Difference between revisions of "2000 CEMC Gauss (Grade 8) Problems/Problem 4"

Latest revision as of 13:41, 20 October 2025

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.

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
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
CEMC Gauss (Grade 8)

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

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+{{Duplicate|[[2000 CEMC Gauss (Grade 8) Problems|2000 CEMC Gauss (Grade 8) #4]] and [[2000 CEMC Gauss (Grade 7) Problems|2000 CEMC Gauss (Grade 7) #6]]}}
 ==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?
@@ Line 16: / Line 17: @@
 ~anabel.disher
+{{CEMC box|year=2000|competition=Gauss (Grade 8)|num-b=3|num-a=5}}
+{{CEMC box|year=2000|competition=Gauss (Grade 7)|num-b=5|num-a=7}}