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

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

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

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2002 AMC 10A Problems/Problem 16

Contents

Problem

Solution

Solution 2

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