Divisors of 10^n- Product of all elements of sets {2^0,2^1…2^n} {5^0,5^1..5^n}. Sum of base 10 logs of these elements = log 2^{[n(n+1)/2]*n+1} * 5^{[n(n+1)/2]*n+1} =log 10^{[n(n+1)/2]*n+1} ={[n(n+1)/2]*n+1} = 792. Substituting options, we get n=11. Not very sure. There must be some better way.

Divisors of 10^n- Product of all elements of sets {2^0,2^1…2^n} {5^0,5^1..5^n}. Sum of base 10 logs of these elements = log 2^{[n(n+1)/2]*n+1} * 5^{[n(n+1)/2]*n+1} =log 10^{[n(n+1)/2]*n+1} ={[n(n+1)/2]*n+1} = 792. Substituting options, we get n=11. Not very sure. There must be some better way.

riyaziqSeptember 29, 2008 at 6:10 pm

absolutely correct !

outtimedSeptember 29, 2008 at 11:16 pm

I’m not satisfied with the solution.Took some time.Is there any better method?

riyaziqSeptember 30, 2008 at 2:17 am

its easy

keep the exponent of 5 constant

and add all of exponents of 2

we get 0+1+2+…n=n(n+1)/2

now exponent of 5 can take any value from 0.n=n+1 values

so total of

n(n+1)^2/2=792

n=11

outtimedSeptember 30, 2008 at 4:09 am