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

Latest revision as of 03:22, 13 November 2022

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

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

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 3

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 Solution1

https://youtu.be/U0jHAadc4MA

~ Education, the Study of Everything

Video Solution by OmegaLearn

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

~ pi_is_3.14

Revision as of 15:36, 12 September 2022 (view source) Thestudyofeverything (talk \| contribs) (→‎Video Solution) ← Older edit		Latest revision as of 03:22, 13 November 2022 (view source) Pi is 3.14 (talk \| contribs) m (→‎Video Solution)
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−	== Video Solution ==	+	== Video Solution by OmegaLearn ==
	https://youtu.be/RdIIEhsbZKw?t=1250		https://youtu.be/RdIIEhsbZKw?t=1250

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