If $N^2 = 12345678987654321$, then how many digits does the number N have?
If $N^2 = 12345678987654321$, then how many digits does the number N have?
8
9
10
11
Solution
The number 12345678987654321 is the well-known square of the repunit $111111111$ (nine 1s). That is, $(111111111)^2 = 12345678987654321$. Hence $N = 111111111$, which has 9 digits.