2008 AMC 12B Problems/Problem 23

Problem 23

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

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

Solutions

Solution 1

Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ . For any factor $2^a \times 5^b$ , there will be another factor $2^n-a \times 5^n-b$ .(This is not true if 10^n is a perfect square.) When these are added, they equal 2^a+n-a \times 5^b+n-b$= 10^n. Log 10^n=n. This means the number of factors divided by 2 times n equals the sum of all the factors, 792.

There are$ (Error compiling LaTeX. Unknown error_msg)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 \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2} $. Plugging in our answer choices into this formula yields 11 (answer choice$ \mathrm{(A)}$) as the correct answer.

=== Solution 2 === We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property$ (Error compiling LaTeX. Unknown error_msg)\log(ab) = \log(a)+\log(b) $now gives <cmath> \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 </cmath> The product of the divisors is (from elementary number theory)$ a^{d(n)/2} $where$ d(n) $is the number of divisors. Note that$ 10^n = 2^n\cdot 5^n $, so$ d(n) = (n + 1)^2 $. Substituting these values with$ a = 10^n $in our equation above, we get$ n(n + 1)^2 = 1584 $, from whence we immediately obtain$ \framebox[1.2\width]{(A)}$as the correct answer.

=== Solution 3 === For every divisor$ (Error compiling LaTeX. Unknown error_msg)d $of$ 10^n $,$ d \le \sqrt{10^n} $, we have$ \log d + \log \frac{10^n}{d} = \log 10^n = n $. There are$ \left \lfloor \frac{(n+1)^2}{2} \right \rfloor $divisors of$ 10^n = 2^n \times 5^n $that are$ \le \sqrt{10^n} $. After casework on the parity of$ n $, we find that the answer is given by$ n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}$.

=== Solution 4 === The sum is <cmath> \sum_{p=0}^{n}\sum_{q=0}^{n} \log(2^p5^q) = \sum_{p=0}^{n}\sum_{q=0}^{n}(p\log(2)+q\log(5)) </cmath> <cmath> = \sum_{p=0}^{n} ((n+1)p\log(2) + \frac{n(n+1)}{2}\log(5)) </cmath> <cmath>= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5) </cmath> <cmath> = \frac{n(n+1)^2}{2} </cmath> Trying for answer choices we get$ (Error compiling LaTeX. Unknown error_msg)n=11$

Alternative thinking

After arriving at the equation $n(n+1)^2 = 1584$ , notice that all of the answer choices are in the form $10+k$ , where $k$ is $1-5$ . We notice that the ones digit of $n(n+1)^2$ is $4$ , and it is dependent on the ones digit of the answer choices. Trying $1-5$ for $n$ , we see that only $1$ yields a ones digit of $4$ , so our answer is $11$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2008 AMC 12B Problems/Problem 23

Contents

Problem 23

Solutions

Solution 1

Alternative thinking

See Also