Difference between revisions of "2002 AMC 10A Problems/Problem 16"
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| − | + | ==Solution 2== | |
Take | Take | ||
<math>a + 1 = a + b + c + d + 5</math> | <math>a + 1 = a + b + c + d + 5</math> | ||
Now we can clearly see: | Now we can clearly see: | ||
<math>-4 = b + c + d</math> | <math>-4 = b + c + d</math> | ||
| − | Continuing this same method with <math>b + 2, c + 3, and d + 4</math> we get altogether | + | Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get altogether |
| − | <math> -4 = b + c + d</math> | + | <math> -4 = b + c + d</math>, |
| − | <math> -3 = a + c + d</math> | + | <math> -3 = a + c + d</math>, |
| − | <math> -2 = a + b + d</math> | + | <math> -2 = a + b + d</math>, |
| − | <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 = \frac{-10}{3}</math>. | Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \frac{-10}{3}</math>. | ||
Revision as of 18:27, 17 September 2014
Contents
Problem
Let 
. What is 
?
Solution
Let 
. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. 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 altogether
,
,
,
,
Adding, we see 
. Therefore, 
.
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. 
