Difference between revisions of "2002 AMC 10A Problems/Problem 16"

(Solution)
(Solution 2)
Line 17: Line 17:
 
<math> -4 = b + c + d</math>,  
 
<math> -4 = b + c + d</math>,  
 
<math> -3 = a + c + d</math>,
 
<math> -3 = a + c + d</math>,
<math> -2 = a + b + d</math>,  
+
<math> -2 = a + b + d</math>, and
 
<math> -1 = a + b + c</math>,
 
<math> -1 = a + b + c</math>,
 
Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \frac{-10}{3}</math>.
 
Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \frac{-10}{3}</math>.

Revision as of 19:28, 17 September 2014

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

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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