Question 661703
Let x = the number
Let n = the hundreds digit = the ones digit
Then the tens digit = 2n
The 3-digit product can be written as
x^2 = 100n + 10(2n) + n = 121n
This gives x = 11{{{sqrt(n)}}}
The product is three digits, and {{{sqrt(n)}}} must be an integer
If n = 1, x = 11 -> x^2 = 121
If n = 4, x = 22 -> x^2 = 484
It seems both of these meet the conditions of the problem, so the possible answers are 11 and 22