Difference between revisions of "2021 AMC 12B Problems/Problem 10"

Revision as of 17:45, 18 February 2021

Problem

Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?

$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution

The sum of the first $37$ integers is given by $n(n+1)/2$ , so $37(37+1)/2=703$ .

Therefore, $703-x-y=xy$

Rearranging, $xy+x+y=703$

$(x+1)(y+1)=704$

Looking at the possible divisors of $704 = 2^6*11$ , $22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those:

$(x+1)(y+1) = 22 * 32$

$x+1=22, y+1 = 32$

$x = 21, y = 31$

Therefore, the difference $y-x=31-21=10$ , choice E). ~ SoySoy4444

Video Solution by Punxsutawney Phil

https://YouTube.com/watch?v=yxt8-rUUosI&t=292s

Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)

https://youtu.be/v8MVGAZapKU

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 <math>(x+1)(y+1)=704</math>
-Looking at the possible divisors of <math>704 = 2^6*11</math>, <math>22</math> and <math>32</math> are within the constraints of <math>0 < x <= y <= 37</math> so we try those:
+Looking at the possible divisors of <math>704 = 2^6*11</math>, <math>22</math> and <math>32</math> are within the constraints of <math>0 < x \leq y \leq 37</math> so we try those:
 <math>(x+1)(y+1) = 22 * 32</math>

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