Difference between revisions of "2002 AMC 10A Problems/Problem 16"
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Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \boxed{\frac{-10}{3}}</math>. | Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \boxed{\frac{-10}{3}}</math>. | ||
− | ==Solution 3(Video solution using | + | ==Solution 3(Video solution using [The Apple Method])== |
https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo | https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo | ||
https://youtu.be/NRdOxPUDngI | https://youtu.be/NRdOxPUDngI | ||
− | == Video Solution == | + | == Video Solution by OmegaLearn == |
https://youtu.be/tKsYSBdeVuw?t=4105 | https://youtu.be/tKsYSBdeVuw?t=4105 | ||
Latest revision as of 15:32, 13 June 2024
Contents
Problem
Let . What is
?
Solution 1
Let . Since one of the sums involves
and
it makes sense to consider
. We have
. Rearranging, we have
, so
. Thus, our answer is
.
Solution 2
Take
.
Now we can clearly see:
.
Continuing this same method with
, and
we get:
,
,
, and
,
Adding, we see
. Therefore,
.
Solution 3(Video solution using [The Apple Method])
https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo https://youtu.be/NRdOxPUDngI
Video Solution by OmegaLearn
https://youtu.be/tKsYSBdeVuw?t=4105
~ pi_is_3.14
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.