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GCD of m and n

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GCD of m and n

by shahab03 » Tue Feb 23, 2010 9:11 pm
Image



I am ok with Statement 1's explanation
1) m is prime

no info on n, insuff

if m=5 and n=15 , GCF=5
if m=5, n=4 then GCF =1

so Not sufficient.

Not following the explanation for statement 2:

2) 2n=7m; n=7/2 m or 1:3.5 ratio

n could be 2, m could be 7, GCF=1
n could be 8, m could be 28, GCF=4
nsuff

together
m has to be 7 for n to be an integer, GCF=1


Why is the answer C?

Can someone please explain.

Thanks
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by thephoenix » Tue Feb 23, 2010 9:26 pm
shahab03 wrote:Image



I am ok with Statement 1's explanation
1) m is prime

no info on n, insuff

if m=5 and n=15 , GCF=5
if m=5, n=4 then GCF =1

so Not sufficient.

Not following the explanation for statement 2:

2) 2n=7m; n=7/2 m or 1:3.5 ratio

n could be 2, m could be 7, GCF=1
n could be 8, m could be 28, GCF=4
nsuff

together
m has to be 7 for n to be an integer, GCF=1


Why is the answer C?

Can someone please explain.

Thanks
for s2)
n=7m/2
since bth r +ve int
m is a multiple of 2
i.e m can be 2,4,6,8........so on
for each value of m the corresponding value of n will be 7,14,21,28......so on a mutiple of 7
now for each set of m and n the value of gcd will differ so the statement is insuff

combining for only m=2 (a prime no. infact the only even prime no. the value of n will be int ,a requirment of qusetion)
so n=7
and gcd of 2 and 7 is 1
hence one value and is therefore suff
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by Pedros » Wed Feb 24, 2010 1:48 am
The 2 statements together says that :

2n = 7 ( prime no.)

2n is even . so 7m must be even as well , the only even prime number is 2 so m is 2 and n is consequantly 7 .

So GCD = 1
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