Difference between revisions of "2023 AMC 10A Problems/Problem 23"
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Consider positive <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have that <math>(a)(a-20) = c</math>. If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than <math>a</math>, and one must be smaller than <math>a-20</math>. We can create two cases and set both equal. We have <math>(a)(a-20) = (a+1)(a-22)</math>, and <math>(a)(a-20) = (a+2)(a-21)</math>. Starting with the first case, we have <math>a^2-20a = a^2-21a-22</math>,or <math>0=-a-22</math>, which gives <math>a=-22</math>, which is not possible. The other case is <math>a^2-20a = a^2-19a-42</math>, so <math>a=42</math>. Thus, our product is <math>(42)(22) = (44)(21)</math>, so <math>c = 924</math>. Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. | Consider positive <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have that <math>(a)(a-20) = c</math>. If there is another pair of two integers that multiply to 30 but have a difference of 23, one integer must be greater than <math>a</math>, and one must be smaller than <math>a-20</math>. We can create two cases and set both equal. We have <math>(a)(a-20) = (a+1)(a-22)</math>, and <math>(a)(a-20) = (a+2)(a-21)</math>. Starting with the first case, we have <math>a^2-20a = a^2-21a-22</math>,or <math>0=-a-22</math>, which gives <math>a=-22</math>, which is not possible. The other case is <math>a^2-20a = a^2-19a-42</math>, so <math>a=42</math>. Thus, our product is <math>(42)(22) = (44)(21)</math>, so <math>c = 924</math>. Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. | ||
-Sepehr2010 | -Sepehr2010 | ||
+ | |||
+ | ==Solution 2 == | ||
+ | We have 4 integers in our problem. Let's call the smallest of them <math>a</math>. <math>a(a+23) = </math> either <math>(a+1)(a+21)</math> or <math>(a+2)(a+22)</math>. So, we have the following: | ||
+ | |||
+ | <math>a^2 + 23a = a^2 + 22a +21</math> or | ||
+ | |||
+ | <math>a^2+23a = a^2 + 24a +44</math>. | ||
+ | |||
+ | The second has negative solutions, so we discard it. The first one 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> => $9 + 2 + 4 = \boxed{\textbf{(C) 15}}#. | ||
+ | |||
+ | ~Arcticturn | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == |
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 has negative solutions, so we discard it. The first one has , and so
. If we check
we get
.
is
times
, and
is
times
, so our solution checks out. Multiplying
by
, we get
=> $9 + 2 + 4 = \boxed{\textbf{(C) 15}}#.
~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.