Aptitude Discussion

Q. |
If $n$ is a positive integer, which one of the following numbers must have a remainder of 3 when divided by any of the numbers 4, 5 and 6? |

✖ A. |
$12n + 3$ |

✖ B. |
$24n + 3$ |

✖ C. |
$90n + 2$ |

✔ D. |
$120n + 3$ |

**Solution:**

Option(**D**) is correct

Let m be a number that has a remainder of 3 when divided by any of the numbers 4, 5 and 6.

Then $m - 3$ must be exactly divisible by all three numbers.

Hence, $ m - 3$ must be a multiple of the Least Common Multiple of the numbers 4, 5 and 6.

The LCM is $3×4×5 = 60$.

Hence, we can suppose $m - 3 =60p$, where $p$ is a positive integer.

Replacing $p$ with $n$, we get $m – 3 = 60n$.

So, $m = 60n + 3$.

Choice (D) is in the same format **120n + 3** $= 60(2n) + 3$

**Dhanyashree**

*()
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Take any positive integer for the place of n and apply it to the option

Ex: 120(1)+3

=123/4 give remainder 3

=123/5 give remainder 3

=123/6 give remainder 3