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

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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 = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n*(n+1)^2}{2} = \frac{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 = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.
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==See Also==
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{{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}

Revision as of 01:52, 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 $0+1+2+3...+n = \frac{n(n+1)}{2}$ total 2's. The total number of 2's is therefore $\frac{n*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}$. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions