Can You Find (CYF)
30 September, 2024:
CYF NO. 61
28 April, 2024:
CYF NO. 60
(Comments by Al Zimmermann dt 2052024 to 10062024)
My program has examined products of three successive integers up to
10,102,125,053,252 × 10,102,125,053,253 × 10,102,125,053,254
= 1,030,951,466,886,907,255,557,038,136,424,474,572,024.
The product is slightly more than 10^{39} and each of the 3 factors are slightly more than 10^{13}.
My program examined all values of (n  1) × n × (n + 1) up to n = 10,102,125,053,253 and
found no palindromes beyond 77 × 78 × 79 = 474,474.
(Comments by Brian Trial dt 11052024 )
A helpful hint, you can use finite differences to create a series of numbers of the form n × (n+1) × ( n+2 ) very rapidly:
a b f
6 6 6
Start with 6 6 6, and then each for each new row,
a = a + 6,
b = b + a.
f = f + b, and f is the next number to check for palindromes:
a b f
6 6 6
12 18 24
18 36 60
24 60 120
The next entry is generated from the last by three additions instead of two multiplies.
This has allowed me to search n up to 8.6 × 10^{12}, so far without success. My program has run for a week so far to reach this.
8691091458776 × ( 8691091458776 + 1 ) × (8691091458776 +2 ) =
656482208188478232061677065881802284656
and notice this is a "near miss" partial palindrome where the first and last 12 digits are palindromes of each other.
31 December, 2023:
CYF NO. 59
(Solution by Al Zimmermann dt 2012024 to 18022024)
There are three solutions up to 10^{20}. These are(Solution by Giorgos Kalogeropoulos dt 6012024 )
I found two more solutions: 2344344434432 and 109514040415901.(Solution by Alessandro Casini dt 10012024 )
Other than the one you mentioned, 2344344434432 = 2772^{3} + 13244^{3} = 7161^{3} + 12551^{3} and12 July, 2023:
CYF NO. 58
(Comments by M F Hasler dt 12072023 )
I checked that there is no solution with n < 2 * 20000. I have submitted this draft : https://oeis.org/draft/A364166 : Indices n such that A002375(n) = A002375(n+1) = number of decompositions of 2n into a sum of two odd primes, which has since been publised A364166
(Comments by Shyam Sunder Gupta dt 31122023 )
I checked that there is no solution with n < 1000000.
12 September, 2022:
CYF NO. 57
(Comments by Alessandro Casini dt 14092022 )
The repunit must have the form p*q^{2} to possess 3 different representation. I looked through the Kamada's factors list up to R_500, but I think I have not found any number like that.
(Comments by M F Hasler dt 14092022 )
I think there cannot be a repunit which is a difference of squares in exactly 3 ways.
Because, for odd numbers n, the number of representations as difference of 2 squares (A100073  OEIS) is numdiv(n) / 2, the number of divisors < sqrt(n).
Repunits R(n) are never squares (except for 1), therefore they have always an even number of divisors.
The following cases are possible:
R(n) is prime => numdiv = 2 => A100073 = 1 (for example, R(2), R(19), R(23)
R(n) is semiprime => numdiv = 4 => A100073 = 2
To have A100073 = 3, one would need to have R(n) with 6 divisors, i.e., in A030515
Such numbers are of the form p^{5} or p^{2}*q where p and q are distinct primes.
I don't think it is possible to have a repunit of that form.
But I admit, I don't have a complete proof yet.
(Comments by Brian Trial dt 18092022 )
Interesting problem.
The problem equates to finding Rn (repunit with n digits) that can be factored in exactly three unique ways since x^{2}  y^{2} = ( x + y )( x  y ) .
In other words, if Rn = f1 * f2, Rn = ( ( f1 + f2 ) / 2 )^{2}  ( ( f1  f2 ) / 2 ) ^{2}.
One possibility is if Rn has the prime factorization p^{2} * q .
This gives exactly three factorization possibilities ( f1 , f2 ) of ( 1, Rn ), ( p , p*q ), and ( q , p^{2} ), which would yield exactly three solutions for x and y.
If Rn has three or more prime factors, we can rule it out since that yields at least 4 factorization possibilities. (1 , Rn), ( p, q*r), ( q , p*r), and (r , p*q).
Exactly three factorizations are also possible if Rn = p^{5} or p^{6}, but I will ignore these cases for now.
I explored the repunits divisible by small values of p^{2} by using modular arithmetic. For example, R9 is divisible by 3^{2}.
If R9 / 9 is prime, we have a solution. If R9 / 9 has a small factor, we can rule it out as a possible solution since that will show it has three or more prime factors.
R9 can be ruled out because it is also divisible by 37, leaving a residual of 333667.
I checked repunits with prime factors up to 20,000,000. For example, R25651 is divisible by 227^{2}.
Unfortunately it also has the "small" factor 12568991, so we can rule it out as a solution possessing a small value of p.
Makoto Kamada maintains the web site https://stdkmd.net/nrr/repunit/ that has complete and partial factorization of repunits up to R300000.
I needed Mr. Kamada's help on R114481. I discovered it is divisible by 479^{2}, but its other factors are all greater than 20,000,000, so I did not find them.
However, he lists 142847911 as a "small" factor, so we can rule this out as well.
The first possible solutions I found are R1865033, R1929139, R2747929 and R3260981, which is as far as I took the search.
They are divisible by 12517^{2}, 3929^{2}, 6203^{2}, and 5711^{2} respectively, but have no other prime factors less than 20,000,000.
Can someone rule these out as solutions by finding other factors, or rule them in as solutions by proving the residual prime?
(Comments by Richard Gosiorovsky dt 17112022 )
Such repunit has to have form r = p1^{2} . p2, where p1, p2 are primes and there is very low probability that such repunit exists. Because:
I/ if repunit r is a prime number there is only way to express it as difference of squares:
r = a^{2}  b^{2} = (a+b).(ab) > ab = 1 because r is prime number
II/ repunit r is composite of two primes: r = p1.p2
There are exactly two ways:
a+b = p1.p2 ab = 1 and
a+b = p1 ab = p2 (p1 > p2)
III/ repunit r is composite of three different primes: r = p1.p2.p3
There are exactly four ways:
a+b = p1.p2.p3 ab = 1
a+b = p1.p2 ab = p3 (p1.p2 > p3)
a+b = p1.p3 ab = p2
a+b = p2.p3 ab = p1
or
a+b = p1.p2.p3 ab = 1
a+b = p3 ab = p1.p2 (p1.p2 < p3)
a+b = p1.p3 ab = p2
a+b = p2.p3 ab = p1
IV/ if repunit r is composite of four or more primes there are lot of combinations of factors
V/ if r = p1^{2} . p2 there are exactly three different ways:
a+b = p1.p1.p2 ab = 1
a+b = p1.p1 ab = p2 (p1.p1 > p2)
a+b = p1.p2 ab = p1
or
a+b = p1.p1.p2 ab = 1
a+b = p2 ab = p1.p1 (p1.p1 < p2)
a+b = p1.p2 ab = p1
(Comments by M F Hasler dt 13072023 )
Reply to the Brian Trials comments :(Comments by Alessandro Casini dt 16072023 )
After the update I've something to say. Let me underline that the p^5 case is ruled out a priori, since a Repunit can't be a perfect power (Mignotte & Bugeaud 1999), so its factorization must be p*q^2. Repunits have a lot in common with Mersenne numbers, whose theory is hugely developed. One of the properties is that each term possesses a primitive prime factor, so forcing the Repunit index to be a prime or a prime squared. Moreover, thanks to (Granville 2012), we know that a Repunit number has a primitive prime factor with an odd exponent, hence after a few tries one shows that the Repunit must have a prime number of digits. This property immediately rules out all Brian Trial's candidates, since their indexes are composite, as confirmed by trial factoring too. Furthermore, as for Mersenne numbers, given that the index is prime, if q^2 divides R_p, then q must be a Wieferich prime base 10. Excluding 3, at now only two Wieferich primes are known less than 10^14 (OEIS A045616) and neither divides a Repunit at all. This means that so far all Repunits with prime index are known squarefree, let alone the particular factorization p*q^2. Summing up, a solution to this problem must be a Repunit with a prime index and p must be a Wieferich prime (very rare) greater than 10^14, not much optimistic. It seems that it will take a long time to give an answer to this problem, at least until a new Wieferich prime base 10 is discovered.14 March, 2022:
CYF NO. 56
(Solution by Alessandro Casini dt 16032022 )
The smallest amicable pair is (2973022993690763475, 3877321503756148525) such that digital root of pair sum is 4.
In his subsequent email, he informed that next two amicable pairs (21048260829161290725, 30819388283587560475) and
(26984027894818400175, 35725020504766879825)
also satisfies the given condition, therefore, together with the previous pair, they are the only 3 pairs less than 10^{20} with such characteristic.
11 November, 2021:
CYF NO. 55
(Comments by Al Zimmermann dt 15112021 to 28112021)
The 153335913313195731171937^{th} triangular number T(n) = 11755951155795937935997379553599771513593751953.(Comments by Maximilian dt 24112021)
For n = 119311115937719393371311137 (27 digits). T(n) = 7117571193151959319913571179373191539575517915771953 (52 digits).(Comments by Giorgos Kalogeropoulos dt 05/07122021)
For n = 19995759119953397335973333 (26 digits). T(n) = n(n+1)/2 = 199915191391599731555773573313573711177939911551111 (51 digits).(Comments by Al Zimmermann dt 7092022 to 9092021)
The Al Zimmermann's Programming Contest (Oddly Triangular) based on CYF No. 55, was run from 7th sept to 8th sept 2022.(Comments by Maximilian dt 11092022)
The sequence contains all numbers of the form 33(9{n}7){k}3{n}, where {x} means to repeat the preceding digitCan you find the largest number(or as large as you can) n such that both n and n^{th} triangular number consists of odd digits only and contain all odd digits i.e., 1,3,5,7 and 9.
12 April, 2021:
CYF NO. 54
(Comments by late shri K D Bajpai dt 01072021 )
I explored the problem for n from 1 to 5.5*10^{8}. No solution was found such that (n*n + n + 41) gives a Strong Pseudoprime (base2).(Comments by Shyam Sunder Gupta)
I tested Eulers trinomial for n up to 10^{11} and no solution was found such that (n*n + n + 41) gives a Strong Pseudoprime (base2).17 June, 2020:
CYF NO. 53
(Solution by Al Zimmermann dt 20112020 )
p = 3 536 123
q = 128 541 727
p * q = 454 539 357 304 421
In his subsequent emails, he informed that:
I solved it by writing a 20line C# program (not counting the
generation of primes, for which I used a library routine), which took 16 seconds to run.
The program considered all prime pairs (p, q) up to q = 982451653 in about 90 seconds.
(982451653 is the 50millionth prime.) The solution I sent was the only pair it found for p > 7.
Given that there were no solutions between (7, 53) and (3536123, 128541727),
I think that the next solution might involve some very large numbers.The program’s run time
is linear with respect to the number of primes considered, so it would take it about 30 minutes to
examine the first 1 billion primes (not including the time it would take to generate those
primes).
In his subsequent email dated 24112020, he informed that:
I tweaked the program to look at all values for q up to 1 trillion =
1,000,000,000,000 = 10^12.It did not find any additional solutions.
(Solution by Brian Trial dt 05042021 )
Primes 3536123 to 128541727 inclusive sum to
454539357304421 and also multiply together to that
result.
This was previously discovered by Robert Munafo
in 2002. See A055233 OEIS
14 December, 2019:
CYF NO. 52
(Comments by late shri K D Bajpai dt 07072021 )
No solution found for the set of 13 consecutive primes which are emirps also. Explored up to (4*10^{7}) th prime,i.e., 776531401.15 September, 2019:
CYF NO. 51
(Comments by Jens Kruse Andersen dt 16092019 )
An upper limit is 10^999+261 = 13 * 87382853 * P990, where P990 is a 990digit prime. There are many smaller candidates.
(Comments by Andreas Bartsch dt 22072024 )
An upper limit is 10^999+19 = 2698217 * 1463229659219525553812101593947619925343 * P954, where P954 is a 954digit prime proven with Primo. The P40 was found after 982 ECM curves at B=3000000 using GMPECM. The only remaining smaller candidate is 10^999+13, see CYF No. 11.
28 November, 2018:
CYF NO. 50
(Solution by Al Zimmermann dt 02122018 )
No solution exists.
Proof by contradiction: Assume that a, b, c are odd integers, that x, y and z are integers, and that b + c = x^{2}, c + a = y^{2} and a + b = z^{2}.
Since a, b and c are odd, we know that x^{2}, y^{2} and z^{2} are even. Therefore x^{2}, y^{2} and z^{2} are multiples of 4. Therefore x^{2}/2, y^{2}/2 and z^{2}/2 are even.
If we solve the original three equations for a, b and c we get:
a = (x^{2} + y^{2} + z^{2}) / 2,
b = (x^{2} – y^{2} + z^{2}) / 2,
c = (x^{2} + y^{2} – z^{2}) /2.
We know that x^{2}/2 and y^{2}/2 and z^{2}/2 are all even, therefore a, b and c must be even. This contradicts our assumption that they are odd.
(Solution by Brian Trial dt 05122018 )
a,b and c cannot all be odd integers.
Proof:
Let a+b = d*d = d^2
a+c = e*e = e^2
b+c = f*f = f^2
Combine equations to get a,b and c in terms of d,e and f.
a = ( d^2 + e^2  f^2) / 2
b = ( d^2  e^2 + f^2) / 2
c = (  d^2 + e^2 + f^2) / 2
Now, d, e and f are either odd or even, so d can be expressed as either 2x or 2x+1
e can be expressed as either 2y or 2y+1 and f can be expressed as either 2z or 2z+1
There are 8 possible combinations of odd and even for d,e and f. If we explore each one, replacing d with either 2x or 2x+1,
e with either 2y or 2y+1, f with either 2z or 2z+1, we always come up with at least one term that is either even, or not an integer.
1. d = 2x, e = 2y , f = 2z
a = ( 4x^2 + 4y^2  4z^2 ) / 2
a = 2x^2 + 2y^2  2z^2 so a is an even number.
2. d = 2x, e = 2y , f = 2z+1
a = ( 4x^2 + 4y^2  (4z^2 + 4z + 1) ) / 2 so a is not an integer.
3. d = 2x, e = 2y+1 , f = 2z
a = ( 4x^2 + (4y^2 + 4y + 1 )  4z^2 ) / 2 so a is not an integer.
4. d = 2x, e = 2y+1 , f = 2z+1
a = ( 4x^2 + (4y^2 + 4y + 1 )  (4z^2 + 4z + 1) / 2
a = 2x^2 + 2y^2 + 2y 2z^2 2z so a is an even number.
5. d = 2x+1, e = 2y , f = 2z
a = ( 4x^2 + 4x + 1 + 4y^2  4z^2 ) / 2 so a is not an integer.
6. d = 2x+1, e = 2y , f = 2z+1
a = ( ( 4x^2 + 4x + 1 ) + 4y^2  (4z^2 + 4z + 1) ) / 2
a = 2x^2 + 2x + 2y^2 2z^2 2z so a is an even number.
7. d = 2x+1, e = 2y+1 , f = 2z
a = ( ( 4x^2 + 4x + 1 ) + (4y^2 + 4y + 1 )  4z^2 ) / 2
a = 2x^2 + 2x + 2y^2 + 2y  2z^2 + 1 so a is an odd number!
Unfortunately, then b is not
b = ( d^2  e^2 + f^2) / 2
b = ( ( 4x^2 + 4x + 1 )  (4y^2 + 4y + 1 ) + 4z^2 ) / 2 so b is an even number.
8. d = 2x+1, e = 2y+1 , f = 2z+1
a = ( ( 4x^2 + 4x + 1 ) + (4y^2 + 4y + 1 )  (4z^2 + 4z + 1) / 2 so a is not an integer.
QED
(Solution by Emmanuel Vantieghem dt 19012019 )
The answer to CYF 50 is obviously negative.
If a+b = x^2
a+c = y^2
b+c =z^2
and a,b,c odd then x^2, y^2, z^2 are divisible by 4
But, summing up the three equalities gives : 2(a+b+c) = x^2 + y^2 + z^2.
The left side is == 2 mod 4 while the right side is 0 mod 4, impossible.
(Solution by Alessandro Casini dt 08032019 )
I think that such numbers do not exist.
In fact, assume that a,b,c exist in order to satisfy the properties.
The perfect squares must be even as sum of two odd numbers and in particular they must be divisible by 4. So:
a+b=0 mod 4
a+c=0 mod 4
b+c=0 mod 4
Subtracting the first with the second equation and rearranging the third one, we get b=c (4) and b=c (4).
The unique solutions are b=c=0 mod 4 and b=c=2 mod 4. In both cases, b and c are even, contrary to our assumption. Therefore, a,b,c can not exist.
17 December, 2017:
CYF NO. 49
(Comments by Daniel Suteu dt 10112019 )
An example for such a pseudoprime is: 3470207934739664512679701940114447720865, which is both an
abundant number and a Fermat pseudoprime to base 2.
However this is not a guaranteed smallest abundant Fermat pseudoprime to base 2. By looking at the record values (based on my
generated list of 22,000+ pseudoprimes), there seems to be quite a big gap between abundacy 1.9967 and
2.0027, which suggests that may be there exists a smaller one:
1.9449 record: 4788772759754985
1.9452 record: 38353281032877674865
1.9452 record: 2638294879881771254145
1.9454 record: 24773130788808935997465
1.9454 record: 429388072625313108651345
1.9459 record: 633708839387221385771985
1.9594 record: 4426632427184293146004185
1.9786 record: 30289748358904212366572385
1.9814 record: 728017010426459878356936705
1.9842 record: 34742187522900518220477982065
1.9856 record: 296874628838164374175009735905
1.9888 record: 282385466973939051351462060010785
1.9893 record: 303874597856107607075523212740305
1.9967 record: 95232935093163427870007414104051665
2.0027 record: 3470207934739664512679701940114447720865
Some other larger base2 pseudoprimes that are also abundant, include:
45737634436431198836267992527251680040385
43502516985853501805353173385040503830615585
35681625351759414447784656906886402710161726145
8360879247062142072476242452391089552717501127905
8912541157773704856248850726153279748167674593185
83608400947968548043246176345350430162640535738785
18324199296528369643710689094829325758885303308048705
28372903573883820144872703713489286337144948101390945
224906436908031007079087447570104271409064434499731585
555385881837969384461386805114747741697881330167516065
1712928022736001657437035367611577294639589883088004705
16068949403392919154075612370341935224248372485244305505
779551322951482550340219275431845821818182317453398081665
3243771449934489563420280731826154811398657190689660346145
3521782835397676946933688469202532605577443098683804760065
4458187585944145020727100588605702609629401058115042012065
4480869812756224982980361246336523683775981809514868335105
16434919376458405148335185917085137785166951881078959368385
43225442828665203343551053810841050234922320859806912515905
83929961868126246820141143600626161173163737039232406516865
839363327864729874359487191942135344679553853090370502721185
1286754527135281077618073992106243955350494847060895462784705
2215351116173516203829705825916100548160379300591878217129985
2768347873560407762689098818902929900263309343470278283249985
12492495099028177216769135406790403675538624687838083630973185
14412192513054899010532954508122570676822692536898161874517985
21735355747738858221724407770659644200025485058748770371779905
27974863795893090960230965223628876028701381681601300706365185
49619164548494076358485360085028000849457165404574906593198145
96575617503741482021640791861376593523200847532295910631907265
149824307412014285797537763203072744032701914138320937509441345
699177242901215844952491647476996222051262265977000379251093505
738403120358970712811833990095379996524800144348145698260302465
856832446705363063968603643425895801652503477562304269424808545
922723577834882097089155855930848203988124904422542752754582465
4941808546075660419264067599369649016479539371630778521435976705
14528634372802949026298064925609466349944920915994190406309741665
16130751161320787755103386812035143794353858112781193905695834945
18571834138590672838737539649602176887168943632625142230078924065
44181041092632451863994500365125906091058066549365702838575250945
47075694600925132483952687753066135110680206374164000678725510145
A related problem (much more difficult) would be: "what is the smallest abundant Carmichael number?"
The analog of A328691 for Carmichael numbers, would begin as:561, 62745, 576480525985, 1886616373665,
3193231538989185, 11947816523586945, 101817952350880305, 171800042106877185(terms below 2^{64}). The
maximal abundancy that we get is sigma(k)/k = 1.893... for k=171800042106877185, which suggests that
abundant Carmichael numbers are possible, but I think the smallest such number would have around
50 digits.
Can you find a smaller solution or prove that Daniel Suteu's solution is the smallest ?.
(Comments by Daniel Suteu dt 16082020 )
Attached a list containing 58 abundant Carmichael numbers, out of which the smallest abundant Carmichael number is 2059832906607460252767290568443059994787898033540634712711845135488141590979778401392385
(Comments by Daniel Suteu dt 31082022 )
Two smaller abundant Fermat pseudoprimes to base 2:
2596282479202818734176082185090403265
12796625128232187655293894634808130945
A somewhat related problem: "What is the smallest abundant LucasCarmichael number?"
An example for an abundant LucasCarmichael number is: 1012591408428327888883952080728349448745451794025524955777432246705535
(Comments by Daniel Suteu dt 10032023 )
Two smaller abundant Fermat pseudoprimes to base 2:
898943937249247967890084629421065
222042825169546323981793629414604065
30 October, 2016:
CYF NO. 48
4 November, 2015:
CYF NO. 47
(Solution by Bernard Fischel dt 28032016 )
Five distincts even numbers : A=32498; B=116498; C=416402; D=1028402; E=2175698;
Verified:
1 A= 32498, B= 116498, A+B= 148996 = 386 * 386
2 A= 32498, C= 416402, A+C= 448900 = 670 * 670
3 A= 32498, D= 1028402, A+D= 1060900 = 1030 * 1030
4 A= 32498, E= 2175698, A+E= 2208196 = 1486 * 1486
5 B= 116498, C= 416402, B+C= 532900 = 730 * 730
6 B= 116498, D= 1028402, B+D= 1144900 = 1070 * 1070
7 B= 116498, E= 2175698, B+E= 2292196 = 1514 * 1514
8 C= 416402, D= 1028402, C+D= 1444804 = 1202 * 1202
9 C= 416402, E= 2175698, C+E= 2592100 = 1610 * 1610
10 D= 1028402, E= 2175698, D+E= 3204100 = 1790 * 1790
(Solution by Brian Trial dt 29082016 )
Five distincts even numbers : 32498, 116498, 416402, 1028402, 2175698.
Generating
386^2, 670^2, 730^2, 1030^2, 1070^2,
1202^2, 1486^2, 1514^2, 1610^2 and 1790^2.
(Comments by Selvam Chidambaram dt 23072016 )
Five distincts even numbers : 685016987580450,
6032343472264570,
110287735000452000,
1019319257458230000 and
6657791941802570000 .
But not necessarily smallest set.
7 July, 2015:
CYF NO. 46
(Solution by Al Zimmermann dt 10072015 )
The following arithmetic progression of five perfect powers has common difference 20073720:
8196769, 28270489, 48344209, 68417929, 88491649
This can be rewritten as: 2863^{2}, 5317^{2}, 6973^{2}, 409^{3}, 9407^{2}
In his subsequent email dated 11072015, he informed that:
I used a brute force approach, which took about 50 lines of code in .NET C#.
It took 40 seconds for the program to find the first solution.
I let the program run for a while longer to see if it would find an arithmetic progression
with larger terms but with a smaller common difference.
When I stopped the program (after 3 hours and 15 minutes) it had generated all perfect powers up to
one trillion (that is, up to 10^12) and had not found a better arithmetic progression.
(Solution by Brian Trial dt 25072015 )
We know you cannot have 4 squares in a row in an A.P., a^2, b^2, c^2, d^2
but what if there is a gap, for example a^2, b^2, c^2, x, d^2 where x = (c^2 + d^2) /2 ?
If you can find this, then (ax)^2, (bx)^2, (cx)^2, x^3, (dx)^2 would be 5 Perfect Powers in A.P.,
with a perfect cube filling the gap. 7^2, 13^2, 17^2, 409, 23^2 fits this criteria,
so that means (7*409)^2, (13*409)^2, (17*409)^2, 409^3, (23*409)^2 is 5 Perfect powers in A.P.
Expanded, this is 8196769, 28270489, 48344209, 68417929, 88491649 with a common difference of 20073720.
I believe this is also the smallest.
6 October, 2014:
CYF NO. 45
25 December, 2013:
CYF NO. 44
(Solution by Fred Schneider dt 29032014 )
You need at least 5 prime factors to get an odd abundant number.
sigma(3*5*7*11) / (3*5*7*11) ~ 1.99481 is the closest you can get.
15 July, 2013:
CYF NO. 43
(Solution by Al Zimmermann dt 11012015 )
For brevity, lets abbreviate “magic square of distinct triangular numbers greater than zero” as MSODTNGTZ.


I am not certain that the above is the smallest MSODTNGTZ, but it has the smallest magic constant (892)
among MSODTNGTZs whose largest entry does not exceed T(31).
13 November, 2012:
CYF NO. 42
(Comments by Shyam Sunder Gupta dt 14072013 )
Bo Gyu Jeong have earlier tested all p^n (p < 100) up to 30000 digits. The largest one he found was: 19^44 = 184144368549628275143663229532787625188711914273876985521. (57 digits)
I have further tested all p^n (p < 10000, 10000 > n > p) . No other solution larger than 19^44 was found. It is quite likely that this is the largest solution.
6 February, 2011:
CYF NO. 41
(Comments by Brian Trial dt 28072015 )
Here are my best efforts:
x^2 + 419x  72 = 28 twin primes x= 1100
x^2 + 929x + 2538 = 148 twin primes x = 11000
x^2  1825x + 984 = 814 twin primes x=110000 and 6002 twin primes x=1100000
(Comments by K D Bajpai dt 18032016 )
The corrected values submitted by Brian Trial are as follows:
x^2 + 419x  72 = 28 twin primes x= 1100
x^2 + 929x + 2538 = 148 twin primes x = 11000
x^2  1825x + 984 = 962 twin primes x=110000 and 6150 twin primes x=1100000
(Comments by Shyam Sunder Gupta dt 31102015 )
My results are:
3x^2 + 27x  14658 = 52 twin primes x = 1100
3x^2 + 231x  2953212 = 259 twin primes x = 11000
3x^2 + 33x  7068948 = 1685 twin primes x = 110000
3x^2 + 33x  9547578 = 9075 twin primes x = 1100000
There are other solutions also with similar records.
Can you beat these records.
26 January, 2010:
CYF NO. 40
(Contribution by Bo Gyu Jeong dt 18032010 )
I have tested all p^n (p < 100) up to 30000 digits. The largest one I found is: 19^44 = 184144368549628275143663229532787625188711914273876985521. (57 digits)
(Contribution by Norman Luhn dt 08082010 )
I start with 100 digits.
p = 349 , n = 39
349^39 = 1479761253386816635938918573914419155994142151
898153246796492828575365682425969779363159565487616149
(Contribution by Shyam Sunder Gupta dt 05022011 )
My record is 143 digit number:
p = 6161527 , n = 21
6161527^21 = 383226448283881161158925442286272657358177586728919276714197431918
81529313713118214228251348587846799583459615559231988282189668125135144387927
(Contribution by Carlos Rivera dt 20092011 )
The prime 92364991 powered to the integer 22 is an integer composed of 176 decimal digits none of which is zero. As a matter of fact none of the digits in 92364991^{22} is zero too.
(Contribution by Maximilian Hasler dt 25092011 )
In fact it is very easy to find such numbers, by considering simply the square of primes such that their square has no zero except for the last digits. Starting from Carlos' solution 92364991^{22}. I found the next larger (I think) to be P = 92364991^{11} + 5839747404697609902097466198015728608858823158478055221217330117206174653951754783434065780808. This one has the same number of digits (176) as 92364991^{22}. To find one with more digits I started at sqrt(10^{178})/6, and already the 5^{th} prime, P' = [10^{89}/6]+1773 is again a solution (i.e. its square has 177 digits which are all nonzero). The least prime larger than 10^{89}/3 is [10^{89}/3]+130 = 33333...33463 (89digits), which is the smallest prime whose square has 178 digits which are all nonzero. This is also the smallest zero less prime power with 178 digits.
Maximilian Hasler in his subsequent email dated 26 sept 2011, added that : The square of the 500 digit prime P=[10^{500}//3]+10210 has 1000 digits, all nonzero. Moreover, it looks very nice: P^2 = [10^{500}/9]*(10^{500}+5) + 6806*10^{500}+104237294 = 11111...11111791755555...55555659792849
He suggested to add some merit function to the puzzle (giving a better score to smaller primes and larger exponents) to make it more interesting
Further details can now be seen on Carlos Riveras page Prime Puzzles and Problems Connection (Puzzle 607. A zeroless Prime power).
(Contribution by Brian Trial dt 24092012 )
I have checked all candidate prime powers with 512 digits or less for (p < 10^{8}) and cold not get any solution except 92364991^{22} with 176 digits.
(In view of large response and suggestion of Maximilian Hasler for some merit function, I have included this problem as new CYF No. 42)
10 May, 2009:
CYF NO. 39
18 January, 2009:
CYF NO. 38
(Comments by Bernard Fischel dt 17122013 )
After many hours of computing the factors of Fibonacci numbers ranging from 1 to 500 (from 1 to 105 digits in length) no solution is found.
However if same number of distinct prime factors are considered then F(3), F(4), F(5), F(6) and F(4), F(5), F(6), F(7) are the only solution found.
Fortunately, after this work I have found a table of a complete factorizations of Fibonacci numbers ranging from 1 to 1000 (from 1 to 209 digits).
The values in the table confirm my previous results. None new sequences appear in the table (http://mersennus.net/fibonacci/f1000.txt).
30 December, 2007:
CYF NO. 37
(Solution by Luke Pebody dt 30122007 )
11968,14366
(Solution by Jens Kruse Andersen dt 01012008 )
x = 11968 = 2^6*11*17. y = 14366 = 2*11*653. s(x^2) = s(y^2) = 191213697.
(Solution by Donovan Johnson dt 31012008 )
Here is the smallest I found:
x=11968,y=14366
sigma(x^2)x^2 = 191213697
sigma(y^2)y^2 = 191213697
(Solution by Frederick Schneider dt 08022008 )
The minimal pair is:
aliquotSum(11968^2) = aliquotSum(14366^2) = 191213697
Using some quadratic tests, I found this low number
aliquotSum(15021^2) = aliquotSum(15843^2) = 111624510
and then checked for a lower solution.
14 July, 2007:
CYF NO. 36
(Comments by Luke Pebody dt 21082007 )
I doubt that this is the smallest solution but the numbers in the
range 36689206836378866312597986802568592633785433489374 to
36689206836378866312597986802568592633785433489377 are abundant.
Here is how I found them: I looked for four perfect/abundant numbers
n_1, n_2, n_3, n_4 such that x=0 (mod n_1) 1 (mod n_2) 2 (mod n_3)
and 3 (mod n_4) were compatible, and then looked for the smallest
solution. I tried to keep the numbers n_k as small as possible, by
using the following procedure:
Initialize N to [6,1,2,3];
Initialize P to [{},{},{2},{3}];
Set p to be 3;
(*)
Let i be the index such that DivisorSigma(N[i])/N[i] is minimal.
If this minimal value is 2 or more, stop;
Otherwise, let q=NextPrime(p);
Let r be the value from P[i] join {q} that maximises
DivisorSigma(N[i]*r)/(N[i]*r);
Multiply N[i] by r and include r in P[i];
If r=q, set p to be q;
Return to (*)
This gives N as [2*3, 5^2*7^2*13*17*19*37*41*43*53*67*73*89*103*109,
2^5*97*107, 3^3*11*23*29*31*47*59*61*71*79*83*101*113].
(Comments by Bruno Mishutka dt 02112007 )
I suspect that this is not the smallest m1 such that (m1, m1+1, m1+2, m1+3) are
all abundant, but it's the smallest set that I could find.
m1 := 141363708067871564084949719820472453374;
It is approximately 1.41*10^38.
(Comments by Shyam Sunder Gupta)
Jens K Andersen found a smaller set of 4 consecutive abundant numbers starting at 172107222115840172372369874858517374 (36 digit number). For details refer Carlos Riveras page Prime Puzzles and Problems Connection (Puzzle 878 Consecutive abundant integers).
Can you find a smaller solution or prove that Jens K Andersen's solution is the smallest set.26 January, 2007:
CYF NO. 35
(Solution by Jaroslaw Wroblewski dt 11022007 )
Let
a=(16n1)(63n1)
b=(18n1)(56n1)
c=(21n1)(48n1)
d=(28n1)(36n1)
Then a,b,c,d form an arithmetic progression and have the same sum of
divisors provided all 8 factors above are prime (as it happens eg. for
n=273168).
Let
a=( 91n1)(400n1)
b=(100n1)(364n1)
c=(112n1)(325n1)
d=(130n1)(280n1)
e=(175n1)(208n1)
Then a,b,c,d,e form an arithmetic progression and have the same sum of
divisors provided all 10 factors above are prime (as it happens eg. for
n=213990).
(Solution by Frederick Schneider dt 24032007 )
I found a solution that is likely minimal. I wrote a program to PARI
to search for all solutions by divisor sum N. Initially, I limited the search to
multiples of 24 but found that the number of a.p.'s of length 3 was very small unless there was a high number of divisors of N. So, I limited the search to numbers
where numDivisors(N)^3 > N. Yesterday, I added a 2nd check to find the
odds of an a.p. of length 4 given there were X solutions between the
maxSolution and the minSolution (and assuming that the solutions were
randomly distributed).
For a candidate, if there was < 1% chance of finding an a.p. , I would
skip over it. Roughly, only about 1 in million numbers are considered
around the 100 million mark.
So, an exhaustive search was not performed through this point but its
likely that this is minimal.
Solution:
119508480 = 2^9 * 3^3 * 5 * 7 * 13 * 19
4 terms in arithmetic progression with difference: 2138892 = 2^2 * 3 * 7 * 25463
34560288 = 2^5 * 3^2 * 7^2 * 31 * 79 (sumDivs = 31 * 13 * 57 * 32
* 80 = 119508480)
36699180 = 2^2 * 3 * 5 * 7 * 59 * 1481 (sumDivs = 7 * 4 * 6
* 8 * 60 * 1482 = 119508480 )
38838072 = 2^3 * 3 * 7 * 13 * 17783 (sumDivs = 15 * 4 * 8 * 14 *
17884 = 119508480)
40976964 = 2^2 * 3^2 * 7 * 113 * 1439 (sumDivs = 7 * 13 * 8 * 114 *
1440 = 119508480)
(Solution by Donovan Johnson dt 30032007 )
Here is the smallest set of four numbers I found:
34560288, 36699180, 38838072, 40976964
AP difference = 2138892
Sigma value = 119508480
Sets of four numbers can be generated from the set of four numbers above by multiplying 34560288 and 2138892 by one or more prime numbers:
d = prime number or the product of two or more prime numbers (use any prime except for the following twelve: 2, 3, 5, 7, 13, 31, 59, 79, 113, 1439, 1481 and 17783)
Smallest of the new set of four numbers = 34560288 * d
AP difference of the new set of four numbers = 2138892 * d
I found one set of four numbers not of the form above:
435831146, 435831566, 435831986, 435832406
AP difference = 420
Sigma value = 653837184
25 October, 2006:
CYF NO. 34
(Comments by Brian Trial dt 05112006 )
I could find no other solutions to CYF 34 in the first
1x10^11 triangular numbers.
This took about 20 hours on a Pentium 1.8 GHz with 2GB ram.
The largest nontrivial number where all the digits except 1
is either 0 or 1 that I found was the 471404th triangular number =
111111101310.
The largest nontrivial example where all the digits except 2
is either 0 or 1 that I found was the
44969278424th triangular number = 1011118001010100601100
(Comments by Nathan Gilbert dt 06112006 )
This is a hard one for sure. There are no others less than 2^64. I'll have to come up with some more robust methods to test any further.
15 January, 2006:
CYF NO. 33
(Solution by Jens Kruse Andersen dt 18012006 )
The maximum is 26 primes for these and their reversals:
165479283, 241798365, 263174598, 283761954, 316547928, 459167328.
The primes for 165479283 are:
2, 3, 5, 7, 29, 47, 61, 79, 83, 97, 283, 479, 547, 829, 4561, 5479,
6547, 8297, 9283, 16547, 65479, 74561, 79283, 165479, 2974561, 65479283.
(Solution by Donovan Johnson dt 18012006 )
Twelve numbers have 26 distinct primes:
165479283:
primes: 2, 3, 5, 7, 29, 47, 61, 79, 83, 97, 283, 479, 547, 829, 4561, 5479, 6547, 8297, 9283, 16547, 65479, 74561, 79283, 165479, 2974561, 65479283
241798365:
primes: 2, 3, 5, 7, 17, 41, 71, 79, 83, 89, 97, 179, 241, 389, 563, 971, 983, 2417, 6389, 8971, 24179, 38971, 417983, 563897, 638971, 2417983
263174598:
primes: 2, 3, 5, 7, 13, 17, 31, 47, 59, 71, 89, 263, 317, 547, 631, 5471, 6317, 7459, 9547, 26317, 54713, 95471, 317459, 895471, 954713, 8954713
283761954:
primes: 2, 3, 5, 7, 19, 37, 59, 61, 67, 73, 83, 167, 283, 619, 673, 761, 2837, 3761, 4591, 37619, 59167, 83761, 91673, 459167, 591673, 837619
316547928:
primes: 2, 3, 5, 7, 13, 29, 31, 47, 61, 79, 97, 479, 547, 613, 829, 4561, 5479, 6547, 8297, 16547, 45613, 65479, 74561, 165479, 2974561, 3165479
382974561:
primes: 2, 3, 5, 7, 29, 47, 61, 79, 83, 97, 283, 479, 547, 829, 4561, 5479, 6547, 8297, 9283, 16547, 65479, 74561, 79283, 165479, 2974561, 65479283
459167328:
primes: 2, 3, 5, 7, 19, 23, 37, 59, 61, 67, 73, 167, 619, 673, 761, 823, 3761, 4591, 8237, 23761, 37619, 59167, 91673, 237619, 459167, 591673
459167382:
primes: 2, 3, 5, 7, 19, 37, 59, 61, 67, 73, 83, 167, 283, 619, 673, 761, 2837, 3761, 4591, 37619, 59167, 83761, 91673, 459167, 591673, 837619
563897142:
primes: 2, 3, 5, 7, 17, 41, 71, 79, 83, 89, 97, 179, 241, 389, 563, 971, 983, 2417, 6389, 8971, 24179, 38971, 417983, 563897, 638971, 2417983
823761954:
primes: 2, 3, 5, 7, 19, 23, 37, 59, 61, 67, 73, 167, 619, 673, 761, 823, 3761, 4591, 8237, 23761, 37619, 59167, 91673, 237619, 459167, 591673
829745613:
primes: 2, 3, 5, 7, 13, 29, 31, 47, 61, 79, 97, 479, 547, 613, 829, 4561, 5479, 6547, 8297, 16547, 45613, 65479, 74561, 165479, 2974561, 3165479
895471362:
primes: 2, 3, 5, 7, 13, 17, 31, 47, 59, 71, 89, 263, 317, 547, 631, 5471, 6317, 7459, 9547, 26317, 54713, 95471, 317459, 895471, 954713, 8954713
(Solution by Carlos Rivera dt 26012006 )
165479283 is the minimal number asked, having 26 primes:
16547 165479 6547 65479 65479283 5 547 5479 47 479 7 79 79283 9283
2 283 83 3 829 8297 29 2974561 97 74561 4561 61
BTW, if you redo the exercise for the pandigitals (numbers having the ten digits 09), the answer is that the minimal number having the maximum distinct primes is... 1654792830 with  again  26 primes (the same than before).
(Solution by Jacques Bernard dt 12032006 )
Here's what I've found.
number 165479283 gives 26 prime
2, 3, 5, 7, 29, 47, 61, 79, 83, 97, 283, 479, 547, 829, 4561,
5479, 6547, 8297, 9283, 16547, 65479, 74561, 79283, 165479, 2974561, 65479283.
(Solution by Nathan Gilbert dt 29032006 )
Having found your site yesterday, I decided to try find the solution to CYF33. I came up with 165479283 as being the answer with 26 primes. Not wishing to stop there, I also found the greatest nine digit number with the maximum number of primes, namely 895471362. The minimum and maximum nine digit numbers containing the least number of primes, 5, are 153694287 and 963518724 respectively. Following is a breakdown of the number of nine digit numbers containing each number of primes.
5 : 34
6 : 192
7 : 800
8 : 2582
9 : 6448
10 : 13036
11 : 21746
12 : 31966
13 : 42202
14 : 49366
15 : 51704
16 : 47534
17 : 38378
18 : 26442
19 : 16140
20 : 8400
21 : 3762
22 : 1488
23 : 454
24 : 158
25 : 36
26 : 12
(Solution by Gaurav Singh dt 06102006 )
I think the answer to the problem will be: 165479283
It contains 26 distinct primes looking forwards and backwards as per the problem. There are some other numbers containing 26 primes but this is the smallest among them. The 26 primes are
2, 3, 5, 7, 29, 47, 61, 79, 83, 97, 283, 479, 547, 829, 4561,
5479, 6547, 8297, 9283, 16547, 65479, 74561, 79283, 165479, 2974561, 65479283.
7 November, 2005:
CYF NO. 32
(Solution by Donovan Johnson dt 13112005 )
I only found one number greater than 15 whose cube consists of digits which are all primes:
1405349897^3 = 2775577757352755375573357273I checked up to 13049558804547.
(Comment by Luke Pebody dt 31122007 )
I think there are infinitely many answers in base 14, starting with 21^3, 27^3, 693^3, 3587^3, 75285^3, 435365^3, 437155^3, 929509^3, 1731765^3, 1902599^3, 1926903^3, et cetera.
My heuristic for base b is that there are (pi(b1))^k numbers up to b^k all of whose baseb digits are prime. Since the probability of a number in this range being a cube is 1/(b^(2/3))^k, it follows that there should be infinitely many if pi(b1)>b^(2/3). This happens for all sufficiently large b, the smallest such being 14.
Note that for b=8, pi(b1)=4=b^(2/3). Thus there are 4^k numbers up to 8^k all of whose base8 digits are prime. The probability of a number in this range being a cube is 1/(4^k). This heuristic says we should only expect finitely many. However, if instead of considering the range [0,8^k] we consider the range [8^{k1},8^k], we get that the expected number in this range is just less than 1, so we expect infinitely many. I used base 14 as we expect exponentially many in this base.
1 May, 2005:
CYF NO. 31
(Solution by B.S.Rangaswamy dt 11052005 )
"Following are the TWO solutions to this interesting problem :
A. 18^3 and 18^4 i.e. 5832 and 104976
B. 69^2 and 69^3 i.e. 4761 and 328509
Apart from the above there are over 3 lakh solutions to this problem, as in
203456789^0 and 203456789^1 i.e. 1 and 203456789
203456798^0 and 203456798^1 i.e. 1 and 203456798
and so on        "
Note: The solution missed above is 2^2 and 2^29 i.e. 4 and 536870912
(Solution by Brian Trial, Ferndale, Michigan, USA dt 17092005 )
"Three solutions to CYF #31.
2 ^ 2 , 2 ^ 29 : 4 , 536870912
18 ^ 3 , 18 ^ 4 : 5832 , 104976
69 ^ 2 , 69 ^ 3 : 4761 , 328509
I also have two solutions where every digit 09 occurs exactly twice:
1638 ^ 2 , 1638 ^ 4 : 2683044 , 7198725105936
6534 ^ 2 , 6534 ^ 3 : 42693156 , 278957081304
And several solutions where every digit 09 occurs exactly three times:
497375 ^ 2 , 497375 ^ 3 : 247381890625 , 123041567849609375
539019 ^ 2 , 539019 ^ 3 : 290541482361 , 156607379280743859
543447 ^ 2 , 543447 ^ 3 : 295334641809 , 160498725087175623
586476 ^ 2 , 586476 ^ 3 : 343954098576 , 201720823916458176
589629 ^ 2 , 589629 ^ 3 : 347662357641 , 204991808273505189
601575 ^ 2 , 601575 ^ 3 : 361892480625 , 217705469031984375
646479 ^ 2 , 646479 ^ 3 : 417935097441 , 270186263858560239
858609 ^ 2 , 858609 ^ 3 : 737209414881 , 632974638501560529
895688 ^ 2 , 895688 ^ 3 : 802256993344 , 718571961854300672
959097 ^ 2 , 959097 ^ 3 : 919867055409 , 882241733241605673
My searches were limited to numbers less than 2^63, so I don't know if solutions exist containing the digits 09 exactly 4 ( or more ) times."
(Solution by Patrick De Geest dt 22102005 )
"I found only three numbers fulfilling your puzzle CYF NO. 31 i.e. 2, 18 and 69.
The numbers, whose two distinct powers contain together all the ten digits (i.e. 0 to 9) only once.
69 ^ 2 and 69 ^ 3 = 4761 and 328509
2 ^ 2 and 2 ^ 29 = 4 and 536870912
18 ^ 3 and 18 ^ 4 = 5832 and 104976"
(Comments by Paul Cleary dt 23012014 )
I have extended the results for No 31 where every digit 09 occurs exactly four times. There are quite a lot 271 to be exact. here's the partial list.
N N^{2} N^{3}
46839081 2193899508924561 102760236804377735568441
47469378 2253341847706884 106964735932016509798152
47693199 2274641230853601 108484916876705732359599
47760623 2281077109348129 108945663853505764924367

99164655 9833628801269025 975148407475906426311375
99424439 9885219070464721 982832360473056354716519
99474426 9895161428029476 984315503230572436180776
99502776 9900802431706176 985157326582314928344576
11 March, 2005:
CYF NO. 30
(Comments by Norman Luhn dt 15032005).
RD(2) and RD(8) are the only repdigit numbers R where R+1 and R1 can be two primes. No solution found up to 5000 digits.
1 January, 2005:
CYF NO. 29
(Comments by Mukesh Kr Jha dt 23072012).
No solution found for this problem for n from 1 to 3,00,00,000 using Maple application.
(Comments by K D Bajpai dt 07012013).
No solution found for this problem for n from 1 to 15,00,00,000 using Maple13 software.
No solution exist for this problem for n from 1 to 10^{12}. For details refer Carlos Riveras page Prime Puzzles and Problems Connection (Puzzle 670 The same sum of divisors).
15 September, 2004:
CYF NO. 28
(Solution by Jens Kruse Andersen dt 18092004 )
n = floor(sqrt(10^999)) + 195 = 31622.....20872 (500 digits). Prp'ed by PrimeForm/GW and proved by Marcel Martin's Primo. The trinomial is famous for consecutive n's giving primes. The first case of two consecutive titanic primes starts at n = floor(sqrt(10^999)) + 7908.
(Solution by Donovan Johnson dt 12102004 )
31622776601683793319988935444327185337195551393252168268575048527925944386392382 21344248108379300295187347284152840055148548856030453880014690519596700153903344 92165717925994065915015347411333948412408531692957709047157646104436925787906203 78086099418283717115484063285529991185968245642033269616046913143361289497918902 66529543612676178781350061388186278580463683134952478031143769334671973819513185 67840323124179540221830804587284461460025357757970282864402902440797789603454398 91633492226526120872I used Mathematica to find the smallest n value to have a 1000 digit result. Then created a file that listed n, n+1, n+2 etc. Then I used PFGW to check this file until it found a n*n+n+41 value that is probable prime. Then I used PRIMO to prove it prime.
4 May, 2004:
CYF NO. 27
(Solution by Jens Kruse Andersen dt 30052004 )
Found by brute force: 9654 * 871203 = 8410593762.
(Solution by Carlos Rivera dt 08052004 and dt 09052004 )
Largest is given by: 9654 * 871203 = 8410593762.
Smallest is given by: 258*3967041 = 1023496578.
2 March, 2004:
CYF NO. 26
(Comments by Mauro Fiorentini dt 05012015 )
If there is any solution other than the three given, it is greater than 10^{9}.
One can find all solutions of the equation sigma(n + k) = sigma(n) + k, with
n < 10^{9}(n composite) and k ≤ 1000 here.
There are very few solutions for some values of k, like 5, 9, 15, 22 and
several for others. I was not able to find any pattern, except that it
seems that there are several solutions when k has many prime factors.
1 February, 2004:
CYF NO. 25
(Solution by Jens Kruse Andersen dt 04022004 )
p^2 = 1000000000.....6894415289, where p = floor(sqrt(10^999))+1840. PrimeForm/GW found the prp and Marcel Martin's Primo proved it. q >2 leads either to a subtitanic or a larger titanic number for all p, prime or not.
(Contribution by Serge Varjabedian dt.15022004).
757^347=
11124235813....................9118170372432285293
find with maple (exactly 1000 digits).

Serge Varjabédian
MPSI Maths (DouaiFrance).
The solution by Serge Varjabédian is not the smallest.
(Contribution by Patrick De Geest dt.23022004).
I believe the answer is (provided I didn't make logical errors... one
can never be too sure) :
100000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000002352698256360096113843651
251602265497838583855923800847707180529498154124968245324183247629037882
419669801646574157531355230799308707840128542599383518090051340599701871
051055747308440431038616929092382828425074193057561525228479805678315972
351037463679215441144433168030762688030946289620535034965489071542184806
352455692246732254479733907881673231756768078938806533458512885000588456
228679628899546278750234188015303240943248595970592728105383326872555593
1369046915503043129202086339020390360187050814757537174426643961
with
p =
316227766016837933199889354443271853371955513932521682685750485279259443
863923822134424810837930029518734728415284005514854885603045388001469051
959670015390334492165717925994065915015347411333948412408531692957709047
157646104436925787906203780860994182837171154840632855299911859682456420
332696160469131433612894979189026652954361267617878135006138818627858046
368313495247803114376933467197381951318567840323124179540221830804587284
46146002535775797028286440290244079778960345440263627749086757651531
of length 500 (certified prime with PRIMO)
and
q = 2
making it a titanic square.
The solution by Patrick De Geest is not the smallest.
1 January, 2004:
CYF NO. 24
(Solution by Jens Kruse Andersen dt 02012004 and Brian Trial dt 12012004).
2^5 = 32, 3+2=5.
(Comments by Shyam Sunder Gupta dt. 31012004).
The one set of other higher values of m^n can be obtained from above solution i.e. 20^5, 200^5, 2000^5 etc. are all solutions.
The second interesting set of higher values are 2^70, 20^70, 200^70 etc.
Can you find any other set.
8 November, 2003:
CYF NO. 23
(Comments by Brian Trial dt. 04122003).
Interesting to note that since 2^n  1 is always a strong pseudoprime base 2 whenever n is a regular pseudoprime, and , 2^n1 is divisible by 3 if n is divisible by 2, then , 2^n1 will be a solution to CYF23 whenever n is an even pseudoprime. So, the lowest solution I've found so far would be: 2^161038  1.
(Solution by Shyam Sunder Gupta dt. 01012004).
The Smallest Strong Pseudoprime (base2) which is a multiple of 3 is 5455590801.
3 August, 2003:
CYF NO. 22
13 June, 2003:
CYF NO. 21
(Solution by Donovan Johnson dt 17102004 ).
I found 37 solutions to:
A prime p such that p + p1 is a cube. Where p1 is reverse of p.
I have checked all primes less than 1 billion. The cube for all solutions is 707^3.
These are:
203993941 + 149399302 = 353393243
205399741 + 147993502 = 353393243
205795741 + 147597502 = 353393243
205993741 + 147399502 = 353393243
207894541 + 145498702 = 353393243
208795441 + 144597802 = 353393243
213399931 + 139993312 = 353393243
214498831 + 138894412 = 353393243
214795831 + 138597412 = 353393243
214894831 + 138498412 = 353393243
217498531 + 135894712 = 353393243
217894531 + 135498712 = 353393243
218993431 + 134399812 = 353393243
219894331 + 133498912 = 353393243
223597921 + 129795322 = 353393243
224399821 + 128993422 = 353393243
225795721 + 127597522 = 353393243
227498521 + 125894722 = 353393243
227597521 + 125795722 = 353393243
228399421 + 124993822 = 353393243
229399321 + 123993922 = 353393243
229894321 + 123498922 = 353393243
229993321 + 123399922 = 353393243
233894911 + 119498332 = 353393243
234399811 + 118993432 = 353393243
236993611 + 116399632 = 353393243
237597511 + 115795732 = 353393243
238399411 + 114993832 = 353393243
239894311 + 113498932 = 353393243
239993311 + 113399932 = 353393243
243399901 + 109993342 = 353393243
243597901 + 109795342 = 353393243
244498801 + 108894442 = 353393243
245696701 + 107696542 = 353393243
246696601 + 106696642 = 353393243
247795501 + 105597742 = 353393243
248795401 + 104597842 = 353393243
(Solution by K D Bajpai dt 11022013 ).
I explored the problem further for all the primes lying in the range of 14000000th prime up to 17000000th prime, using Maple and found 82 solutions . The cube for all solutions is 1019^3.
These are here
(Solution by Shyam Sunder Gupta dt 14072013 ).
I extended the computations to 1000000000th prime and found additional solutions. Infact more than 130 more solutions for the same cube founded by K D Bajpai was noted. Also two more cube were found resulting in number of solutions. These cubes are 2321^3 and 4891^3.
21 May, 2003:
CYF NO. 20
(Comments by Jens Kruse Andersen dt. 15062003).
Regarding Brian's observations:
It is actually also true for 4's. 1's, 4's and 7's are just cases of a more
general result I found:
Theorem: Let x be any given ndigit number with leading 0's allowed. There is
exactly one ndigit number a, leading 0's allowed, such that a^3 ends in x.
Proof by induction on n.
Case n=1: x and a are digits. Inspect the 10 digits and check their cubes end
in distinct digits.
Case n+1: Let the given (n+1)digit x start with digit d and end with ndigit
y: x=d*10^n+y.
By the induction hypothesis, there is exactly one ndigit b such that b^3 ends
in y. Consider (n+1)digit numbers a(e)=e*10^n+b where e is a digit. The last
n digits of a(e)^3 are the same as for b^3, i.e. the ndigit y. According to
case n=1, e^3 can end in any digit and there is exactly one e such that e^3
ends in a given digit. And e^3 has to end in some given digit to make
(e*10^n+b)^3 end in x, which only requires that the digit before the last n
digits is d.
The theorem tells us that there are exactly 3^n ndigit numbers a, for which
a^3 ends in n digits taken from {1,4,7}.
This does not change the heuristics I made earlier and it still looks like
there are probably no solutions.
Another way to view the theorem: If a and c are distinct ndigit numbers then
the last n digits of a^3 and c^3 are not the same.
(Comments by Jens Kruse Andersen dt. 24052003).
There are no solutions n^3 for n <10^9. It is highly unlikely there are any
solutions at all.
Some crude heuristics for ddigit n:
There are around 10^d ddigit n and n^3 has around 3d digits. Let us assume
the digits are random. Each digit has 3/10 chance of being 1,4 or 7. The
chance all 3d digits are 1,4 or 7 is 0.3^(3d). The expected number of ddigit
solutions is then candidates*probability = 10^d*0.3^(3d) = 0.27^d. The total
expected number of solutions with more than 9 digits is the sum for d=10 to
infinity of 0.27^d ~= 1/350000. The chance of a solution seems to be around
1/350000.
With those odds, I will not search further.
To at least give some result, here is the first cube ending in seventeen 1,4 or 7:
1029036564^3 = 1089663539514111741717774144
The expected distance between such numbers is 1/(0.3^17) ~= 775*10^6 so the
heuristics seem plausible.
(Comments by Brian dt. 4062003).
I was unable to find a solution,
but here are some numbers that
come close:
1923^3 = 7111117467, all but 1 digit
36311371^3 = 47877111441171171117811, all but 2 digits
4928924264^3 = 119744737146141407441471711744, all but 4 digits.
One thing I find remarkable, and I don't know why this should
be, is for every integer n there exists a number with n digits
whose cube ends in a string of n 1's.
368288471^3 = 49953321 584048960 111111111
858716637368288471^3 = 633212722 189717735 780029541 304784194 111111111 111111111
172789893 835778279 858716637 368288471 =.... 111111111 111111111 111111111 111111111
This implies that it's possible (though unlikely!) a titanic number out there solves CYF #20.
This seems to be true of 7's as well, though I haven't investigated them
as far.
And it does NOT seem to be true of 4's, which also makes me wonder
why.
29 March, 2003:
CYF NO. 19
(Solved by Brian), Ferndale, Michigan U.S.A. (email dated 9052003)
Brian Trial says in his (email dated
9052003) that "5657 ^ 3 = 5399 ^ 3 + 2731 ^ 3 + 1487 ^ 3
For digits 0  9, the lowest solution I found is
14083 ^ 3 = 13259 ^ 3 + 7129 ^ 3 + 4639 ^ 3
"
1 March, 2003:
CYF NO. 18
(Comments by Brian dt. 29032003).
(Comments by Shyam Sunder Gupta dt. 29032003).
The online integer sequences A000954 and A056636 may not give the solutions. The question asked contain the condition of distinct primes which is not considered in the above sequences. The best known solution are 38 (31+7, as 19+19 is not to be considered as these are no distinct primes)and 68 (7+61, 31+37).
1 February, 2003:
CYF NO. 17
(Comments by Brian Trial vide his email dt. 09042003 ).
Brian Trial says in his email that " 1117111 is the 86969th prime. 86969 is also prime but alas,1 away from being palindromic! ."
1 January, 2003:
CYF NO. 16
(Solved by Jan Munch Pedersen vide his email dt. 2012003 ).
Jan Munch Pedersen says in his email that "
For CYF 16 my best guess for smallest prime p >11
with no solutions to sigma(x)=2^p is p=487.
Let us look at the more general equation sigma(x)=2^n.
If 2^p1 is prime, then sigma(2^p1)=2^p, so, Mersenne
primes are solutions. If N=(2^p11)*(2^p21)*...*(2^pm1)
is a product of distinct Mersenne primes, then
sigma(N)=2^p1*2^p2*..*2^pm=2^(p1+p2+...+pm) so these numbers
are also solutions.
I am pretty sure there are no other solutions to sigma(x)=2^n
but I don't have the skills to prove it. It is suffiecient
to prove that any solution x is squarefree.
I wrote a small program that for each 2 <= n <= 1000 tried to
write n as a sum of distinct powers of Mersenne primes. It
found that the following n's doesn't have any solutions:
4, 6, 11, 470, 475, 477, 480, 482, 483, 484, 485, 486, 487, 488, 489, 490,
491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505,
506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520,
522, 525, 527, 532
With my assumption this gives that the next prime solution
after 11 is 487.
(Solved by David Terr vide his email dt. 3012003 ).
David Terr says in his email that" If sigma(k)=2^n then k must be a product of distinct Mersenne primes of the form 2^p_i1 with Sum(p_i)=n. There are no numbers k with sigma(k)=2^n for n between 482 and 520 since the sum of the Mersenne prime exponents up to 127 is 481 and 521 is the next Mersenne prime exponent. Since 487 is a prime in this range, there is no k with sigma(k)=2^487. The same holds for 491, 499, 503, 509, and all primes between 704339 and 756838.
1 December, 2002:
CYF NO. 15
(Solved by Brian Trial vide his email dt. 03122002 ).
Brian Trial says in his email that"
I have to thank Christian Zeller, who in 1883 published this concise
formula giving the day of the week for any given month, day, and year.
My "C" implementation of his formula and the answer to this problem is
below.
I got an answer of 172."
int Zeller(int month, int day, int year)
{
int year_hi = year / 100;
int year_lo = year % 100;
/* Reverend Zeller's formula */
int DayOfTheWeek = day + (int)(((13 * month)  1) / 5) + year_lo +
(int)(year_lo / 4) + (int)(year_hi / 4)  year_hi * 2;
while ( DayOfTheWeek < 0 )
{
DayOfTheWeek +=7;
}
DayOfTheWeek %= 7;
return DayOfTheWeek;
}
void main(void)
{
int month, year;
int count = 0;
/* January, 2000 is considered 11th month of 1999 under Zeller's
formula, February is considered 12th month. */
/* Sunday = 0, Friday = 5 */
if ( Zeller(11,13,1999) == 5) count++;
if ( Zeller(12,13,1999) == 5) count++;
for (year = 2000; year ≤ 2098; year++)
for (month = 1; month ≤ 12; month++)
if ( Zeller(month,13,year) == 5)
count++;
/* Go through March  December of year 2099 */
for (month = 1; month ≤ 10; month++)
if ( Zeller(month,13,2099) == 5)
count++;
printf("%d \n",count);
}
(Solved by Robin Suri vide his email dt. 04122002 ).
Robin Suri says in his email that" there are 172 fridays in this century which fall on 13th."
(Comments by Mauro Fiorentini dt 05012015 )
For a given a random 13th, it is more likey to be a Friday than any other day.
In 400 years (a complete cycle of leap years) there are:
685 Mondays,
685 Tuesdays,
687 Wednesdays,
684 Thursdays,
688 Fridays,
684 Saturdays,
and 687 Sundays.
As far as I know, B. H. Brown was the first to notice this (American Mathematical Monthly, n. 40, 1933),
but is is quite likely that somebody else noticed it before.
11 November, 2002:
CYF NO. 14
(Solved by Jens Kruse Andersen vide his email dt. 12112002 ).
Jens Kruse Andersen says in his email that " The smallest values of prime p and q such that the sum of digits of p^q is p. If p+q must be minimized then it is trivial. 30 seconds with a pocket calculator gave: 17^3 = 4913. I have not searched exhaustively for solutions where one of p and q is minimized. However it looks like they are both minimized in 17^3."
(Solved by Brian Trial vide his email dt. 19112002 ).
Brian Trial says in his email that" It seems likely that 17^3 is lowest solution.In fact, I conjecture that there exists a value r such that if q > r p^q is ALWAYS pandigital.." Can anyone Prove this conjecture?
(Solved by Robin Suri vide his email dt. 29112002 ).
Robin Suri says in his email that" the smallest pair of primes 'p' and 'q' is p=17 and q=3."
5 October, 2002:
CYF NO. 13
(Comments by adam stinchcombe vide his email dt. 05052003 ).
adam stinchcombe says in his email that " I have obtained: A slight modification in this polynomial yields a slightly higher count of primes. In searching for primes of the form ax^2+bx+c, it is good to put the axis of symmetery at b/2a=50000.5 so that when x yields a prime value then 100001x also yields a prime value (what we Americans call a "twofer," for two for one sales). The general principle is sound but one should realize that, once a candidate formula is found, the value at x=1 might be composite while the value at x=100001 might be prime, so that shifting x by 1 would up the prime count. In this fashion I identified a shift of 5357 giving 16 more primes. The polynomial x^289287*x+1991687729 gives 44500 Prp on the range x=1..100000. "
(Comments by Brian Trial vide his email dt. 02112002 ).
Brian Trial says in his email that " x^2  99999x + 1116427201 generates 44285 primes for x = 1 to 100,000 ."
(Comments by Jens Kruse Andersen vide his email dt. 17102002 ).
Jens Kruse Andersen says in his email that " I have found primes for 44484 values of x in x^2100001x+2498695637. I chose A=1 to limit the numbers and B=100001 to make the polynomial symmetric around the middle of the interval: 50000.5 (= B/2A). I tried odd C between 10^10 and 10^10. For each C, I used modulo arithmetic to test whether at least 85% of the numbers would have no prime factor below 250. In these cases I counted the number of prp's. This is far from an exhaustive search for any of the coefficients. Actually the C value with the most primes only just cleared the first test with 85.16%. The runner up was C=1116527201 with 88.91% and 44284 primes. I never got above 91%. I could use the chinese remainder theorem to get 100% but C would become very big and primes less likely, even completely without prime factors below 250. There is still a long way to 50000 primes."
21 September, 2002:
CYF NO. 12
(Solved by Jens Kruse Andersen vide his email dt. 25092002 ).
Jens Kruse Andersen says in his emails that " I have found the smallest nontrivial solution in the first second of my search run: A=90 and n=19.We can see that 90^19, 90^20, 90^21 and 90^22 all have digitsum 90. Note digitsum(90^x) = digitsum(9^x). Later I have checked there are no solutions for 1 < A < 90 and n ≤9996. I have not proved there are no solutions for n < 9996 but it seems extremely unlikely.".
(Comments by Brian Trial vide his email dt. 23112002 ).
Brian Trial says in his email that " Jen's Kruse Andersen's solution may be unique. Checking A < 8000, limiting to numbers of 2000 digits or less, revealed no more solutions, and only 1 solution where only SOD( A ^ n ) = SOD( A ^ ( n + 1 ) ) = SOD( A ^ ( n + 2 ) ) = A SOD( 181 ^ 18) = SOD( 181 ^ 19 ) = SOD( 181 ^ 20 ) = 181 I found an interesting number in my search, A, with exponents m > 1 , n > 1, such that SOD ( A ^ n ) = SOD ( A ^ m ) = A, and SOD ( (A+1) ^ (n+1) ) = SOD( (A+1) ^ (m+1) ) = (A+1). Can anyone find it?"
6 September, 2002:
CYF NO. 11
(Comments by Patrick De Geest vide his email dt. 17092002 & 18092002).
Patrick De Geest says in his email that "I ran Ubasic and soon my program gave the following smallest Titanic EVEN Semiprime. 10^999+5026 = (2) * (5*10^998 +2513). In his subsequent email , he says that the smallest titanic semiprime MIGHT be 10^999 +139 , provided I can 'exclude' the following two numbers: 10^999 + 13 and 10^999 + 69, Both numbers have smaller factor larger than 10^8. I guess that this is the hard part of the puzzle".
(Comments by Jens Kruse Andersen vide his email dt. 18092002 ).
Jens Kruse Andersen says in his emails that " The smallest titanic semiprime is either 10^999+13 or 10^999+139. 10^999+139 is a proven semiprime with prime factor 31 and cofactor (10^999+139)/31 proved prime with Primo and validated by Cert_Val. 10^999+13 is composite but I have found no factors and don't know how many there are. After running GMPECM for several hours, the smallest factor is probably at least 25 digits and hard to find. I have not heard of any semiprime test without a known factor. 10^999+7 is prime. According to Primeform/GW all remaining numbers below 10^999+139 have a relatively small prime factor with a composite cofactor. The only numbers without a factor below 500000 are 10^999+19 (factor: 2698217) and 10^999+69 (factor: 3067890809). Of course the question remains: Can anyone determine whether 10^999+13 is a semiprime?
(Comments by Paul Zimmermann vide his email dt. 12112002 ).
Paul Zimmermann says in his emails that " I performed 1800 ecm curves with 1st stage bound B1=1000000 with 10^999+13 without any success. So it seems that its smallest prime factor is larger than 35 digits. I also tried a large P1 run and it failed too."
(Comments by David Harrison vide his email dt. 20052010 ).
David Harrison says in his email that "I have put some effort into attempting to factor 10^999+13 using the ECM for Can You Find problem number 11.
Using Prime95 for stage 1, and GMPECM for stage 2, and the optimal parameters specified here:
http://www.loria.fr/~zimmerma/records/ecm/params.htm
The following work has been done:
2,351 curves at B1=3M, B2=6,610,263,676
4,482 curves at B1=11M, B2=36,007,711,756
7,557 curves at B1=43M, B2=198,654,756,318
I have heard that the exact value of B2 doesn't really matter, just so long as it's in the right general area to
match the number of curves run. Nevertheless I include it for completions sake, and also because I happen to have it.
As a complete shot in the dark, I also ran 4 curves with each of these bounds (16 curves in total):
B1=110M, B2=970,090,831,726
B1=260M, B2=3,134,108,727,766
B1=850M, B2=15,793,458,981,318
B1=2600M, B2=86,489,653,750,516
Sadly, and as you may have guessed, no factor was found during all this which means that all of its factors are most
likely over 50 digits. I wish good luck to whomever decides to take this number out for the next few thousand curves".
(Update by David Harrison vide his email dt. 15022011 ).
David Harrison says in his email that just to let you know I decided to do some more work on it. So far I have done 5400 curves with B1=26e7 and B2~=3e12.
(Solution by David Harrison vide his email dt. 05102014 ).
It is now solved! After 7000 ECM curves at B1=2.9e9, I have factored 10^{999} + 13 into a p65 and c935, therefore it is NOT a semiprime.
10^{999} + 13 = 26757495525262452858412972724728927157469061956724253045372193707 * c935
Here is its factordb page: http://factordb.com/index.php?query=10^999%2B13
Since there are no more remaining 1000 digit unfactored numbers below 10^{999} + 139,
therefore 10^{999} + 139 is proven to be the smallest titanic semiprime!
14 April, 2002:
CYF NO. 10
(Comments by Brian Trial vide his email dt. 14062002 and dt. 25062002).
Brian Trial says in his emails that " The smallest number not divisible by 10 expressible in 3 ways is 1115892288 and in 4 ways is 104645899008. I have not yet found one expressible in 5 ways. However the smallest number expressible in 6 ways is 11158922880 and in 8 ways is 1046458990080. The fact that they are divisible by 10 can double the number of ways expressible.".
(Solution by Brian Trial vide his email dt. 04102002 ).
Brian Trial says in his email that " EP0RN degree 5:
10119126106147652568 =
2478732804 * 4082378742 =
2278590444 * 4440958722 =
2071654884 * 4884561702 =
1257864408 * 8044687521 =
1143628488 * 8848263411 =
2^3 * 3^4 * 13 * 17^4 * 53 * 61 * 67^2 * 911
This took several days on a Pentium 4, 1.8 GHz. On my request for confirmation that the above is the smallest such number, Brian Trial says in his subsequent email that "It is possible that this is not the smallest solution.I used a couple of short cuts to speed my search through more likely candidates, but an unlikely candidate might have slipped through my net."
Can anyone confirm that 10119126106147652568 is a smallest EP0RN of degree 5?
10 March, 2002:
CYF NO. 9
(Comments by Norman Luhn vide his email dt. 11032002).
Norman Luhn says in his email that "I have tested all Fibonacci numbers up to 30000 digits. I mean , your problem is perhaps unsoluble.I be looking forward to solve it !"
(Comments by Brian Trial vide his email dt. 12032002 and dt. 13032002).
Brian Trial says in his emails that " if my program is free of serious bugs then I have checked up to F18733167( a number containing 3915000 digits) and I think it is nearly certain that CYF 9 has no solutions > 233. Based on his computations ( 36 hours on a Pentium 4 1.8 KHz), Brian Trial also conjectures that " All fibonacci numbers > F285 are Pandigital i.e. containing all digits from 0 to 9 ." This is really Great. Lets call this Brian's conjecture.
Can you prove or disprove Brian's Conjecture?.
Other Observations by Brian Trial: The largest number I can find where not all the digits are represented is F285,where no 5 exists among it's 60 digits. Before this is F258, 54 digits with no 6, and F224, 47 digits with no 7. The first number where all 10 digits are represented is F61.
We can see that
F61 = 2504730781961
F224 = 29090180355503362256910111038089984964854261893
F258 = 370959230771131880927453318055001997489772178180790104
F285 = 162926777992448823780908130212788963731840407743629812913410
F286 = 263621064469290555679241849789653324393054271110084140201023
(Comments by Mauro Fiorentini 08052021 )
I noticed an argument, just of theoretical interest, about Brian's conjecture (CFY 9): if it is true, it is very likely to be computable: remainders of Fibonacci numbers modulo 10^{n} form a cycle of length 15 * 10^{(n1)}, so if for some n we can show that the last n digits are pandigital for all modules in the cycle, the conjecture is proved. Unfortunately n is likely to be quite large (I think about 60), so actual computation is out of question, unless one can find a clever trick.
10 February, 2002:
CYF NO. 8
(Solved by Brian Trial, Ferndale, Michigan U.S.A.). He found not one but two solutions (email dated 8032002)
Brian Trial says in his (email dated
8032002) that " Not one but two solutions: With my revised program I'm pretty confident these are the smallest solutions too.
97524222465 = 3 * 5 * 42751 * 152081
97524222466 = 2 * 11 * 19 *
233311537
97524222467 = 7 * 29 * 149 * 3224261
97524222468 = 2 * 2 * 3 *
8127018539
97524222469 = 73 * 73 * 251 * 72911
97524222470 = 2 * 5 * 67
* 145558541
97524222471 = 3 * 3 * 13 * 833540363
97524222472 = 2 * 2 * 2
* 12190527809
97524222473 = 17 * 17 * 3499 * 96443
97524222474 = 2 * 3 *
7 * 2322005297
97524222475 = 5 * 5 * 18493 * 210943
97524222476 = 2 * 2
* 23029 * 1058711
97524222477 = 3 * 11 * 18457 * 160117
97524222478 = 2
* 23 * 151 * 14040343
97524222479 = 47 * 181 * 181 * 63337
212220020305
 212220020319 is the second solution "
(Comments by Prof. François Bergeron vide his email dt.21102002).
Prof. François Bergeron from Canada says in his email that " I have since learned that people have tried to look for k successive kprimes, but this is asking for the maximum possible sequence of kprimes. I was happy to learn of the sequence found by Brian Trial for 15 consecutive 4primes. The next problem in this line is thus to find 31 consecutive 5primes."
Fifteen years ago, when I turned 35, I observed that for three
consecutive years my age had been the product of two primes. At the time, I was led to
consider the problem below.
" Is it always possible two find 2^k1 consecutive integers that have exactly
k prime factors, taking into account their multiplicities? "
More precisely,for an integer $n$ we set
Omega(n):=a_1+a_2+...+a_s,
whenever the decomposition of $n$, into distinct prime factors, is
n = p_1^{a_1}p_2^{a_2} ... p_s^{a_s}$
The problem is to find 2^k$ consecutive integers (arranged here around
their mean n):
na,... ,,n1,n,n+1,... , n+a, (with a=2^(k1)1)
for each of which Omega takes the value k.
It is easy to see that this is best possible, since any list of 2^k
consecutive integers contains one integer that is divisible by 2^k. Except
for the special case 2^k (easily dealt with), this particular member of the
list is clearly divisible by some other integer, hence its Omega value is
larger thank.
A little more thought leads to the observation that the central value n
(the
mean) of this list of 2^k1 integers should be of the form 2^(k1) p for
some prime p. In this case, we say that p is knice. It is easy to see that
if p is knice, it is automatically jnice for all j less or equal to k.
This can be used to look for solutions in small cases.
An ancillary question is to characterize the density distribution of knice
primes. Among the first 10,000 primes, there are 679 primes that are 2nice,
3 that are 3nice (the smallest being 52919), and none that are 4nice.
Apparently, 12190527809 (Brian Trial, Ferndale, Michigan, see
http://www.shyamsundergupta.com/canyoufind.htm) is the smallest 4nice
prime.
Francois Bergeron
Professor, Departemnt de Mathematics,
Universite du Quebec a Montreal,
C.P. 8888, Succursale CentreVille,
Montreal, H3C 3P8,
Canada.
20 January, 2002:
CYF NO. 7
(Solved by Brian Trial, Ferndale, Michigan
U.S.A.). He found that 3 consecutive abundant numbers with 2 of them odd are
27523728059933744479478128774219545978059153664131775,
27523728059933744479478128774219545978059153664131776,
27523728059933744479478128774219545978059153664131777 (email
dated 14022002)
Brian Trial explains in his (email dated 15022002) that "I realized I could find a solution by choosing 3 abundant numbers with no common factors, a0, a1, a2, then solving the simultaneous linear Diophantine equations. a1 * x1  a0 * x0 = 1 , a2 * x2  a1 * x1 = 1. I figured it might be easier to choose a0 and a1 very small, and let a2 be large. I originally chose a0 = 945 (the smallest odd abundant) and a1 = 2^3 * 11 (smallest even abundant relatively prime to a0) but this made a2 disastrously large. As a compromise I chose a0 = 3^5 * 5^2 * 13 = 78975 , a2 = 7^2 * 11^2 * (product of primes 17 ... 97), a1 = 2^6 * 101 = 6464, Solving 6464 * x1  78975 * x0 = 1 was relatively easy, yielding a1 = 6464 * (78975t + 25034). The second equation became a2 * x2  6464 * (78975t + 25034) = 1 , a2 * x2  510493300t = 161819777 , And the smallest solution of x2 = (510494400  136838699) * 161819777 = 373655701 * 161819777 , So the solutions are a2 * x2, a2 * x2  1, a2 * x2  2 "
(Comments by Mauro Fiorentini dt 05012015 and 04052021 )
For CFY 7 the best answer so far is contained in the answer to CFY 36:
141363708067871564084949719820472453374 + 1, but if you ask for a
sequence of exactly 3 consecutive abundant number, first being odd, then
27523728059933744479478128774219545978059153664131775 is not an answer,
as 27523728059933744479478128774219545978059153664131775 is the start of 4 consecutive abundant numbers
It is possible to find a solution, but it is for sure outrageously
large, as the two odd numbers cannot be multiple of 3. We do not know sequences
of exactly 3 consecutive abundant numbers starting with an odd
number; if they exist, their numbers are greater than 10 ^{19}. Refer Abundant Numbers by Mauro Fiorentini
6 January, 2002:
CYF NO. 6
(Comments by Jan Munch Pedersen. vide his email dt. 08012002).
Jan Munch Pedersen says in his email that "I have scanned the 2,500,000+ amaciable pairs in my database and I found none with one or both members a palindrome, so, it would be interesting just to find such a pair. Since all pairs with smaller member below 10^12 are already known we are looking for a palindrome with at least 13 digits."
31 December, 2001:
CYF NO. 5
(Solved by Jan Munch Pedersen. ). He found that 6789783 is the smallest Odd Abundant Number not ending in 5 and also not divisible by 9. (email dated 03012002)
Jan Munch Pedersen says in his email that "6789783 is the answer to CYF #5. I got the result two ways. It is easily seen that the number must be divisble by 3 (but as required not by 3^2). So the question can be reformulated to find the smallest a coprime to 30 with (4*sigma(a))/(3*a)>2, which is the same as sigma(a)>(3/2)*a. I had the answer to this (7^2*11*13*17*19) in a table already. To be absolutely sure I wrote a small program that confirmed 6789783 as the result in a few minutes."
23 December, 2001:
CYF NO. 4
(Solved by Carlos Rivera ). He found that 9876542103 is the largest ten digit Pandigital Smith number. (email dated 14022002)
Carlos Rivera says in his email that "The answer can be found extremely easily using the following method that may be interesting. Starting from the largest pandigital 987643210 and diminishing 9 recursively. This way generate all the pandigitals required, and other no pandigital ofcourse. The first pandigital that is also a Smith one comes very very fast( at the step 123): 9876542103 = 9876543210  123*9 = 3*3*53*20705539 and (9+8+7+6+5+4+3+2+1+0) = 45 = (3+3+5+3+2+0+7+0+5+5+3+9)."
17 December, 2001:
CYF NO. 3
(Solved by Patrick De Geest ). He found that n = 3088803 is the smallest multidigit Palindromic number such that 10^{999} + n is prime. (email dated 22122001)
9 December, 2001:
CYF NO. 2
(Comments by Patrick De Geest vide his email dt. 03022002).
Patrick De Geest says in his email that "the smallest palindrome is larger than 4294994924."
(Comments by Shyam Sunder Gupta dt. 10032002).
Smallest palindromic StrongPseudoprime(base2) is larger than 10^18."
30 November, 2001:
CYF NO. 1
(Comments by Jens Kruse Andersen vide his email dt. 30052004).
Jens Kruse Andersen says in his email that "Let k = floor(sqrt(10^999/72))+485639. Then the 2psp by Phil Carmody is (6k+1)(12k+1). I rediscovered it without knowing Carmody must have used the same method. Giving the formula will prevent others from spending time rediscovering this particular psp (which is almost guaranteed nonminimal)."
(Comments by Phil Carmody vide his email dt. 06052003).
Phil Carmody says in his email that "I bet you thought CYF #1 was put to bed! However a good puzzle never dies,
so here's my contribution to it:
(03:04) gp > qqq
%35 =
100000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000026062130069041934295325878339598116808871909955708929
075268462986657243448273375538862028079314199309373447323355405595025708250125
575976835479977670790297422796447379915957169467638392937427217997348488396724
607196777440812487844011103187290077356938929525351851634478363859615268867388
857507487090191278861574111603012803523287643459926969940981392939144553212883
952371848826529730771243099461724655482124452159329498033661766893511540808198
3665080675821437094604032791536044323044793660282591888101871081
(03:04) gp > Mod(2,qqq)^(qqq1)
%36 = Mod(1,
100000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000026062130069041934295325878339598116808871909955708929
075268462986657243448273375538862028079314199309373447323355405595025708250125
575976835479977670790297422796447379915957169467638392937427217997348488396724
607196777440812487844011103187290077356938929525351851634478363859615268867388
857507487090191278861574111603012803523287643459926969940981392939144553212883
952371848826529730771243099461724655482124452159329498033661766893511540808198
3665080675821437094604032791536044323044793660282591888101871081)
Not a 2SPSP, but still a 2PSP, as requested.
"
(Comments by Patrick De Geest vide his email dt. 03022002).
Patrick De Geest says in his email that "A hard puzzle it is indeed. With Ubasic I prepare a list of titanic pseudoprimes to base 2 and then I check with PRIMO is they are prime. Up till now all candidates turned out to be prime so the search continues.10^999 + d = prime for d =7, 663, 2121, 2593, 3561, 4717, 5863, 9459 and 11239."
(Comments by Shyam Sunder Gupta dt. 10022002).
The upper limit of the solution is (2^33261)/3 , as this the Titanic Pseudoprime (base2) consisting of 1001 digits. So if solution smaller than this is not found than (2^3326 1)/3 is the answer."
(Comments by Carlos Rivera vide his email dt.10022002).
Carlos rivera says in his email that " I have found something interesting about the question " What is the probability that a composite number of K digits is pseudoprime base 2? " and then something interesting for the puzzle CYF1 in the Shyam pages. In the page 28 of the R.K.Guy well known book (UPiNT, Vol.1, 2nd ed.) we may read that the quantity of numbers pseudoprimes base 2 less than x, P2(x) is bound by the following formulas due to C. Pomerance: e^[ln x^(5/14)] < P2(x)< x.e^(ln x.lnlnln x)/2lnln x).(Pomerance) has a heuristic argument that the true estimate is the upper bound with the 2 ommited . This (an estimative formula for the population of pseudoprimes base 2) is what we need to estimate the probability that a random number of K digits is psp(2).According with the upper bound the probability that a (any) random number is psp(2) is approximately P2(2)/x, or e^(ln x.lnlnln x)/lnln x) (following the indication of Pomerance about eliminate the 2 of the upper bound). Consequently the average quantity of numbers that need to be tested before we get ONE SINGLE psp(2) is 1/e^(ln x.lnlnln x)/lnln x) or e^(ln x.lnlnln x)/lnln x). Well, when the number we are looking for in CYF1 is of 1000 digits we may equate x=10^999. For this x then e^(ln x.lnlnln x)/lnln x) approximately equal to 1.3x10^264. Consequently this target (CYF1) is nowadays a target unaffordable IF the the Pomerance formulas are right. Even numbers really smaller that 1000 digits are practically & nowdays unaffordable (i.e. 50 digits, as you can verify calculating e^(ln x.lnlnln x)/lnln x) for x=10^49."
(Comments by Carlos Rivera vide his email dt.12022002).
Carlos rivera says in his email that "I have produced the following smaller composite 1000 digits psp(2) than yours: 2^3319 1 . Needles to say this hardly can be the smallest titanic, but a smaller upper limit, without any doubt. I guess I can find smaller candidates of that kind."
(Comments by Carlos Rivera vide his email dt.13022002).
Carlos rivera says in his email that " here is an another interesting titanic psp(2) number with 1002 digits; F8*F10*F11, where Fn = 2^(2^n)+1. I calculated it using a theorem of Cipolla found in the A12 of R. K.Guy, p.28."
(Comments by Gerbicz Robert vide his email dt.08072002).
Gerbicz Robert from Hungary says in his email that I have found a smaller composite Titanic pseudoprime (base 2) than 2^33191:
N =
100000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000000000000000000000000000000000000000000000000000
000000000000185017449758229220443851253776129894319687315600262882074787718636
495656409455047338875076375761449987618955122262411693001773217422124857363682
726860768539754082900336842988430308161524666290049940172774507740995673691808
603018327385739947515790358472902692773403047880640767316143525884645758947227
727920415160370563059844256424954764342100369939756013445112654207275438518088
669787306043735471206941348415357323391002320332526716198901659328563926038635
757430581534458759023575114811443901333697210682885124397427691209420775575137
016625535158009885991758112475590076207503610014087176295675374669798201512953
5418285896411998662217493237822384486461302995204680976314083569.
N is also a Carmichael number. Every Carmichael number is odd it implies that every Carmichael number is also a pseudoprime in base 2. If N=P*Q*R where P,Q and R are prime numbers and P=6*k+1,Q=12*k+1,R=18*k+1 then N is a Carmichael number(k is an integer). Using this fact I have searched the smallest Carmichael number of this kind: N>10^999,N=(6*k+1)*(12*k+1)*(18*k+1)>6*k*12*k*18*k=1296*k^3, so it includes that k>integer part of(cube root of 10^999/1296) this number is: C=int(10^333/(1296^(1/3)))=
91720201358184074552072498678030137690191240119958148249255(use ubasic with word 100 and point 100 to calculate this integer) We can found that if D=C+56566126 then P=6*D+1,Q=12*D+1,R=18*D+1 are prime numbers,so N=(6*D+1)*(12*D+1)*(18*D+1) is a Carmichael number.It is not need to check that P,Q,R are primes because it is easy to verify that N is divisible by 6*D+1 so N is composite and it is easy to see that 2^(N1)=1 mod N so N is a pseudoprime in base 2.
(Visit Patrick De Geest for smallest n digit length pseudoprimes(base2) "
Can You reduce the upper limit of the smallest titanic Pseudoprime(base2)?.

email your answer