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

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

Revision as of 23:21, 8 November 2022

Problem

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 $2^{n-a} \times 5^{n-b}$. Note this is not true if $10^n$ is a perfect square. When these are added, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$. $\log 10^n=n$ 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 (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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