Problems & Puzzles: Puzzles

 Puzzle 645. Generalized Dudeney Numbers Perhaps you know already what a Dudeney number is. If not, see this reference.In short, a Dudeney number is any number n^3 such that SOD(n^3)=n; SOD means here "sum of digits". Then it's natural generalize the concept: A Generalized Dudeney Number is any number n^m such that SOD(n^m)=n. Steffen Jakob has the record: SOD(547210^25662) = 547210 where 547210^25662 is a number of 147253 digits. Q1. Perhaps you would like to improve such record (more digits). Two Sunday additions: Q2. Perhaps you will success beating the record using a prime base n?Q3. Perhaps you will find useful to redefine the n-searching space for each exponent m, as was done here.

Contributions came from Steffen Jakob, Jan van Delden, Giovanni Resta.

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Steffen wrote:

1001983^37099 is a new prime based record (222626 digits). :-)

I just beat my old hiscore. The new largest number is 653230^30192
(175569 digits). I already updated my web page.

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Jan wrote:

Q1: Large solutions can be found with less effort if one uses: DigitSum(v^p)=DigitSum((v*10^k)^p)=v*10^k.

Instead of trying the powers of v*10^k the numbers can be kept relatively small by searching for v*10^k in the digit sum of v^p.
I used v=9 because we always have DigitSum(9^p)=0 mod 9, which hopefully increases our luck.

Solution: 900000^209412 Digits 1246890

Q3: If we assume that the average digit is 4.5 (the digits are uniformly distributed over 0..9), and the digit sum equals v, we should require v/4.5 digits approximately.
The power required can be approximated by using 9^p=10^(v/4.5), i.e. p*log[10](9)=v/4.5 hence p=v/(4.5*log[10](9)). If we substitute v=9*10^5 we get p about 209590.
So the given estimate is not bad at all.

Or more sophisticated: 9^p=4.5*[10^(v/4.5)-1]/9 assuming all those digits are 4.5 (which is not possible in real life..), which generates essentially the same p.

The real trick is to calculate an interval in which p probably lies. Each digit can be described as drawn from a uniform distribution (*) with mean 4.5 and variance 99/12.
If we draw n digits the average value of a digit will follow a normal distribution with mean 4.5 and variance 99/(12*n), if n is large enough. The variance decreases rapidly with n.
If n is approx 200000 this would give us an 95% confidence interval for the average digit [4.475,4.525]. Which would give an estimate for p: [208432,210761] <skewed>.

(*) This might not be true, because the digits are not drawn independently, but calculated using v^p. However the predictions for v=9*10^k, for k<=5, seem to work fine...

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Resta wrote:

My best result for puzzle 645 is
35900000^3122353 which has 23589672 digits whose sum is 35900000.

I also searched for examples of n^e with d digits such
that n,e, and d are prime numbers.

One such example is  5059937^167677 with 1124131 digits.

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