I guess it should be tens digit.sud21 wrote:If x is unit digit, y is tenth digit of a certain number, what x and y of (123,456,789)^2?
As the number ends with 9, the square of the number will end with 1. Hence, x = 1.
Now, we have to calculate the tens digit of 123456789².
123456789² = (123456780 + 9)² = (123456780² + 2*123456780*9 + 9²)
Now the underlined term in the above expression will end with 00. Hence, it won't contribute in determining the tens digit or y of the given term.
Hence, y of 123456789² = y of (2*123456780*9 + 9²) = y of (18*123456780 + 81) = y of (...40 + 81) = y of (...21) = 2
The correct answer is D.
