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2002 AMC 10A Problems/Problem 16

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

See Also

2002 AMC 10A (Problems • 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

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