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

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==Problem==
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Two distinct numbers are selected from the set <math>\{1,2,3,4,\dots,36,37\}</math> so that the sum of the remaining <math>35</math> numbers is the product of these two numbers. What is the difference of these two numbers?
  
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<math>\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10</math>
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==Solution==
 
The sum of the first <math>37</math> integers is given by <math>n(n+1)/2</math>, so <math>37(37+1)/2=703</math>.
 
The sum of the first <math>37</math> integers is given by <math>n(n+1)/2</math>, so <math>37(37+1)/2=703</math>.
  
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Therefore, the difference <math>y-x=31-21=10</math>, choice E).
 
Therefore, the difference <math>y-x=31-21=10</math>, choice E).
 
~ SoySoy4444
 
~ SoySoy4444
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==See Also==
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{{AMC12 box|year=2021|ab=B|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 20:05, 11 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 $0 < x <= y <= 37$ 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

See Also

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

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