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

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

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Say that $a + 1 = a + b + c + d + 5$

With substitution we get

Following this pattern with $b + 2$ , $c + 3$ , and $d + 4$ we can get: $-4 = b + c + d$ $-3 = a + c + d$ $-2 = a + b + d$ $-1 = a + b + c$ Adding, we can get $-10 = 3a + 3b + 3c + 3d$ , therefore $a +b + c + d$ = $-10/3$

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2002 AMC 10A Problems/Problem 16

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