$\large N=\underbrace{11111 \cdots 11111}_{\text{Number of 1s }= 2018}$

What is the digit sum of $N^2$ ?

The answer is 18148.

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Lets first try to develop a pattern:

$1^{2}$ = $1$ | $1$ = $1^{2}$

$11^{2}$ = $121$ | $1$ + $2$ + $1$ = $2^{2}$

$111^{2}$ = $12321$ | $1$ + $2$ + $3$ + $2$ + $1$ = $3^{2}$

$1111^{2}$ = $1234321$ | $1$ + $2$ + $3$ + $4$ + $3$ + $2$ + $1$ = $4^{2}$

...$1111111111^{2}$ = $12345678987654321$ | $1$ + $2$ + $3$ + $4$ + $5$ + $6$ + $7$ + $8$ + $9$ + $8$ + $7$ + $6$ + $5$ + $4$ + $3$ + $2$ + $1$ = $9^{2}$

$11111111111^{2}$ = $12345678900987654321$ | $1$ + $2$ + $3$ + $4$ + $5$ + $6$ + $7$ + $8$ + $9$ + $0$ + $0$ + $9$ + $8$ + $7$ + $6$ + $5$ + $4$ + $3$ + $2$ + $1$ = $1^{2}$ + $9^{2}$

The pattern that appears is every time $9$ ones are added, the sum of its digits increase by $9^{2}$ (after the number is squared). While every individual $1$ added, increases the value being squared by $1$ (after the number is squared). Using this, we can divide the number of ones $N$ has, by $9$ . This leaves us with: $2018$ = ( $224$ x $9$ ) + $2$ . So the final equation will have to include the $224$ sets of $9^{2}$ and also $2^{2}$ for the remainder of $2$ .