# Easy Number System Solved QuestionAptitude Discussion

 Q. If each of the three nonzero numbers  a, b and  c  is divisible by 3, then  abc  must be divisible by which one of the following the numbers?
 ✖ A. 8 ✔ B. 27 ✖ C. 81 ✖ D. 121

Solution:
Option(B) is correct

Since each one of the three numbers $a$, $b$, and $c$ is divisible by 3, the numbers can be represented as $3p,3q$ and $3r$, respectively, where $p, q$ and $r$ are integers.

The product of the three numbers is $3p×3q×3r =27(pqr)$.
Since $p, q$ and $r$ are integers, $pqr$ is an integer and therefore $abc$ is divisible by 27.

## (3) Comment(s)

Datta
()

Number should be divisible by 9 and 3. such least number is 27.

Vivek
()

If it would have been 3 distinct numbers a,b and c divisible by 3. Then the answer would necessarily have been 81.

Taking just the first 3 multiples of 3.

$a=3, b=6, d=9$,

$a.b.c = 162$

$81 divides 162$

Ruhi
()

the question is poorly framed. The solution implies that the divisor of product of a,b,c was being asked. however, the question doesn't make this clear.