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Let n~ be defined for all positive integers n as the remainder when (n

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Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 05 Nov 2015, 04:57
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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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New post 05 Nov 2015, 06:14
2
2
n~ = (n-1)!

so 32~ = (32-1)! = 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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Re: Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 15 Apr 2017, 12:39
1
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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New post 16 Apr 2017, 00:27
1
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?
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Re: Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 16 Apr 2017, 18:09
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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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Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 16 Apr 2017, 22:51
1
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) 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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New post 27 Jun 2017, 15:09
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
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Re: Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 09 Jul 2017, 03:12
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



\(= \frac{(32-1)!}{32}\)

\(= \frac{31!}{32}\)

Numbers \(8 * 4 = 32\), will cancel out the denomintor and hence the reminder will be ZERO.

Hence, Answer is A
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Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 10 Nov 2017, 03:20
Bunuel VeritasPrepKarishma

32! has various multiples of 32 built in say 8 * 4, 16* 2. Will not various no of multiples affect my remainder?
For eg, even if 8*4 will cancel out 32 in denominator leaving remainder = 0, I still have one multiple of 16*2, correct?
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Re: Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 14 Nov 2017, 06:08
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


32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

Since we can safely say that there are at least five 2s in 31! (for example, 31! has the factors 16 = 2^4 and 8=2^3), the remainder is zero.

Alternate Solution:

32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

We want to know the remainder when 31! is divided by 2^5. If we can establish that there are at least 5 factors of 2 in 31!, then we will know that 2^5 evenly divides into 31!, which means that the remainder would be 0. Let’s determine if we can find at least 5 twos in 31!:

31! = 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x … x 1

31! = 31 x (2 x 15) x 29 x (2 X 2 x 14) x 27 x (2 x 13) x 25 x (2 x 2 x 2 x 3) x … x 1

Notice that we have found seven 2s, which is two more than what we needed. Thus, we know that 2^5 will evenly divide into 31!, leaving a remainder of 0.

Answer: 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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New post 30 Nov 2017, 22:24
Hi,

So basically as long as there are enough prime factor powers in the numerator to cancel out all the prime factor powers in the denominator, the remainder is 0 correct?
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Re: Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 30 Nov 2017, 23:04
calappa1234 wrote:
Hi,

So basically as long as there are enough prime factor powers in the numerator to cancel out all the prime factor powers in the denominator, the remainder is 0 correct?


Yes. If the numerator contains the same (or more) primes as denominator and the (positive integer) powers of primes in the numerator are at least as big as the powers of the same primes in the denominator, then the result of the division will be an integer, so the remainder will be 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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New post 06 Feb 2018, 11:12
Hi All,

This prompt is an example of a 'Symbolism' question - the prompt 'makes up' a math symbol, tells you what it means and asks you to perform a calculation using it. Based on the information in the prompt, we're told that....

32~ is the remainder when (32-1)! is divided by 32.

Now, there's no way that the GMAT would expect you to calculate the value of 31!, so we have to think in terms of what 31! actually is.

Here's an example that's a bit easier to deal with:

4! = (4)(3)(2)(1) = 24

What numbers divide EVENLY into 24?

1, 2, 3, 4, 6, 8, 12 and 24

You can clearly see why 1, 2, 3 and 4 divide in - they're in the 'chain' of numbers that are multiplied together.
6 divides in because (2)(3) = 6 - and you can see the (2) and the (3) in the 'chain'
Similarly, 8, 12 and 24 are also 'combinations' of the numbers in the 'chain', so they divide evenly in too.

31! has LOTS of numbers in it, so it's evenly divisibly by LOTS of different integers. If you were to write out 31!, you would see a (2) and a (16). This means that (2)(16) = 32 divides evenly into 31!, so there will be a remainder of 0.

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Let n~ be defined for all positive integers n as the remainder when (n  [#permalink]

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New post 07 Feb 2018, 08:29
Bunuel wrote:
calappa1234 wrote:
Hi,

So basically as long as there are enough prime factor powers in the numerator to cancel out all the prime factor powers in the denominator, the remainder is 0 correct?


Yes. If the numerator contains the same (or more) primes as denominator and the (positive integer) powers of primes in the numerator are at least as big as the powers of the same primes in the denominator, then the result of the division will be an integer, so the remainder will be 0.



Hi Bunuel, long time no hear no see :)

Here is my solution.

to define if \(31!\) is divisible by \(32\) make prime factorization of \(32\) so we get this --- > \(2^4\)

\(31!\) certainly will have four \(2\) as prime factors, hence \(\frac{31!}{32}\) we get integer.

if this solution correct. What if similar question would have remainder - how could i find a remainder ? :?

i would appreciate your explanation :)

thank you :)
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Let n~ be defined for all positive integers n as the remainder when (n &nbs [#permalink] 07 Feb 2018, 08:29
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