BMAD 65 Posted June 1, 2013 Report Share Posted June 1, 2013 Herman knows how old he is turning this birthday; you don't. He is as many years old as the largest number of divisors of any integer N less than or equal to 20,000. How old is Herman turning, and what's the smallest such N? Quote Link to post Share on other sites

0 Solution Barcallica 2 Posted June 4, 2013 Solution Report Share Posted June 4, 2013 for some reason my factors is half of yours but even still i am finding a higher divisible number N= 13860, Herman turning 144?? Maybe I understood wrong N=15120 and he is turning 80. seems more realistic. in my first answer it was 72 not 144. my mistake Quote Link to post Share on other sites

0 Barcallica 2 Posted June 2, 2013 Report Share Posted June 2, 2013 N= 13860, Herman turning 144?? Maybe I understood wrong Quote Link to post Share on other sites

0 dark_magician_92 4 Posted June 3, 2013 Report Share Posted June 3, 2013 135? Quote Link to post Share on other sites

0 dark_magician_92 4 Posted June 3, 2013 Report Share Posted June 3, 2013 17640 Quote Link to post Share on other sites

0 BMAD 65 Posted June 3, 2013 Author Report Share Posted June 3, 2013 for some reason my factors is half of yours but even still i am finding a higher divisible number N= 13860, Herman turning 144?? Maybe I understood wrong Quote Link to post Share on other sites

0 dark_magician_92 4 Posted June 5, 2013 Report Share Posted June 5, 2013 was the answer i posted, wrong? Age - 135, number-17640 Quote Link to post Share on other sites

0 BMAD 65 Posted June 5, 2013 Author Report Share Posted June 5, 2013 was the answer i posted, wrong? Age - 135, number-17640 yes. it was incorrect. Quote Link to post Share on other sites

0 bonanova 85 Posted June 5, 2013 Report Share Posted June 5, 2013 N=15120 and he is turning 80. seems more realistic. in my first answer it was 72 not 144. my mistake I agree with this solution. Nice puzzle. if n = Sum_{1}^{r} (p_{i}^{ei}) where p_{i} are primes and e_{i} are exponents, then f = Prod_{1}^{r} (e_{i}+1) gives the number of factors of n and is maximized when the e_{i} are largest. Thus, use the smallest r primes and ensure that e_{i} are non-increasing. For any number N, we can write N = Sum_{1}^{r} (p_{i}^{xi}) where x_{i} assume real values. x_{i} = [log N + Sum(log p_{i})]/[r + log p_{i}] - 1. Use integer e_{i} that are close to the real x_{i} to maximize factors of n < N. This makes the search lightning fast, doable by hand. Here are the calculated x_{i} and the best results for two, three and four primes: Primes x_{i} e_{i} n (factors) 2 3 5 7 11 13 > 3.86 2.07 1.09 .73 .40 .31 2 3 5 7 11 > 4.09 2.21 1.19 .81 .47 2 3 5 7 > 4.50 2.47 1.37 .96 4 3 1 1 gives 15120 (80) 2 3 5 > 5.40 3.04 1.76 7 3 1 gives 17280 (64) 2 3 > 7.44 4.32 9 3 gives 13824 (40) also 7 4 gives 10368 (40) Quote Link to post Share on other sites

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