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Write a program that reads N and then continuously read integers and check if there are odd or even until odd and even number equals N

This is what I tried:
using System;
namespace OddAndEvenNumbers
{
class Program
{
static void Main(string[] args)
{
int requiredOddAndEvenNumbers = Convert.ToInt32(Console.ReadLine());
int oddNumbersCount = 0;
int evenNumbersCount = 0;
int n = Convert.ToInt32(Console.ReadLine());
while (oddNumbersCount <= requiredOddAndEvenNumbers || evenNumbersCount <= requiredOddAndEvenNumbers)
{
if (n % 2 == 0)
{
evenNumbersCount++;
}
else if (n % 2 != 0)
{
oddNumbersCount++;
}
}
Console.WriteLine(evenNumbersCount + " " + oddNumbersCount);
Console.ReadLine();
}
}
}
I don't know why it doesn't work.
The program should stop when odd numbers and even number equals N.

You never update n so the loop will never exit
So evenNumbersCount and oddNumbersCount will never count up
using System;
namespace OddAndEvenNumbers
{
class Program
{
static void Main(string[] args)
{
int requiredOddAndEvenNumbers = Convert.ToInt32(Console.ReadLine());
int oddNumbersCount = 0;
int evenNumbersCount = 0;
int n = Convert.ToInt32(Console.ReadLine());
while (oddNumbersCount <= requiredOddAndEvenNumbers || evenNumbersCount <= requiredOddAndEvenNumbers)
{
if (n % 2 == 0)
{
evenNumbersCount++;
}
else if (n % 2 != 0)
{
oddNumbersCount++;
}
n++;
}
Console.WriteLine(evenNumbersCount + " " + oddNumbersCount);
Console.ReadLine();
}
}
}

It seems you've forgotten to get new number inside the loop. It keeps checking the first number
ItThe while loop works until the Odd + Even numbers are equal to n.
int requiredOddAndEvenNumbers = Convert.ToInt32(Console.ReadLine());
int oddNumbersCount = 0;
int evenNumbersCount = 0;
int n = 0;
while ((oddNumbersCount + evenNumbersCount) < requiredOddAndEvenNumbers)
{
n = Convert.ToInt32(Console.ReadLine());
if (n % 2 == 0)
{
evenNumbersCount++;
}
else if (n % 2 != 0)
{
oddNumbersCount++;
}
}
Console.WriteLine("evenNumbersCount: {0}", evenNumbersCount);
Console.WriteLine("oddNumbersCount: {0}", oddNumbersCount);

Problem with prime numbers not printed correctly [duplicate]

This question already has answers here:
Check if number is prime number
(31 answers)
Closed 2 years ago.
I have a problem with my code and i don't know how to solve it. Basically this program prints prime numbers based on the user input and at the end it prints their sum. This works perfectly until a certain amount, example: if i input 10, it shows ten correct prime numbers, but if i input 100, it also prints a number that is not prime, in this case 533. I don't know where i'm wrong.
Thanks for the support.
EDIT: I solved it on my own. Basically there was an error in the first "If" inside the for loop, i've simply added "c = n - 1;" after n++. Now it works perfectly.
Console.Write("How many prime numbers?: ");
int l = Convert.ToInt32(Console.ReadLine());
int n = 2;
int sum = 0;
sum += n;
Console.WriteLine(n);
n++;
int i = 1;
l++;
while (i < l)
{
for (int c = n - 1; c > 1; c--)
{
if (n % c == 0)
{
n++;
}
else if (n % c != 0 && c == 2)
{
sum += n;
Console.WriteLine(n);
n++;
i++;
}
}
}
Console.WriteLine("Sum: " + sum);

Let's start from extracting method:
public static bool IsPrime(int value) {
if (value <= 1)
return false;
if (value % 2 == 0)
retutn value == 2;
int n = (int) (Math.Sqrt(value) + 0.5);
for (int d = 3; d <= n; d += 2)
if (value % d == 0)
return false;
return true;
}
Having this method implemented you can easily compute the sum of the first N primes:
int N = 100;
long s = 0;
for (int p = 1; N > 0; ++p) {
if (IsPrime(p)) {
s += p;
N -= 1;
}
}
Console.Write(s);
Another (a bit more complex) possibility is prime numbers enumeration:
public static IEnumerable<long> Primes() {
yield return 2;
List<int> knownPrimes = new List<int>();
for (int p = 3; ; p += 2) {
int n = (int) (Math.Sqrt(p) = 0.5);
bool isPrime = true;
foreach (int d in knownPrimes) {
if (d > n)
break;
if (p % n == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
knownPrimes.Add(p);
yield return p;
}
}
}
Then you can query enumeration with a help of Linq:
using System.Linq;
...
long s = Primes()
.Take(N)
.Sum();

Is a given number a prime or not, if not output all its divisors

The main part of my code is working, the only thing that doesn't work is the output of all its divisors. My result if it's not a prime should be like this:
Input -> 4
Output -> false 1 2 4
Console.WriteLine("Type your number: ");
int n = Convert.ToInt32(Console.ReadLine());
int a = 0, i;
for (i = 1; i <= n; i++)
{
if (n % i == 0)
{
a++;
}
}
if (a == 2)
{
Console.WriteLine("true");
}
else
{
Console.WriteLine("false" + i);
}
Console.ReadLine();

To print all the divisors, you'll need to gather them up in a collection of some sort – a list of integers, here.
In addition, all integers are divisible by 1, so you don't want to start there; neither do you want to end at n, since n % n == 0.
var divisors = new List<int>();
for (var i = 2; i < 2; i++)
{
if (n % i == 0)
{
divisors.Add(i);
}
}
if (divisors.Count == 0)
{
Console.WriteLine("true");
}
else
{
Console.WriteLine("false " + String.Join(" ", divisors));
}

Here is a working solution. You basically have to store your divisors somewhere or print them directly:
public static void Method(int n)
{
if (IsPrime(n))
{
Console.WriteLine($"{n} is a prime");
return;
}
var divisors = new List<int>();
for(var i = 1; i <= n; i++)
{
if (n % i == 0)
divisors.Add(i);
}
Console.WriteLine($"{n} isn't a prime");
Console.WriteLine($"The divisors are: {string.Join(", ", divisors)}");
}
public static bool IsPrime(int n)
{
for(var i = 2; i < n; i++)
{
if (n % i == 0)
return false;
}
return true;
}

From a brief inspection, there are two ways to generate the output. So far, you count the number of divisors, but neither store them nor write them to the output. You could replace
if (n % i == 0)
{
a++;
}
by
if (n % i == 0)
{
Console.WriteLine(i);
a++;
}
or store the divisors in a
List<int>
to generate the output afterwards.

A recursion related issue in c#

This is the background to this question:
Background
Take any integer n greater than 1 and apply the following algorithm
If n is odd then n = n × 3 + 1 else n = n / 2
If n is equal to 1 then stop, otherwise go to step 1
The following demonstrates what happens when using a starting n of 6
6 - 3 - 10 - 5 - 16 - 8 - 4 - 2 - 1
After 8 generations of the algorithm we get to 1.
It is conjectured that for every number greater than 1 the repeated application of this algorithm will
eventually get to 1.
The question is how can I find a number that takes exactly 500 generations to reduce to 1?
The code below is my version but appearntly got some wrong logic. Could you help me correct this? Thanks in advance.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Sequence1
{
class Program
{
static void Main(string[] args)
{
int start = 1;
int flag = 0;
int value;
while(true){
int temp = (start - 1) / 3;
string sta = temp.ToString();
if (Int32.TryParse(sta, out value) )
{
if (((start - 1) / 3) % 2 == 1)
{
start = (start - 1) / 3;
flag++;
if (flag == 500)
{
break;
}
}
else
{
start = start * 2;
flag++;
if (flag == 500)
{
break;
}
}
}
else
{
start = start * 2;
flag++;
if (flag == 500)
{
break;
}
}
}
Console.WriteLine("result is {0}", start);
Console.ReadLine();
}
}
}

Since your question's title is "A recursion related issue", I will give you a recursive solution.
int Process(int input, int maxRecursionDepth)
{
// condition to break recursion
if (maxRecursionDepth == 0 || input == 1)
return input;
if (input % 2 == 1) // odd case
return Process(input * 3 + 1, maxRecursionDepth - 1);
else // even case
return Process(input / 2, maxRecursionDepth - 1);
}
Now to find all number in a specified range, that return 1 after exactly 500 recursions:
int startRange = 1, endRange = 1000;
int maxDepth = 500;
List<int> resultList = new List<int>();
for (int i = startRange; i <= endRange; i++)
{
if (Process(i, maxDepth) == 1)
resultList.Add(i);
}

Your problem is a part of Collatz conjecture (about recursively defined function) which has not been solved yet:
http://en.wikipedia.org/wiki/Collatz_conjecture
so I think brute force is a good way out:
public static int GetMinNumber(int generations) {
if (generations < 0)
throw new ArgumentOutOfRangeException("generations");
// Memoization will be quite good here
// but since it takes about 1 second (on my computer) to solve the problem
// and it's a throwaway code (all you need is a number "1979515")
// I haven't done the memoization
for (int result = 1; ; ++result) {
int n = result;
int itterations = 0;
while (n != 1) {
n = (n % 2) == 0 ? n / 2 : 3 * n + 1;
itterations += 1;
if (itterations > generations)
break;
}
if (itterations == generations)
return result;
}
}
...
int test1 = GetMinNumber(8); // <- 6
int test2 = GetMinNumber(500); // <- 1979515

Observing the problem,
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
In the third iteration we hit the number 10, which is smaller than 13
So instead of calculating the sequence count every time we can use a cache.
static int GetMinCollatz(int maxChain)
{
const long number = 1000000;
int minNumber = 0;
// Temporary values
int tempCount = 0;
long temp = 0;
// Cache
int[] sequenceCache = new int[number + 1];
// Fill the array with -1
for (int index = 0; index < sequenceCache.Length; index++)
{
sequenceCache[index] = -1;
}
sequenceCache[1] = 1;
for (int index = 2; index <= number; index++)
{
tempCount = 0;
temp = index;
// If the number is repeated then we can find
// sequence count from cache
while (temp != 1 && temp >= index)
{
if (temp % 2 == 0)
temp = temp / 2;
else
temp = temp * 3 + 1;
tempCount++;
}
sequenceCache[index] = tempCount + sequenceCache[temp];
if (sequenceCache[index] == maxChain)
{
minNumber = index;
}
}
return minNumber;
}
For more details refer project euler and this.

A recursive solution
private void ReduceTo1(int input, ref int totalCount)
{
totalCount++;
if (input % 2 == 0)
{
input = input / 2;
}
else
{
input = input * 3 + 1;
}
if (input != 1)
ReduceTo1(input, ref totalCount);
}
to test
int desireValue = 0;
for (int i = 1; i < 100000; i++)
{
int totalCount = 0;
ReduceTo1(i, ref totalCount);
if (totalCount >= 500)
{
desireValue = i;
break;
}
}

Input an Integer and get its prime factors and duplicates [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Prime Factors In C#
I am trying to get this coding to give me all the prime factors of the integer that is inputted, including its duplicates. I have this current code and it seems to be working sort of but it isn't showing all of it's prime factors and duplicates. Any help would be appreciated.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace _1_Numeric_Problem
{
class Program
{
static void Main(string[] args)
{
string myInput;
int myInt;
int p;
Console.WriteLine(("Please input an integer"));
myInput = Console.ReadLine();
myInt = Int32.Parse(myInput);
{
for (int i = 2; i > 1; i++)
{
if (i == 100000)
break;
if (myInt % i == 0)
{
if (i <= 3)
{
Console.Write("{0} ", i);
Console.ReadLine();
continue;
}
else
{
for (p = 2; p < i; p++)
if (i % p != 0)
{
Console.Write("{0} ", i);
Console.ReadLine();
return;
Console.ReadLine();
}
else
{
p++;
continue;
}
}
}
}
}
}
}
}

Try replacing the following code:
for (p = 2; p < i; p++) {
if (i % p != 0) {
Console.Write("{0} ", i);
Console.ReadLine();
return;
Console.ReadLine();
} else {
p++;
continue;
}
}
With this instead:
bool isPrime = true;
for (p = 2; p <= Math.Sqrt(i); p++) {
if (i % p == 0) {
isPrime = false;
break;
}
if (isPrime) {
Console.Write("{0} ", i);
Console.ReadLine();
}
}

Can't you just make a for loop like this?
for (int i = 2; i < myInt; i++)
{
if(myInt % i == 0)
//Do something with it.
}

The basic algorithm for integer factorization using trial division tries each potential factor starting from 2, and if it divides n, outputs the factor, reduces n, and searches for the next factor; note that f is not incremented if it divides n, since it might again divide the reduced n. The loop stops when f is greater than the square root of n, since at that point n must be prime. Here's the pseudocode:
function factors(n)
f := 2
while f * f <= n
if n % f == 0
output f
n := n / f
else
f := f + 1
output n
There are better ways to factor integers, but that should get you started. When you're ready for more, I modestly recommend this essay on my blog.