$5$ and $11$ divides a perfect square $abc0ac$. What is the number?

I started this way - expressing the number as $(10^5+10)a+(10^4)b + (10^3+1)c$
Which fooled me. How can I start?
P.S: This is a problem from BdMO-2016 regionals.

  • $\begingroup$ 5 is a divisor, so the last digit must be 0 or 5. If c is 0, a must be 0, which is false, so c must be 5. Hence, a must be 2. So your task is to find out b. $\endgroup$ – Huang Dec 26 '16 at 12:40
  • $\begingroup$ @Huang Why a must be 0 if c = 0? Didn't get this.. :| $\endgroup$ – Rezwan Arefin Dec 26 '16 at 12:41
  • $\begingroup$ since $5$ is a divisor of this number $\endgroup$ – Dr. Sonnhard Graubner Dec 26 '16 at 12:42
  • $\begingroup$ @RezwanArefin since it's a perfect square. It the ones digit is 0, the ones digit of its root must be 0, so the tens and ones digits are both 0 of that number. $\endgroup$ – Huang Dec 26 '16 at 12:45
  • 1
    $\begingroup$ A natural number is a multiple of $\;11\;$ iff (the sum of its digits in even poisition) minus (the sum of its digits in odd position) is a multiple of $\;11\;$ . With this and knowing $\;c=5\implies a=2\;$ you can solve this at once. $\endgroup$ – DonAntonio Dec 26 '16 at 13:10

Because $25$ divides N it follows $ac \in \{ 00, 25, 50, 75\}$. Because $11$ divides N it follows that the alternating sum of the digits in the number is divisible by 11, therefore $11 | 2a -b$.

Now just take each possibility for $ac$

  • $\begingroup$ Already got solution from Huang's comment. This is also a nice solution :D $\endgroup$ – Rezwan Arefin Dec 26 '16 at 12:50
  • $\begingroup$ @RezwanArefin No, you didn't...and you should have checked that supposed solution is wrong! $\endgroup$ – DonAntonio Dec 26 '16 at 13:05
  • $\begingroup$ @DonAntonio Yep. ! I also got the wrong solution .. Thanks $\endgroup$ – Rezwan Arefin Dec 26 '16 at 13:09

I will post an answer for completion.
If the number is divisible by $5$ then $c=5$ or $c=0$. It cannot equal $0$ by Huang's comment. Therefore $c=5$ and because a perfect square ending in $5$ necessarily ends in $25$ we know that $a=2$.
If a number is divisible by $11$ then the alternating sum of its digits: $a-b+c-0+a-c=2a-b=4-b=11n$ (i.e. the sum must be a multiple of $11$). Because $0\le b \le 9$, then $n$ can only equal $0$, from which we know that $b=4$ and the number equals $$245025=495^2$$


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.