Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 CEMC Gauss (Grade 8) 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...")
 
 
Line 63: Line 63:
  
 
~anabel.disher
 
~anabel.disher
 +
{{CEMC box|year=2009|competition=Gauss (Grade 8)|num-b=18|num-a=20}}

Latest revision as of 21:28, 19 October 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

2009 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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)
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_CEMC_Gauss_(Grade_8)_Problems/Problem_19&oldid=263646"