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Mathematics Question on Binomial theorem

The remainder obtained when (⌊1)2+(⌊2)2+(⌊3)2+...+(⌊100)2(\lfloor 1)^2 + (\lfloor 2)^2 + (\lfloor 3)^2 + ... + (\lfloor 100)^2 is divided by 10210^2 is ;

A

14

B

17

C

28

D

27

Answer

17

Explanation

Solution

Terms greater than 5! i.e., (5!)2,(6!)2,...,(100!)2(5 !)^2, ( 6 !)^2, ..., (100!)^2 is divisible by 100 ∴\therefore For terms (5!)2,(6!)2, (5 !)^2, ( 6 !)^2, ..., (100!)2(100!)^2 remainder is 0. Now consider (1!)2+(2!)2+(3!)2+(4!)2(1 !)^2 + (2 !)^2 + (3 !)^2 + (4 !)^2 = 1 + 4+ 36+ 576 = 617 When 617 is divided by 100, its remainder is 17. ∴\therefore Required remainder is 17.