There is a $3$ digit number such that the product of the digits at tens place and hundreds place is $17$ greater than a perfect square number and the sum of the $3$ digits is a perfect square number. Again the digit at tens place is the average of a square number and a cubic number. What is the minimum value of this 3 digit number?

This is a problem from BdMO 2006 Regionals.. I cant find a way to start with... Any hint will be helpful :)


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Take the meaning of perfect square and cube numbers to exclude zero.

First, considder the tens digit. What digits (numbers from $1$ to $9$) can be the average of a square and a cube? The cube must be $1$ or $8$; $27$ is too big. For $8$, the only case that works is $\frac{2^3+2^2}{2}=6$. For $1$, the answer could be $1$ or $5$. So $t$ is $1, 5,$ or $6$.

Now list all the numbers up to $81$ (which is as big as the product of two digits can get) that are of the form $n^2+17: \{18,21,26,33,42,53,66,81\}$. We immediately see that $t$ can't be one, because $h$ would have to be at least $18$, and $t$ can't be $5$ because for this range, $n^2+17$ is never a multiple of $5$. Then $t=6$.

But only three of the numbers in our list are divisible by $6$. The first to try is $18$. That would give a number of the form $36\cdot$ and you need to choose the last digit to make the sum $3+6+\cdot$ a perfect square. You can take it from there.


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