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\ 06-Jan-2025 Surendra Prajapati Proof-6 Reader’s Sign _______________________ Date __________

1. Express 1234 in its expanded form.

2. Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
3. Choose the correct option:
(a) Which of the following numbers is divisible by 4?
(i) 15742 (ii) 64728 (iii) 34769 (iv) 57894

(b) Which of the following numbers is not divisible by 3?
(i) 45276 (ii) 3794814 (iii) 47857 (iv) 67584
4. Write ‘T’ for true and ‘F’ for false statement.
(a) If a number is divisible by 3, it must be divisible by 9.
(b) A number is divisible by 18, if it is divisible by both 3 and 6.

(c) All numbers which are divisible by 4 must also be divisible by 8.
(d) If a number exactly divides two numbers separately, it must exactly divide their sum.
5. Replace * by a smallest digit in the following numbers so that the numbers are divisible by 8.

(a) 865* (b) 88*0 (c) 9*68 (d) 1*128
Introduction

In previous classes, we have learnt tests of divisibility. In this chapter, we shall explore the reasons
behind the validity of these tests. Additionally, we will learn how to create puzzles involving numbers.

Numbers in General Form

In earlier classes, we learnt to write a number in expanded form by using the place value of its
digits. For example,
(a) 89 = 8 × 10 + 9 (b) 55 = 5 × 10 + 5

Hence, we observe that, in general, any 2-digit number ab, having a and b as tens and units digits
respectively, can be written as:
ab = 10 × a + b = 10a + b
In a 2-digit and 3-digit number
On reversing the digits, we get Note: ab and abc respectively, a, b and

ba = 10 × b + a = 10b + a c are whole numbers and a ≠ 0.
Now consider the following examples for 3-digit numbers:
(a) 543 = 5 × 100 + 4 × 10 + 3 (b) 983 = 9 × 100 + 8 × 10 + 3

(c) 345 = 3 × 100 + 4 × 10 + 5 (d) 389 = 3 × 100 + 8 × 10 + 9
Thus, in general, any 3-digit number abc made up of digits a, b, and c denote hundreds, tens and
ones digit of a number respectively, can be written as:
abc = 100a + 10b + c a

And by changing the place of its digits in cyclic order, we obtain numbers b
cab = 100c + 10a + b, and

bca = 100b + 10c + a c

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