Difference between revisions of "2002 AMC 10A Problems/Problem 16"
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==Solution 2== | ==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</math>, and <math>d + 4</math> we get | + | Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get: |
<math> -4 = b + c + d</math>, | <math> -4 = b + c + d</math>, | ||
<math> -3 = a + c + d</math>, | <math> -3 = a + c + d</math>, | ||
Revision as of 18:29, 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:
,
,
, and
,
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. 
