We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1?
(1) The least common multiple of a and b is 42
(2) The greatest common factor of a and b is 2
I am struggling in this type of questions ..Any help is greatly appreciated.
From F.S 1 , for a=6,b=7, we have the remainder as 1, which is not more than 1. Again, for a=3,b=14, we have a remainder as 2, which is more than 1. Insufficient.
From F.S 2, the integers a and b are of the form 2x and 2y, where x and y are co-primes.
Also, 2y = 2x*q + R, where q is a non-negative integer.
or R = 2(y-qx). As x and y are co-primes, thus y is not equal to qx, for any integral value of q. Thus, (y-qx) will never be zero. And as R is always positive, the value of R will always be more than 1. Sufficient.
Note that as a>2, we can have the first value of a only as 4.
All that is equal and not-Deep Dive In-equality
Hit and Trial for Integral Solutions