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  1. A $10$-digit number has one $1$, two $2$'s, three $3$'s and four $4$'s as its digits in some order. Prove that it can never be a perfect square.
Martin Sleziak
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Anish Bhattacharya
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1 Answers1

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The digit sum $30$ is divisible by $3$ but not by $9$. So our number is divisible by $3$ but not by $9$, and therefore cannot be a perfect square.

André Nicolas
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  • why can't it be a perfect square if it's divisible by 3 and not by 9? Is there a proof to that statement? – Anish Bhattacharya Dec 28 '13 at 07:16
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    @AnishBhattacharya,Just think about it for a while.Since the number is divisible by 3,it must be of the form $3k$. What happens when you square it? – rah4927 Dec 28 '13 at 07:17
  • Here is a proof. Suppose that $n=m^2$ and $3$ divides $n$. Since $3$ is prime, and $3$ divides the product $m\times m$, it follows that $3$ divides $m$. But then $9$ divides $m^2$. – André Nicolas Dec 28 '13 at 07:19
  • Thanks a lot!! I get it now. that's a rigorous proof. I hope I didn't waste both of your time. Thanks again! – Anish Bhattacharya Dec 28 '13 at 07:28
  • You are welcome. It is a complete proof if (as is likely) you have proved already in your course that if $n$ is a non-negative integer written in decimal form, then the remainder when you divide $n$ by $9$ is the same as the remainder when you divide the digit sum of $n$ by $9$. Or more weakly $n$ is divisible by $3$ if and only if the digit sum is, and is divisible by $9$ if and only if the digit sum is. – André Nicolas Dec 28 '13 at 07:32
  • Anish, this may help: Write out any number with 1 ones, 2 twos, 3 threes and 4 fours. Verify that it is divisible by 3 and not 9. Now since the number is divisible by 3, its square root should also be divisible by 3. But if the square root is divisible by 3, the number should be divisible by 9 but you know it is not. Hope this long explanation helps – user44197 Dec 28 '13 at 07:33