Number System
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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

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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.