Number Play Class 8 Notes
easyNumber Play Class 8 Notes
Number Play Class 8 Notes
Consecutive Numbers
Numbers that follow each other in order. Example : 3, 4, 5, 6 are consecutive numbers.
Patterns in Sums of Consecutive Numbers
The sum of any four consecutive numbers is always the even.
Even and Odd Sums Properties
Odd ± Odd = Even : For example, (3 - 1 = 2).
Even ± Even = Even : For example, (4 + 2 = 6).
Odd ± Even = Odd : For example, (3 + 2 = 5).
Conclusion : Adding or subtracting four consecutive numbers will always yield an even result.
Identifying Even Numbers
An even number is any integer that can be divided by 2 without a remainder.
Last Digit Rule : A number is even if it ends in 0, 2, 4, 6, or 8.
Adding Two Numbers to Result in an Ever
Number
Combinations:
Even + Even = Even : Example : (2 + 4 = 6).
Odd + Odd = Even : Example : (3 + 5 = 8).
Even + Odd = Odd : Example : (2 + 3 = 5).
Divisibility Rules
Divisibility by 4:
Rule : Adding two even numbers that are multiples of 4 gives a multiple of 4.
Example : (4p + 4q = 4(p + q)).
Divisibility by 8 :
Rule : The sum of two multiples of 8 is also a multiple of 8.
Example : (8a + 8b = 8(a + b)).
Divisibility by 7 :
Rule : A number is divisible by 7 if it can be expressed as (7j).
Divisibility by 12 :
Rule : A number is a multiple of 12 if it can be expressed as (12m).
Checking Divisibility Quickly
Divisibility by 10 : Last digit must be 0.
Divisibility by 5 : Last digit must be 0 or 5.
Divisibility by 2 : Last digit must be even (0, 2, 4, 6, 8).
Digital Roots
Definition : The digital root is found by repeatedly summing the digits until a single-digit number is obtained.
Example : For 489710 :
First sum :(4 + 8 + 9 + 7+ l+ 0 = 29)
Second sum : (2 + 9 = 11)
Third sum : (1 + 1 = 2)
Conclusion : The digital root is 2.
Cryptarithms
Cryptarithms are puzzles where letters represent digits. Each letter stands for a unique digit from 0 to 9.
Example : In the equation SEND + MORE = MONEY, each letter represents a different digit.
Uniqueness : No two letters can represent the same digit. This rule is crucial for solving the puzzle correctly.
Example : If S = 9, then no other letter can equal 9.
Leading Digit Rule : In most cases, the first letter of a number cannot be zero, as it would not be a valid number.
Example : In SEND, S cannot be 0.
Solving Method : To solve a cryptarithm, you can use logical reasoning and trial-and-error to find which digits correspond to each letter.
Example : For SEND + MORE = MONEY, you might start by guessing values for S and M based on their positions.
Verification : After assigning digits to letters, verify the solution by substituting back into the original equation to ensure it holds true.
Example : If you find S = 9, E = 5, N = 6, D = 7, M = 1,0 = 0, R=8, Y = 2, check if 9567 + 1085 = 10652.
Is This a Multiple Of ?
Sum of Consecutive Numbers : Consecutive numbers are numbers that follow each other in order. For example, 3, 4, 5, and 6 are consecutive numbers.
Patterns in Sums of Consecutive Numbers
When you choose any four consecutive numbers and add them together, you will notice that the result is always the same, regardless of which four numbers you choose. This is because the sum of any four consecutive numbers can be expressed in a general form.
Even and Odd Sums : When you place '+' and '-' signs between four consecutive numbers, the resulting expressions will always yield an even number. This is due to the properties of even and odd numbers :
- Odd ± Odd = Even : For example, 3 - 1 = 2.
- Even ± Even = Even : For example, 4 + 2 = 6.
- Odd ± Even = Odd : For example, 3 + 2 = 5.
Identifying Even Numbers
Without Computing:
- An even number is any integer that can be divided by 2 without leaving a remainder.
- You can identify even numbers by their last digit: if a number ends in 0, 2, 4, 6, or 8, it is even.
Adding Two Numbers to Result in an Even Number
- Even + Even = Even : For example, 2 + 4 = 6.
- Odd + Odd = Even : For example, 3 + 5 = 8.
- Even + Odd = Odd : For example, 2 + 3 = 5.
Divisibility Rules
Divisibility by 4
1. Adding two (even) numbers that are multiples of 4 will always give a multiple of 4.
4p and 4q.
4p + 4q = 4 (p + q). (which is divisible by 4)
2. Adding two even numbers that are not multiples of 4 will always give a multiple of 4 because their remainders of 2 add up to 4.
(4p + 2) and (4q + 2).
(4p + 2) + (4q + 2)
= 4p + 4q + 4
= 4 (p + q + 1) (which is divisible by 4)
4p and (4q + 2)
= 4p + (4q + 2)
= 4p + 4q + 2
= 4 (p + q) + 2.. (which is not divisible by 4)
Divisibility by 8
1. The two numbers have 8 as a factor; in other words, the two numbers are multiples of 8.
8a and 8b.
2. As multiples of 8 are obtained by repeatedly adding 8, the sum of two multiples of 8 will also be a multiple of 8.
8a + 8b = 8 (a + b).
If a divides M and a divides N, then a divides M + N and a divides M - N. In other words, if M and N are multiples of a, then M + N and M - N will also be multiples of a.
3. A number divisible by 8 can be expressed as a sum of two multiples of 8 or sum of two non-multiples of 8.
8m = 8a + 8b
8m = p + q (p, q not multiples of 8)
Divisibility by 7
1. Numbers divisible by 7 will have 7 as a factor 7j.
2. This contains a total of mj rows. So this is also a multiple of 7.
(7j) × m
We can say : If A is divisible by k, then all multiples of A are divisible by k.
Divisibility by 12
1. A number divisible by 12 is a multiple of 12.
12m
2. Factors of multiples of 12 will include factors of 12.
12m =2 × 6 × m
= 3 × 4 × m
If A is divisible by k, then A is divisible by all the factors of k.
In general, if A is divisible by k and A is also divisible by m, then A is divisible by the LCM of k and m.
Checking Divisibility Quickly
Divisibility by 10, 5, and 2
1. Divisibility by 10: A number is divisible by 10 if its units digit (the last digit) is 0.
Why it works : Any number can be expressed in terms of its place values. For example, in the number 4075, the last digit is 5, and the other digits (4, 0, 7) contribute multiples of 10. Thus, a number is divisible by 10 only if the last digit is 0.
2. Divisibility by 5 : A number is divisible by 5 if its units digit is either 0 or 5.
Why it works : Similar to divisibility by 10, the last digit determines whether the number can be grouped into sets of 5. If the last digit is 0 or 5, it can be evenly divided by 5.
3. Divisibility by 2 : A number is divisible by 2 if its units digit is even (i.e., 0, 2, 4, 6, or 8).
Why it works : Even numbers can be divided into two equal parts, while odd numbers cannot. Thus, the last digit being even indicates that the entire number can be divided by 2.
A Shortcut for Divisibility by 9
- A number is divisible by 9 if the sum of its digits is divisible by 9.
- Example : For the number 123, we find the sum of its digits : (1 + 2 + 3 = 6). Since 6 is not divisible by 9, 123 is not divisible by 9.
- Why it works : This rule is based on the properties of numbers in base 10. Each digit’s place value contributes to the overall value of the number, and when summed, if the total is divisible by 9, the original number is as well.
A Shortcut for Divisibility by 3
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- Example : For the number 456, the sum of its digits is (4 + 5 + 6 = 15). Since 15 is divisible by 3, 456 is also divisible by 3.
- Why it works : Similar to the rule for 9, this is because of how numbers are structured in base 10. Each digit contributes to the total in a way that maintains divisibility by 3.
A Shortcut for Divisibility by 11
To check if a number is divisible by 11, alternate the signs of the digits starting from the rightmost digit (units place). Then, sum these values. If the result is divisible by 11, the original number is also divisible by 11.
Example : For the number 328105 :
Start from the right:
(-5 + 0 - 1 + 8 - 2 + 3 = 3)
Since 3 is not divisible by 11, 328105 is not divisible by 11. .
Why it works : This method is based on the properties of the number system and how remainders behave when divided by 11.
When a Number is Divisible by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
Example : For the number 24 :
Check divisibility by 2 : The last digit is 4 (even), so it is divisible by 2.
Check divisibility by 3: The sum of the digits (2 + 4 = 6) is divisible by 3.
Since both conditions are met, 24 is divisible by 6.
Why it works : Since 6 is the product of 2 and 3, a number must meet both criteria to be divisible by 6.
Digital Roots
The digital root of a number is found by repeatedly summing its digits until a single-digit number is obtained.
Example : For the number 489710 :
First sum :(4 + 8 + 9 + 7 + 1 + 0 = 29)
Second sum : (2 + 9 = 11)
Third sum : (1 + 1 = 2)
Thus, the digital root is 2.
Digits in Disguise
Cryptarithms : Cryptarithms are fascinating puzzles where letters represent digits. Each letter corresponds to a unique digit,-and the goal is to find out what digit each letter stands for. This type of problem encourages logical reasoning and problem¬solving skills, making it a fun way to engage with numbers.
Key Features of Cryptarithms :
- Unique Representation : Each letter represents a different digit (0-9). No two letters can represent the same digit.
- Leading Digits : In a multi-digit number, the first digit cannot be zero. For example, if a letter represents the first digit of a number, it must be between 1 and 9.
- Mathematical Operations : Cryptarithms often involve basic arithmetic operations such as addition, subtraction, multiplication, or division.
Example of a Cryptarithm
Solve UT × 3 = PUT
The equation can be rewritten in terms of digits :
Step 1 : Let (UT = 10U + T) (where (U) is the tens digit and (T) is the units digit).
Let (PUT = 100P + 10U + T) (where (P) is the hundreds digit).
Thus, the equation becomes :
(10U + T) × 3 = 100P + 10U + T
Step 2 : Simplifying the equation
Expanding the left side gives us :
30U + 3T = 100P + 10U + T
Now, we can rearrange this to isolate terms, involving (U) and (T):
30U + 3T - 10U - T = 100P
This simplifies to:
20U + 2T = 100P
Step 3 : Further Simplification
Dividing the entire equation by 2 gives us :
10U + T = 50P
Step 4 : Analysing the Equation
From this equation, we can see that (10U + T) must be a multiple of 50. Since (U) and (T) are digits (0 - 9), (10U + T) can take values from 0 to 99. The multiples of 50 within this range are 0, 50.
Step 5 : Finding Possible Values Case 1: (10U + T = 50)
This gives us (U = 5) and (T = 0).
Therefore, (UT = 50) and (PUT = 150).
Checking the equation : (50 × 3 = 150), which holds true.