Saurav's Square #3

Let $a$ and $b$ be the two 7-digit numbers in the problem. Then $10^7 a + b = b^2$ , so $10^7 a = b(b-1)$ . Exactly one of $b$ and $b-1$ is even, so whichever one it is has to be divisible by $2^7$ . Exactly one of $b$ and $b-1$ is divisible by $5$ , so whichever one it is has to be divisible by $5^7$ . Since $b < 10^7$ , we can't have chosen the same factor for both divisibility requirements.

So there are two cases: $2^7 | b$ and $5^7 | b-1$ , or $2^7 | b-1$ and $5^7 | b$ . These are simple Chinese Remainder Theorem calculations. We get $b \equiv 7109376$ mod $10^7$ and $b \equiv 2890625$ mod $10^7$ respectively. I don't know if $T$ was allowed to be $0$ (since $2890625^2 = 08355712890625$ ), but in any event $7109376$ is the larger of the two. We can check that $7109376^2 = 50543227109376.$

There are 2 numbers satisfying the condition : 2890625 and 7109376. So, largest number is 7109376.