comon factor, an official question

My guess will be A.

1. tells you that k = j + 1. If this is the case, you can safely say that j and k are consecutive numbers. The greatest common factor for consecutive numbers is always 1. Test this for any pair of consecutive integers.

2. is insufficient, IMO. jk is divisible by 5 = at least one of the two integers is divisible by 5. However, there are many possible cases:
a. j = 5 and k = 10 - greatest common factor 5
b. j = 6 and k = 30 - greatest common factor 6
c. j = 20 and k = 100 - greatest common factor 20

only one q , if j and k are + integers

can we take j =0 ?

coz if j> 0 then GCD is always 2 for two consecutive integers and A is Suff.

Please let me know.

j and k are positive integers... '0' is neither +ve nor -ve...so consider natural numbers i.e. 1 onwards..

Humm A only IMO:

As people have shown for jk to be divisible by 5: gcd(1,5) =1
gcd(5, 15) =5

etc.

Can somebody use steps to solve how consecutive integers will have 1 as gcd?

Take any two +ve consecutive integers:

eg. 17 and 18

17 = 17*1
18 = 2*3*3*1

HCF of (a,b) is the biggest number that can divide both a and b

In this case, the only common factor of 1.

You can test the same for any 2 consecutive +ve integers

1,2
100,101
1399,1400 etc...

A for me as well.

Thanks Dana.i will bear this tip in mind.