Difference between revisions of "2013 AIME II Problems/Problem 2"
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+ | ==Problem 2== | ||
Positive integers <math>a</math> and <math>b</math> satisfy the condition | Positive integers <math>a</math> and <math>b</math> satisfy the condition | ||
<cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath> | <cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath> |
Revision as of 16:50, 6 April 2013
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
.
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 |