Difference between revisions of "2023 AMC 10A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | If the positive integer <math> | + | If the positive integer <math>n</math> has positive integer divisors <math>a</math> and <math>b</math> with <math>n = ab</math>, then <math>a</math> and <math>b</math> are said to be <math>\textit{complementary}</math> divisors of <math>n</math>. Suppose that <math>N</math> is a positive integer that has one complementary pair of divisors that differ by <math>20</math> and another pair of complementary divisors that differ by <math>23</math>. What is the sum of the digits of <math>N</math>? |
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> | <math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> |
Revision as of 13:04, 11 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 ~Mathkiddie
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 be and
as well as
and
. 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
+
=129,
-
=1. We finish by getting
65 and
. So
Adding the digits, we have
.
~averageguy
Solution 6
can be written
with a positive integer
and
with a positive integer
.
The above equations can be reorganized as
The only solution is and
.
Thus,
.
Therefore,
.
So the sum of the digits of
is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 7
We can write as
or
where
and
are divisors of
Since
we know that
, and we can view this as a quadratic in
Since the solution for must be an integer, the discriminant for this quadratic must be a perfect square and therefore
so
Since the discriminant of this quadratic in must also be a perfect square we know that
which we can simplify as
Since they are both positive integers
and
are factors of
so
and
or
and
These systems of equations give us and
respectively, if we plug our values for
into the equation for
we get
and
respectively. The first equation gives us
or
and the second gives us
or
, since
is positive we know that
and
, therefore the sum of the digits of
is
~SailS
Video Solution by epicbird08
~EpicBird08
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.