Remainder Theorem: Concepts, and Tricks

Remainder Theorem (शेषफल प्रमेय) is one of the fundamental and frequently tested topics in the Number System section of various competitive examinations, especially for the Staff Selection Commission (SSC) General Duty (GD) exam. Mastering the remainder of the concepts, shortcuts, and related theorems is important for quickly solving complex division problems. Here, we provide a guide to the Remainder Theorem, its core concepts, and how to apply them effectively to solve SSC GD Previous Year Questions. 

What is the Remainder Theorem?

In the context of competitive exams, the Remainder Theorem is primarily based on the Division Algorithm. It establishes a relationship between the Dividend, Divisor, Quotient, and Remainder. The fundamental concept is that when an integer is divided by another integer, the result is a quotient and a remainder, where the remainder must be less than the divisor.

The Basic Division Algorithm

The relationship between the four components of division can be mathematically expressed by the formula:

Dividend = (Divisor × Quotient) + Remainder

Where:

• Dividend (भाज्य): The number being divided.

• Divisor (भाजक): The number by which the dividend is divided.

• Quotient (भागफल): The number of times the divisor goes into the dividend.

• Remainder (शेषफल): The number left over after the division. The remainder (R) must always satisfy the condition 0 ≤ R < Divisor

Key Concepts and Tricks for Remainder Theorem

Aspirants need to become familiar with shortcut methods and advanced concepts related to the Remainder Theorem in order to excel in the SSC GD Mathematics section. Here are some concepts:

1. Concept of Positive and Negative Remainders

The remainder is always a non-negative integer by definition. However, for ease of calculation in complicated questions involving large numbers and exponents, the concept of a Negative Remainder is used.

• Positive Remainder: The standard, non-negative remainder.

• Negative Remainder: The deficiency needed in the dividend to make it perfectly divisible by the divisor.

Conversion: To find the actual positive remainder from a negative remainder, simply add the divisor to the negative remainder.

Positive Remainder = Divisor + Negative Remainder

Example:

When 15 is divided by 4:

• Positive Remainder:  15 = 4 × 3 + 3, remainder = 3

• Negative Remainder:  15 is 1 less than the next multiple of 4 (16), remainder = -1

• Conversion Check:  4 + (-1) = 3

2. Remainder of Sums and Products

The most important trick for complex arithmetic problems is finding the remainder of individual components first:

Remainder of a Sum:
Remainder of (A + B) ÷ N = Remainder of [Rem(A ÷ N) + Rem(B ÷ N)] ÷ N

Remainder of a Product (Multiplicative Property):
Remainder of (A × B) ÷ N = Remainder of [Rem(A ÷ N) × Rem(B ÷ N)] ÷ N 

This property is key to solving multiplication problems efficiently without calculating the full product.

3. Special Theorem: Divisibility by (a + b) or (a - b) Forms

These formulae are frequently tested in SSC GD PYQs:

• (aⁿ + bⁿ) is always divisible by (a + b) when n is an odd number.

• (aⁿ − bⁿ) is always divisible by (a + b) when n is an even number.

• (aⁿ − bⁿ) is always divisible by (a − b) for every natural number n.

Remainder Theorem Practice Questions (SSC GD PYQs)

Below are examples of the problem types commonly found in the SSC GD exam, which require the application of the above concepts.

Type 1: Basic Division Algorithm Problems

These questions test your understanding of the relationship between the four components: Dividend, Divisor, Quotient, and Remainder.

Example 1:

In a division problem, the divisor is 8 times the quotient and 4 times the remainder. If the remainder is 12, find the dividend.

Solution:

Given Remainder (R): 12

Calculate Divisor (d):
The divisor is 4 times the remainder.
d = 4 × R = 4 × 12 = 48

Calculate Quotient (q):
The divisor is 8 times the quotient.
d = 8 × q
48 = 8 × q
q = 48 ÷ 8 = 6

Calculate Dividend (D):
Use the Division Algorithm.
Dividend = (Divisor × Quotient) + Remainder
D = (48 × 6) + 12
D = 288 + 12
D = 300

Answer: The dividend is 300.

Type 2: Finding Remainder with Exponents (Negative Remainder Trick)

These problems are best solved using the negative remainder concept to handle large powers.

Example 2:

Find the remainder when 2^103  is divided by 9.

Solution:

Find a power of 2 that is close to a multiple of 9.
2^3 = 8

Express the dividend using this power:
2^103 = 2^102 × 2^1
= (2^3)^34 × 2
= 8^34 × 2

Apply the remainder property:
Remainder of (2^103 ÷ 9)
= Remainder of (8^34 × 2 ÷ 9)

Use the negative remainder trick:
Remainder of (8 ÷ 9) = -1

So, remainder of [(-1)^34 × 2 ÷ 9]

Simplify:
(-1)^34 = 1

Remainder of (1 × 2 ÷ 9)
= Remainder of (2 ÷ 9)

Answer: The remainder is 2.

Remainder Theorem Video

Watch the detailed Remainder Theorem video to understand key concepts, shortcut tricks, and SSC GD–level questions. The lecture explains division rules, negative remainder methods, and exponent-based problems step by step for better clarity.