1
$\begingroup$

$6, 96, 996, 9996, ... , 99999...996, ...$

Consider the above number sequence. Here in the $n$ $^{th}$ term $n-1$ digits are $9$. How can we tell about the existence of a perfect square number in this sequence?

$\endgroup$
  • 3
    $\begingroup$ If the perfect square were $10^{2n}-4$, then its factors would be $10^n+2,10^n-2$. So the perfect square must be of the form $10^{2n-1}-4$. $\endgroup$ – abiessu Mar 4 '16 at 17:57
16
$\begingroup$

The sum of the digits of the $n^{th}$ term of this sequence is $$\text{Sum}=9(n-1)+6=9n-3=3(3n-1)$$ Thus, every number in the sequence is divisible by $3$ but not by $9$. Hence, no number in the sequence can be a perfect square.

$\endgroup$
5
$\begingroup$

HINT: consider divisibility by a suitable small number $n$ and also by $n^2$

$\endgroup$

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.