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The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10

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Joined: 11 Dec 2013
Posts: 120
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)
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The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10  [#permalink]

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New post 14 Feb 2019, 23:13
6
00:00
A
B
C
D
E

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
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Status: Learning stage
Joined: 01 Oct 2017
Posts: 991
WE: Supply Chain Management (Energy and Utilities)
Re: The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10  [#permalink]

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New post 14 Feb 2019, 23:30
1
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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The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10  [#permalink]

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New post 02 Mar 2019, 12:49
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.
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The remainder obtained when $(1 !)^2 + (2!)^2 + (3 !) ^2+ \dotsc + (10   [#permalink] 02 Mar 2019, 12:49
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