# Concept: Successive Division

CONTENTS

INTRODUCTION

When we divide N successively by 'a' and 'b', it means we first divide N by a, and then divide the quotient of this division by b.

Remember:
• Dividend = Divisor × Quotient + Remainder

Example: Divide 100 successively by 2 and 5.

Solution:
We first divide 100 by 2. The quotient will be 50

Now, we divide 50 by 5 and the final answer will be 10.

Example: What are the remainder when 100 is successively divided by 3 and 5.

Solution:
We first divide 100 by 3. The quotient will be 33 while the remainder will be 1.

Now, we divide 33 by 5. The quotient will be 6 while the remainder will be 3.

∴ When 100 is successively divided by 3 and 5, the remainders are 1 and 3 respectively.

SMALLEST POSSIBLE NUMBER

Example: Find the smallest possible number which when successively divided by 3 and 4 leaves remainder of 1 and 3 respectively.

Solution:
In such examples we need to start from the last of the successive divisions.

When divided by 4 the remainder is 3. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 4 leaves remainder 3 is 3 itself.

Now, 3 must have been the quotient of previous division.

We need to find the number which when divided by 3 gives quotient as 3 and leaves remainder of 1.

∴ Original number = 3 × 3 + 1 = 10

The smallest number which when successively divided by 3 and 4 leaves remainder of 1 and 3 respectively is 10.

Example: Find the smallest possible number which when successively divided by 2 and 5 leaves remainder of 1 and 4 respectively.

Solution:
In such examples we need to start from the last of the successive divisions.

When divided by 5 the remainder is 4. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 5 leaves remainder 4 is 4 itself.

Now, 4 must have been the quotient of previous division.

We need to find the number which when divided by 2 gives quotient as 4 and leaves remainder of 1.

∴ Original number = 2 × 4 + 1 = 9

The smallest number which when successively divided by 2 and 5 leaves remainder of 1 and 4 respectively is 9.

Example: Find the smallest possible number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively.

Solution:
In such examples we need to start from the last of the successive divisions.

When divided by 5 the remainder is 3. Since we need to find the smallest possible such number, we will assume the quotient for this division to be 0.
∴ The smallest possible number which when divided by 5 leaves remainder 3 is 3 itself.

Now, 3 must have been the quotient of previous division.

We need to find the number which when divided by 3 gives quotient as 3 and leaves remainder of 2.

∴ Number = 3 × 3 + 2 = 11

Now, 11 must have been the quotient of previous division.

We need to find the number which when divided by 2 gives quotient as 11 and leaves remainder of 1.

∴ Initial Number = 2 × 11 + 1 = 23

The smallest number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively is 23.

GENERAL FORM

A number (N) when successively divided by a, b and c leaving remainders p, q and r respectively can be written as

N = (a × b × c) × k + n

where k is a whole number and n is the smallest such number.

Example: Find the highest three-digit number possible number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively.

Solution:
In such example we need to first find the smallest possible such number.

The smallest number which when successively divided by 2, 3 and 5 leaves remainder of 1, 2 and 3 respectively is 23. (from previous example)

Now, N = 2 × 3 × 5 × k + 23

⇒ N = 30k + 23

Now, N = 30k + 23 < 1000

⇒ k < 977/30
∴ highest possible value of k = 32

∴ N = 30 × 32 + 23 = 983

Example: Find the smallest three-digit number possible number which when successively divided by 3 and 5 leaves remainder of 1 and 2 respectively.

Solution:
In such example we need to first find the smallest possible such number.

The smallest number which when successively divided by 3 and 5 leaves remainder of 1 and 2 respectively is 7.

Now, N = 3 × 5 × k + 7

⇒ N = 15k + 7

Now, N = 15k + 7 > 99

⇒ k > 92/15
∴ least possible value of k = 7

∴ N = 15 × 7 + 7 = 112

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