How to Solve: Prime Numbers

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Prime Numbers.pdf 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 https://www.youtube.com/watch?v=WSx0QaGYv8M 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 img-1.JPG Sample Problems Q1. Check if a 123 is prime number or not Solution: img-2.JPG ¤ 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: img-3.JPG ¤ 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!

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Sieve of Eratosthenes - Primes between 1 and 100.pdf Hi All, I have posted a video on YouTube to discuss about Sieve of Eratosthenes Method: To Find Prime Numbers between 1 to 100 https://www.youtube.com/watch?v=lD3Yp_h2E8A 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 1.JPG STEP 1 : Cancel out 1 as it is not prime 2.JPG 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 : 3.JPG STEP 3.2 : : Cancel out all multiples of 3 except 3. We will get : 4.JPG STEP 3.3 : : Cancel out all multiples of 5 except 5. We will get : 5.JPG STEP 3.4 : : Cancel out all multiples of 7 except 7. We will get : 6.JPG STEP 4 : Remaining numbers are prime numbers 7.JPG 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!

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Even and Odd Numbers.pdf How to Solve: Even and Odd Numbers Hi All, I have posted a video on YouTube to discuss Even and Odd Numbers https://www.youtube.com/watch?v=6WX-7UwJUnw 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 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 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 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 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 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 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 Hope it helps!

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Positive and Negative Numbers.pdf How to Solve: Positive and Negative Numbers Hi All, I have posted a video on YouTube to discuss Positive and Negative Numbers https://www.youtube.com/watch?v=XVyIyAC6m1s Attached pdf of this Article as SPOILER at the top! Happy learning!

Following is Covered in the Video ⁍ Types of Numbers ⁍ Zero ⁍ Properties of Positive & Negative Numbers ⁍ Solved Problems Types of Numbers Three types of numbers are there in a number line ⁍ Positive Numbers – All numbers to the right of 0 on the number line ⁍ Negative Numbers – All numbers to the left of 0 on the number line ⁍ Zero (neither +ve nor –ve) img-1.jpg Zero ⁍ Zero is neither positive nor negative ⁍ Zero is an even number ⁍ For all x ≠ 0, x^0 = 1 ⁍ Division by zero is not defined Properties of Positive and Negative Numbers ⁍ Addition (P is Positive and N is Negative) P + P = P, Ex: 2 + 3 = 5 P + N = P or N or 0, Ex1: 3 + (-2) = 1, Ex2: 2 + (-3) = -1, Ex3: 3 + (-3) = 0 N + P = P or N or 0, Ex: Same as above N + N = N, Ex: -2 + (-3) = = -5 ⁍ Subtration P – P = P or N or 0, Ex1: 3 - 2 = 1, Ex2: 2 - 3 = -1, Ex3: 3 - 3 = 0 P – N = P, Ex: 3 - (-2) = 5 N – P = N, Ex: -3 - 2 = -5 N – N = P or N or 0, Ex1: -3 - (-4) = 1, Ex2: -3 - (-2) = -1, Ex3: -3 - (-3) = 0 ⁍ Division P / P = P, Ex: 6/2 = 3 P / N = N, Ex: 6/-2 = -3 N / P = N, Ex: -6/2 = -3 N / N = P, Ex: -6/-2 = 3 ⁍ Multiplication P * P = P, Ex: 2 * 3 = 6 P * N = N, Ex: 2 * -3 = -6 N * P = N, Ex: -2 * 3 = -6 N * N = P, Ex: -2 * -3 = 6 ⁍ IF we are multiplying ODD number of Negative numbers then we will get a NEGATIVE number (Assuming they are not getting multiplied with zero) ⁍ IF we are multiplying EVEN number of Negative numbers then we will get a POSITIVE number (Assuming they are not getting multiplied with zero) ⁍ If Product/Ratio of two numbers is positive, then both the numbers will have the SAME sign. Example : xy > 0 or x/y > 0 => x and y have the SAME sign => Either x > 0 and y > 0 Or x < 0 and y < 0 ⁍ If Product/Ratio of two numbers is negative, then both the numbers will have DIFFERENT signs. Example : xy < 0 or x/y < 0 => x and y have DIFFERENT signs => Either x > 0 and y < 0 Or x < 0 and y > 0 Solved Problems Q1. Which of the following cannot be the value of x if y/(x-2) = 5 ? A. 1 B. 2 C. 3 D. 4 E. 5 Sol : Denominator cannot be equal to 0 => x - 2 ≠ 0 => x ≠ 2 So, Answer will be B Q2. Given that a and b are positive, and c and d are negative, which of the following will be positive for sure? (select all possible) A. ab + cd B. ab + acd C. acd + bcd D. abc + d E. c + cd Sol : A. ab + cd = P*P + N*N = P + P = P = TRUE B. ab + acd = P*P + P*N*N = P + P = P = TRUE C. acd + bcd = P*N*N + P*N*N = P + P = P = TRUE D. abc + d = P*P*N + N = N + N = N = FALSE E. c + cd = N + N*N = N + P = N or P or 0 = FALSE So, Answer will be A, B, C Q3. Is bd>0 ? A. ab > 0 B. cd > 0 Sol: Stat A : ab > 0 There are two cases a>0 and b>0 a<0 and b<0 In both the cases we don’t know anything about the sign of d so NOT sufficient Stat B: cd > 0 There are two cases c>0 and d>0 c<0 and d<0 In both the cases we don’t know anything about the sign of b so NOT sufficient Combining both the statements we will have four cases (1) a>0 b>0 c> 0 d>0 (2) a>0 b>0 c< 0 d<0 (3) a<0 b<0 c>0 d>0 (4) a<0 b<0 c<0 d<0 In case 1 and 4 bd > 0 and in case 2 and 3 bd < 0 So, Together also NOT sufficient . So, Answer will be E Q4. Is bd>0 ? A. ab > 0 B. ad > 0 Sol: Stat A : ab > 0 There are two cases a>0 and b>0 a<0 and b<0 In both the cases we don’t know anything about the sign of d so NOT sufficient Stat B: ad > 0 There are two cases a>0 and d>0 a<0 and d<0 In both the cases we don’t know anything about the sign of b so NOT sufficient Combining both the statements we will have two cases (Since we have a common variable “a” in both the statements so we will combine the two statements based on the sign of the common variable) First case of STAT A will be combined with the first case of Stat B and Second case of STAT A will be combined with the second case of Stat B (1) a>0 b>0 d>0 (2) a<0 b<0 d<0 In both the cases bd > 0 So, Together the two statements are SUFFICIENT . So, Answer will be C Hope it helps!

How to Solve: Prime Numbers

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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