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

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==Problem 23==
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== Problem ==
 
The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>?
 
The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>?
  
 
<math>\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math>
 
<math>\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math>
  
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__TOC__
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== Solutions ==
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==Solutions==
 
 
=== Solution 1 ===
 
=== Solution 1 ===
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Every factor of <math>10^n</math> will be of the form <math>2^a \times 5^b , a\leq n , b\leq n</math>. Using the logarithmic property <math>\log(a \times b) = \log(a)+\log(b)</math>, it suffices to count the total number of 2's and 5's running through all possible <math>(a,b)</math>. For every factor <math>2^a \times 5^b</math>, there will be another <math>2^b \times 5^a</math>, 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 <math>\log(2)+\log(5) = \log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>.
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Every factor of <math>10^n</math> will be of the form <math>2^a \times 5^b , a\leq n , b\leq n</math>. 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 <math>\log(a \times b) = \log(a)+\log(b)</math>. For any factor <math>2^a \times 5^b</math>, there will be another factor <math>2^{n-a} \times 5^{n-b}</math>. Note this is not true if <math>10^n</math> is a perfect square. When these are multiplied, they equal <math>2^{a+n-a} \times 5^{b+n-b} = 10^n</math>. <math>\log 10^n=n</math> so the number of factors divided by 2 times n equals the sum of all the factors, 792.  
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There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice <math>\mathrm{(A)}</math>) as the correct answer.
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There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>(n+1)^2</math> total factors. <math>\frac{(n+1)^2}{2}*n = 792</math>. We then plug in answer choices and arrive at the answer <math>\boxed {11}</math>
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===Solution 2===
 
  
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=== Solution 2 ===
 
We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property <math>\log(ab) = \log(a)+\log(b)</math> 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) <math>a^{d(n)/2}</math> where <math>d(n)</math> is the number of divisors. Note that <math>10^n = 2^n\cdot 5^n</math>, so <math>d(n) = (n + 1)^2</math>. Substituting these values with <math>a = 10^n</math> in our equation above, we get <math>n(n + 1)^2 = 1584</math>, from whence we immediately obtain <math>\framebox[1.2\width]{(A)}</math> as the correct answer.
 
We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property <math>\log(ab) = \log(a)+\log(b)</math> 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) <math>a^{d(n)/2}</math> where <math>d(n)</math> is the number of divisors. Note that <math>10^n = 2^n\cdot 5^n</math>, so <math>d(n) = (n + 1)^2</math>. Substituting these values with <math>a = 10^n</math> in our equation above, we get <math>n(n + 1)^2 = 1584</math>, from whence we immediately obtain <math>\framebox[1.2\width]{(A)}</math> as the correct answer.
  
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<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} \log(2) + \frac{n(n+1)^2}{2} \log(5) </cmath>
 
<cmath> = \frac{n(n+1)^2}{2} </cmath>
 
<cmath> = \frac{n(n+1)^2}{2} </cmath>
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Try for answer choices we get <math>n=11</math>
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Trying for answer choices we get <math>n=11</math>
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== 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. <math>(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>
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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>
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== Alternative thinking ==
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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==
 
==See Also==

Latest revision as of 17:02, 17 January 2023

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 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 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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