Difference between revisions of "2022 AMC 8 Problems/Problem 3"
Happydolphin (talk | contribs) (→Solution 1) |
MRENTHUSIASM (talk | contribs) |
||
Line 19: | Line 19: | ||
Together, the numbers <math>a,b,</math> and <math>c</math> can be chosen in <math>\boxed{\textbf{(E) } 4}</math> ways. | Together, the numbers <math>a,b,</math> and <math>c</math> can be chosen in <math>\boxed{\textbf{(E) } 4}</math> ways. | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
Line 39: | Line 41: | ||
In total, there are <math>3+1=\boxed{\textbf{(E) } 4}</math> ways to choose distinct positive integer values of <math>a,b,c</math>. | In total, there are <math>3+1=\boxed{\textbf{(E) } 4}</math> ways to choose distinct positive integer values of <math>a,b,c</math>. | ||
− | ==Video Solution | + | ~MathFun1000 |
+ | |||
+ | ==Video Solution== | ||
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142 | https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=142 | ||
+ | ~Interstigation | ||
==Video Solution 2== | ==Video Solution 2== | ||
https://youtu.be/LHnC_Wz6fOU | https://youtu.be/LHnC_Wz6fOU | ||
− | ==Video Solution | + | ~savannahsolver |
+ | |||
+ | ==Video Solution== | ||
https://youtu.be/Q0R6dnIO95Y?t=98 | https://youtu.be/Q0R6dnIO95Y?t=98 | ||
+ | |||
+ | ~STEMbreezy | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=2|num-a=4}} | {{AMC8 box|year=2022|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:24, 20 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.