2024 AMC 8 Problems/Problem 5
Contents
- 1 Problem
- 2 Solution 1
- 3 Video Solution (A Clever Explanation You’ll Get Instantly)
- 4 Video Solution 1 (easy to digest) by Power Solve
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution 2 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution by Interstigation
- 10 Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)
- 11 See Also
Problem
Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of . Which of the following integers cannot be the sum of the two numbers?
Solution 1
First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:
is possible:
is possible:
is possible:
is possible:
The only integer that cannot be the sum is
-ILoveMath31415926535 & countmath1 & Nivaar
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=eqTGEN2doXnn-oZE&t=443
~hsnacademy
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/HE7JjZQ6xCk?si=2M33-oRy1ExY3_6l&t=239
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=W8N4vwOx84KauOA6&t=1108
~Math-X
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=Ylw-kJkSpq8
~NiuniuMaths
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=51pqs80PUnY
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=296
Video Solution by Daily Dose of Math (Understandable, Quick, and Speedy)
https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR
~Thesmartgreekmathdude
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.