Difference between revisions of "2013 AIME II Problems/Problem 2"
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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. | + | 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 (also the problem specifies that <math>a</math> and <math>b</math> are positive). |
==Solution 2== | ==Solution 2== |
Latest revision as of 23:19, 10 January 2024
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 (also the problem specifies that and are positive).
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.