+812 A four-digit hexadecimal integer is written on a napkin such that the units digit is illegible. The first three digits are 2, $F$, and 1. If the integer is a multiple of $19_{10}$, what is the units digit?
+8 Let x be the units digit. The value of the hexadecimal integer in base 10 is \(2 \cdot 16^3 + 15 \cdot 16^2 +1 \cdot 16 +x = 12048+x\)
The smallest multiple of 19 greater than 12048 is 12065. In order for 12048 + x to be a multiple of 19, x has to be 17 which is not possible in hexadecimal. Therefore, it is not possible.
