A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
86
53
49
46
33
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Divisibility & Prime Number
This topic has expert replies
-
DanaJ
- Site Admin
- Posts: 2567
- Joined: Thu Jan 01, 2009 10:05 am
- Thanked: 712 times
- Followed by:550 members
- GMAT Score:770
I'm thinking: go for the multiples of 4, add 1 and then check to see if that number fits the second condition:
4 + 1 = 5 - NO
8 + 1 = 9 - NO
12 + 1 = 13 - YES - first one
16 + 1 = 17 - NO
20 + 1 = 21 - NO
24 + 1 = 25 - NO
28 + 1 = 29 - NO
32 + 1 = 33 - YES - second one
This makes the sum, IMHO, 46.
4 + 1 = 5 - NO
8 + 1 = 9 - NO
12 + 1 = 13 - YES - first one
16 + 1 = 17 - NO
20 + 1 = 21 - NO
24 + 1 = 25 - NO
28 + 1 = 29 - NO
32 + 1 = 33 - YES - second one
This makes the sum, IMHO, 46.
-
franciskyle
- Senior | Next Rank: 100 Posts
- Posts: 32
- Joined: Wed Mar 11, 2009 9:56 pm
- Location: Whitler, BC, Canada
For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.
A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
-
cramya
- Legendary Member
- Posts: 2467
- Joined: Thu Aug 28, 2008 6:14 pm
- Thanked: 331 times
- Followed by:11 members
Good solutions!
n = 4k+1 and n=5q+3
Find the smallest number that satisfies this i.e. 13
Therefore the family of numbers will be
a*LCM(4,5) + 13
20a+13
When a=0 the number is 13 which u obtained above
When a = 1 the number will be 33
So 33+13 = 46
Hope this helps!
Regards,
CR
n = 4k+1 and n=5q+3
Find the smallest number that satisfies this i.e. 13
Therefore the family of numbers will be
a*LCM(4,5) + 13
20a+13
When a=0 the number is 13 which u obtained above
When a = 1 the number will be 33
So 33+13 = 46
Hope this helps!
Regards,
CR
GMAT/MBA Expert
-
Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Ah, note that the correct answer shouldn't give the same remainders that are mentioned in the question. The question asks for the sum of two different numbers with certain remainders. We know each of those two numbers will have a remainder of 3 when divided by 5; when we add those two numbers we should now have a remainder of 1 when we divide by 5 (because 3+3 has a remainder of 1). We also want a number with a remainder of 2 when divided by 4, since the two numbers we're adding give a remainder of 1 when divided by 4.franciskyle wrote:For this one, I would start with deciding which answers gave a remainder of 3 when divided by 5 (the units digit would have to be either 3 or 8 because to be divisible by the units would have to be 0 or 5... add 3 to get 3 and 8 respectively). That leaves us with B & E.
A quick look and you know that 4 goes into 33 eight times with a remainder of 1. Therefore E.
Still, two answer choices remain, unfortunately, so some other method would be needed to choose between 46 and 86.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com
-
franciskyle
- Senior | Next Rank: 100 Posts
- Posts: 32
- Joined: Wed Mar 11, 2009 9:56 pm
- Location: Whitler, BC, Canada
