-2
$\begingroup$

$abc0a$c is a six digit perfect square number which is divisible by 5 and 11. Find out the number.

Source: Bangladesh Math Olympiad 2016 Junior Category

I tried but failed to find any relation between 5, 11 and a square number. I have also seen a similar question with 13 and 15. I found that last digits of the square number are 0 or 5. Can anyone give me necessary hints to solve this problem.

$\endgroup$
  • 2
    $\begingroup$ There is an easy test for multiples of $11$. There are limited options for the final two digits of a square divisible by $5$. $\endgroup$ – Mark Bennet Jan 24 at 17:22
3
$\begingroup$

We need to use divisibility rules.

First, because the number is divisible by 5, we know that $c=0,5$.

But, if $c=0$, then $a=0$, since that implies that the 6 digit number is divisible by 10, implying that it also divides 100. But, this would make the number not a 6 digit number. Hence, $c=5$.

But, if $c=5$, $a=2$, since that means that the number divides 5 but not 2, implying that the last two digits of the number must be 25 (any odd number squared is congruent to 1 mod 4). So, we know our number is of the form $2b5025$.

Next, since the number is divisible by 11, $a+c+a-b-c=2a-b$ is divisible by 11. So, 11 divides $4-b$, which implies $b=4$. So, our final answer is $$245025$$

$\endgroup$
  • $\begingroup$ $$(10d+5)^2=100d(d+1)+25$$ $\endgroup$ – lab bhattacharjee Jan 24 at 17:47
4
$\begingroup$

Since it's a perfect square divisible by $5$ and $11$, it is divisible by $25$ and $121$. This immediately tells you that the last two digits $ac$ is one of $00$, $25$, $50$ or $75$.

Using divisibility test for $11$:

Case $00$: $0b0000$ is not divisible by $11$ for any $b$.

Case $25$: $2b5025$ gives you $b = 4$ if it is divisible by $11$ and is a perfect square.

Case $50$: $5b0050$ is not divisible by $11$ for any $b$.

Case $75$: $7b5075$ gives you $b = 3$ if it is divisible by $11$ but it is not a perfect square.

So the answer is 245025.

$\endgroup$
  • $\begingroup$ Nice answer! My only critique is that you don't really use the fact that it divides 121, but only 11. The last two digits being 00, 25, 50, or 75 comes exclusively from dividing 25. $\endgroup$ – Don Thousand Jan 24 at 17:25
  • $\begingroup$ Also, $b=0$ satisfies your Case 00 trivially :) $\endgroup$ – Don Thousand Jan 24 at 17:34
  • $\begingroup$ Case 00 should be ruled out by saying “$0b0000$ is not a six-digit number”. $\endgroup$ – MJD Jan 24 at 17:35
  • $\begingroup$ @MJD Yea, that's the argument I used. $\endgroup$ – Don Thousand Jan 24 at 17:38
0
$\begingroup$

The number is a perfect square which is divisible by 55, so it is divisible by $55^2=3025$

We look at the square multiples of $3025$ and find $81*3025 =245025$ to fit the desired format.

$\endgroup$
-1
$\begingroup$

Hint. If square number is divisible by prime $p$, it's divisible by $p^2$

$\endgroup$
  • 3
    $\begingroup$ 14 is not a square, so the hint of Vasily is not injured by your correct observation. $\endgroup$ – Daniele Tampieri Jan 24 at 17:58
  • 1
    $\begingroup$ @DanieleTampieri Regardless, this isn't really an answer. $\endgroup$ – Don Thousand Jan 24 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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