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 It is currently 05 Dec 2019, 11:13 ### GMAT Club Daily Prep

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6 00:00

Difficulty:   55% (hard)

Question Stats: 60% (02:06) correct 40% (02:01) wrong based on 67 sessions

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The remainder obtained when $$(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (100 !)^2$$ is divided by $$10^2$$ is
A) 14
B) 17
C) 7
D) 27
E) 24

Source : Ignus
Director  D
Status: Learning stage
Joined: 01 Oct 2017
Posts: 991
WE: Supply Chain Management (Energy and Utilities)

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10^2= 100= 2^2*5^2
So, if a number contains 2^2 and 5^2 we will get remainder as 0 for that number which is true for numbers above 5!^2
So, remainder will be = (1+4+36+576)/100
We get remainder as 17. B is the correct choice. The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10 [#permalink] 02 Mar 2019, 12:49 Display posts from previous: Sort by # The remainder obtained when$(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10  