Multiples the 6 are the commodities of 6 and natural numbers. In various other words, a lot of of 6 is a number that deserve to be split by 6 and pipeline the remainder zero. Interestingly, multiples the 6 have a difference of 6 between each other. In this mini-lesson, we will certainly calculate the multiples the 6 and learn exciting facts about these multiples. In this mini-lesson, let"s learn an ext about multiples that 6 in the tabular kind with examples.

You are watching: What are the first five multiples of 6

**First 5 multiples that 6**: 6, 12, 18, 24, 30

**Prime factorization of 6:**2 x 3

1. | What space the Multiples that 6? |

2. | First 20 Multiples of 6 |

3. | Important Notes |

4. | FAQs ~ above Multiples that 6 |

The product of 6 with any number is a multiple of 6.

6×1 = 6

6×2 = 12

6×3 = 18

We get successive multiples of 6 by skip counting.

If we skip count 128 times by 6, we will acquire the 128th multiple of 6. I.e., 128×6 = 768.

We achieve the very first 20 multiples the 6 by recognize the product of 6 and the very first 20 natural numbers.

ProductMultiples6 × 1 | 6 |

6 × 2 | 12 |

6 × 3 | 18 |

6 × 4 | 24 |

6 × 5 | 30 |

6 × 6 | 36 |

6 × 7 | 42 |

6 × 8 | 48 |

6 × 9 | 54 |

6 × 10 | 60 |

6 × 11 | 66 |

6 × 12 | 72 |

6 × 13 | 78 |

6 × 14 | 84 |

6 × 15 | 90 |

6 × 16 | 96 |

6 × 17 | 102 |

6 × 18 | 108 |

6 × 19 | 114 |

6 × 20 | 120 |

To understand the concept of recognize multiples, let united state take a few more examples.

**Important Notes:**

**Example 1 **

Andrea decides to make arm bands one by one in together a way that there room 6 beads in the an initial bracelet, 12 in the second, 18 in the third, and so on. How numerous beads would certainly she must make the 45th bracelet?

**Solution**

In the 4th bracelet, Andrea needs 4×6 = 24 beads

In the 8th bracelet, she needs 8×6 = 48 beads.

Hence for nth bracelet, she needs n×6 beads.

Here she looks because that the 45th bracelet.

Thus, Andrea needs 45×6 = 270 beads.

**Example 2 **

Mike is playing a video clip game. For every jump over the obstacle, he collects 6 coins. He desires to understand how plenty of coins he would have collected if he the cross 112 such jumps. Deserve to you assist him to calculate the score?

**Solution**

In one jump, Mike it s okay 6 coins.

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In 112 jumps, the would have actually collected 112×6 = 672 coins.

**Example 3 **

Mia gets a score that 6 points native 8 judges at every level. What will be her score in ~ the finish of the 66th level?