2026/027) Let n be a positive integer such that $12n^2+12n+11$ is a 4-digit number with all 4 digits equal. Determine the value of n.

All 4 digits are same

Let it be 1111x . Add 1 in both sides to get

$12n^2 + 12n + 12 = 1111x + 1$

Work mod 12 to get 

$7x \equiv = 0 \pmod {12}$

So x is odd and 

Trying x 1 3 5 9 11 we get x is 5 .

So $12n^2 + 12n + 12 = 5556$

Or $n^2 + n + 1 = 463$

Or $n = 21$