Difference between revisions of "2008 AMC 12B Problems/Problem 23"
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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. | 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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=== Solution 3 === | === Solution 3 === |
Revision as of 12:22, 2 February 2015
Problem 23
The sum of the base- logarithms of the divisors of
is
. What is
?
Solution
Solution 1
Every factor of will be of the form
. Using the logarithmic property
, it suffices to count the total number of 2's and 5's running through all possible
. For every factor
, there will be another
, 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
, the final sum will be the total number of 2's occurring in all factors of
.
There are choices for the exponent of 5 in each factor, and for each of those choices, there are
factors (each corresponding to a different exponent of 2), yielding
total 2's. The total number of 2's is therefore
. Plugging in our answer choices into this formula yields 11 (answer choice
) as the correct answer.
Solution 2
We are given The property
now gives
The product of the divisors is (from elementary number theory)
where
is the number of divisors. Note that
, so*
. Substituting these values with
in our equation above, we get
, from whence we immediately obtain $\framebox{11 \,\mathrm{(A)}}$ (Error compiling LaTeX. Unknown error_msg) as the correct answer.
Solution 3
For every divisor of
,
, we have
. There are
divisors of
that are
. After casework on the parity of
, we find that the answer is given by
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.