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

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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>?
  
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<math>\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15</math>
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<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__
 
==Solution==
 
==Solution==
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Every factor of <math>10^n</math> will be of the form <math>2^a * 5^b , a\leq n , b\leq n</math>. Logarithmically, addition and multiplication are interchangeable (i.e. <math>\log(a*b) = \log(a)+\log(b)</math>), so we need only count the number of 2's and 5's occurring in total. For every factor <math>2^a * 5^b</math>, there will be another <math>2^b * 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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=== 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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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*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) 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>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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=== Solution 2 ===
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For every divisor <math>d</math> of <math>10^n</math>, <math>d \le \sqrt{10^n}</math>, we have <math>\log d + \log \frac{10^n}{d} = \log 10^n = n</math>. There are <math>\left \lfloor \frac{(n+1)^2}{2} \right \rfloor</math> divisors of <math>10^n = 2^n \times 5^n</math> that are <math>\le \sqrt{10^n}</math>. After casework on the parity of <math>n</math>, we find that the answer is given by <math>n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}</math>.  
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2008|ab=B|num-b=22|num-a=24}}
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[[Category:Introductory Algebra Problems]]

Revision as of 19:57, 1 December 2008

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

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

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 $n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}$.

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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