Difference between revisions of "2019 AMC 12A Problems/Problem 15"
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==Solution== | ==Solution== | ||
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+ | 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 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>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>). | ||
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+ | Only a couple 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 (by seeing whether <math>\sqrt{x} + \sqrt{b} + \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) } 100^{164}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 20:05, 9 February 2019
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
Since all four terms on the left are positive integers, from , we know that both
has to be 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 couple possible sums are seen: ,
,
,
, and
. By testing each of these (by seeing whether
), only the pair
and
work. Therefore,
and
are
and
, and our answer is
.
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.