Question

How many four digit whole numbers ‘n’ are possible such that the last four digits of $n^2$ are in fact the original number ‘n’?

• A. 1
• B. 2
• C. 3

Answer 1

The correct option is A 1

Option (b)

Looking at the last digit, the last digit must be either 0, 1, 5 or 6. Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76. Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376. Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376. Out of those, only 9376 is a 4 digit number.