Difference between revisions of "2023 AMC 10A Problems/Problem 23"
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== Solution 5 == | == Solution 5 == | ||
− | Since we are given that some pairs of divisors differ by 20 and 23 we can do <math>(x-10)(x+10)=(y-</math>\frac{23}{2}) | + | Since we are given that some pairs of divisors differ by 20 and 23 we can do <math>(x-10)(x+10)=(y-</math>\frac{23}{2})(y+<math>\frac{23}{2})</math>, <math>y=\frac{A}{2}</math> <math>x^2-100=\frac{(a^2-529)}{4}</math>. where x-10 and x-10 are factors are the ones that differ by 20 and y-<cmath>\frac{23}{2} and y+</cmath>\frac{23}{2} are the ones that differ by 23. Since both divisors are integers y must be in the form of \frac{A}{2} where A is an odd integer. After solving and substituting we get that <math>y=\frac{A}{2}</math> <math>x^2-100=\frac{(A^2-529)}{4}</math>. Multiplying by 4 by both sides and simplifying we get that A^2-4x^2=129<math>. We use difference of squares to get that A+2x=129, A-2x=1. So </math>A=65 and <math>x=32. Plugging back x into the original equation we get that c=</math>(42)(22) so c=924<math>. The answer is </math>\boxed{\textbf{(C) 15}}$. |
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == |
Revision as of 15:51, 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
Solution 4
Say one factorization is The two cases for the other factorization are
and
We know it must be the first because of AM-GM intuition: lesser factors are closer together. Thus,
and we find that
meaning the answer is
~DouDragon
Solution 5
Since we are given that some pairs of divisors differ by 20 and 23 we can do \frac{23}{2})(y+
,
. where x-10 and x-10 are factors are the ones that differ by 20 and y-
\frac{23}{2} are the ones that differ by 23. Since both divisors are integers y must be in the form of \frac{A}{2} where A is an odd integer. After solving and substituting we get that
. Multiplying by 4 by both sides and simplifying we get that A^2-4x^2=129
A=65 and
(42)(22) so c=924
\boxed{\textbf{(C) 15}}$.
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.