Difference between revisions of "2013 AIME II Problems/Problem 2"
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==Solution== | ==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 <math>\log_{2^a}(\log_{2^b}(2^{1000}))=1</math> (because <math>2^0=1</math>). Doing this again, we get <math>\log_{2^b}(2^{1000})=2^a</math>. Doing the process one more time, we finally eliminate all of the logs, getting <math>{(2^{b})}^{(2^a)}=2^{1000}</math>. Using the property that <math>{a^ | + | 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 <math>\log_{2^a}(\log_{2^b}(2^{1000}))=1</math> (because <math>2^0=1</math>). Doing this again, we get <math>\log_{2^b}(2^{1000})=2^a</math>. Doing the process one more time, we finally eliminate all of the logs, getting <math>{(2^{b})}^{(2^a)}=2^{1000}</math>. Using the property that <math>{(a^x)^{y}}=a^{xy}</math>, we simplify to <math>2^{b\cdot2^{a}}=2^{1000}</math>. Eliminating equal bases leaves <math>b\cdot2^a=1000</math>. The largest <math>a</math> such that <math>2^a</math> divides <math>1000</math> is <math>3</math>, so we only need to check <math>1</math>,<math>2</math>, and <math>3</math>. When <math>a=1</math>, <math>b=500</math>; when <math>a=2</math>, <math>b=250</math>; when <math>a=3</math>, <math>b=125</math>. Summing all the <math>a</math>'s and <math>b</math>'s gives the answer of <math>\boxed{881}</math>. |
Revision as of 16:49, 5 April 2013
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
.