2000 CEMC Gauss (Grade 8) Problems/Problem 4

The following problem is from both the 2000 CEMC Gauss (Grade 8) #4 and 2000 CEMC Gauss (Grade 7) #6, so both problems redirect to this page.

Problem

In the addition shown, a digit, either the same or different, can be placed in each of the two boxes. What is the sum of the two missing digits?

$\text{ (A) }\ 9 \qquad\text{ (B) }\ 11 \qquad\text{ (C) }\ 13 \qquad\text{ (D) }\ 3 \qquad\text{ (E) }\ 7$

Let $x$ be the digit that is missing in the tens place, and $y$ be the other missing digit.

$3 + 1 + 8 = 12$ , which is bigger than $10$ , so we must add $1$ to the tens place.

$6 + 9 + x + 1 = 16 + x$ must end in $8$ , which happens when $x = 2$ . There is also a 1 carried to the hundreds place.

$8 + 7 + y + 1 = 16 + y$ must end in $1$ , which happens when $y = 5$ .

We can now see that $x + y = 2 + 5 = \boxed {\textbf {(E) } 7}$ .

~anabel.disher

2000 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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CEMC Gauss (Grade 8)

2000 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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CEMC Gauss (Grade 7)