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

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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 $2^a \times 5^b , a\leq n , b\leq n$ . 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 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 $11$ .

@@ Line 28: / Line 28: @@
 ==Alternative thinking==
-After arriving at the equation <math>n(n+1)^2 = 1584</math>, notice that all of the answer choices are in the form <math>10+k</math>, where <math>k</math> is <math>1-5</math>. We notice that the ones digit of <math>n(n+1)^2</math> is <math>4</math>, and it is dependent on the ones digit of the answer choices. Trying <math>1-5</math> for <math>n</math>, we see that only <math>1</math> yields a ones digit of <math>4</math>.
+After arriving at the equation <math>n(n+1)^2 = 1584</math>, notice that all of the answer choices are in the form <math>10+k</math>, where <math>k</math> is <math>1-5</math>. We notice that the ones digit of <math>n(n+1)^2</math> is <math>4</math>, and it is dependent on the ones digit of the answer choices. Trying <math>1-5</math> for <math>n</math>, we see that only <math>1</math> yields a ones digit of <math>4</math>, so our answer is <math>11</math>.
 ==See Also==

2008 AMC 12B (Problems • Answer Key • Resources)
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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