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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^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>.
 
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>.
 +
 +
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.
  
 
== 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 01:37, 16 July 2021

Problem 2

Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$.

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 $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (because $2^0=1$). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$. Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$. Using the property that ${(a^x)^{y}}=a^{xy}$, we simplify to $2^{b\cdot2^{a}}=2^{1000}$. Eliminating equal bases leaves $b\cdot2^a=1000$. The largest $a$ such that $2^a$ divides $1000$ is $3$, so we only need to check $1$,$2$, and $3$. When $a=1$, $b=500$; when $a=2$, $b=250$; when $a=3$, $b=125$. Summing all the $a$'s and $b$'s gives the answer of $\boxed{881}$.

Note that $a$ cannot be $0,$ since that would cause the $\log_{2^a}$ to have a $1$ in the base, which is not possible.

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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