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How many three-digit even integers 'n' are there which are divisible by 9 so that both n and \(n^2\) leave the same remainder when divided by 10?
(A) 20
(B) 21
(C) 22
(D) 23
(E) 24
**The source did not provide an OA
Since n and n^2 leaves the same remainder when divided by 10 the units digit of n and n^2 has to be same. This implies that units digit of n should be either 0 or 6 since n can only be even.
Sum of the digits of n has to be divisible by 9.
Therefore considering
Case i) when the units digit 0
the sum of the hundreds digit and tens digit must be either 9 or 18
- Cases for the sum to be 9 => 9,0; 8,1; 7,2; 6,3; 5,4; 4,5; 3,6; 2,7; 1,8 - Total 9
- Cases for the sum to be 18 => 9,9,0 - Only 1
Case ii) when the units digit is 6
the sum of the hundreds digit and tens digit must be either 3 or 12
- Cases for the sum to be 3 => 3,0; 2,1; 1,2; - Total 3
- Cases for the sum to be 12 => 9,3; 8,4; 7,5; 6,6; 5,7; 4,8; 3,9 - Only 7
Total possible combinations = 20
Answer - A