Veritas Prep Test
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 84
- Joined: Sat Jun 18, 2011 9:50 pm
- Location: New Delhi
- Thanked: 35 times
- Followed by:3 members
- GMAT Score:800
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~ ?
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me for free GMAT Learning!
Contact me for free GMAT Learning!
- theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
(n-1)!/n
For 32,
31!/32
32 = 4 x 8 --> both of these numbers exist in 31!.. remainder should be 0
What is the OA? Also, please post the answer choices along with Questions
For 32,
31!/32
32 = 4 x 8 --> both of these numbers exist in 31!.. remainder should be 0
What is the OA? Also, please post the answer choices along with Questions
R A H U L
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
So, 32~ = the remainder when 31! is divided by 32.CSASHISHPANDAY 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~ ?
31! = (1)(2)(3)(4)(5)(6)(7)(8)(9)....(30)(31)
Since (4)(8) = 32, we can conclude that 31! is a multiple of 32, which means the remainder is 0 when we divide 31! by 32
Cheers,
Brent
-
- Senior | Next Rank: 100 Posts
- Posts: 84
- Joined: Sat Jun 18, 2011 9:50 pm
- Location: New Delhi
- Thanked: 35 times
- Followed by:3 members
- GMAT Score:800
Thanks OA is 0
If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
Contact me for free GMAT Learning!
Contact me for free GMAT Learning!
GMAT/MBA Expert
- Jeff@TargetTestPrep
- GMAT Instructor
- Posts: 1462
- Joined: Thu Apr 09, 2015 9:34 am
- Location: New York, NY
- Thanked: 39 times
- Followed by:22 members
32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5CSASHISHPANDAY 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
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
Jeffrey Miller
Head of GMAT Instruction
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews