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, $a + b + c + d = \boxed{\frac{-10}{3}}$ .

Solution 3(Video solution using The Apple Method)

https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo https://youtu.be/NRdOxPUDngI

Video Solution

https://youtu.be/tKsYSBdeVuw?t=4105

~ pi_is_3.14

2002 AMC 10A (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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

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