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15 Jul 2022, 06:48
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Show SpoilerDOWNLOAD PDF: How To Solve: Prime Numbers
Attachment:
How To Solve: Prime Numbers
Attached pdf of this Article as SPOILER at the top! Happy learning!
Hi All,
I have posted a video on YouTube to discuss about Prime Numbers
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
Theory
- ¤ What is a Prime Number?
¤ Generic form of a Prime Number
¤ How to check if a number is prime?
¤ Sample Problems
What is a Prime Number?
Prime numbers are the numbers which have only two factors
¤ The number itself
¤ And 1
Example of Prime numbers are : 2, 3, 5, 7, 11, 13
2 is the only even prime number
Only positive numbers can be prime numbers
Generic Form of Prime Numbers
All Prime Numbers (apart from 2 and 3) can be written as
¤ 6n + 1 or
¤ 6n - 1
Example
7 = 6 +1 (of the form 6n+1)
11 = 12 – 1 (of the form 6n-1)
13 = 12 + 1 (of the form 6n+1)
17 = 18 -1 (of the form 6n-1)
How to check if a number (n) is prime?
¤ Divide the number by 6 and check if the remainder is 1 or 5
¤ If the remainder is not 1 or 5 then it is NOT a prime number
¤ If the remainder is 1 or 5 then it CAN be a prime number
¤ Divide the number by all prime numbers from 2 to Sqrt(n) and check if it is divisible by any of these numbers
¤ If the number is divisible by ANY of the prime numbers from 2 to Sqrt(n) then it is NOT a prime number
¤ If the number is NOT divisible by ALL the prime numbers from 2 to Sqrt(n) then it is a prime number
Attachment:
img-1.JPG [ 66.19 KiB | Viewed 3334 times ]
Sample Problems
Q1. Check if a 123 is prime number or not
Solution:
Attachment:
img-2.JPG [ 64.58 KiB | Viewed 3293 times ]
¤ 123 divided by 6 gives 3 as remainder which is NOT 1 or 5
¤ 123 is NOT a prime number
Q2. Check if a 127 is prime number or not
Solution:
Attachment:
img-3.JPG [ 62.36 KiB | Viewed 3405 times ]
¤ 127 divided by 6 gives 1 as remainder
¤ 127 CAN be a prime number
¤ We need to divide 127 by all prime numbers from 2 to Sqrt(127) and check if it is divisible by any of these numbers
¤ Closest integer to Sqrt(127) is Sqrt(121) = 11. So prime numbers from 2 to 11. 2,3,5,7,11
¤ 127 is NOT divisible by ALL the prime numbers 2,3,5,7,11.
¤ 127 is a prime number
Hope it Helps!
23 Jul 2022, 02:29
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Attachment:
Hi All,
I have posted a video on YouTube to discuss about Sieve of Eratosthenes Method: To Find Prime Numbers between 1 to 100
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
- ¤ Sieve of Eratosthenes Method to find all prime numbers ≤ a number
¤ Find Prime numbers between 1 to 100
Sieve of Eratosthenes Method
Sieve of Eratosthenes Method is a method to find all the prime numbers ≤ a number.
Following are the steps involved in finding all the prime numbers ≤ n
STEP 1 : Cancel out 1 as it is not prime
STEP 2 : Find all the prime numbers from 2 to √𝑛
STEP 3 : Take each prime number from 2 to √𝑛 and cancel out all the multiples of these prime numbers except the number itself.
STEP 4 : Remaining numbers are prime numbers
Find Prime numbers between 1 to 100
Attachment:
1.JPG [ 62.46 KiB | Viewed 2847 times ]
STEP 1 : Cancel out 1 as it is not prime
Attachment:
2.JPG [ 62.63 KiB | Viewed 2879 times ]
STEP 2 : Find all the prime numbers from 2 to √𝑛
We need to find all the prime numbers from 2 to √100
=> Primes between 2 to 10, which are 2, 3, 5, 7 [ Watch this video to know about prime numbers ]
STEP 3 : Take each prime number from 2 to √𝑛 and cancel out all the multiples of these prime numbers except the number itself.
So, we will take all prime numbers 2, 3, 5 and 7 and cancel out their multiples, except the numbers 2, 3, 5, 7
STEP 3.1 : : Cancel out all multiples of 2 except 2. We will get :
Attachment:
3.JPG [ 66.11 KiB | Viewed 2921 times ]
STEP 3.2 : : Cancel out all multiples of 3 except 3. We will get :
Attachment:
4.JPG [ 67.17 KiB | Viewed 2824 times ]
STEP 3.3 : : Cancel out all multiples of 5 except 5. We will get :
Attachment:
5.JPG [ 68.18 KiB | Viewed 2837 times ]
STEP 3.4 : : Cancel out all multiples of 7 except 7. We will get :
Attachment:
6.JPG [ 70.42 KiB | Viewed 2839 times ]
STEP 4 : Remaining numbers are prime numbers
Attachment:
7.JPG [ 26.4 KiB | Viewed 2849 times ]
There are 25 Prime Numbers between 1 and 100, which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Hope it Helps!
05 Jan 2024, 22:22
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Attachment:
How to Solve: Even and Odd Numbers
Hi All,
I have posted a video on YouTube to discuss Even and Odd Numbers
Attached pdf of this Article as SPOILER at the top! Happy learning!
Following is Covered in the Video
Theory of Even and Odd Numbers
- ⁍ What are Even Numbers ?
⁍ Even Number Problem ?
⁍ What are Odd Numbers ?
⁍ Odd Numbers Problem ?
⁍ Properties of Even and Odd Numbers ?
⁍ Solved Problems
What are Even Numbers
A number which gives 0 as remainder when divided by 2 is an even number.
⁍ Example: 18
⁍ Even numbers end with a units’ digit of 0, 2, 4, 6, 8
⁍ An even number “n” is represented as
n = 2k , where k is an integer
⁍ Example: Consecutive even numbers can be taken as
2k-4, 2k-2, 2k, 2k+2, 2k+4
Even Numbers Problem: Sum of three consecutive even numbers is 60. Find the numbers?
Sol: Let the three numbers be 2k-2, 2k, 2k+2
=> 2k-2 + 2k + 2k+2 = 60
=> 6k = 60
=> k = 10
=> Numbers are 18, 20, 22
=> 2k-2 + 2k + 2k+2 = 60
=> 6k = 60
=> k = 10
=> Numbers are 18, 20, 22
What are Odd Numbers
A number which gives 1 as remainder when divided by 2 is odd number.
⁍ Example: 19
⁍ Odd numbers end with a units’ digit of 1, 3, 5, 7, 9
⁍ An odd number “n” is represented as
n = 2k + 1 or 2k-1 , where k is an integer
⁍ Example: Consecutive odd numbers are
2k-5, 2k-3, 2k-1, 2k+1, 2k+3, 2k+5
Odd Numbers Problem: Sum of 4 consecutive odd numbers is 80. Find the numbers?
Sol: Let the three numbers be 2k-3, 2k-1, 2k+1, 2k+3
=> 2k-3 + 2k-1 + 2k+1 + 2k+3 = 80
=> 8k = 80
=> k = 10
=> Numbers are 17, 19, 21, 23
=> 2k-3 + 2k-1 + 2k+1 + 2k+3 = 80
=> 8k = 80
=> k = 10
=> Numbers are 17, 19, 21, 23
Properties of Even and Odd Numbers
| Addition and Subtraction | |
| -Addition E + E = E E + O = O O + E = O O + O = E Adding Odd number of Odds will give us O Adding even number of Odds will give us E where E -> Even, O -> Odd | -Subtraction E – E = E E – O = O O – E = O O – O = E Subtracting odd number of Odds will give us O Subtracting even number of Odds will give us E where E -> Even, O -> Odd |
| Division and Multiplication | |
| -Division E / E = E or F or O E / O = E or F O / E = F O / O = O or F where E -> Even, O -> Odd, F -> Fraction | -Multiplication E * E = E => \(E^{+ve Integer}\) = E E * O = E O * E = E O * O = O => \(O^{+ve Integer}\) = O where E -> Even, O -> Odd |
⁍ Product of numbers will be even when there is at least one number is Even.
Example: 3*3*2 = 18 = Even as there was one even number 2 on the left side
⁍ Product of numbers will be odd ONLY when all numbers are odd.
Example: 3*3*3 = 27 = Odd as all the numbers on the left side were odd
Solved Problems
Q1. If x, y, and z are integers and x + yz is odd, then which of the following must be true?
I. x + z is even
II. x + y is odd
III. y + z is odd
IV. xy is even
V. yz is even
VI. xz is odd
VII. xyz is even
Sol: x + yz = O
=> We will have four cases
Either x is E and yz is O.
=> Only one case possible
Case 1: x = E, y = O, z = O
Or x is O and yz is E
=> Three cases possible
Case 2: x = O, y = E, z = O
Case 3: x = O, y = O, z = E
Case 4: x = O, y = E, z = E
I. x + y is odd
We have to check for x + y in all the 4 possible cases
For x + y to be odd, one has to even and other has to be odd
But in 3rd case both x and y are odd. So, not possible
II. y + z is odd
Similar logic, in case 1 and case 4 its not possible
III. x + z is even
Similar logic, in case 1, 3,4 its not possible
IV. xy is even
At least one has to even. Not possible in case 3
V. yz is even
At least one has to even. Not possible in case 1
VI. xz is odd
Both have to odd. Not possible in case 1,3,4
VII. xyz is even
At least one has to even. Possible in all the cases
Answer VII
=> We will have four cases
Either x is E and yz is O.
=> Only one case possible
Case 1: x = E, y = O, z = O
Or x is O and yz is E
=> Three cases possible
Case 2: x = O, y = E, z = O
Case 3: x = O, y = O, z = E
Case 4: x = O, y = E, z = E
I. x + y is odd
We have to check for x + y in all the 4 possible cases
For x + y to be odd, one has to even and other has to be odd
But in 3rd case both x and y are odd. So, not possible
II. y + z is odd
Similar logic, in case 1 and case 4 its not possible
III. x + z is even
Similar logic, in case 1, 3,4 its not possible
IV. xy is even
At least one has to even. Not possible in case 3
V. yz is even
At least one has to even. Not possible in case 1
VI. xz is odd
Both have to odd. Not possible in case 1,3,4
VII. xyz is even
At least one has to even. Possible in all the cases
Answer VII
Q2.If x is even, y is odd, z is even, then whether the following are odd or even
I. x + yz
II. x + y + yz
III. xy + z
IV. (x+1)*(y+1)*(z+1)
V. xy*(z+1)
VI. \((x+1)^2*y*(z+1)^3\)
Sol:
I. E + O*E = E + E = E
II. E + O + O*E = E + O + E = O
III. E*O + E = E + E = E
IV. (E + O) * (O + O) * (E + O) = O ( E * O = E
V. E *O * (E + O) = E
VI. \((E+O)^2*O*(E+O)^3\) = O * O * O = O
I. E + O*E = E + E = E
II. E + O + O*E = E + O + E = O
III. E*O + E = E + E = E
IV. (E + O) * (O + O) * (E + O) = O ( E * O = E
V. E *O * (E + O) = E
VI. \((E+O)^2*O*(E+O)^3\) = O * O * O = O
Q3. Product of 4 consecutive numbers will be divisible by all of the following EXCEPT?
A. 6
B. 8
C. 12
D. 24
E. 48?
Sol: Let's take values to solve this. Lets take 4 numbers as 1, 2, 3, 4
Their product = 1*2*3*4 = 24 and will be divisible by all numbers except E
Answer E
Their product = 1*2*3*4 = 24 and will be divisible by all numbers except E
Answer E
Theory: Product of n consecutive numbers will always be divisible by n!
Q4. Sum of three consecutive even numbers is divisible by all of the following EXCEPT
A. 1
B. 2
C. 3
D. 4
E. 6
Sol: Let numbers be 2k-2, 2k, 2k+2
=> Sum = 6k
=> Will be divisible by 6 and all factors of 6
=> Divisible by 1, 2 , 3 and 6
Answer D
=> Sum = 6k
=> Will be divisible by 6 and all factors of 6
=> Divisible by 1, 2 , 3 and 6
Answer D
Q5. Sum of three consecutive odd numbers is divisible by which of the following
A. 2
B. 3
C. 4
D. 5
E. 6
Sol: Let numbers be 2k+1, 2k+3, 2k+5
=> Sum = 6k + 9 = 3*(2k + 3)
=> Will be divisible by 3
Answer B
=> Sum = 6k + 9 = 3*(2k + 3)
=> Will be divisible by 3
Answer B
Hope it helps!
