GMATPrep - Another trap

This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 277
Joined: Sun Jun 17, 2007 2:51 pm
Location: New York, NY
Thanked: 6 times
Followed by:1 members

GMATPrep - Another trap

by givemeanid » Mon Sep 03, 2007 9:09 am
I came across this while taking a GMATPrep test.

If k is a positive integer, then 20k is divisible by how many different positive integers?

1. k is prime
2. k = 7
So It Goes

Master | Next Rank: 500 Posts
Posts: 101
Joined: Tue Aug 07, 2007 1:45 am
Followed by:1 members

by agps » Mon Sep 03, 2007 9:22 am
I would answer D (either)

2) (easier) 20k = 20*7 = you know the number then you can determine all possible factors, sufficient.

1) any prime is divisible only by itself and 1. any prime multiplied by 20 will be divisible by it's factors and the factors of 20 (2, 4, 5, 10) so it's sufficient.

OA?

User avatar
Master | Next Rank: 500 Posts
Posts: 277
Joined: Sun Jun 17, 2007 2:51 pm
Location: New York, NY
Thanked: 6 times
Followed by:1 members

by givemeanid » Mon Sep 03, 2007 9:37 am
That's why its a trap son :)
Try it again. D is not the OA
So It Goes

Master | Next Rank: 500 Posts
Posts: 101
Joined: Tue Aug 07, 2007 1:45 am
Followed by:1 members

by agps » Mon Sep 03, 2007 9:48 am
hmmm,
is it because of 2? 2 is a prime, but also a factor of 20, then, not knowing what prime nuber it is you can't be sure if you have 4 different factors (for 2) or 5 (for any other prime number)

it that case, the answer would be B.

User avatar
Master | Next Rank: 500 Posts
Posts: 277
Joined: Sun Jun 17, 2007 2:51 pm
Location: New York, NY
Thanked: 6 times
Followed by:1 members

by givemeanid » Mon Sep 03, 2007 12:47 pm
Now you got it. You have to consider 2 (or 5) as a value of k!
So It Goes