4d wrote:
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
\((1!)^2=1\)
\((2!)^2=4\)
\((3!)^2=36\)
\((4!)^2=576\)
\((5!)^2=14400\)
All the square terms after 5! yield last two digits(00). Hence. all the terms after \((5!)^2\)yield a remainder zero when divided by 100.
Now the remainder of the given expression becomes:
\(Rem(\frac{1+4+36+76}{100})=Rem(\frac{617}{100})=17\)
Ans. (B)
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PKN
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59% (02:12) correct 
