Difference between revisions of "2002 AMC 10A Problems/Problem 16"
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<math> -1 = a + b + c</math>, | <math> -1 = a + b + c</math>, | ||
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 [The Apple Method])== | ||
| + | https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo | ||
==See Also== | ==See Also== | ||
Revision as of 12:25, 28 April 2020
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
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. 
