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Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 19"

(Created page with "==Problem== In the addition shown, <math>P</math>, <math>Q</math>, and <math>R</math> each represent a single digit, and the sum is <math>2009</math>. <cmath>\begin{array}{r...")
 
 
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From the numbers that we have obtained, <math>P + Q + R = 4 + 5 + 1 = \boxed {\textbf {(B) } 10}</math>.
 
From the numbers that we have obtained, <math>P + Q + R = 4 + 5 + 1 = \boxed {\textbf {(B) } 10}</math>.
 +
 +
We can also verify our answer is correct by calculating <math>1555 + 454</math>. This ends up to be <math>2009</math>, which is the same as the sum in the problem.
  
 
~anabel.disher
 
~anabel.disher

Latest revision as of 12:56, 19 June 2025

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

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