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Question

Need help on this...

heman · Answer

Cant twist the problem more than this GMAC Animals !!

Qstem m,p={I , >2}
2<m<p

p/m is not an Int

p=mx+ r where x is any Integer

gonna take (2) first
(2) m=5 p=6 LCM is 30
6/5 --> r=1 No

m=3 p=10 LCM is 30
10/3 --> r=1 No

BD

(1) GCF is 2

m=4 p=6 :6/4 r=2 Yes

m=8 p=10: 10/8 --> r=2 Yes

m=20 p=22: 22/20 --> r=2 Yes

m=4 p=10: 10/4--> r=2 Yes

Hence D??

Heman

Professor · Answer

only A.

what about if the integers are 5 and 6?

heman · Answer

For that r=1 

Do you have a case for (2) when r!= 1 ?

Heman

haas_mba07 · Answer

Professor,
 Can you please provide some more insight into problems like these? I get hopelessly lost at such problems.

yezz · Answer

I think the question want us to prove they are not consecutive

any two cnsecutives are relatively prime to each other

ie : GCF = 1

FROM ONE

GCF =2 THUS THEY ARE NOT CONSECUTIVE

FROM TWO

5,6 LCM = 30 BUT THEY ARE CONSECUTIVE AND 5 IS NOT A FACTOR OF 6 AND 5<6 ........INSUFF

I WILL GO FOR A ON THIS ONE

cicerone · Answer

Folks the answer is A.
Look at this.
Statement 1: GCD is 2.
mean m=2a and p=2b, where a and b do not have any common factor (if they have a common factor then the GCD will not be 2)i.e a and b are coprimes.

Since p>m , b>a and remember b and a are +ve numbers and coprimes.
Take the least values for a and b ie
a=2 and b=3,
So m=4 and p=6 The remainder is 2>1
Since b-a =1 the remainder =2x1=2
If b-a=2 the remainder will be 2X2=4.
If b-a=3 the remainder will be 2x3=6.If this is > m this will be further divided , but the minimum remainder possible is 2.

So always the remainder is >1.
So it is A.

It is already explainde that Statement 2 alone is not sufficient.
For exmaple 
m=5 and p=6 Their LCM = 30 and remainder is 1.
m=15 and p=30 their LCM=30 and remainder is 0.
m=6 and p=10 their LCM=30 and remainder is 4.

So it is not sufficent.

Hence A.

Professor · Answer

from i, m and p are even integers and p cannot be 2m. so in that case, r >1 when p is divided by m. try plugging some integers.

from ii, LCM of m and p = 30 means m amd p can be any integers that are > 1 and has 30 LCM. for ex: 3 and 10, 5 and 6, 6 and 10, 10 and 15, 6 and 15 or so on..... some of these pairs have r =1 and some have r>1. so insufff............................... 

HTH..

Futuristic · Answer

this was tricky.....
agree with A

haas_mba07 · Answer

Thanks Cicerone for the great explanation.

I guess when we have such problems referring to GCD, we should try to represent the two numbers as multiples of the GCD.

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