Write a java program that finds and displays all the prime numbers less than 1000?

Answer 1

CLS
FOR n = 2 TO 100
FOR k = 2 TO n / 2
flag = 0
r = n MOD k
IF r = 0 THEN
flag = 1
EXIT FOR
END IF
NEXT k
IF flag = 0 THEN PRINT n,
NEXT n
END

Answer 2

// Semi-optimized Sieve of Eratosthenes implementation.

// num - determine prime numbers from 0 to num-1

// Returns a boolean array. a[n] = true iff n is prime

public static final boolean[] primesSieveE(final int num) {

// We're going to be performing this particular division a lot,

// so let's do the math once and store the answer for later.

final int num_2 = num / 2;

final boolean a[] = new boolean[num];

// Initially set each value to "true"

Arrays.fill(a, true);

// By definition, 0 and 1 are not prime.

a[0] = false;

a[1] = false;

// To cut down some of the work later on, let's also mark off all multiples of 2.

for (int i = 4; i < num; i += 2) {

a[i] = false;

}

// Now the actual algorithm...

// Loop through each Prime number in our array and mark off all multiples.

// Note that we only need to iterate up through num / 2 because once we get

// past the halfway mark, all multiples of i will be outside the bounds of our array.

for (int i = 3; i < num_2; i += 2) {

// Only mark off multiples if i is prime.

if (a[i]) {

// Mark all multiples of i as composites.

for (int j = i + i; j < num; j += i) {

a[j] = false;

}

}

}

return a;

}

Answer 3

count=1

i=3

n=i-1

while count<1000:

if i%n == 0:

i= i+2

n= i-1

else:

n=n-1

while n==1:

print i

count=count+1

i= i+2

n= i-1

while count==1000:

print (i-2)

break

print "1000th prime is ", i-2

Answer 4

A for loop is simpler in this case; though of course, any for loop can be emulated with a while loop. Actually, you need two nested for (or while) nested loops. The outer one loops through all the numbers from 1 to 1000; the inner one is used to check all the factors, from 1 to (number being tested minus 1). For illustration purposes, when you are checking the number 6, you check whether 1, 2, 3, 4 and 5 are factors, and if they are, you add them to the sum of factors, which you must of course initialize to zero. In this example, the factors are 1, 2, and 3, so the sum is 6.

If the sum is equal to the number being tested, you print it out. This requires an if.

You didn't specify what computer language you want to use; the pseudocode would look something like the following - adapt it to your favorite language (I am using underlines for indentation, otherwise the indentation would be lost):

for number = 1 to 1000

_ sum = 0

_ for factor = 1 to number - 1

_ _ if number % factor = 0

_ _ _ sum += factor

_ if number = sum

_ _ print number

To change the for loops to while loops (not convenient, except as an academic exercise), initialize the variable before the while, and don't forget to increment it at the end. Thus, the outside loop would be:

number = 1

while number <= 1000

_ (main body of loop here)

_ number++

Answer 5

If you wanted to output numbers in PHP to get less than a thousand and display all those numbers you could put it in a while loop like this.