Difference between revisions of "2022 AMC 8 Problems/Problem 3"
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The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath> | The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath> | ||
− | We | + | We apply casework to <math>a</math>: |
If <math>a=1</math>, then there are <math>3</math> cases: | If <math>a=1</math>, then there are <math>3</math> cases: |
Latest revision as of 00:00, 18 January 2024
Contents
Problem
When three positive integers ,
, and
are multiplied together, their product is
. Suppose
. In how many ways can the numbers be chosen?
Solution 1
The positive divisors of are
It is clear that
so we apply casework to
- If
then
- If
then
- If
then
- If
then
Together, the numbers and
can be chosen in
ways.
~MRENTHUSIASM
Solution 2
The positive divisors of are
We apply casework to
:
If , then there are
cases:
If , then there is only
case:
In total, there are ways to choose distinct positive integer values of
.
~MathFun1000
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=tkBYOey2NioTPPPq&t=221
~Math-X
Video Solution 2 (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution 3
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142
~Interstigation
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/Q0R6dnIO95Y?t=98
~STEMbreezy
Video Solution 6
https://www.youtube.com/watch?v=KkZ95iNlFyc
~harungurcan
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.