2014 CEMC Gauss (Grade 7) Problems/Problem 20

Problem

The product of three consecutive odd numbers is $9177$ . What is the sum of the numbers?

$\textbf{(A)}\ 51 \qquad\textbf{(B)}\ 57 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 63 \qquad\textbf{(E)}\ 69$

Solution

Let $x$ be the second largest number. We then have $x - 2$ as the smallest number and $x + 2$ as the largest because the three numbers are consecutive odd numbers.

Their product is then:

$(x - 2) \times x \times (x + 2) = 9177$

Factoring $9177$ , we have:

$(x - 2) \times x \times (x + 2) = 3 \times 7 \times 19 \times 23$

We can see that $(x - 2) = 19$ , $x = 3 \times 7 = 21$ , and $x + 2 = 23$ from this. We could have also obtained this result by seeing that $21 \times 23 \times 25 = 483 \times 25 = 48300 \div 4 = 12075$ is too large but the number must be around $20 \times 20 \times 20 = 8000$ , and then seeing $19 \times 21 \times 23 = 9177$ .

The sum is then:

$(x - 2) + x + (x + 2) = 3x = 3 \times 21 = \boxed {\textbf {(D) } 63}$

~anabel.disher

2014 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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)

Page

Toolbox

Search

2014 CEMC Gauss (Grade 7) Problems/Problem 20

Problem

Solution