The base-10 integers 36, 64, and 81 can be converted into other bases so that their values are represented by the same digits The base-10 integers 36, 64, and 81 can be converted into other bases so that their values are represented by the same digits $\triangle\Box\Box$, where $\triangle$ and $\Box$ are two distinct digits from 0-9. What is the value of $\triangle\Box\Box$?
This problem is very interesting to me. I don't have a lot of experience with changing the base. Any hints are greatly appreciated.
 A: Hint:
$$
36=6^2 \qquad 64=8^2 \qquad 81=9^2
$$
A: Let us call the number you get $xyy$.  In base $b,xyy_b$ represents $xb^2+yb+1$.  We know that $b^2 \le 36 \lt b^3$ for its base to have three digits, so $3 \lt b \le 6$ and we have only three possibilities.  We can compute $36_{10}=210_4=121_5=100_6$ and only the last can be $xyy$.  This means each number is the square of its base, so $64=100_8,81=100_9$
A: Cute.
Note: $36, 64, 81$ are all perfect squares so they are all in the form $1*b^2 + 0*b + 0 = 100_b$. So $\triangle = 1; \Box = 0$ is certainly a solution.  But is it the only solution?
$\triangle\Box\Box_b = b^2*\triangle + b*\Box + \Box = 36$.
If we assume $\triangle \ne 0$ we have $36 = 6^2 + 0 + 0$; So the lowest possible base can be $6$  $36 = 100_6=121_5;210_4;1100_3$ which is more than $3$ digits. Only $100$ is of the form $\triangle\Box\Box$.
If we allow for $\triangle = 0$ we have $\triangle\Box\Box = 0\Box\Box = \Box*b + \Box = \Box(b+1)$ for some base $b$.
Thus we have $\Box(b+1) = 36;\Box(c+1) = 64; \Box (d+1) = 81$ so $\Box$ is a common divisor of $36, 64$ and $81$.  $1$ is the only common factor.
So $\triangle\Box\Box = 011$ will work for $011_{35} = 36; 011_{63}=64;011_{80} = 81$.  But those answers are ... perverse.
I am certain, that $\triangle = 0$ was not to be considered.
So I think we can safely so $\triangle = 1; \Box = 0$ for bases $6,8,9$.
