2008 AMC 12B Problems/Problem 23
Problem 23
The sum of the base- logarithms of the divisors of
is
. What is
?
Solutions
Solution 1
Every factor of will be of the form
. 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
. For any factor
, there will be another factor
.(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+1n+1
0+1+2+3...+n = \frac{n(n+1)}{2}
\frac{n \cdot(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}
\mathrm{(A)}$) as the correct answer.
=== Solution 2 ===
We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property$ (Error compiling LaTeX. Unknown error_msg)\log(ab) = \log(a)+\log(b)a^{d(n)/2}
d(n)
10^n = 2^n\cdot 5^n
d(n) = (n + 1)^2
a = 10^n
n(n + 1)^2 = 1584
\framebox[1.2\width]{(A)}$as the correct answer.
=== Solution 3 ===
For every divisor$ (Error compiling LaTeX. Unknown error_msg)d10^n
d \le \sqrt{10^n}
\log d + \log \frac{10^n}{d} = \log 10^n = n
\left \lfloor \frac{(n+1)^2}{2} \right \rfloor
10^n = 2^n \times 5^n
\le \sqrt{10^n}
n
n \times \frac{(n+1)^2}{2} = 792 \Longrightarrow n = 11\ \mathrm{(A)}$.
=== Solution 4 === The sum is <cmath> \sum_{p=0}^{n}\sum_{q=0}^{n} \log(2^p5^q) = \sum_{p=0}^{n}\sum_{q=0}^{n}(p\log(2)+q\log(5)) </cmath> <cmath> = \sum_{p=0}^{n} ((n+1)p\log(2) + \frac{n(n+1)}{2}\log(5)) </cmath> <cmath>= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5) </cmath> <cmath> = \frac{n(n+1)^2}{2} </cmath> Trying for answer choices we get$ (Error compiling LaTeX. Unknown error_msg)n=11$
Alternative thinking
After arriving at the equation , notice that all of the answer choices are in the form
, where
is
. We notice that the ones digit of
is
, and it is dependent on the ones digit of the answer choices. Trying
for
, we see that only
yields a ones digit of
, so our answer is
.
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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