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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
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Number Theory : Level 1
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CAT 2025 Lesson : Number Theory - Arithmetic Operations

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2.4 Arithmetic operations

While performing mathematical calculations, we must operate in order.

Eg., 3+42=3+23 + \dfrac{4}{2} = 3 + 23+24​=3+2 and not 3+42\dfrac{3 + 4}{2}23+4​

Eg., 3×5−2=15−2=133 \times 5 - 2 = 15 - 2 = 133×5−2=15−2=13 and not 3×3=93 \times 3 = 93×3=9.

The rules of which mathematical operator comes first are denoted by BODMAS

2.4.1 BODMAS

BODMAS is an acronym, where each letter (as stated below) represents an arithmetic operation.

B = Brackets
O = Orders (i.e. Powers and Square Roots, Cube Roots, etc.)
D = Division
M = Multiplication
A = Addition
S = Subtraction

BODMAS represents the order in which arithmetic operations are to be performed in an expression, from left to right.

In other words, operations involving Brackets are completed first, followed by those involving Orders (Powers), Division, Multiplication, Addition and finally Subtraction.

We should always follow this BODMAS order while solving questions. If not our answers might go wrong.

The order in which the operations within different types of brackets are to be done is as follows

1) 123‾\overline{\color{white} 123}123 called line bracket. Eg, 4×5+3‾=4×8=324 \times \overline{5 + 3} = 4 \times 8 = 324×5+3​=4×8=32
2) () called parenthesis or common bracket
3) {} called curly bracket
4) [] called rectangular bracket

Example 8

8+3×(4×5−6‾)×{15−12}=?8 + 3 \times (4 \times \overline{5 - 6}) \times \{15 - 12 \}= ?8+3×(4×5−6​)×{15−12}=?

Solution

8+3×(4×5−6‾)×{15−12}8 + 3 \times (4 \times \overline{5 - 6}) \times \{15 - 12 \}8+3×(4×5−6​)×{15−12}
=== 8+3×(4×−1)×38 + 3 \times (4 \times -1) \times 38+3×(4×−1)×3
=== 8+3×−4×38 + 3 \times -4 \times 38+3×−4×3
=== 8−368-368−36
=== −28-28−28
Answer: −28-28−28


2.4.2 Basics of Remainders

When a number say nnn (also called dividend) is divided by divisor (ddd), it leaves a quotient (qqq) and remainder (rrr).



So, nnn can be expressed as n=dq+rn = dq + rn=dq+r

For example, when 373737 is divided by 999, we get a quotient of 444 and remainder of 111.

37=9×4+137 = 9 \times 4 + 137=9×4+1


Example 9

148714871487, when divided by xxx, leaves a remainder of 555. What is the largest three-digit number that xxx can be?

Solution

The dividend is 148714871487 and remainder is 555. xxx is the divisor. Let the quotient be qqq.

1487=xq+51487 = xq + 51487=xq+5

⇒ xq=1482xq = 1482xq=1482

⇒ x=1482qx = \dfrac{1482}{q}x=q1482​

Value of xxx is highest when qqq is lowest.

When q=1q = 1q=1, we get a 444-digit number.

When q=2q = 2q=2, x=14822=741x = \dfrac{1482}{2} = 741x=21482​=741, which is the largest possible 333-digit number.

Answer: 741741741

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