February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST We don’t care what your relationship status this year  we love you just the way you are. AND we want you to crush the GMAT! February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST Buy "AllInOne Standard ($149)", get free Daily quiz (2 mon). Coupon code : SPECIAL
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37 GMAT 2: 650 Q43 V36 GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
05 Nov 2015, 04:57
Question Stats:
70% (01:22) correct 30% (01:45) wrong based on 294 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
KUDO me plenty




Intern
Joined: 21 Aug 2013
Posts: 3
Concentration: International Business, Leadership
GPA: 2.9
WE: Programming (Computer Software)

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
05 Nov 2015, 06:14
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




Current Student
Joined: 12 Aug 2015
Posts: 2621

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
16 Mar 2016, 00:57



Director
Joined: 12 Nov 2016
Posts: 725
Location: United States
GPA: 2.66

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
15 Apr 2017, 12:39
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52938

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
16 Apr 2017, 00:27



Director
Joined: 12 Nov 2016
Posts: 725
Location: United States
GPA: 2.66

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



Math Expert
Joined: 02 Sep 2009
Posts: 52938

Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
16 Apr 2017, 22:51
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) or Excell, use Wolframalpha: https://www.wolframalpha.com/Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 12 Nov 2016
Posts: 725
Location: United States
GPA: 2.66

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
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 "A"



Retired Moderator
Joined: 19 Mar 2014
Posts: 934
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
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 \(= \frac{(321)!}{32}\) \(= \frac{31!}{32}\) Numbers \(8 * 4 = 32\), will cancel out the denomintor and hence the reminder will be ZERO. Hence, Answer is A
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/howtoget60awamyguide64327.html#p470475



Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1299
Location: India
WE: Engineering (Other)

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



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2827

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
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
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 18 Jun 2017
Posts: 49
GMAT 1: 660 Q39 V40 GMAT 2: 700 Q45 V41

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52938

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
30 Nov 2017, 23:04



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13551
Location: United States (CA)

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
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 (321)! 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. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



VP
Joined: 09 Mar 2016
Posts: 1285

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



Director
Joined: 09 Mar 2018
Posts: 986
Location: India

Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
Show Tags
10 Feb 2019, 18:51
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~ = 31!/31 Now if 31 from denominator is consumed completely this means that remainder is 0. A
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.
Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up.




Re: Let n~ be defined for all positive integers n as the remainder when (n
[#permalink]
10 Feb 2019, 18:51






