Both n and n^2 to leave the same remainder when divided by 10, both n and n^2 should have the same units digit (the units digit of a number defines the remaider upon division by 10. For example, 17 leaves the remainder of 7, 23 leaves the remainder of 3, ...). This means that the units digit of n must be 0, 1, 5, or 6. Since given that n is even, then only the units digit of 0 and 6 are valid.

So, we need to find the number of multiples of 9, with the units digit of 0 or 6.

xy0 --> x + y must be a multiple of 9 (18 or 9): 99, 90, 81, 72, 63, 54, 45, 36, 27, or 18. Ten numbers.

xy6 --> x + y must be 3 or 12.
x + y = 3: 30, 21, or 12. Three numbers.
x + y = 12: 93, 84, 75, 66, 57, 48, or 39. Seven numbers.

10 + 3 + 7 = 20.

Answer: A.
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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

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

three digit even numbers can end in 0,2,4,6 and 8.
even numbers where n and \(n^2\) have same digit can only end in 0 and 6.

let the three digit number = \(100*a+10*b +c\)
c = [0,6]

case A: when c ends in 0
a + b = 9 or 18
(a,b) = (1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2)(8,1)(9,0),(9,9); Total = 10

case B: when c ends in 6
a+b = 3
(a,b) = (1,2),(2,1),(3,0); total = 3
or
a+b = 12
(a,b) = (3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3); total = 7
Total numbers = 10+3+7

Alternative:

between [1000 - 100], numbers divisible by 9 = 100
in these 100 numbers, which are divisible by 9, even numbers = 100/2 = 50
even numbers can only end in 0,2,4,6 and 8 and even numbers where n and \(n^2\) are same can only end in 0 and 6.
so out of 50 even numbers only 20 even numbers will end in 0 or 6.
Ans:20

Number of 3 digit number div by 9 = ((999-108)/2)+1=100.
Even 3 digit number div by 9 = 100/2 = 50.

For n and n^2 to leave the same remainder, the even number must end with 0 or 6.
Multiples of 9 - 18,36,54,72,90.
Therefore 2/5 multiples end with 6 or 0. So, Ans is 2/5*50=20.
Bunuel, is my approach correct?