Problem - 2382
Let $A$ and $B$ be two positive integers and $A=B^2$. If $A$ satisfies the following conditions, find the value of $B$:
- $A$'s thousands digit is $4$
- $A$'s tens digit is $9$
- The sum of all $A$'s digits is $19$
Answer
64
First, let's consider the last two digits of $A$. Because its tens digit is odd, its units digit must be $6$ (see # 4151). Now, the sum of its digits is already $19$. Hence, its hundreds digit must be $0$. In fact, we find $4096 = 64^2$ indeed is a square. Hence, $B=\boxed{64}$.