Carcass wrote:
Suppose n is an integer such that \(2 < n^2 <100\). If the units digit of \(n^2\) is 6 and the units digit of \((n —1)^2\) is 5, what is the units digit of \((n +1)^2\)?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 9
We can solve this question by using the concept of CYCLICITY!
The Cyclicity (repeating pattern) of 6 is 0 i.e. any number with last digit as 6 when raised to any power - the last digit will always be 6
The Cyclicity (repeating pattern) of 5 is again 0 i.e. any number with last digit as 5 when raised to any power - the last digit will always be 5
Using this;
If \(n^2\) = _ 6, \(n\) must have units digit as 6
If \((n - 1)^2\) = _ 5, \(n\) must have units digit as 5
Therefore, In \((n + 1)^2\), \((n + 1)\) must have units digit as 7
(_\(7)^2\) = _9
Hence, option E