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Problem Of The Week 33

with 4 comments


The sum of base-10 logarithms of divisors of 10^n is 792. what is n?
(A) 11  (B) 12  (C) 10  (D) 13 (E) 14

Written by Implex

September 27, 2008 at 11:08 pm

4 Responses

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  1. 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.

    riyaziq

    September 29, 2008 at 6:10 pm

  2. absolutely correct !

    outtimed

    September 29, 2008 at 11:16 pm

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

    riyaziq

    September 30, 2008 at 2:17 am

  4. 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

    outtimed

    September 30, 2008 at 4:09 am


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