Xavier writes down a three-digit positive integer. Youssef writes a in front of the digits of Xavier's number, so that the resulting number has four digits. If this four-digit number is times as large as Xavier's original number, what number did Xavier write down?
So here's my approach:
we know xaviers number is in the range of 100-999
and we know it has to be a multiple of 13
and the number we are trying to find is between 1000-9999
and it has to start with a 3.
"If this four-digit number is times as large as Xavier's original number, what number did Xavier write down"? How many times larger?
Let Xavier's three-digit number be represented as $ABC$, where $A$, $B$, and $C$ are digits representing the hundreds, tens, and ones digits, respectively. Then, Youssef's four-digit number is $3ABC$.
Since the four-digit number is 13 times as large as the original number, we have:
3ABC=13⋅ABC
Simplifying this equation, we get:
3000+100A+10B+C=13(100A+10B+C)
Expanding the right side, we get:
3000+100A+10B+C=1300A+130B+13C
Simplifying this equation further, we get:
1170A+120B=2999
Since $A$ and $B$ are digits, the left-hand side of this equation is an integer between 0 and 999. Therefore, the only possible value for $A$ is 2, and the only possible value for $B$ is 4. Substituting these values into the equation, we get:
1170(2)+120(4)=2520+480=2999
Therefore, Xavier's original number is $ABC = \boxed{240}$.