Difference between revisions of "2008 AMC 12B Problems/Problem 23"
(→Solution 1) |
|||
Line 1: | Line 1: | ||
− | == Problem | + | == Problem == |
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>? | ||
Revision as of 13:47, 15 February 2021
Contents
Problem
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
. Note this is not true if
is a perfect square. When these are added, they equal
.
so the number of factors divided by 2 times n equals the sum of all the factors, 792.
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 factors.
. We then plug in answer choices and arrive at the 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
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
.
Solution 4
The sum is
Trying for answer choices we get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.