## Skip counting by 3’s

Skip counting is the method of skipping certain numbers in the number sequence. To skip count by 3’s start with the number 3 and mark every 3rd number from the given number.

Let us skip count by 3 for the numbers up to 100.

1 | 2 | 3 |
4 | 5 | 6 |
7 | 8 | 9 |
10 |

11 | 12 | 13 | 14 | 15 |
16 | 17 | 18 |
19 | 20 |

21 |
22 | 23 | 24 |
25 | 26 | 27 |
28 | 29 | 30 |

31 | 32 | 33 |
34 | 35 | 36 |
37 | 38 | 39 |
40 |

41 | 42 |
43 | 44 | 45 |
46 | 47 | 48 |
49 | 50 |

51 |
52 | 53 | 54 |
55 | 56 | 57 |
58 | 59 | 60 |

61 | 62 | 63 |
64 | 65 | 66 |
67 | 68 | 69 |
70 |

71 | 72 |
73 | 74 | 75 |
76 | 77 | 78 |
79 | 80 |

81 |
82 | 83 | 84 |
85 | 86 | 87 |
88 | 89 | 90 |

91 | 92 | 93 |
94 | 95 | 96 |
97 | 98 | 99 |
100 |

The skip counting here also gives the multiples of 3 up to 100.

## Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 …

The multiples of **3** are all the values that we get by multiplying any number by **3**.

The multiples of **3** are the same as the values we get by skip counting by **3**.

**Example:**

The first five multiples of three are obtained by multiplying the numbers up to **5** by **3**.

*1 x 3 = 3*

*2 x 3 = 6*

*3 x 3 = 9*

*4 x 3 = 12*

*5 x 3 = 15*

We can find any number of multiples in a similar manner.

## The nth multiple of 3

We know that we can find any number of multiples of **3** by multiplying the given number by **3**. The value of the multiple of any nth value is given by **3n.**

**Example:**

**Find the 16th multiple of 3.**

The nth multiple of **3** is given by **3n**

The 16th multiple of **3** is given by **3 x 16 = 48**

## Did you know?

**Difference between factors and multiples of 3**

3 is a prime number. The prime numbers are the numbers that are only divisible by 1 and the number itself. Hence the factors of 3 are 1 and 3. The factors are the numbers that a given value can be split into, whereas, the multiples are the values that are obtained by the product of 3 and any other number.

Note that factors of a number are smaller than or equal to the number itself. The multiples of a number are either equal to or bigger than a given number.

Every number is both a factor and a multiple of itself.

### Test of divisibility for 3

The test of divisibility of 3 is the test that we can use to find if a given number is a multiple of 3 or not. In other words, the test of divisibility of 3 helps to determine whether a given number is divisible by 3.

To find if a given number is divisible by 3, find the sum of the digits of the number. If the sum is divisible by 3, then the given number is also divisible by 3.

**Example:**

**Conduct the test of divisibility to find if the given value is divisible by 3.**

**7542**

Here the digits of the number are **7, 5, 4,** and **2**.

The sum of the digits are **7 + 5 + 4 + 2 = 18**

The sum 18 is divisible by **3**, hence the number **7542** is also divisible by **3**.

The test of divisibility is helpful in identifying the multiples of 3 for large numbers

without actually dividing the number by 3.

### Least common multiple:

The least common multiple is the first common multiple of two numbers. To find the least common multiple of two numbers write down the multiples of both the numbers until you find one common multiple.

Let us take the example of two numbers 9 and 12 to find the least common mulitple.

Multiples of 9: 9, 18, 27, **36**, 45, 54, 63, 72, 81,90

Multiples of 12: 12, 24, **36**, 48, 60, 72

We can see that 36 is the first common multiple that appears in the multiples

sequence.

#### Hence 36 is the first common multiple of 9 and 12.

Note that there will be other multiples that are common to the multiples sequence

of both the numbers. But the multiple that is common and occurs first is called

the least common multiple.

The least common multiples are helpful in finding the common denominator in

fraction addition and subtraction.

## Check what I know?

Write down the next 6 consecutive multiples of 3 that come after 30.

**The next 6 consecutive multiples of 3 that come after 30 is 33, 36, 39, 42,****45, 48.**

The next consecutive multiple is obtained by adding 3 to 30

30+3=33

33+3=36

36+3=39

39+3=42

42+3=45

45+3=48

Alternatively, we know that 30 is the 10th multiple of 3

The next consecutive multiples are

11×3=33

12×3=36

13×3=39

14×3=42

15×3=45

16×3=48

Find the 30th and 40th multiple of 3 and find their sum.

**The sum of the 30th and the 40th multiples of 3 is 90 + 120 = 210**

The 30 th multiple of 3 is 30×3=90

The 40 th multiple of 3 is 40×3=120

John has a ribbon that he has to cut to pieces to decorate his room. Each piece of ribbon has to be 3 cm long. If the roll of ribbon is 161 cm long, find the number of pieces that he can cut from the roll. Also, state the length of ribbon left out.

**John can cut out 53 pieces of 3-cm ribbons and will have left a piece****of 2 cm.**

The length of the ribbon is 161 cm.

If John cuts the ribbon into 3 cm pieces, the length of the ribbon would reduce in multiples of 3.

Hence we can find the number of multiples of 3 in the number 161.

Divide 161 by 3

161÷3=53 R 2

There are 53 threes in the number 161 and a remainder of 2

**Find the multiples of 3 from the list of numbers using division**

**36, 78, 94, 46, 59, 39**

**The multiples of 3 are 36, 78 and 39.**

To find the multiples of 3 from the given list of numbers, divide each of the

numbers by 3.

If the result of the division gives an answer with remainder 0, the given number is

divisible by 3. Else, the number is not divisible by 3.

Let us find by dividing each of the numbers

36÷3=12 R 0

78÷3=26 R 0

94÷3=31 R 1

46÷3=15 R 1

59÷3=19 R 2

39÷3=13 R 0

The values for which we get the remainder 0 are the multiples of 3.

**List out the multiples of 3 without actually dividing.**

**The values 4578, 6435 and 456 are divisible by 5.**

Recollect the divisibility test for 3!

The divisibility test for 3 helps to find the multiples of 3 without actually dividing the number.

Start by adding all the digits in a number.

If the sum is divisible by 3, then the number is also divisible by 3.

4+5+7+8=24

2+8+7+6=23

7+0+0+1=8

6+4+3+2=15

1+4+7+9+8=29

4+5+6=15

5+8+9=22

The sums 24,15 and 15 are divisible by 3.

**Find the least common multiple of 6 and 15.**

**30 is the least common multiple of the numbers 6 and 15.**

The least common multiple of two numbers is the first multiple that is common to

both the numbers.

To find the common multiple, write the multiples of both the numbers.

The multiples of 6 : 6, 12, 18, 24, 30, 36

The multiples of 15: 15, 30, 45, 60, 75

The first common value that appears in the multiples sequence is the number 30.

## Frequently Asked Questions

**59781098678978259**

**Solution**

To find if a very large number is divisible by 3, add the digits to find their sum.

If the sum of the numbers is divisible by 3, the number is divisible by 3

Here we get the sum as

*5+9+7+8+1+0+9+8+6+7+8+9+7+8+2+5+9= 108*

**Now how do you find if 108 is divisible by 3 without dividing?**

We do the same divisibility test again.

**1 + 0 + 8 = 9**

Since **9** is divisible by **3**, the given number is also divisible by **3**.

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