BdMO National Higher Secondary 2011/9

[Masum]

No one didn't post a full solution!
So,here is one.
Let $N$ has $k$ digits.Then the repeat of $N,\overline{N}=(10^k+1)N$.
Note that if $10^k+1$ is square-free,then we need $10^k+1$ to have a square factor.
From the tricky lemma,$10^k+1$ is not square-free for $k=11$
Then $11^2|10^{11}+1$
Now note that $ 100000000001=11^{2}*23*4093*8779 $
$\overline{N}=11^{2}*23*4093*8779*N$ is a square.This suggests us to take a number divisible by $23*4093*8779$ which is a square.Now we need to check by hand to verify that $4093,8779$ are primes.Then a convenient choice for $ N=a^2*23*8779*4093$ for some $a$
But $N$ must be a $11$ digit number,as we derived.Check with some smaller values of $a$.We get $a=10$ because $23*4093*8779=826446281$ which is a $9$ digit number.Thus,$N=82644628100$ is such a number that satisfies the condition.
Try to find more such numbers if you can.

[sm.joty]

Can any participant of our national Olympiad solve this problem. This problem is really so much hard for us.

[Nadim Ul Abrar]

But $N$ must be a $11$ digit number,as we derived.Check with some smaller values of $a$.We get $a=10$ because $23*4093*8779=826446281$ which is a $9$ digit number.Thus,$N=82644628100$ is such a number that satisfies the condition.

but
$26446281002644628100 = 2^2.3.5^2.101.3541.27961.8815427 $

And what about this one's ??
$1322314049613223140496,2975206611629752066116,5289256198452892561984$