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If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh

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S
Joined: 18 Jul 2018
Posts: 42
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
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If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh  [#permalink]

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New post 05 Aug 2018, 08:04
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E

Difficulty:

  45% (medium)

Question Stats:

60% (01:07) correct 40% (02:03) wrong based on 25 sessions

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If \(P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2\). Then what is the remainder when \(5^{2P}\) is divided by 13

1) 0
2) 1
3) 4
4) 8
5) 12
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Joined: 01 Oct 2017
Posts: 507
WE: Supply Chain Management (Energy and Utilities)
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Re: If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh  [#permalink]

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New post 05 Aug 2018, 08:52
1
Afc0892 wrote:
If P= \(1!^2\) + \(2!^2\) + \(3!^2\) + .....+ \(10!^2\). Then what is the remainder when \(5^{2P}\) is divided by 13

1) 0
2) 1
3) 4
4) 8
5) 12


What is the pattern of \(Rem(\frac{5^{2p}}{13})\)?

1. If P=1, \(Rem(\frac{5^2}{13})=12\)
2. If P=2, \(Rem(\frac{5^4}{13})=1\)
3. If P=3, \(Rem(\frac{5^6}{13})=12\)
4. If P=4, \(Rem(\frac{5^8}{13})=1\)

So, when p is odd, remainder is 12 otherwise remainder is 1.

You know, factorial of any positive integer greater than or equal to 5 has unit digit '0'.

Now. P= \(1!^2\) + \(2!^2\) + \(3!^2\) + .....+ \(10!^2\)
P=1+4+6+4+0..............+0=15 (Note:- only unit digits added, because unit digit of a integer decides whether it is even or odd)
Or, unit digit of P=5(odd)

Therefore, the remainder when \(5^{2*{odd}}\) is divided by 13 is 12.

Ans. (E)
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Regards,

PKN

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Re: If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh &nbs [#permalink] 05 Aug 2018, 08:52
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