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2008 AMC 12B Problems/Problem 23

Revision as of 12:50, 25 July 2020 by Theasian (talk | contribs) (Alternative solution)

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 $2^a \times 5^b , a\leq n , b\leq n$. Using the logarithmic property $\log(a \times b) = \log(a)+\log(b)$, it suffices to count the total number of 2's and 5's running through all possible $(a,b)$. For every factor $2^a \times 5^b$, there will be another $2^b \times 5^a$, 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 $\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}$. Plugging in our answer choices into this formula yields 11 (answer choice $\mathrm{(A)}$) as the correct answer.

Solution 2

We are given \[\log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792\] 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 $n=11$

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 that 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 $4$.

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

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