# NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.2

NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.2 (Ex. 16.2) Playing with Numbers in Hindi and English Medium free PDF file for CBSE Session 2021-2022. Videos related to all questions of chapter 16 of Class VIII Maths are solved step by step.

Class 8 mathematics exercise 16.2 contains the questions based on divisibility concepts. In this exercise questions are easy to solve but somehow tricky. Video solutions will help in all the doubts.

## Class 8 Maths Chapter 16 Exercise 16.2 Solution

Class: 8 | Mathematics |

Chapter: 16 | Playing with Numbers |

Exercise: 16.2 | Hindi and English Medium Solutions |

### CBSE NCERT Class 8 Maths Chapter 16 Exercise 16.2 Solution in Hindi and English Medium

### Class 8 Maths Chapter 16 Exercise 16.2 Solution in Videos

#### Test of Divisibility

##### 1. Test of Divisibility by 2

Consider a 2-digit number having b and c as the tens digit and units digit respectively.

Then, this number is (10b + c). Clearly, this number is divisible by 2, when c is divisible by 2.

This happens when c = 0, 2, 4, 6 or 8.

Again, consider a 3-digit number having a, b and c as the hundreds digit, tens digit and one’s digit respectively.

Then, this number is (100a + 10b + c).

Clearly, this number is divisible by 2, when c is divisible by 2.

This happens when c = 0, 2, 4, 6 or 8.

Rule: In general, a given number is divisible by 2 only when its units digit is 0, 2, 4, 6 or 8.

Example:

Each of the numbers 60, 72, 84, 96, 108 is divisible by 2.

##### 2. Test of Divisibility by 3

Consider a 2-digit number having b and c as the tens digit and units respectively.

Then, this number is (10b + c) = 9b + (b + c)

Clearly, this number is divisible by 3 only when (b + c) is divisible by 3

Thus, a 2-digit number is divisible by 3 only when the sum of its digits is divisible by 3

Again, consider a 3-digit number having a, b and c as the hundreds digit, tens digit and ones digit respectively.

Then, this number is (100a + 10b + c) = 99a + 9b + (a + b + c)

Clearly, this number is divisible by 3 only when the sum of its digit is divisible by 3. In general, we have the following rule:

Rule: A given number is divisible by 3 only when the sum of its digits is divisible by 3.

##### Test the divisibility of each of the following numbers by 3. (i) 16785, (ii) 976485

(i) The given number is 16758.

Sum of its digits = (1 + 6 + 7 + 8 + 5) = 27, which is divisible by 3

So, 16785 is divisible by 3

(ii) The given number is 976485

Sum of its digits = (9 + 7 + 6 + 4 + 8 + 5) = 39, which is divisible by 3

So, 976485 is divisible by 3

##### 3. Test of Divisiblity by 5

Consider a 2-digit number having b and c as tens digit and units digit respectively

Then, this number is (10b + c) Clearly, this number is divisible by 5 only when c is divisible by 5.

This happens when c = 0 or c = 5. Thus, a 2-digit number is divisible by 5 when its unit digit is 0 or 5. Again, consider a 3-digit number having a, b and c as hundreds digit, tens and units digit respectively.

Then, this number is (100a + 10b + c). Clearly, this number is divisible by 5 only when c is divisible by 5. This happens when c = 0 or c = 5. Thus, a 3-digit number is divisible by 5 only when its units digit is 0 or 5.

In general, we have the following rule:

Rule: A given number is divisible by 5 only when the units digit is 0 or 5.

Example:

Each of the numbers 67230 and 83715 is divisible by 5. And, none of the numbers 136, 247, 158, 373, 419, 241, 514, 632 is divisible by 5.

##### 4. Test of Divisibility by 9

Consider a 2-digit number having b and c as tens digit and units digit respectively.

Then, this number is (10b + c) = [9b + (b + c)].

Clearly, this number is divisible by 9 only when (b + c) is divisible by 9.

Thus, a 2-digit number is divisible by 9 only when the sum of its digits is divisible by 9. Again, consider a 3-digit number having a, b and c as hundreds digit, tens and units digit respectively.

Then, this number is (100a + 10b + c) = 99a + 9b + (a + b + c).

Clearly, this number is divisible by 9 only when (a + b + c) is divisible by 9. Thus, a 3-digit number is divisible by 9 only when the sum of its digit is divisible by 9. In general, we have the following rule:

Rule: A given number is divisible by 9 only when the sum of its digits is divisible by 9.

Example:

Test the divisibility of numbers 27891 by 9.

The given number is 27891

Sum of its digits = (2 + 7 + 8 + 9 + 1) = 27. which is divisible by 9

So, 27891 is divisible by 9.

##### Letters for Digits (Cry Ptarithms)

Example:

Replace A, B, C, D by suitable numerals in the following:

6 A B 5

+ D 5 8 C

9 3 5 1

Clearly, C = 6, Now (5 + 6) = 11

So, 1 is carried over

(1 + 8 + 6) = 15,

So, B = 6 and 1 is carried over.

(1 + 5 + 7) = 13,

So, A = 7 and 1 is carried over.

(1 + 6 + 2) = 9,

So, D = 2. . . A = 7, B = 6, C = 6 and D = 2.

#### Magic Squares and Triangles

The squares containing 9, 16, 25, ….. boxes with numbers in each box in such a way that the sum of the numbers in each row, each column and each diagonal remains the same are called magic squares. We shall discuss magic square of the order 3 x 3 having 3 rows and 3 columns i.e., nine boxes.

Let us consider a magic square of the order 3 x 3 containing nine square boxes

We find that:

(i) Numbers from 1 tp 9 have been put in each box.

(ii) Numbers in the horizontal arrangement constitute the rows R, R and R.

(iii) Row R has numbers 6, 7, 8.

1 Row R has number 1, 5, 9.

2 Row R has number 8, 3, 4. 3

(iv) Number in the vertical arrangement.

Constitute the columns C, C and C. 1 2 3

(v) Column C has numbers 6, 1, 8.

1 Column C has number 7, 5, 3.

2 Column C has number 2, 9, 4.

(vi) The numbers written in the boxes from the top left box to the bottom right box to the bottom right box constitute the numbers (6, 5, 4) in the main diagonal.

(vii) The numbers written in the top right box to the bottom left box constitute the number the number (2, 5, 8) in the other diagonal.

(viii) The sum of the numbers in each row, in each column and in each diagonal is equal to 15.

The constant term 15 is called the magic constant or magic sum.

##### Complete the magic square with numbers 5 to 13 such that the sum of the numbers in each row, each column and both the diagonals is equal to 27.

(i) In the first row the given numbers are 12 and 8. Hence, middle box must have 7 in it. (Since, 12 + 8 + 7 = 27)

(ii) In the third column two boxes contain the numbers 8 and 6. Hence, the middle box should have 13. (Since, 6 + 8 + 13 = 27)

(iii) In the main diagonal, two extreme boxes have numbers 12 and 6. Hence, the middle box should have 9. (Since, 12 + 6 + 9 =27)

(iv) Now, consider the other diagonal. In other diagonal the given numbers are 8 and 9. Hence, to make the sum 27, third box should have number 10 in it. (Since, 8 + 9 + 10 = 27)

(v) Now, in the first column, first and last box have numbers 12 and 10. Hence, for the sum to be 27, it should have 5 in it.

(vi) Now, in the last row, given numbers are 10 and 6. Hence, middle box should have number 11 to make the sum 27. Thus, in each row, in each column and in both the diagonals the numbers sum up to 27.

##### Test the divisibility of numbers 865917 by 9.

The given number is 865917

Sum of its digits = (8 + 6 + 5 + 9 + 1 + 7) = 36. which is divisible by 9

865917 is divisible by 9

##### Find all possible values of y for which the 4-digit number 51y3 is divisible by 9. Also, find each such number.

The given number is 51y3

Sum of its digits = (5 + 1 + y + 3) = (9 + y)

which must divisible by 9

This happens when y = 0 or y = 9

So, the required numbers are 5103 and 5193.

##### Find the values of A, B and C in the following: 3 5 A – C B 8 = 1 8 3

Since (11 – 8) = 3, so, A = 1 and 1 is taken away from 5,

Now, (14 – 6) = 8, So, B = 6 and 1 is taken away from 3

Now, 2 – 1 = 1, So, C = 1

A = 1, B = 6 and C = 1