Difference between revisions of "2013 AIME II Problems/Problem 2"

Latest revision as of 22:19, 10 January 2024

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 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 $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (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 (also the problem specifies that $a$ and $b$ are positive).

Solution 2

We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$ , then simplify to get $\dfrac{1000}b=2^a$ .

At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$ . To help us see the next step, we factorize $1000$ : $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$ ; when $a=1$ , $b=500$ ; when $a=2$ , $b=250$ ; and finally, when $a=3$ , $b=125.$ We sum all the pairs to get $\boxed{881}.$

~Technodoggo

Video Solution

https://youtu.be/zf9ld5KL_g4

~Lucas

@@ Line 1: / Line 1: @@
+==Problem 2==
 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>
 Find the sum of all possible values of <math>a+b</math>.
+==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 <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 (also the problem specifies that <math>a</math> and <math>b</math> are positive).
+==Solution 2==
+We proceed as in Solution 1, raising <math>2</math> to both sides to achieve <math>\log_{2^a}(\log_{2^b}(2^{1000})) = 1.</math> We raise <math>2^a</math> to both sides to get <math>\log_{2^b}(2^{1000})=2^a</math>, then simplify to get <math>\dfrac{1000}b=2^a</math>.
+At this point, we want both <math>a</math> and <math>b</math> to be integers. Thus, <math>2^a</math> can only be a power of <math>2</math>. To help us see the next step, we factorize <math>1000</math>: <math>\dfrac{2^35^3}b=2^a.</math> It should be clear that <math>a</math> must be from <math>1</math> to <math>3</math>; when <math>a=1</math>, <math>b=500</math>; when <math>a=2</math>, <math>b=250</math>; and finally, when <math>a=3</math>, <math>b=125.</math> We sum all the pairs to get <math>\boxed{881}.</math>
+~Technodoggo
+==Video Solution==
+https://youtu.be/zf9ld5KL_g4
+~Lucas
+== See also ==
+{{AIME box|year=2013|n=II|num-b=1|num-a=3}}
+{{MAA Notice}}

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