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.
ReplyDeleteabsolutely correct !
ReplyDeleteI'm not satisfied with the solution.Took some time.Is there any better method?
ReplyDeleteits easy
ReplyDeletekeep 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