Difference between revisions of "2001 CEMC Gauss (Grade 8) Problems/Problem 9"

Latest revision as of 17:52, 20 October 2025

The following problem is from both the 2001 CEMC Gauss (Grade 8) #9 and 2001 CEMC Gauss (Grade 7) #14, so both problems redirect to this page.

In the square shown, the numbers in each row, column, and diagonal multiply to give the same result. The sum of the two missing numbers is

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$\text{ (A) }\ 28 \qquad\text{ (B) }\ 15 \qquad\text{ (C) }\ 30 \qquad\text{ (D) }\ 38 \qquad\text{ (E) }\ 72$

Let $X$ and $Y$ be the missing numbers, like this:

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From the middle row and the middle column, we see:

$1 \times 6 \times Y = 9 \times 6 \times 4$

This means $Y = 9 \times 4 = 36$

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From the bottom row and middle column, we have:

$12 \times 9 \times X = 1 \times 6 \times 36$

$X = \frac{1 \times 6 \times 36}{12 \times 9} = \frac{6 \times 9 \times 4}{6 \times 2 \times 9} = \frac{4}{2} = 2$

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We then see that $X + Y$ , like the problem asked for, is $2 + 36 = \boxed {\textbf {(D) } 38}$ .

~anabel.disher

2001 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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)

2001 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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|[[2001 CEMC Gauss (Grade 8) Problems|2001 CEMC Gauss (Grade 8) #8]] and [[2001 CEMC Gauss (Grade 7) Problems|2001 CEMC Gauss (Grade 7) #11]]}}
+{{Duplicate|[[2001 CEMC Gauss (Grade 8) Problems|2001 CEMC Gauss (Grade 8) #9]] and [[2001 CEMC Gauss (Grade 7) Problems|2001 CEMC Gauss (Grade 7) #14]]}}
 ==Problem==
 In the square shown, the numbers in each row, column, and diagonal multiply to give the same result. The sum of the two missing numbers is