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

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== Solutions ==
 
== 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>. 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>.(This is not true if 10^n<math> is a perfect square.) When these are added, they equal 2^(a+n-a)</math> \times 5^(b+n-b)<math> = </math>10^n. Log <math>10^n=n. This means the number of factors divided by 2 times n equals the sum of all the factors, 792.  
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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 </math>5^(n-b)<math>.(This is not true if </math>10^n<math> is a perfect square.) When these are added, they equal 2^(a+n-a)</math> \times 5^(b+n-b)<math> = </math>10^n. Log <math>10^n=n. This means the number of factors divided by 2 times n equals the sum of all the factors, 792.  
 
   
 
   
 
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 factors. So we have \frac{n(n+1)}{2}</math> divided by 2 times n = 792
 
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 factors. So we have \frac{n(n+1)}{2}</math> divided by 2 times n = 792

Revision as of 11:19, 14 December 2020

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 $2^a \times 5^b , a\leq n , b\leq n$. 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$5^(n-b)$.(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. This means the number of factors divided by 2 times n equals the sum of all the factors, 792.

There are$ (Error compiling LaTeX. Unknown error_msg)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 factors. So we have \frac{n(n+1)}{2}$ divided by 2 times n = 792

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 $11$.

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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