Difference between revisions of "2022 AMC 8 Problems/Problem 13"
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By substitution, we have <math>(2n+d)+n=28,</math> from which <math>d=28-3n.</math> Note that <math>n=1,2,3,\ldots,9</math> each generate a positive integer for <math>d,</math> so there are <math>\boxed{\textbf{(D) } 9}</math> possible values for <math>d.</math> | By substitution, we have <math>(2n+d)+n=28,</math> from which <math>d=28-3n.</math> Note that <math>n=1,2,3,\ldots,9</math> each generate a positive integer for <math>d,</math> so there are <math>\boxed{\textbf{(D) } 9}</math> possible values for <math>d.</math> | ||
| − | ~ | + | ~MRENTHUSIASM |
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/Ij9pAy6tQSg?t=1110 | https://youtu.be/Ij9pAy6tQSg?t=1110 | ||
Revision as of 19:34, 16 January 2023
Problem
How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is 
.”
Solution
Let 
and 
be positive integers such that 
and 
It follows that 
for some positive integer 
We wish to find the number of possible values for
By substitution, we have 
from which 
Note that 
each generate a positive integer for 
so there are 
possible values for
~MRENTHUSIASM
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1110
~Interstigation
Video Solution
https://youtu.be/p29Fe2dLGs8?t=139
~STEMbreezy
Video Solution
~savannahsolver
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
