# Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 6"

## Problem

$A$ is a $4$-digit number $abcd$. $B$ is a $5$-digit number formed by augmenting $A$ with a $3$ on the right, i.e. $B=abcd3$.

$C$ is another $5$-digit number formed by placing a $2$ on the left $A$, i.e. $C=2abcd$. If $B$ is three times $C$, what is the number $A$?

## Solution

What this problem is basically saying is that $10000a + 1000b + 100c + 100d + 3 = 50000 + 1000a+100b+10c+d$. This can be simplified to $9000a + 900b + 90c + 9d = 49997 \Rightarrow 9(1000a+100b+10c+d) = 49997 \Rightarrow 1000a+100b+10c+d = \frac{49997}{9}$. This has no integer solution.