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

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Director
Joined: 18 Jul 2018
Posts: 503
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh  [#permalink]

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05 Aug 2018, 07:04
4
00:00

Difficulty:

55% (hard)

Question Stats:

58% (01:26) correct 42% (02:24) wrong based on 33 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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When you want something, the whole universe conspires in helping you achieve it.

Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
Re: If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh  [#permalink]

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05 Aug 2018, 07:52
2
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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Re: If P = 1!^2 + 2!^2 + 3!^2 + ... + 10!^2. Then what is the remainder wh &nbs [#permalink] 05 Aug 2018, 07:52
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