0
$\begingroup$

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 :)

$\endgroup$

closed as off-topic by heropup, user133281, TMM, KonKan, Alex Mathers Oct 1 '16 at 1:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, user133281, TMM, KonKan, Alex Mathers
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.