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This makes use of the fact that every square is of the form 4k or 4k+1.
Remember that the digit in the units place of a perfect square would belong to {0,1,4,5,9}
If the last digit is 0, then the last 4 digits are 1010 and hence the remainder on division by 4 is 2.
In this way we see that the remainder set is {2,3}. Hence no perfect squares exist such that number 101 appears before the units digit
you have considered only four digit numbers but the question is general
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