2013 AIME II Problems/Problem 2
Problem 2
Positive integers and
satisfy the condition
Find the sum of all possible values of
.
Solution 1
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.
Solution 2
We proceed as in Solution 1, raising to both sides to achieve
We raise
to both sides to get
, then simplify to get
.
At this point, we want both and
to be integers. Thus,
can only be a power of
. To help us see the next step, we factorize
:
It should be clear that
must be from
to
; when
,
; when
,
; and finally, when
,
We sum all the pairs to get
~Technodoggo
Video Solution
~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 |
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