Aptitude 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**.

**Datta**

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**Vivek**

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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**

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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.

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