2025 CSAT — Q40

The right-hand side is the famous repunit-square palindrome: a string of $1$s squared produces $1, 121, 12321, 1234321, \dots$ rising to a peak digit equal to the number of $1$s, then descending. The target $12345678987654321$ peaks at $9$, so it is

Hence $N = 111111111$, which has $9$ digits.

(Confirm by length: $N^2$ has $17$ digits; a number with $d$ digits has a square with $2d-1$ or $2d$ digits, so $2d - 1 = 17 \Rightarrow d = 9$.)

Answer: (b) 9.

Two independent routes both land instantly. Pattern recognition: the palindrome $123\ldots n\ldots321$ is exactly $(\underbrace{1\ldots1}_{n})^2$, and its peak digit names the count of $1$s — here the peak is $9$, so $N$ has $9$ ones. Digit-length sanity: a $d$-digit number squares to a $(2d-1)$- or $2d$-digit number; $N^2$ shows $17$ digits, and $2d-1 = 17 \Rightarrow d = 9$ pins it without recognising the pattern at all. Carrying both methods means you answer in seconds and self-check for free — the hallmark of a prepared CSAT solver on a “freebie” item.