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Number Theory

Number Theory

MODULES

Basics of Numbers
Types of Numbers
Fractions
Arithmetic Operations
Other Numerical Operations
Algebraic Expansion
Prime Numbers
Counting Integers
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Number Theory 1
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Number Theory 2
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Algebraic Expansion 1
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PRACTICE

Number Theory : Level 1
Number Theory : Level 2
Number Theory : Level 3
ALL MODULES

CAT 2025 Lesson : Number Theory - Basics of Numbers

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1. Introduction

Numbers are used to measure and count. The number system used globally is the decimal number system, with the base 10. This just means that there are 10 digits in this number system. They are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Unless specified otherwise, the base 10 number system will be used for all calculations.

The base 9 number system will use the first 9 digits mentioned above which are 0, 1, 2, 3, 4, 5, 6, 7, 8. Likewise, the base 8 number system will not include the digits 8 and 9. So, the base 2 number system will include only the digits 0 and 1. This is called the binary number system.

In number systems of base higher than 10, additional digits need to be created. The widely agreed form is to use letters for the digits after 9. For example, the hexadecimal (base 16) representation uses the letters A, B, C, D, E and F for the numbers 10, 11, 12, 13, 14 and 15 respectively. We will discuss these in detail in the Number Systems lesson.

1.1 Decimal Number System

We follow a positional number system, wherein the position of a digit has a certain value called the Place Value. The actual value of the digit in that position is called Face Value.

In the number 34.25, the face values are 3, 4, 2 and 5, while their respective place values are 101,100,10−110^1, 10^0, 10^{-1}101,100,10−1 and 10−210^{-2}10−2.

When we multiply the face values with their respective place values and add them up, we get the value of the number

34.25=3×101+4×100+2×10−1+5×10−234.25 = 3 \times 10^1 + 4 \times 10^0 + 2 \times 10^{-1} + 5 \times 10^{-2}34.25=3×101+4×100+2×10−1+5×10−2

34.25=3×10+4×1+210+510034.25 = 3 \times 10 + 4 \times 1 + \dfrac{2}{10} + \dfrac{5}{100}34.25=3×10+4×1+102​+1005​

Note that digits to the left of the decimal point are multiplied with 10010^0100, 10110^1101, 10210^2102 and so on, while the place values to the right of the decimal point are 10−110^{-1}10−1, 10−210^{-2}10−2, and so on.

Therefore, digits in the places to the left of the decimal point are called units digit, tens digit, hundreds digit, thousands digit and so on. The digits in places to the right of the decimal point are called tenths digit, hundredths digit, thousandths digit and so on.

In the above example, 333 is tens digit, 444 is the units digit, 222 is the tenths digit and 555 is the hundredths digit.

1.2 Application in Questions

Questions in MBA tests might require us to find 222-digit or 333-digit numbers with certain conditions around the digits. In these questions, we assume the digits to be variables and form the numbers.

A 222-digit number can be written as 10a+b10a + b10a+b, where aaa and bbb are the tens and units digit respectively.

Likewise, a 333–digit number can be written as 100a+10b+c100a + 10b + c100a+10b+c where aaa, bbb and ccc are the hundreds, tens and units digit respectively. 444–digit numbers can be written 1000a+100b+10c+d1000a + 100b + 10c + d1000a+100b+10c+d , where aaa, bbb, ccc and ddd are the thousands, hundreds, tens and units digit respectively.

Example 1

When a 333-digit number is reversed, we get another 333-digit number. If the difference between these two numbers is 297297297, then what is the difference between the hundreds and the units digit of the number?

Solution

Let aaa, bbb and ccc be the hundreds, tens and units digit of the number. Therefore, the number is 100a+10b+c\bm{100a} + \bm{10b} + \bm{c}100a+10b+c.

When the digits are reversed, the number formed is 100c+10b+a\bm{100c} + \bm{10b} + \bm{a}100c+10b+a.

100a+10b+c−(100c+10b+a)=297100 a + 10 b + c - (100 c + 10 b + a) = 297100a+10b+c−(100c+10b+a)=297

⇒ 99a−99c=297 99 a - 99c = 29799a−99c=297

⇒ a−c=3 a - c = 3a−c=3

Answer: 333

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