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Difference between revisions of "2008 AMC 12B Problems/Problem 23"
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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.
 
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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===Solution 2 (number-theoretic bash)===
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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{11 \,\mathrm{(A)}}</math> as the correct answer.
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 +
 
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*This is by use of a well-known formula for <math>d(n)</math>.
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=== Solution 3 ===
 
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>.  
 
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>.  
  

Revision as of 10:07, 17 February 2014

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 (number-theoretic bash)

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 $\framebox{11 \,\mathrm{(A)}}$ (Error compiling LaTeX. Unknown error_msg) as the correct answer.


  • This is by use of a well-known formula for $d(n)$.

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

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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