Difference between revisions of "2022 AMC 8 Problems/Problem 3"
Mathfun1000 (talk | contribs) m |
MRENTHUSIASM (talk | contribs) (→Solution 2: Made the start of both solutions the same. Minor stylistic change.) |
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==Solution 2== | ==Solution 2== | ||
− | The positive divisors of <math>100</math> are < | + | The positive divisors of <math>100</math> are <cmath>1,2,4,5,10,20,25,50,100.</cmath> |
+ | We can do casework on <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: |
Revision as of 03:31, 17 January 2023
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 can do casework on
:
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
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142 ~Interstigation
Video Solution 2
~savannahsolver
Video Solution
https://youtu.be/Q0R6dnIO95Y?t=98
~STEMbreezy
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.