How many three-digit even integers 'n' are there which are divisible b

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?

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