How can you print prime numbers in basic from 1 to 100?

Answer 1

Two solutions immediately spring to mind:

1. Run over all the numbers in the range and if the number is even print it;
2. run over the numbers starting with the first even number in the range using a step of 2 (so that only even numbers are considered) and print the numbers.

Answer 2

Copy this function in your javascript file and call it somewhere

function determine_Odd_Even()

{

var a;

var b;

for(a=1;a<=100;a++)

{

b = a%2;

if(b==0)

{

document.write(a+" is even<br/>");

}

else

{

document.write(a+ " is odd<br/>");

}

}

}

Answer 3

In pseudo-code:

let num := 0;

while( num < 100 )

{ num := num + 1;

print( num );

print( " is " );

if( num % 2 = 1 ) // remainder of division by 2.print( "odd" );

elseprint( "even" );

print( 0x0a ); // newline character

}

In C++, the above pseudo-code can be interpreted as follows:

#include <iostream>

int main()

{ for( int num=1; num<= 100; ++num )

std::cout << num << " is " << (num%2?"odd":"even") << std::endl;

return( 0 );

}

Answer 4

Count from 1 to 100 and test each number to see if it is prime.

To test if a number is prime, use the following algorithm:

If the number is negative, multiply by -1 to make it positive.

If the number is less than 2 then it is not prime.

Otherwise, if the number is even then it is prime if it is 2 (all other even numbers are composite).

Otherwise, if the number has any odd factor in the range 3 to its square root, then it is not prime.

Otherwise it is prime.

Answer 5

Firstly, let's be clear about your interval (the range of values). As written, the whole numbers between 1 and 100 excludes both 1 and 100. That is, the set of whole numbers n such that 1 < n < 100. This may or may not be what you meant, hence it is important to be clear.

When we define algorithms that operate upon a range of values, we always use half-closed range. That is, given two endpoints a and b, we define the range of n such that a <= n < b. It is half-closed because b is excluded from the range (it is one-past-the-end of the range). We denote a half-closed ranges as [a:b). Note the use of parenthesis to close the range, as opposed to the square bracket that opens the range. If we denote the range as [a:b] then both endpoints will be included in the range, thus creating a closed range.

In C Programming, we can implement a half-closed range [a:b) using a for loop:

for (n=a; n

// use n...

}

This is (slightly) more efficient than using a closed range [a:b] because that would require us to check for equality with the second endpoint:

for (n=a; n<=b; ++n) { // closed range [a:b] - includes both a and b // use n...

}

Note that to convert a closed range to a half-closed range, we simply add 1 to b. Thus [a:b] becomes [a:b+1).

To print all even numbers in the half-closed range [min:max) we can use the following algorithm:

// print even numbers in the half-closed range [min:max)

void print_evens (int min, int max) {

if (min%2) ++min; // if min is odd, add 1 to make it even

while (min

printf ("%d ", min);

min += 2;

}

printf ("\n");

}

Note that we start by making min even. This ensures efficiency because we know that odd and even numbers in a range will alternate, so there's no need to check every number, we simply start with the first even number, which is either min itself, or min+1.

Depending on which closed range you actually meant, we can call this function as follows:

print_evens (1,100); // means [1,99]

print_evens (1, 101); // means [1,100]

print_evens (2, 100); // means [2,99]

print_evens (2, 101); // means [2,100]

Answer 6

You write a for loop. Either increment the variable by 2 at a time, or increment it by 1 at a time, but inside the loop, test whether the number is divisible by 2.

Answer 7

for(int i = 1; i < 100; i+=2) {

System.out.println(i);

}

Answer 8

You can always write it out by hand...

int i = 1;

printf("%d\n", (i++));

printf("%d\n", (i++));

printf("%d\n", (i++));

...copy-paste 97 more times...

Answer 9

for(int i = 0; i <= 100; ++i) {

System.out.println(i);

}

Answer 10

Pseudocode:

for (int i = 8; i < 100; i = i + 2)

{

print("The value is ", i + 2);

}