The digit at the unit place of $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} . . . . . . (2015!)^{2}$

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Solution

The correct option is B
7

This question does not require binomial expansion.The first four terms are 1,4,36 and 576.Terms after that will have at least two zeros at the end.So the terms which decide the digit at the unit place are the first four terms only.We will consider only the first four terms.

The last digits of $(1!)^{2} + (2!)^{2} + (3!)^{2} + (4!)^{2} . . . . . . (2015!)^{2}$ = Last digit of (1+4+36+576) = 7

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