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# prove that in any perfect square the digits immediately to the left of units digit cannot be 101

mycroft holmes
272 Points
8 years ago

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

ashwen venkatesh
19 Points
7 years ago

you have considered only four digit numbers but the question is general