Difference between revisions of "2002 AMC 10A Problems/Problem 16"
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<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math> | <math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>. | Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 15:12, 27 November 2015
Contents
Problem
Let 
. What is 
?
Solution 1
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. 
