Divisor

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Divisor

by billyr » Sun Mar 01, 2009 12:04 am
Is there an easy way to approach problem like this/special pattern such as repharse the equation or picking a number?


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by sureshbala » Sun Mar 01, 2009 2:18 am
Given x = 8y+12, where x and y are +ve integers and we need to find the GCD of (x,y)

Statement I: x = 12u, where u is an integer.

Now y= (x-12)/8 = 3(u-1)/2.

Since y is an integer, u-1 must be divisible by 2.

So for u = 3, we have x = 36 and y = 3 and GCD will be 3.
For u = 5, we have x =60 and y = 6 and GCD will be 6.

So statement I is not sufficient.

Statement II: y = 12z, where z is an integer.

Since x = 8y+1, we have x =12(8z+1)

Now GCD of x and y = GCD of 12(8z+1) and 12z.

Clearly the GCD of z and 8z+1 =1.

So GCD of x and y = 12

Hence II is sufficient