Difference between revisions of "2019 AMC 12A Problems/Problem 15"
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Only a few possible sums are seen: <math>16+36=52</math>, <math>36+64=100</math>, <math>64+100=164</math>, <math>100+100=200</math>, and <math>4+196=200</math>. By testing each of these (seeing whether <math>\sqrt{x} + \sqrt{y} + \frac{x}{2} + \frac{y}{2} = 100</math>), only the pair <math>x = 64</math> and <math>y=100</math> work. Therefore, <math>a</math> and <math>b</math> are <math>10^{64}</math> and <math>10^{100}</math>, and our answer is <math>\boxed{\textbf{(D) } 10^{164}}</math>. | Only a few possible sums are seen: <math>16+36=52</math>, <math>36+64=100</math>, <math>64+100=164</math>, <math>100+100=200</math>, and <math>4+196=200</math>. By testing each of these (seeing whether <math>\sqrt{x} + \sqrt{y} + \frac{x}{2} + \frac{y}{2} = 100</math>), only the pair <math>x = 64</math> and <math>y=100</math> work. Therefore, <math>a</math> and <math>b</math> are <math>10^{64}</math> and <math>10^{100}</math>, and our answer is <math>\boxed{\textbf{(D) } 10^{164}}</math>. | ||
− | ==Solution | + | ==Solution 3== |
Given that <math>\sqrt{\log{a}}</math> and <math>\sqrt{\log{b}}</math> are both integers, <math>a</math> and <math>b</math> must be in the form <math>10^{m^2}</math> and <math>10^{n^2}</math>, respectively for some positive integers <math>m</math> and <math>n</math>. Note that <math>\log \sqrt{a} = \frac{m^2}{2}</math>. By substituting for a and b, the equation becomes <math>m + n + \frac{m^2}{2} + \frac{n^2}{2} = 100</math>. After multiplying the equation by 2 and completing the square with respect to <math>m</math> and <math>n</math>, the equation becomes <math>(m + 1)^2 + (n + 1)^2 = 202</math>. Testing squares of positive integers that add to <math>202</math>, <math>11^2 + 9^2</math> is the only option. Without loss of generality, let <math>m = 10</math> and <math>n = 8</math>. Plugging in <math>m</math> and <math>n</math> to solve for <math>a</math> and <math>b</math> gives us <math>a = 10^{100}</math> and <math>b = 10^{64}</math>. Therefore, <math>ab = \boxed{\textbf{(D) } 10^{164}}</math>. | Given that <math>\sqrt{\log{a}}</math> and <math>\sqrt{\log{b}}</math> are both integers, <math>a</math> and <math>b</math> must be in the form <math>10^{m^2}</math> and <math>10^{n^2}</math>, respectively for some positive integers <math>m</math> and <math>n</math>. Note that <math>\log \sqrt{a} = \frac{m^2}{2}</math>. By substituting for a and b, the equation becomes <math>m + n + \frac{m^2}{2} + \frac{n^2}{2} = 100</math>. After multiplying the equation by 2 and completing the square with respect to <math>m</math> and <math>n</math>, the equation becomes <math>(m + 1)^2 + (n + 1)^2 = 202</math>. Testing squares of positive integers that add to <math>202</math>, <math>11^2 + 9^2</math> is the only option. Without loss of generality, let <math>m = 10</math> and <math>n = 8</math>. Plugging in <math>m</math> and <math>n</math> to solve for <math>a</math> and <math>b</math> gives us <math>a = 10^{100}</math> and <math>b = 10^{64}</math>. Therefore, <math>ab = \boxed{\textbf{(D) } 10^{164}}</math>. | ||
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== Video Solution == | == Video Solution == |
Revision as of 10:29, 29 September 2021
Problem
Positive real numbers and
have the property that
and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is
?
Solution 1
Since is a positive integer, we get
for some integer
; since
is a positive integer, we get
. Thus
; similarly
. Substituting, we get
, i.e.
. It follows that
. The values of
for
are
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Two of those values must add up to and we see that
, so
and
, and our answer is
.
Solution 2
Since all four terms on the left are positive integers, from , we know that both
has to be a perfect square and
has to be a power of ten. The same applies to
for the same reason. Setting
and
to
and
, where
and
are the perfect squares,
. By listing all the perfect squares up to
(as
is larger than the largest possible sum of
and
of
from answer choice
), two of those perfect squares must add up to one of the possible sums of
and
given from the answer choices (
,
,
,
, or
).
Only a few possible sums are seen: ,
,
,
, and
. By testing each of these (seeing whether
), only the pair
and
work. Therefore,
and
are
and
, and our answer is
.
Solution 3
Given that and
are both integers,
and
must be in the form
and
, respectively for some positive integers
and
. Note that
. By substituting for a and b, the equation becomes
. After multiplying the equation by 2 and completing the square with respect to
and
, the equation becomes
. Testing squares of positive integers that add to
,
is the only option. Without loss of generality, let
and
. Plugging in
and
to solve for
and
gives us
and
. Therefore,
.
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.