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Difference between revisions of "2019 AMC 12A Problems/Problem 15"

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==Solution 1==
 
==Solution 1==
  
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Since <math>\sqrt{\log{a}}</math> is a positive integer, we get <math>\log a = x^2</math> for some integer <math>x</math>; since <math>\log \sqrt{a} = \tfrac 12 \log a</math> is a positive integer, we get <math>x=2m</math>. Thus <math>a=10^{4m^2}</math>; similarly <math>b=10^{4n^2}</math>. Substituting, we get <math>2(m+n+m^2+n^2)=100</math>, i.e. <math>m(m+1) + n(n+1) = 50</math>. It follows that <math>m,n \le 6</math>. The values of <math>m(m+1)</math> for <math>m=1,\ldots , 6</math> are
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{| style="border:0px solid black"
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|- <!-- Start of a new row -->
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| style =”width:85%” | <math>m</math> || <math>1\qquad</math> || <math>2\qquad</math> || <math>3\qquad</math> || <math>4\qquad</math> || <math>5\qquad</math> || <math>6\qquad</math>
 +
|-
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| <math>m(m+1)\qquad</math>
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|<math>2</math>
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|<math>6</math>
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|<math>12</math>
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|<math>20\qquad</math>
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|<math>30</math>
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|<math>42</math>
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|}
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Two of those values must add up to <math>50</math> and we see that <math>20+30=50</math>, so <math>m=4, n=5</math> and  <math>ab=10^{4(m^2+n^2)}=10^{4(4^2+5^2)}</math>, and our answer is <math>\boxed{\textbf{(D) } 10^{164}}</math>.
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 +
==Solution 2==
 
Since all four terms on the left are positive integers, from <math>\sqrt{\log{a}}</math>, we know that both <math>\log{a}</math> has to be a perfect square and <math>a</math> has to be a power of ten. The same applies to <math>b</math> for the same reason. Setting <math>a</math> and <math>b</math> to <math>10^x</math> and <math>10^y</math>, where <math>x</math> and <math>y</math> are the perfect squares, <math>ab = 10^{x+y}</math>. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <math>y</math> of <math>200</math> from answer choice <math>\text{E}</math>), two of those perfect squares must add up to one of the possible sums of <math>x</math> and <math>y</math> given from the answer choices (<math>52</math>, <math>100</math>, <math>144</math>, <math>164</math>, or <math>200</math>).  
 
Since all four terms on the left are positive integers, from <math>\sqrt{\log{a}}</math>, we know that both <math>\log{a}</math> has to be a perfect square and <math>a</math> has to be a power of ten. The same applies to <math>b</math> for the same reason. Setting <math>a</math> and <math>b</math> to <math>10^x</math> and <math>10^y</math>, where <math>x</math> and <math>y</math> are the perfect squares, <math>ab = 10^{x+y}</math>. By listing all the [https://artofproblemsolving.com/wiki/index.php/Perfect_square perfect squares] up to <math>14^2</math> (as <math>15^2</math> is larger than the largest possible sum of <math>x</math> and <math>y</math> of <math>200</math> from answer choice <math>\text{E}</math>), two of those perfect squares must add up to one of the possible sums of <math>x</math> and <math>y</math> given from the answer choices (<math>52</math>, <math>100</math>, <math>144</math>, <math>164</math>, or <math>200</math>).  
  

Revision as of 10:29, 29 September 2021

Problem

Positive real numbers $a$ and $b$ have the property that \[\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100\] and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is $ab$?

$\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200}$

Solution 1

Since $\sqrt{\log{a}}$ is a positive integer, we get $\log a = x^2$ for some integer $x$; since $\log \sqrt{a} = \tfrac 12 \log a$ is a positive integer, we get $x=2m$. Thus $a=10^{4m^2}$; similarly $b=10^{4n^2}$. Substituting, we get $2(m+n+m^2+n^2)=100$, i.e. $m(m+1) + n(n+1) = 50$. It follows that $m,n \le 6$. The values of $m(m+1)$ for $m=1,\ldots , 6$ are

$m$ $1\qquad$ $2\qquad$ $3\qquad$ $4\qquad$ $5\qquad$ $6\qquad$
$m(m+1)\qquad$ $2$ $6$ $12$ $20\qquad$ $30$ $42$

Two of those values must add up to $50$ and we see that $20+30=50$, so $m=4, n=5$ and $ab=10^{4(m^2+n^2)}=10^{4(4^2+5^2)}$, and our answer is $\boxed{\textbf{(D) } 10^{164}}$.

Solution 2

Since all four terms on the left are positive integers, from $\sqrt{\log{a}}$, we know that both $\log{a}$ has to be a perfect square and $a$ has to be a power of ten. The same applies to $b$ for the same reason. Setting $a$ and $b$ to $10^x$ and $10^y$, where $x$ and $y$ are the perfect squares, $ab = 10^{x+y}$. By listing all the perfect squares up to $14^2$ (as $15^2$ is larger than the largest possible sum of $x$ and $y$ of $200$ from answer choice $\text{E}$), two of those perfect squares must add up to one of the possible sums of $x$ and $y$ given from the answer choices ($52$, $100$, $144$, $164$, or $200$).

Only a few possible sums are seen: $16+36=52$, $36+64=100$, $64+100=164$, $100+100=200$, and $4+196=200$. By testing each of these (seeing whether $\sqrt{x} + \sqrt{y} + \frac{x}{2} + \frac{y}{2} = 100$), only the pair $x = 64$ and $y=100$ work. Therefore, $a$ and $b$ are $10^{64}$ and $10^{100}$, and our answer is $\boxed{\textbf{(D) } 10^{164}}$.

Solution 2

Given that $\sqrt{\log{a}}$ and $\sqrt{\log{b}}$ are both integers, $a$ and $b$ must be in the form $10^{m^2}$ and $10^{n^2}$, respectively for some positive integers $m$ and $n$. Note that $\log \sqrt{a} = \frac{m^2}{2}$. By substituting for a and b, the equation becomes $m + n + \frac{m^2}{2} + \frac{n^2}{2} = 100$. After multiplying the equation by 2 and completing the square with respect to $m$ and $n$, the equation becomes $(m + 1)^2 + (n + 1)^2 = 202$. Testing squares of positive integers that add to $202$, $11^2 + 9^2$ is the only option. Without loss of generality, let $m = 10$ and $n = 8$. Plugging in $m$ and $n$ to solve for $a$ and $b$ gives us $a = 10^{100}$ and $b = 10^{64}$. Therefore, $ab = \boxed{\textbf{(D) } 10^{164}}$.


Video Solution

https://youtu.be/RdIIEhsbZKw?t=1250

~ pi_is_3.14

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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