Herman's Happy Birthday

[BMAD]

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?

[Barcallica]

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

[Barcallica]

N= 13860, Herman turning 144?? Maybe I understood wrong

[dark_magician_92]

135?

[dark_magician_92]

17640

[BMAD]

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

[dark_magician_92]

was the answer i posted, wrong? Age - 135, number-17640

[BMAD]

was the answer i posted, wrong? Age - 135, number-17640

yes. it was incorrect.

[bonanova]

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 = Sum₁^r (p_i^e_i) where p_i are primes and e_i are exponents,

then f = Prod₁^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 = Sum₁^r (p_i^x_i) 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)