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

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==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=p4iCAZRUESs
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2021|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:53, 12 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

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

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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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