Difference between revisions of "2023 AMC 10A Problems/Problem 23"
Technodoggo (talk | contribs) (→Solution 3) |
Technodoggo (talk | contribs) (→Solution 3) |
||
Line 36: | Line 36: | ||
Since both <math>(2y+2x+43)</math> and <math>(2y-2x+3)</math> must be integer, we get two equations. | Since both <math>(2y+2x+43)</math> and <math>(2y-2x+3)</math> must be integer, we get two equations. | ||
<cmath>\begin{align} | <cmath>\begin{align} | ||
− | 129 &= (2y+2x+43)\\ | + | 129 or 43 &= (2y+2x+43)\\ |
− | 1 &= 2y-2x+3\\ | + | 1 or 3&= 2y-2x+3\\ |
\end{align}</cmath> | \end{align}</cmath> | ||
+ | 43 & 1 yields (0,0) which is not what we want. | ||
+ | 129 & 1 yields (22,21) which is more interesting. | ||
+ | |||
Simplifying the equations, we get: | Simplifying the equations, we get: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 01:09, 10 November 2023
Contents
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
Solution 3
From the problems, it follows that
Since both
and
must be integer, we get two equations.
43 & 1 yields (0,0) which is not what we want.
129 & 1 yields (22,21) which is more interesting.
Simplifying the equations, we get:
So, the answer is .
~Technodoggo
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.