Difference between revisions of "2002 AMC 10A Problems/Problem 16"

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==Alternate Solution==
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<math>a + 1 = a + b + c + d + 5</math>
 
<math>a + 1 = a + b + c + d + 5</math>

Revision as of 19:25, 17 September 2014

Problem

Let $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. What is $a + b + c + d$?

$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$

Solution

Let $x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have $4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20$. Rearranging, we have $3(a+b+c+d)=-10$, so $a+b+c+d=\frac{-10}{3}$. Thus, our answer is $\boxed{\text{(B)}\ -10/3}$.


Also: Take $a + 1 = a + b + c + d + 5$ Now we can clearly see: $-4 = b + c + d$ Continuing this same method with $b + 2, c + 3, and d + 4$ we get altogether $-4 = b + c + d$ $-3 = a + c + d$ $-2 = a + b + d$ $-1 = a + b + c$ Adding, we see $-10 = 3a + 3b + 3c + 3d$. Therefore, $a + b + c + d = \frac{-10}{3}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
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

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