# 2002 AMC 10A Problems/Problem 16

## 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 1

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 $\boxed{\text{(B)}\ -10/3}$.

## Solution 2

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: $-4 = b + c + d$, $-3 = a + c + d$, $-2 = a + b + d$, and $-1 = a + b + c$, Adding, we see $-10 = 3a + 3b + 3c + 3d$. Therefore, $a + b + c + d = \boxed{\frac{-10}{3}}$.

## Video Solution by OmegaLearn

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