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Difference between revisions of "2008 AMC 12B Problems/Problem 23"

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==Solution==
 
==Solution==
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Every factor of <math>10^n</math> will be of the form <math>2^a * 5^b , a\leqn , b\leqn</math>. Logarithmically, addition and multiplication are interchangeable (i.e. <math>log(a*b) = log(a)+log(b)</math>), so we need only count the number of 2's and 5's occurring in total. For every factor <math>2^a * 5^b</math>, there will be another <math>2^b * 5^a</math>, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since <math>log(2)+log(5) = log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>.
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Every factor of <math>10^n</math> will be of the form <math>2^a * 5^b , a\leq n , b\leq n</math>. Logarithmically, addition and multiplication are interchangeable (i.e. <math>log(a*b) = log(a)+log(b)</math>), so we need only count the number of 2's and 5's occurring in total. For every factor <math>2^a * 5^b</math>, there will be another <math>2^b * 5^a</math>, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since <math>log(2)+log(5) = log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>.
  
 
There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = (n(n+1))/2</math> total 2's. The total number of 2's is therefore <math>(n*(n+1)^2)/2 = (n^3+2n^2+n)/2</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.
 
There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = (n(n+1))/2</math> total 2's. The total number of 2's is therefore <math>(n*(n+1)^2)/2 = (n^3+2n^2+n)/2</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.

Revision as of 00:50, 2 March 2008

Problem 23

The sum of the base-$10$ logarithms of the divisors of $10^n$ is $792$. What is $n$?

$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$

Solution

Every factor of $10^n$ will be of the form $2^a * 5^b , a\leq n , b\leq n$. Logarithmically, addition and multiplication are interchangeable (i.e. $log(a*b) = log(a)+log(b)$), so we need only count the number of 2's and 5's occurring in total. For every factor $2^a * 5^b$, there will be another $2^b * 5^a$, so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since $log(2)+log(5) = log(10) = 1$, the final sum will be the total number of 2's occurring in all factors of $10^n$.

There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $0+1+2+3...+n = (n(n+1))/2$ total 2's. The total number of 2's is therefore $(n*(n+1)^2)/2 = (n^3+2n^2+n)/2$. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.

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