Find the two-digit positive integer

Let the 2 digit number be 10t + u, where t is the tens digit, u is the units digit. We then have the following 2 equations: (1) t^2 + u^2 = 41, which tells us that t,u are 4 and 5 in some order. (2) 10t + u = 6(t + u) = 6t + 6u. Then 4t = 5u, and it is clear that t = 5, u = 4; so 10t + u = 54.

The only way to write 41 as the sum of two squares is $41 = 4^2 + 5^2 = 5^2 + 4^2$ .

So the number either is 45 or 54.

We can simply check if these satisfy the second condition and in fact, $54 = 6 \cdot (5 + 4)$ while $45 \neq 6 \cdot (4 + 5)$ .

So the answer is $\boxed{54}$ .