Difference between revisions of "1999 AMC 8 Problems/Problem 3"
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By adding each triplet, we can see that <math>\boxed{(D)}</math> gives us <math>0</math>, not <math>1</math>, as our sum. | By adding each triplet, we can see that <math>\boxed{(D)}</math> gives us <math>0</math>, not <math>1</math>, as our sum. | ||
+ | |||
+ | ==Video(By YippieMath)== | ||
+ | https://www.youtube.com/watch?v=nMNIR7PfsG0&t=15s | ||
==See Also== | ==See Also== |
Latest revision as of 17:31, 17 May 2025
Problem
Which triplet of numbers has a sum NOT equal to 1?
Solution
By adding each triplet, we can see that gives us
, not
, as our sum.
Video(By YippieMath)
https://www.youtube.com/watch?v=nMNIR7PfsG0&t=15s
See Also
1999 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.