E.g 176Γ176
Diff- 24- square - 576
Substarct 24 from 176 = 152
152Γ2 = 304
Concate 304 and 576 such that 5 will carry
Ans: 30976
Diff- 24- square - 576
Substarct 24 from 176 = 152
152Γ2 = 304
Concate 304 and 576 such that 5 will carry
Ans: 30976
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I think you can now easily find the squares of any number by understanding the above shortcuts. From tomorrow onwards, we will start learning how to identify whether a number is a perfect square or not, along with the rules for it.
Excited?
Excited?
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Basic Mathematicsπ€
Base 200 : i.e for numbers between 175-225. 1. Get the difference between 200 and given number and square diff(Write it on right side such that it will consume last two digits and carry 100th digit. 2. Subtract the difference( if number < 200) or Add the differenceβ¦
Here, we find square of number 212 which is 44944.
This form of number are called palindrome that is from middle they are at same distance on both the sides and similar also.
This form of number are called palindrome that is from middle they are at same distance on both the sides and similar also.
More examples of Palindrome are:
β’ 1^2 = 1
β’ 11^2 = 121
β’ 111^2 = 12321
β’ 1111^2 = 1234321
β’ 11111^2 = 123454321
β’
β’
And so on upto 9 times 1.
β’ 1^2 = 1
β’ 11^2 = 121
β’ 111^2 = 12321
β’ 1111^2 = 1234321
β’ 11111^2 = 123454321
β’
β’
And so on upto 9 times 1.
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How to identify given number is perfect square or not?
β’ Last digit of a perfect square is NEVER 2,3,7 or 8 And it's always 0,1,4,5,6 and 9.
e.g: 100, 81, 64, 25, 36, 529...
β’ If last digit of square is 5, then the 2nd last digit is always 2.
e.g: 25, 625, 1225...
β’ If last digit of square is 6, then the 2nd last digit is always ODD.
e.g: 36, 256...
β’ If last digit of a square is 1,4 and 9 then 2nd last digit is always EVEN.
e.g: 81, 121, 144, 484, 49, 1089...
β’ The number of zero's at the end of a square is always EVEN & the non zero part should be a perfect Square.
e.g: 900, 25600, 10000...
β’ Last digit of a perfect square is NEVER 2,3,7 or 8 And it's always 0,1,4,5,6 and 9.
e.g: 100, 81, 64, 25, 36, 529...
β’ If last digit of square is 5, then the 2nd last digit is always 2.
e.g: 25, 625, 1225...
β’ If last digit of square is 6, then the 2nd last digit is always ODD.
e.g: 36, 256...
β’ If last digit of a square is 1,4 and 9 then 2nd last digit is always EVEN.
e.g: 81, 121, 144, 484, 49, 1089...
β’ The number of zero's at the end of a square is always EVEN & the non zero part should be a perfect Square.
e.g: 900, 25600, 10000...
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β€2
β’ 7744 is the Only Perfect square number of the form/type XXYY.
i.e 88^2 = 7744.
i.e 88^2 = 7744.
π2
Cube And Cube Roots:
You should know the Cubes of numbers from 1-10.
1^3- 1
2^3- 8
3^3- 27
4^3- 64
5^3- 125
6^3- 216
7^3- 343
8^3- 512
9^3- 729
10^3- 1000
You can easily find cube of any number by using formulas given:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3
(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
You should know the Cubes of numbers from 1-10.
1^3- 1
2^3- 8
3^3- 27
4^3- 64
5^3- 125
6^3- 216
7^3- 343
8^3- 512
9^3- 729
10^3- 1000
You can easily find cube of any number by using formulas given:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3
(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
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Example:
23^3 = (20 + 3)Β³
Step 1 β Apply Formula
(20 + 3)^3 = 20^3 + 3(20^2Γ3) + 3(20Γ3^2) + 3^3
Step 2 β Solve
= 8000 + 3(400Γ3) + 3(20Γ9) + 27
= 8000 + 3600 + 540 + 27 ξ
= 12,167
23^3 = (20 + 3)Β³
Step 1 β Apply Formula
(20 + 3)^3 = 20^3 + 3(20^2Γ3) + 3(20Γ3^2) + 3^3
Step 2 β Solve
= 8000 + 3(400Γ3) + 3(20Γ9) + 27
= 8000 + 3600 + 540 + 27 ξ
= 12,167
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7Β³ = (10 β 3)Β³
Step 1 β Apply Formula
(10 - 3)^3 = 10^3 - 3(10^2Γ3) + 3(10Γ3^2) - 3^3
Step 2 β Solve
= 1000 - 3(100Γ3) + 3(10Γ9) - 27
= 1000 - 900 + 270 - 27 ξ
= 343
Step 1 β Apply Formula
(10 - 3)^3 = 10^3 - 3(10^2Γ3) + 3(10Γ3^2) - 3^3
Step 2 β Solve
= 1000 - 3(100Γ3) + 3(10Γ9) - 27
= 1000 - 900 + 270 - 27 ξ
= 343
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How to identify cube of number: There is trick:
Check last digit of number and Check last digit of answer also
Number Cube Last digit
1 1 1
2 8 8
3 27 7
4 64 4
5 125 5
6 216 6
7 343 3
8 512 2
9 729 9
10 1000 0
By checking only last digits you can eliminate at least 1 or 2 or 3 options.
You will be close to answer.
Check last digit of number and Check last digit of answer also
Number Cube Last digit
1 1 1
2 8 8
3 27 7
4 64 4
5 125 5
6 216 6
7 343 3
8 512 2
9 729 9
10 1000 0
By checking only last digits you can eliminate at least 1 or 2 or 3 options.
You will be close to answer.
β€3π3π―1
π4β€3
Basic Mathematicsπ€
Find the cube of 54?
Here it is expected that you should check the last digit...
As 4 is there at last in 54... According to shortcut...
There should be 4Γ4Γ4 = 16Γ4 i.e. 6Γ4= 24 i.e. 4 should be there at last digit in answer.
As there are two such options.
We will go for formula....
54Β³= (50+4)Β³
=12500+3Γ2500Γ4+3Γ50Γ16+64
=125000+30000+2400+64
=1,57,464
As 4 is there at last in 54... According to shortcut...
There should be 4Γ4Γ4 = 16Γ4 i.e. 6Γ4= 24 i.e. 4 should be there at last digit in answer.
As there are two such options.
We will go for formula....
54Β³= (50+4)Β³
=12500+3Γ2500Γ4+3Γ50Γ16+64
=125000+30000+2400+64
=1,57,464
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LCM (Least Common Multiple)
Definition βThe smallest number that is divisible by all given numbers.
Steps to Find LCM
Method 1 β Prime Factorization
β’ Factorize each number into prime numbers.
β’ Take each prime factor only once with its highest power.
β’ Multiply them together.
Example:
Find LCM of 12 and 18.
12 = 2Β² Γ 3ΒΉ
18 = 2ΒΉ Γ 3Β²
LCM = 2Β² Γ 3Β² = 36
---
Method 2 β Division Method (Shortcut)
β’ Divide the numbers by prime numbers till all become 1.
β’ Multiply all divisors together.
Example:
Find LCM of 15, 20, 30.
Divisor: 15 , 20, 30
2- 15 10 15
2- 15 5 15
3- 5 5 5
5- 1 1 1
LCM = 2 Γ 2 Γ 3 Γ 5 = 60
Definition βThe smallest number that is divisible by all given numbers.
Steps to Find LCM
Method 1 β Prime Factorization
β’ Factorize each number into prime numbers.
β’ Take each prime factor only once with its highest power.
β’ Multiply them together.
Example:
Find LCM of 12 and 18.
12 = 2Β² Γ 3ΒΉ
18 = 2ΒΉ Γ 3Β²
LCM = 2Β² Γ 3Β² = 36
---
Method 2 β Division Method (Shortcut)
β’ Divide the numbers by prime numbers till all become 1.
β’ Multiply all divisors together.
Example:
Find LCM of 15, 20, 30.
Divisor: 15 , 20, 30
2- 15 10 15
2- 15 5 15
3- 5 5 5
5- 1 1 1
LCM = 2 Γ 2 Γ 3 Γ 5 = 60
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β€2
HCF (Highest Common Factor)
It is also called GCD (Greatest Common Divisor).
#Definition
The HCF of two or more numbers is the largest positive integer that divides each of the numbers completely (without leaving any remainder).
Steps to Find HCF:
Method 1: Prime Factorization
1. Find the prime factors of each number.
2. Identify the common prime factors.
3. Multiply them to get the HCF.
Example:
Find HCF of 18 and 24.
Prime factors of 18 β 2Γ3Γ3
Prime factors of 24 β 2Γ2Γ2Γ3
Common factors β 2Γ3-> HCF = 6
It is also called GCD (Greatest Common Divisor).
#Definition
The HCF of two or more numbers is the largest positive integer that divides each of the numbers completely (without leaving any remainder).
Steps to Find HCF:
Method 1: Prime Factorization
1. Find the prime factors of each number.
2. Identify the common prime factors.
3. Multiply them to get the HCF.
Example:
Find HCF of 18 and 24.
Prime factors of 18 β 2Γ3Γ3
Prime factors of 24 β 2Γ2Γ2Γ3
Common factors β 2Γ3-> HCF = 6
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Method 2: Division Method (Euclidean Algorithm)
1. Divide the larger number by the smaller number.
2. Take the remainder.
3. Divide the smaller number by the remainder.
4. Repeat until the remainder becomes 0.
The last divisor is the HCF.
Example:
Find HCF of 36 and 60:
60/36=1 remainder 24
36/24=1 remainder 12
24/12=2 remainder 0
HCF = 12
1. Divide the larger number by the smaller number.
2. Take the remainder.
3. Divide the smaller number by the remainder.
4. Repeat until the remainder becomes 0.
The last divisor is the HCF.
Example:
Find HCF of 36 and 60:
60/36=1 remainder 24
36/24=1 remainder 12
24/12=2 remainder 0
HCF = 12
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Another way to find HCF when Two numbers are there and LCM is Given :
Formula: HCF Γ LCM = N1 Γ N2
e.g: N1 = 15, N2 = 20, LCM = 60, HCF= ?
HCF = 15Γ20/60
= 300/60
HCF = 5
Note: This formula is valid for only two numbers.
Formula: HCF Γ LCM = N1 Γ N2
e.g: N1 = 15, N2 = 20, LCM = 60, HCF= ?
HCF = 15Γ20/60
= 300/60
HCF = 5
Note: This formula is valid for only two numbers.
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