Difference between revisions of "2023 AMC 10A Problems/Problem 23"
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<math>a^2+23a = a^2 + 24a +44</math>. | <math>a^2+23a = a^2 + 24a +44</math>. | ||
− | The second has negative solutions, so we discard it. The first | + | The second equation has negative solutions, so we discard it. The first equation has <math>a = 21</math>, and so <math>a + 23 = 44</math>. If we check <math>(a+1)(a+21)</math> we get <math>22 \cdot 42 = 21 \cdot 44</math>. <math>44</math> is <math>2</math> times <math>22</math>, and <math>42</math> is <math>2</math> times <math>21</math>, so our solution checks out. Multiplying <math>21</math> by <math>44</math>, we get <math>924</math> => <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}</math>. |
~Arcticturn | ~Arcticturn |
Revision as of 23:20, 9 November 2023
Problem
If the positive integer has positive integer divisors
and
with
, then
and
are said to be
divisors of
. Suppose that
is a positive integer that has one complementary pair of divisors that differ by
and another pair of complementary divisors that differ by
. What is the sum of the digits of
?
Solution 1
Consider positive with a difference of
. Suppose
. Then, we have that
. If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than
, and one must be smaller than
. We can create two cases and set both equal. We have
, and
. Starting with the first case, we have
,or
, which gives
, which is not possible. The other case is
, so
. Thus, our product is
, so
. Adding the digits, we have
.
-Sepehr2010
Solution 2
We have 4 integers in our problem. Let's call the smallest of them .
either
or
. So, we have the following:
or
.
The second equation has negative solutions, so we discard it. The first equation has , and so
. If we check
we get
.
is
times
, and
is
times
, so our solution checks out. Multiplying
by
, we get
=>
.
~Arcticturn
Video Solution 1 by OmegaLearn
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.