A Copy-and-Paste Square

Suppose that n is the number sought, and that it has k digits. This means that $10^{k-1}\le n < 10^k$ . According to our assumption, $10^kn+n=(10^k+1)n$ is a perfect square.

I claim that $10^k+1$ must have a perfect square factor (other than 1). If this were not the case, then $10^k+1$ would be a product of distinct primes $p_1p_2\cdots p_m$ . Since $(10^k+1)n$ is a perfect square, all of these $p_i$ 's must also divide n . This would mean that $n \ge 10^k+1$ , which is impossible.

So we need to find a number of the form $10^k+1$ that has a perfect square factor. The smallest such number is $10^{11}+1$ , which is divisible by $11^2=121$ .

In order for $(10^{11}+1)n=11^2\left(\frac{10^{11}+1}{121}\right)n$ to be a perfect square, $\left(\frac{10^{11}+1}{121}\right)n$ must be a perfect square. So n must be equal to $\frac{10^{11}+1}{121}\cdot a^2$ for some natural number a . But we can't simply let $a=1$ , the smallest natural number, because n needs to have $k=11$ digits, and $\frac{10^{11}+1}{121}=826446281$ has nine digits. The smallest natural number a for which $\frac{10^{11}+1}{121}\cdot a^2$ has eleven digits is 4. So $n=\frac{10^{11}+1}{121}\cdot 4^2=13223140496$ , and our answer is $\boxed{496}$ .

(For the record, I adapted both this problem and this solution from Mathematical Olympiad Challenges by Andreescu & Gelca.)