2009 CEMC Gauss (Grade 8) Problems/Problem 19

Problem

In the addition shown, $P$ , $Q$ , and $R$ each represent a single digit, and the sum is $2009$ .

$\begin{array}{rccccc} & & P & Q & P & \\ + & R & Q & Q & Q & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

The value of $P + Q + R$ is

$\text{ (A) }\ 9 \qquad\text{ (B) }\ 10 \qquad\text{ (C) }\ 11 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 13$

Solution

We can notice that from the tenths place, for $Q + Q$ to result in a number ending in $0$ , either $Q$ is $0$ , or it is $5$ .

Let's consider the case where $Q = 0$ . $\begin{array}{rccccc} & & P & 0 & P & \\ + & R & 0 & 0 & 0 & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

From the ones place, that would mean $P = 9$ , since there is no other way for $P + Q$ to result in a number ending in $9$ . $\begin{array}{rccccc} & & 9 & 0 & 9 & \\ + & R & 0 & 0 & 0 & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

However, from the hundreds place, there is no way for $P + Q$ to end in a number ending with $0$ when $P = 9$ and $Q = 0$ without carrying. This means that $Q$ cannot be $0$ , and must be $5$ .

$\begin{array}{rccccc} & & P & 5 & P & \\ + & R & 5 & 5 & 5 & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

From the ones place, $P$ must be $4$ for $P + Q$ to end in a number ending with $9$ :

$\begin{array}{rccccc} & & 4 & 5 & 4 & \\ + & R & 5 & 5 & 5 & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

From the thousandths place, either $R = 1$ or $R = 2$ . However, since there is carrying, $R = 1$ .

$\begin{array}{rccccc} & & 4 & 5 & 4 & \\ + & 1 & 5 & 5 & 5 & \\ \hline & 2 & 0 & 0 & 9 & \\ \end{array}$

From the numbers that we have obtained, $P + Q + R = 4 + 5 + 1 = \boxed {\textbf {(B) } 10}$ .

We can also verify our answer is correct by calculating $1555 + 454$ . This ends up to be $2009$ , which is the same as the sum in the problem.

~anabel.disher

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2009 CEMC Gauss (Grade 8) Problems/Problem 19

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