function is_prime (n) such that n>=0
if n<2 return false // 0 and 1 are non-prime
if n%2=0 return n=2 // 2 is the only even prime
max_div = sqrt (n) + 1
for div=3 to max_div step 2 // odd factors (3, 5, 7, 9, etc)
if n%div==0 then return false // div is a factor of n, thus n is non-prime
next div
return true // if we get this far, n has no factors, thus n is prime
function main // test the algorithm
for n=2 to 100 step 1 // inclusive if is_prime (n) = true then output n next n
Note: % is the modulo operator, returning the remainder after division. E.g., n%d returns 0 when d is a factor of n, and non-zero if d is not a factor of n
function prime (input: number) output: true if number is prime, otherwise false
{
// Test even values.
if number modulo 2 true
{
print number
}
number = number + 1
}
}
function main (input: none) output: none
{
print primes (2, 100)
}
Loop through some numbers - for example, 2 through 100 - and check each one whether it is a prime number (write a second loop to test whether it is divisible by any number between 2 and the number minus 1). If, in this second loop, you find a factor that is greater than 1 and less than the number, it is not a prime, and you can print it out.
Yes. All non-prime numbers are composite numbers, which simply means a number has one or more prime factors other than 1 and the number itself.
I am providing a succinct and easy to understand version of the program. I have run it in 3-4 compilers and it works perfect. Mind you, you should not enter a number more than 2147483647 (which is the largest number a variable can process in C!). If you do, no problem, but it will display all numbers above it, including the even numbers to be prime. So here you are:#include#includemain(){long int a,b,c;printf("Enter the number: ");scanf("%ld",&a);for (b=2;b
The algorithm is fairly straightforward. For any integer i greater than or equal to zero:i is non-prime if i is less than 2.otherwise, i is prime if i is 2.otherwise, i is non-prime if i is greater than 2 but is divisible by 2.otherwise, i is non-prime if it has any prime factors less than or equal to its square root.otherwise, i is prime.Step 4 is easier to implement by testing all odd divisors from 3 to the square root of i.
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
Develop an algorithm to display all prime numbers from 2 to 100. Give both the pseudocode version and the flowchart version. Convert your pseudocode into a Java program.
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29 is a Prime number. There are no prime numbers between 29 and
NO. There are more prime numbers between 1 and 100 than the prime numbers between 101 and 200.number of prime numbers between 1 and 100 = 25number of prime numbers between 101 and 200 = 20
Yes
There are no prime numbers between 114 and 126.
There are no prime numbers between 33 and 36.
A) Here's an example of a flowchart and pseudocode that could be used to display the prime numbers between 1 and 10000: Flowchart: START Set up an array of numbers from 1 to 10000 Set an empty array to store the prime numbers Set i = 2, the first prime number For each number in the array, check if it is divisible by i If it is divisible by i, it is not a prime number and move to the next number in the array If it is not divisible by i, it is a prime number and add it to the prime numbers array Increase i by 1 and go back to step 4 Repeat steps 4 through 7 until i is greater than the square root of 10000 Display the prime numbers array END
The prime numbers between 30 and 40 are 31 and 37.The prime numbers between 30 and 40 are 31 and 37.
The prime numbers between 31 and 50 are 37,41,43,47.
The prime numbers between 12 and 48 are 13,17,19,23,29,31,37,41,43,47.
The prime numbers between 41 and 54 are 43,47,53.