Difference between revisions of "2021 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | The sum of the first <math> | + | The sum of the first <math>n</math> integers is given by <math>\frac{n(n+1)}{2}</math>, so <math>\frac{37(37+1)}{2}=703</math>. |
Therefore, <math>703-x-y=xy</math> | Therefore, <math>703-x-y=xy</math> | ||
| Line 13: | Line 13: | ||
<math>(x+1)(y+1)=704</math> | <math>(x+1)(y+1)=704</math> | ||
| − | Looking at the possible divisors of <math>704 = 2^6 | + | Looking at the possible divisors of <math>704 = 2^6\cdot11</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 | + | <math>(x+1)(y+1) = 22\cdot32</math> |
<math>x+1=22, y+1 = 32</math> | <math>x+1=22, y+1 = 32</math> | ||
| Line 21: | Line 21: | ||
<math>x = 21, y = 31</math> | <math>x = 21, y = 31</math> | ||
| − | Therefore, the difference <math>y-x=31-21=10</math> | + | Therefore, the difference <math>y-x=31-21=\boxed{\textbf{(E) }10}</math>. |
| + | |||
~ SoySoy4444 | ~ SoySoy4444 | ||
| + | |||
| + | ~MathFun1000 (<math>\LaTeX</math> help) | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
Revision as of 16:38, 7 January 2022
Contents
Problem
Two distinct numbers are selected from the set 
so that the sum of the remaining 
numbers is the product of these two numbers. What is the difference of these two numbers?
Solution
The sum of the first 
integers is given by 
, so 
.
Therefore, 
Rearranging, 
. We can factor this equation by SFFT to get
Looking at the possible divisors of 
, 
and 
are within the constraints of 
so we try those:
Therefore, the difference 
.
~ SoySoy4444
~MathFun1000 (
help)
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=yxt8-rUUosI&t=292s
Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
Video Solution by TheBeautyofMath
https://youtu.be/kuZXQYHycdk?t=1227
~IceMatrix
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
