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 |
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/oUEa7AjMF2A?si=q1GMT9DB8-AY0Ld5&t=1882 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/YnRJ1Q0GxgU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/yT-9qbMw8y8 | ||
+ | |||
+ | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=12|num-a=14}} | {{AMC8 box|year=2022|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:34, 2 September 2024
Contents
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 by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=q1GMT9DB8-AY0Ld5&t=1882
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1110
~Interstigation
Video Solution
https://youtu.be/p29Fe2dLGs8?t=139
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
~harungurcan
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.