2008 AMC 12B Problems/Problem 23


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

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


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 $2^{n-a} \times 5^{n-b}$. Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$. $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.

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 $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$. We then plug in answer choices and arrive at the answer $\boxed {11}$

Solution 2

We are given \[\log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792\] The property $\log(ab) = \log(a)+\log(b)$ now gives \[\log_{10}(d_1 d_2\cdot\ldots d_k) = 792\] 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 $d$ 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 \[\sum_{p=0}^{n}\sum_{q=0}^{n} \log(2^p5^q) = \sum_{p=0}^{n}\sum_{q=0}^{n}(p\log(2)+q\log(5))\] \[= \sum_{p=0}^{n} ((n+1)p\log(2) + \frac{n(n+1)}{2}\log(5))\] \[= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5)\] \[= \frac{n(n+1)^2}{2}\] Trying for answer choices we get $n=11$

Solution 5

Let integer $d$ be the number of divisors $10^n$ has. Then, we set up $\frac{d}{2}$ pairs of divisors such that each pair $(a,b)$ satisfies $ab = 10^n$ -- ex. $(1, 10^n), (2, 5*10^{n-1})$, etc. Then the sum of the base-$10$ logarithms is $\sum_{j=1}^{\frac{d}{2}} \log_{10} a_j + \log_{10} b_j = \sum_{j=1}^{\frac{d}{2}} \log_{10} a_j b_j = \sum_{j=1}^{\frac{d}{2}} \log_{10} 10^n = \sum_{j=1}^{\frac{d}{2}} n = \frac{nd}{2}$ We can use the property that only perfect squares have an odd number of factors, as for perfect square $p^2$, we have ordered pair $(p,p)$ that works. For even $n$, then, $10^{\frac{n}{2}}$ can be multiplied by itself to get $10^n$, so $d$ is odd. But, in our summation, $\frac{d}{2}$ does not exist for even $n$ as $d$ is then odd, so $d$ must be even. And, since $\frac{nd}{2}$ = $792$, We want to find a $d$ for our odd options $n$ such that $nd$ = $1584$. For $n=11$, $d=144$ works, and integer $d$ can not be found for other odd $n$. So, we get $\framebox[1.2\width]{(A) 11}$

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 $11$.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions

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