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Patterns in Mathematics - Patterns in Numbers

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A number pattern is a sequence of numbers that follows a specific rule. For example, in the sequence 5,10,15,20,…5, 10, 15, 20, \dots, the rule is to add 55 to the previous term.

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Patterns in Addition: Adding numbers like 9,99,9999, 99, 999 can be simplified using the logic n+9=n+10−1n + 9 = n + 10 - 1 and n+99=n+100−1n + 99 = n + 100 - 1.

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Patterns in Multiplication: Multiplying by 5,25,1255, 25, 125 can be simplified by using powers of 1010 and 22. For example, a×5=a×102a \times 5 = a \times \frac{10}{2} and a×25=a×1004a \times 25 = a \times \frac{100}{4}.

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Triangular Numbers: These are numbers that can be arranged in the shape of an equilateral triangle. The sequence is 1,3,6,10,15,…1, 3, 6, 10, 15, \dots. The nthn^{th} triangular number is given by the sum of first nn natural numbers.

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Square Numbers: These are numbers that can be arranged in a square shape. The sequence is 1,4,9,16,25,…1, 4, 9, 16, 25, \dots. The nthn^{th} square number is n×nn \times n or n2n^2.

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Even and Odd Patterns: The sum of two even numbers is always even (2+4=62 + 4 = 6). The sum of two odd numbers is always even (3+5=83 + 5 = 8). The sum of an even and an odd number is always odd (2+3=52 + 3 = 5).

📐Formulae

n+9=n+10−1n + 9 = n + 10 - 1

n+99=n+100−1n + 99 = n + 100 - 1

a×5=a×102a \times 5 = \frac{a \times 10}{2}

a×25=a×1004a \times 25 = \frac{a \times 100}{4}

nth Square Number=n×n=n2n^{th} \text{ Square Number} = n \times n = n^2

Sum of first n odd numbers=n2\text{Sum of first } n \text{ odd numbers} = n^2

💡Examples

Problem 1:

Observe the pattern and find the next two terms: 1,3,6,10,…1, 3, 6, 10, \dots

Solution:

15,2115, 21

Explanation:

These are triangular numbers. The difference between terms increases by 11 each time: 3−1=23 - 1 = 2, 6−3=36 - 3 = 3, 10−6=410 - 6 = 4. So, the next differences will be 55 and 66. Term 5: 10+5=1510 + 5 = 15. Term 6: 15+6=2115 + 6 = 21.

Problem 2:

Solve 123+999123 + 999 using number patterns.

Solution:

11221122

Explanation:

We can write 999999 as (1000−1)(1000 - 1). Therefore, 123+999=123+1000−1=1123−1=1122123 + 999 = 123 + 1000 - 1 = 1123 - 1 = 1122.

Problem 3:

Find the product of 64×2564 \times 25 using the division-multiplication pattern.

Solution:

16001600

Explanation:

Using the rule a×25=(a×100)÷4a \times 25 = (a \times 100) \div 4, we get: 64×25=64004=160064 \times 25 = \frac{6400}{4} = 1600.

Problem 4:

Find the sum of the first 66 odd numbers without actual addition.

Solution:

3636

Explanation:

The sum of the first nn odd numbers is n2n^2. Here n=6n = 6, so the sum is 62=6×6=366^2 = 6 \times 6 = 36.

Problem 5:

Calculate 45×9945 \times 99 using the subtraction pattern.

Solution:

44554455

Explanation:

We know 45×99=45×(100−1)=4500−4545 \times 99 = 45 \times (100 - 1) = 4500 - 45. Calculating the subtraction: 4500−454455\begin{array}{r} 4500 \\ - 45 \\ \hline 4455 \end{array}

Patterns in Numbers Class 6 Notes & Examples | CBSE Maths