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Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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05 Nov 2015, 05:57
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Let n~ be defined for all positive integers n as the remainder when (n  1)! is divided by n. What is the value of 32~ ? A. 0 B. 1 C. 2 D. 8 E. 31
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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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05 Nov 2015, 07:14
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n~ = (n1)!
so 32~ = (321)! = 31!
when 31!/32 we have 16*2 inside 31!
hence 32 gets cancelled and we get remainder as 0

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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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32 = 8*4 which are both present in 31! so Remainder =0 hence A
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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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15 Apr 2017, 13:39
Ok so I was initially not able to do this problem because I didn't understand how to divide 31!/32...
Now that you guys have made the problem more clear I have a question
5!/15 must be an integer? Because 5x4x3x2x1 contains the factors of 15 (5 x 3)
When I do 17!/30 in a calculator the result is not an integer?

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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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16 Apr 2017, 01:27

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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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16 Apr 2017, 19:09
Bunuel wrote: Nunuboy1994 wrote: Ok so I was initially not able to do this problem because I didn't understand how to divide 31!/32...
Now that you guys have made the problem more clear I have a question
5!/15 must be an integer? Because 5x4x3x2x1 contains the factors of 15 (5 x 3)
When I do 17!/30 in a calculator the result is not an integer? First of all, 17!/30 = 11856247603200 = integer, but what 17!/30 has to do with this problem? I trying to demonstrate the concept in this problem, which is new to me, in a different example I think the calculator's notation just makes it appear as 1.something^e even though that doesn't necessarily mean it's not an integer. It's clear now.

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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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16 Apr 2017, 23:51
Nunuboy1994 wrote: Bunuel wrote: Nunuboy1994 wrote: Ok so I was initially not able to do this problem because I didn't understand how to divide 31!/32...
Now that you guys have made the problem more clear I have a question
5!/15 must be an integer? Because 5x4x3x2x1 contains the factors of 15 (5 x 3)
When I do 17!/30 in a calculator the result is not an integer? First of all, 17!/30 = 11856247603200 = integer, but what 17!/30 has to do with this problem? I trying to demonstrate the concept in this problem, which is new to me, in a different example I think the calculator's notation just makes it appear as 1.something^e even though that doesn't necessarily mean it's not an integer. It's clear now. For big numbers do not user calculator (unless it's not advanced on) or Excell, use Wolframalpha: https://www.wolframalpha.com/Hope it helps.
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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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27 Jun 2017, 16:09
shasadou wrote: Let n~ be defined for all positive integers n as the remainder when (n  1)! is divided by n.
What is the value of 32~ ?
A. 0 B. 1 C. 2 D. 8 E. 31 Simple all this question is asking is if you have 31!/32 then what is the remainder? You don't necessarily have to expand 31! 32 fits in because you have 16 and 2 so there is no remainder Thus "A"

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Re: Let n~ be defined for all positive integers n as the remainder when (n [#permalink]
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09 Jul 2017, 04:12
shasadou wrote: Let n~ be defined for all positive integers n as the remainder when (n  1)! is divided by n.
What is the value of 32~ ?
A. 0 B. 1 C. 2 D. 8 E. 31 \(= \frac{(321)!}{32}\) \(= \frac{31!}{32}\) Numbers \(8 * 4 = 32\), will cancel out the denomintor and hence the reminder will be ZERO. Hence, Answer is A
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