Difference between revisions of "2024 AMC 10A Problems/Problem 15"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | == Solution 2 == | ||
+ | Let <math>M+1213=x^2</math> and <math>M+3773=y^2</math>. Subtracting the two equations will give <cmath>y^2-x^2=2560.</cmath> Factoring the left hand side as a difference of squares gives <cmath>(y-x)(y+x)=2560.</cmath> The goal in the problem was to maximize <math>M</math>, which therefore means maximizing both <math>x</math> and <math>y</math> and maximizing <math>y+x</math>. | ||
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+ | <math>y-x</math> and <math>y+x</math> are two factors of <math>2560</math>, therefore maximizing <math>y+x</math> means making <math>y-x</math> and <math>y+x</math> as far as possible from each other. We start with <math>2560=1\cdot2560</math>, the most obvious choice that makes <math>y+x</math> as large as possible, but solving <math>y-x=1</math> and <math>y+x=2560</math> will give non-integer solutions, therefore this case does not work. | ||
+ | |||
+ | The next most obvious choice would be <math>2560=2\cdot1280</math>, which does give integer solutions to <math>x</math> and <math>y</math>. <cmath>y-x=2</cmath> | ||
+ | <cmath>y+x=1280</cmath> gives the solutions <math>y=641,x=639</math>. | ||
+ | |||
+ | Since we're only focusing on the units digit of <math>M=x^2-1213=y^2-3773</math>, then we only have to focus on the units digits of <math>x^2</math> and <math>1213</math> or <math>y^2</math> and <math>3773</math>. <math>1^2</math> and <math>9^2</math> give the same units digit, and <math>1213</math> and <math>3773</math> also have the same units digit. So the units digit of <math>M</math> is <math>10-|1-3|=\boxed{\text{(E) }8}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:09, 8 November 2024
Contents
[hide]Problem
Let be the greatest integer such that both
and
are perfect squares. What is the units digit of
?
Solution
Let and
for some positive integers
and
We subtract the first equation from the second, then apply the difference of squares:
Note that
and
have the same parity, and
We wish to maximize both and
so we maximize
and minimize
It follows that
from which
Finally, we get so the units digit of
is
~MRENTHUSIASM
Solution 2
Let and
. Subtracting the two equations will give
Factoring the left hand side as a difference of squares gives
The goal in the problem was to maximize
, which therefore means maximizing both
and
and maximizing
.
and
are two factors of
, therefore maximizing
means making
and
as far as possible from each other. We start with
, the most obvious choice that makes
as large as possible, but solving
and
will give non-integer solutions, therefore this case does not work.
The next most obvious choice would be , which does give integer solutions to
and
.
gives the solutions
.
Since we're only focusing on the units digit of , then we only have to focus on the units digits of
and
or
and
.
and
give the same units digit, and
and
also have the same units digit. So the units digit of
is
See also
2024 AMC 10A (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 10 Problems and Solutions |
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