Did you recognize from the number 132, if you take the amount of every the 2-digit numbers, you get 12 + 13 + 21 + 23 + 31 + 32 = 132 making that the smallest number among 264, 396, and 35964 with this property? In this lesson, we will discover the determinants of 132, prime components of 132, and factors that 132 in pairs in addition to solved examples for a far better understanding.

You are watching: What are the factors of 132

**Factors that 132:**1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132

**Prime administrate of 132:**132 = 2 × 2 × 3 × 11

1. | What space the factors of 132? |

2. | How come Calculate components of 132? |

3. | Factors that 132 by prime Factorization |

4. | Factors that 132 in Pairs |

5. | FAQs on factors of 132 |

## What room the factors of 132?

Factors of a number are the numbers that division the given number precisely without any remainder. Factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132.**So,132 is a composite number together it has more factors various other than 1 and also itself.**

**How to calculation the determinants of 132?**

**We have the right to use various methods favor divisibility test, prime factorization, and also the upside-down department method to calculate the components of 132. Us express 132 as a product the its prime components in the prime factorization method as well as we divide 132 through its divisors in the division method. Let us see which numbers divide 132 exactly there is no a remainder.**

**Hence, the factors that 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132.**

**Explore components using illustrations and interactive examples.**

## Factors of 132 by prime Factorization

We have the right to do element factorization of any kind of number by:

Upside-down department methodFactor tree method### Prime administrate by division Method

The upside-down department got that is name since the division symbol is flipped upside down.

STEP 1: By utilizing divisibility rules, we uncover out the smallest specific prime divisor (factor) of the provided number. Here, 132 is an also number. So the is divisible by 2. In other words, 2 divides 132 with no remainder. Therefore 2 is the the smallest prime factor of 132.STEP 2: We division the given number by its smallest variable other 보다 1 (prime factor), 132/2 = 66STEP 3: we then uncover the prime factors of the derived quotient.Repeat step 1 and Step 2 it rotates we gain a element number as a quotient. In the critical step, 33 is the quotient, 33/3= 11. 11 is the quotient here, therefore we avoid the procedure as 11 is a element number.Therefore 132 = 2 × 2 × 3 × 11.

### Prime administer by element Tree Method

First, we determine the two factors that provide 132. 132 is the root of this factor tree.

132 = 2 × 66

Here, 66 is an even composite number, therefore it can be more factorized.

66 = 2 × 66

We proceed this procedure until we space left with just prime numbers, i.e., till we cannot further variable the derived numbers.

Basically, we branch the end 132 into its element factors. Variable tree is not distinctive for a offered number.

Instead of express 132 as 2 × 66, we deserve to express 132 as 4 × 33 or 3 × 44

Here is a simple activity to shot it on your own, rather of 2 × 66, if we had provided 3 × 44, perform you think we will obtain the same factor tree?

Can you draw the aspect tree v 3 and 44 as the branches?

## Factors that 132 in Pairs

Factor pairs room the two numbers that, as soon as multiplied, give the number 132.

Factor bag of 132Notation1 ×132 | (1, 132) |

2 × 66 | (2, 66) |

3 × 44 | (3, 44) |

4 × 33 | (4, 33) |

6 × 22 | (6, 22) |

11 × 12 | (11, 12) |

**We deserve to have an adverse factors likewise for a given number.**As the product of two an adverse numbers is positive (- × - = +), therefore (-1, -132), (-2, -66), and (-3, -44), etc., are likewise factor pairs of 132

But, for now, allow us focus on the positive components in this article. V factors, us are only looking for whole numbers that room equal to or much less than the initial number.

**Important Notes:**

**Think Tank:**

**Example 1: **Huia provides 132 bran biscuits because that the institution gala. She wants to placed them in packets the each have the same variety of biscuits, however she no want any kind of left over. What number of equal packets can Huia make up?

**Solution:**

We need to discover different feasible ways that packing.Total variety of biscuits = variety of packets × variety of biscuits in each packetHence, 132 = p × bFor that, we need to list out the aspect pairs of 132

132 = 1 × 132, i.e, 132 packets with 1 biscuit in each.

132 = 2 × 66, i.e, 66 packets with 2 biscuits in each.

132 = 3 × 44, i.e, 44 packets with 3 biscuits in each.

132 = 4 × 33, i.e, 33 packets with 4 biscuits in each.

132 = 6 × 22, i.e, 22 packets with 6 biscuits in each.

132 = 11 × 12, i.e, 12 packets through 11 biscuits in each.

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This would work-related the other way round too, like 11 packets through 12 biscuits in each. Hence, there room 12 ways of pack possible.