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Playing With Numbers

    Introduction

    Factors and Multiples

    Prime and Composite Numbers

    Even and Odd numbers

    Tests for Divisibility of Numbers

    Prime Factorisation

    Common Factors and Common Multiples

    Exercise 3.1

    Exercise 3.2

    Exercise 3.3

    Exercise 3.4

    Exercise 3.5

    Exercise 3.6

    Exercise 3.7

    What Have We Discussed?

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

    Enhanced Curriculum Support

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Playing With Numbers > Exercise 3.1

Exercise 3.1

1. Which of the following numbers are divisible by 2, by 3 and by 6? (Enter Yes/No)

NumberDivisible by 2Divisible by 3Divisible by 6
(i) 321729
(ii) 197232
(iii) 972132
(iv) 1790184
(v) 312792
(vi) 800552
(vii) 4335
(viii) 726352

2. Determine which of the following numbers are divisible by 5 and by 10. Check whether the numbers that are divisible by 10 are also divisible by 2 and 5. (Enter Yes/No)

NumberDivisible by 5Divisible by 10Divisible by 2 (if divisible by 10)
(i) 25
(ii) 125
(iii) 250
(iv) 1250
(v) 10205
(vi) 70985
(vii) 45880
  1. Fill the table using divisibility test for 3 and 9:
NumberSum of the digits in the numberDivisible by 3Divisible by 3
72 + =
197 + + =
4689 + + + =
79875 + + + + =
9889749 + 8 + 8 + 9 + 7 + 4 = 45YesYes

4. Make 3 different 3 digit numbers using 1, 9 and 8, where each digit can be used only once. Check which of these numbers are divisible by 9.

Instructions

A Number formed by using the digits 1,9,8 without repetition
981,918,819,891,189,
If a number is divisible by 9 then sum of the digits also divisible by
Therefor all these numbers formed by using digits 1,9,8 are divisible by

5. Which numbers among 2, 3, 5, 6, 9 divides 12345 exactly? Write 12345 in reverse order and test now which numbers divide it exactly?

1235

Divisibility by 2 : Units place should have ,,,,

Therefore, 12345 divisible by 2.

Divisibility by 3 : The of its digits is a multiple of

1 + 2 + 3 + 4 + 5 =

Therefore, 12345 divisible by 3

Divisibility by 5 : Units place - or

Therefore, 12345 divisible by 5.

Divisibility by 6 : should be divisible by both and

Therefore, 12345 divisible by 6.

Divisibility by 9 : the of digits of the number is divisible by 9

1 + 2 + 3+ 4 + 5 =

Therefore, 12345 divisible by 9.

Reverse Order :

Checking the divisiblity rules

Therefore:

54321 divisible by 2.

54321 divisible by 3.

54321 divisible by 5.

54321 divisible by 6.

54321 divisible by 9.

6. Write different 2 digit numbers using digits 3, 4 and 5. Check whether these numbers are divisible by 2, 3, 5, 6 and 9?

The Number formed by using the digits 3, 4, 5 are 34, , 43, , 53,

Divisible by 2: Units place - 0, 2, 4, 6 and 8

Therefore , are divisible by 2

Divisible by 3: of the digits is multiple of 3

34 = 3 + 4 =

35 = 3 + 5 =

43 = 4 + 3 =

45 = 4 + 5 =

53 = 5 + 3 =

54 = 5 + 4 =

Therfore and are divisible by 3.

Divisible by 5 : Units place - 0 or 5

Therefore , are divisible by 5

Divisibility by 6 : should be divisible by both 2 and 3

Therefore is divisible by 6

Divisibility by 9 : the sum of digits of the number is divisible by 9

34 = 3 + 4 = 7

35 = 3 + 5 = 8

43 = 4 + 3 = 7

45 = 4 + 5 = 9

54 = 5 + 4 = 9

53 = 5 + 3 = 8

Therefore, and are divisible by 9.

7. Write the smallest digit and the greatest possible digit in the blank space of each of the following numbers so that the number formed are divisible by 3.

(i) ? 6724

(ii) 4765 ? 2 (iii) 7221 ? 5

Instructions

(a) ... 6724

Sum of the digits = 4 + 2 + 7 + 6 =
Thus, The smallest digit to be placed is blank space = 2.
Then the sum = 19 + 2 = , which is divisible by 3.
The greatest digit to be placed in blank space = .
Then, the sum = 19 + 8 = , which is divisible by 3
Therefore, the required digits are and .

(b) 4765 ... 2

Sum of the digits = 2 + 5 + 6 + 7 + 4 =
Thus, The smallest digit to be placed in blank space = .
Then, sum = 24 + 0 = , which is divisible by 3.
The greatest digit to be placed in blank space = .
Then, the sum = 24 + 9 = , which is divisible by 3.
Therefore, the required digits are and .

(iii) 7221...5

Sum of the digits = 7 + 2 + 2 + 1 + 5 =
Thus, The smallest digit to be placed in blank space = .
Then, sum = 17 + 1 = , which is divisible by 3.
The greatest digit to be placed in blank space = .
Then, the sum = 17 + 7 = , which is divisible by 3.
Therefore, the required digits are and .

8. Find the smallest number that must be added to 123, so that it becomes exactly divisible by 5?

A number is divisible by 5 if its last digit is 0 or 5.

The given number is 123. The next multiples of 5 after 123 are:

125 (since 125 is the next number ending in 5).

To get from 123 to 125, we add:

125 − 123 =

Thus, the smallest number that must be added to 123 to make it exactly divisible by 5 is .

9. Find the smallest number that has to be subtracted from 256, so that it becomes exactly divisible by 10?

A number is exactly divisible by 10 if its last digit is 0.

The given number is 256. To make it divisible by 10, we need to change the last digit to 0.

The nearest lower multiple of 10 is .

To get from 256 to 250, we subtract: 256 − 250 =

Thus, the smallest number that must be subtracted from 256 to make it exactly divisible by 10 is .