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 | + | Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair. We also know the product of both the complementary divisors give the same number so <math>(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})</math> . |
− | + | Now we let <math>y=\frac{a}{2}</math>. Then we substitute and get <math>x^2-100=\frac{(a^2-529)}{4}</math>. Finally we multiply by 4 and get <math>4x^2-a^2=-129, a^2-4x^2=129</math>. | |
− | + | Then we use differences of squares and get a+2x=129, a-2x=1. We finish by getting <math>a=</math>65 and <math>x=32</math>. So <math>(42)(22) = 924</math> Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. | |
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == |
Revision as of 16:42, 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 and we can let the pair. We also know the product of both the complementary divisors give the same number so .
Now we let
. Then we substitute and get
. Finally we multiply by 4 and get
.
Then we use differences of squares and get a+2x=129, a-2x=1. We finish by getting
65 and
. So
Adding the digits, we have
.
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.