The "connection" between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b. For instance, the smallest common multiple of 8 and 12 is 24, and the product of 8 and 12 is 96, so the connection between 8 and 12 is 24/96 = 1/4
The positive integer y is less than 20 and the connection between y and 6 is equal to 1/1 . How many possible values of y are there?
A 7
B 8
C 9
D 10
E 11
Kaplan: multiples and ratios
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 45
- Joined: Wed Jun 11, 2008 4:25 pm
- Thanked: 4 times
- GMAT Score:730
-
- Master | Next Rank: 500 Posts
- Posts: 399
- Joined: Wed Apr 15, 2009 3:48 am
- Location: india
- Thanked: 39 times
A
value of y will be 1,5,7,11,13,17,19
the value of y has to be a no. which is when multiplied by 6 should give a smallest common mutiple.
then only the connection b/n y and 6 is 1
just see denominator is 6*y
so numerator has to be a smallest common multiple and divisble by 6y
value of y will be 1,5,7,11,13,17,19
the value of y has to be a no. which is when multiplied by 6 should give a smallest common mutiple.
then only the connection b/n y and 6 is 1
just see denominator is 6*y
so numerator has to be a smallest common multiple and divisble by 6y
It does not matter how many times you get knocked down , but how many times you get up
-
- Master | Next Rank: 500 Posts
- Posts: 128
- Joined: Thu Jul 30, 2009 1:46 pm
- Thanked: 1 times
I dont understand the logic.. why should I look for numbers that do not have 2 or 3 as factors?praky_rules wrote:Find all pos. numbers <20 that do not have 2 or 3 as factors. 1,5,7,11,13,17,19.
7.
somebody explain!!!
-
- Legendary Member
- Posts: 727
- Joined: Sun Jun 08, 2008 9:32 pm
- Thanked: 8 times
- Followed by:1 members
Because, 2 and 3 are factors of 6.fruti_yum wrote:I dont understand the logic.. why should I look for numbers that do not have 2 or 3 as factors?praky_rules wrote:Find all pos. numbers <20 that do not have 2 or 3 as factors. 1,5,7,11,13,17,19.
7.
somebody explain!!!
What we think, we become
-
- Junior | Next Rank: 30 Posts
- Posts: 11
- Joined: Tue Aug 04, 2009 12:12 pm
- Thanked: 1 times
We know that LCM(x,y)*HCF(x,y) = Product(x,y)---(1)
From the problem we have LCM(x,y)/Product (x,y) = 1
from (1) it means 1/HCF(x,y) = 1 => HCF(x,y) = 1.
x,y will have HCF 1.
so we will look for all numbers between 1 and 20 which has HCF 1 with 6.
list has {1,5,7,11,13,17,19} = 7 elements
From the problem we have LCM(x,y)/Product (x,y) = 1
from (1) it means 1/HCF(x,y) = 1 => HCF(x,y) = 1.
x,y will have HCF 1.
so we will look for all numbers between 1 and 20 which has HCF 1 with 6.
list has {1,5,7,11,13,17,19} = 7 elements