Difference between revisions of "2013 AIME II Problems/Problem 2"
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Note that <math>a</math> cannot be <math>0,</math> since that would cause the <math>\log_{2^a}</math> to have a <math>1</math> in the base, which is not possible. | Note that <math>a</math> cannot be <math>0,</math> since that would cause the <math>\log_{2^a}</math> to have a <math>1</math> in the base, which is not possible. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/zf9ld5KL_g4 | ||
+ | ~Lucas | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=II|num-b=1|num-a=3}} | {{AIME box|year=2013|n=II|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:21, 19 September 2022
Contents
Problem 2
Positive integers and
satisfy the condition
Find the sum of all possible values of
.
Solution
To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means (because
). Doing this again, we get
. Doing the process one more time, we finally eliminate all of the logs, getting
. Using the property that
, we simplify to
. Eliminating equal bases leaves
. The largest
such that
divides
is
, so we only need to check
,
, and
. When
,
; when
,
; when
,
. Summing all the
's and
's gives the answer of
.
Note that cannot be
since that would cause the
to have a
in the base, which is not possible.
Video Solution
https://youtu.be/zf9ld5KL_g4 ~Lucas
See also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.