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Kaplan: multiples and ratios

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by Nermal » Sat Sep 12, 2009 8:51 am
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
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by praky_rules » Sat Sep 12, 2009 9:17 am
Find all pos. numbers <20 that do not have 2 or 3 as factors. 1,5,7,11,13,17,19.

7.
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IMO

by xcusemeplz2009 » Sat Sep 12, 2009 9:31 am
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
It does not matter how many times you get knocked down , but how many times you get up
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by fruti_yum » Tue Sep 15, 2009 5:37 pm
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.
I dont understand the logic.. why should I look for numbers that do not have 2 or 3 as factors?
somebody explain!!!
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by umaa » Tue Sep 15, 2009 6:00 pm
fruti_yum wrote:
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.
I dont understand the logic.. why should I look for numbers that do not have 2 or 3 as factors?
somebody explain!!!
Because, 2 and 3 are factors of 6.
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by blr_gmat_prep » Tue Sep 15, 2009 10:55 pm
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
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