Consider all 2 digit numbers N such that the last 2 digits of N^2 are N itself. What is the sum of all these 2 digit numbers?

Answer 1

the answer is 76

Answer 2

sorry the answer should be 101

I do like this:
first consider the possible last digit of the number, then you have 0,1,5,6

Let N be ab

For number end with 0,
the last two digits of N will be 00
no numbers can be formed as 00 is not a
2 digit numbers

For number end with 1,
it must satisfy the conditions:
last digit of 2a=a
no numbers can be formed

For number end with 5,
it must satisfy the conditions:
last digit of 10a+2=a
the number 25 can be formed

For number end with ,
it must satisfy the conditions:
last digit of 12a+3=a
the number 76 can be formed

Sum of N=76+25=101