1. ## Quiz

Hello everbosy.
I have been dpoing alot of there cprogramming tutorials
Excellent Work
And well i also did there quizes bu found them quite easy.
So maybe you guys can Make up a quick Quiz.
The tuts i have done are.
Basics
If Statements.
Loops
Thats all so far
So maybe quickly Make a quick test As hard as you can think.

2. Sure, why not? I'll make you a little quiz thingee. Can you tell me what this does:
Code:
```#include <iostream>
using namespace std;

void myFunc()
{
cout << "Hello!" << endl;
}

int main(void)
{
for (int x = 0 ;x < 100 ; x++)
myFunc();
return 0;
}```

3. This is one that appears on MANY homeworks!

Request a number from the user, and print to the screen a pyramid using *'s. Like so:

Code:
```Enter number --> 4

*
**
***
****
***
**
*

Enter number --> 2

*
**
*```
To make it trickier, try a diamond, or make the pyramid right-side up.

4. Krak
Im guessing it Multiplies the Word Hello and stops at 100 or 99

PJYelton
You i have no idea after all i sad i only did Basics If statments and loops.
Unless im missing somthing i better go readthem al again.
By the way im only 13.Im doing pretty good.

5. what PJ posted could be done with the basics, if statements, and loops.

sounds great that you are gaining interest now. hope you enjoy it.

6. Yeah, it is possible using nothing but for loops. To start you off, try doing a rectangle instead which should be a lot easier and can be done with two for loops, one nested within the other.
Code:
```What is the width --> 6
What is the height --> 4

******
******
******
******```

7. Make a prime number generator that tests for and lists all the prime numbers between two inputed values!

8. ## prime number generator?!

Finding prime numbers between 1 and 10 is easy, even so for 1 and 100. But say I want the prime numbers between 131400 and 131500 ?
The program would have to modulus test all prime numbers under 131400 - 131500 progressively. Is it just me or does that sound abit too hard for a beginner task?

9. Not really, the problem would run exactly the same way, although it will obviously take a little longer to execute. Theres no reason unless you want to get fancy to check all of the numbers below the starting point.

10. Originally posted by PJYelton
Theres no reason unless you want to get fancy to check all of the numbers below the starting point.
You want to break out Fermat's lesser theorem on a 13 year old kid who just barely finished loops? I approve.

11. ## ~

A test for coolazz0:

If c is an integer, what would c have to equal for the final condition to be true:

Code:
```if (c > 0 && c < 5) {
if (c - 1 > (c - 2)) {
if (c++ - ++c < 4) {
if (c / (c++) > 0.5) {
// final condition
}
}
}
}```