The numbers \(\{a, b,c\}\) are three positive integers. If \(\dfrac{abc}{14}\) equals an integer, and \(\dfrac{bc}4\)

[Gmat_mission]

The numbers \(\{a, b,c\}\) are three positive integers. If \(\dfrac{abc}{14}\) equals an integer, and \(\dfrac{bc}4\) equals an integer, what is the smallest possible integer value of \(a?\)

A. 1
B. 2
C. 4
D. 7
E. 14

Answer: A

Source: Magoosh

[deloitte247]

$$\frac{abc}{14}and\ \frac{bc}{4}$$
For the smallest positive integer value of a =>
$$\frac{9}{1}\cdot\frac{bc}{14}$$
So, bc must be a multiple of 14 and 4 for it to be completely divisible by 14 and 4 without a remainder. And by doing so, the value of a will always be 1.
E.g if bc = 28 and a = 1
abc/14 is an integer and 6c/4 is an integer.
If bc=56 and a=1, then abc/14 is an integer, and bc/14 is an integer also.
So, therefore, the possible integer value of a = 1. Hence, A is the correct answer.

[Scott@TargetTestPrep]

The numbers \(\{a, b,c\}\) are three positive integers. If \(\dfrac{abc}{14}\) equals an integer, and \(\dfrac{bc}4\) equals an integer, what is the smallest possible integer value of \(a?\)

A. 1
B. 2
C. 4
D. 7
E. 14

Answer: A

Solution:

We are told that abc/14 is an integer and bc/4 is an integer. Let’s write abc/14 = k and bc/4 = s, where k and s are integers. It follows that abc = 14k and bc = 4s. Substituting bc = 4s in abc = 14k, we obtain:

a(4s) = 14k

a = 14k/4s = 7k/2s

We observe that the expression 7k/2s can be made equal to 1 simply by choosing k = 2 and s = 7. Since 1 is the smallest positive integer, it is also the smallest possible value of a.

Answer: A