1. ## C++ program.....

This is the program....

An integer is said to be prime if it is divisible by only 1 and itself. For example 2,3,5 and 7 are prime, but 4,6,8 and 9 are not.

---A) Write a function that determines weather a number is prime.

---B) Use this function in a program that determines and prints all the prime numbers between 1 and 10,000. HOw many of these 10,000 numbers do you really have to test before being sure that you have found all the primes.

---C) Initially, you might think that n/2 is the upper limit for which you must test to see whether a number is prime, but you need only go as high as the square root of n. WHY? Rewrite a program and run it both ways. Estimate the performance improvement.

........PLEASE HELP!!!!!!!! I was able to do every other program for the exception of this one........and this is the most important of them all..............HELP!!!!!!!!!!!!!!!!!!!!!!

2. This really isn't that hard. You should give it a shot yourself! Here's some tips to get started though...

a) if it's 2, it's prime. If it's divisible by anything between 2 and itself, it's not prime. You can use a loop to check that.
b) Use a loop to print out all the primes from 1 to 1000.
for(int i = 1; i <= sqrt(10000); ++i)
if(isPrime(i))
cout << i;

You should at least have figured out (b). I answered, just to be nice for a change, but in the future, don't expect answers to homework if you haven't demonstrated that you've at least tried to figure it out.

The answer to part 2 of (b) is part of question (c), by the way. Just thought I'd point that out, since you didn't seem to have noticed it. And for part C, a useful point to ponder is the fact that x*y = y*x. I won't go farther than that, otherwise I'd have done your entire homework for you. Oh yeah, for the last part of (c), try using a profiler if you have one. It'll greatly improve your estimate's accuracy. Good luck anyways, and don't try to mooch off of us again

3. [groan]
Not another prime number question. Do your own homework.
[/groan]

4. Just for the record, any non prime number is divisible by a prime number that is < itself. Also checking to sqrt(n) rather than n/2 to figure out if it's a prime ups speed by a LOT.

5. Also checking to sqrt(n) rather than n/2 to figure out if it's a prime ups speed by a LOT.
He could have figured that out just by looking at question (C)

6. Originally posted by Hunter2
He could have figured that out just by looking at question (C)
That he could, that he could. Maybe I should finish reading before I hit reply next time.

7. Oh well, it might do some good to spam dumb answers at people who come with dumb questions. Keep up the good work!