Use the if with arrays

We are going to simplify your problem to the maximum, because you are taking a totally complex course.

For this, we just have to know what kind of coins you have and how much is the change that you have to return.

Your function, will receive the amount of change you need (which you said works and well), and will be adding coins until you have more change to return.

I will respect your change array (it seems perfect to me if you still can not use classes, lists or dictionaries in the uni)

teruggave[0, 0] = 2;
teruggave[1, 0] = 1;
teruggave[2, 0] = 0.50;
teruggave[3, 0] = 0.20;
teruggave[4, 0] = 0.10;

And I'm going to use it in this class. Keep in mind, that before entering the amount of coins in the should be at 0

Therefore, you should have some function or something to clean it beforehand (make teruggave[0, X] = 0; for all X)

public void DevolverMonedas(decimal ValorACalcular)
{
    int MonedaMaxima = 0;
    While (ValorACalcular > 0)
    {
        if (teruggave[MonedaMaxima, 0] <= ValorACalcular)
        {
            teruggave[MonedaMaxima, 1]++;
            ValorACalcular -= teruggave[MonedaMaxima, 0];
        }
        else
        {
            MonedaMaxima++;
        }
    }
    for(int i = 0; i<5; i++)
    {
        Console.WriteLine($"Cantidad de monedas de {teruggave[i, 0]} = {teruggave[i, 1]}");
    }
}

Let's analyze the function

Notice that if for example, enter 0.8, what happens is the following

• in the first round, it compares 0.8 with 2, it gives it bigger so it only makes MonedaMaxima ++
• second round, compare 0.8 with 1, give it bigger so just go here MonedaMaxima ++
• third round, compare 0.8 with 0.5, give it less, add 1 to the position of 0.5, subtract 0.5 from 0.8, you have 0.3
• fourth round, compare 0.3 with 0.5, give higher, make MonedaMaxima ++
• fifth round, compare 0.3 with 0.2, give it less, add 1 to the position of 0.2, subtract 0.2 to 0.3, you have 0.1

And you can already give an idea of how it goes.

Your way of trying to solve the problem is extremely complex and does not adapt well to changes (for example, adjust to foreign currencies with different division). The easiest thing is to write a routine, let's call it ProducirCambio() , which takes from parameters the sum of the change to be returned and a field with the different available currencies. The routine will fill a dimension of the field with the amount of each coin to be returned and it does so using a for loop, discounting the change to return the value of each coin that is returned until there is nothing left.

// Parametros:
//  - cambio: monto a devolver en monedas.
//  - monedas: array de dos dimensiones de tamaño
//      [n, 2] donde n es la cantidad de monedas
//      distintas; [i, 0] debe contener el valor
//      de la moneda, [i, 1] deber ser 0 y será
//      cambiado a tener la cantidad de monedas
//      del valor especificado a devolver.
public static void
ProducirCambio(decimal cambio, decimal[,] monedas)
{
    // Ya no se usan monedas de €0,01 y €0,02 en
    // Holanda, por eso ajustamos el cambio a un
    // valor divisible por €0,05.
    int ajustado = (int)(cambio * 20M);
    cambio = (decimal)ajustado / 20M;

    // Pasar por cada valor de moneda y agregar
    // a la devolución las cantidades necesarias.
    for (int i = 0; i < monedas.GetLength(0); i++)
    {
        decimal m = monedas[i, 0];
        for (; cambio >= m; cambio -= m)
            monedas[i, 1]++;
        if (cambio == 0M) break;
    }
}

This routine could be called like this:

decimal[] monedas = new decimal[]
    { 2M, 1M, 0.5M, 0.2M, 0.1M, 0.05M };
decimal[,] devolucion =
    new decimal[monedas.Length, 2];
for (int i = 0; i < monedas.Length; i++)
    devolucion[i, 0] = monedas[i];
ProducirCambio(cambio, devolucion);
for (int i = 0; i < monedas.Length; i++)
    Console.WriteLine("{1} monedas de {0}",
                      devolucion[i, 0],
                      devolucion[i, 1]);

EDIT: To demonstrate the ease of introducing changes to this solution ( refactoring ), I altered the ProducirCambio() routine so that it takes only the values of the change and of the coins as parameters and returns the two-dimensional field as a result:

// Parametros:
//  - cambio: monto a devolver en monedas.
//  - monedas: array de los valores de las monedas
//      disponibles.
public static decimal[,]
ProducirCambio(decimal cambio, decimal[] monedas)
{
    // Ya no se usan monedas de €0,01 y €0,02 en
    // Holanda, por eso ajustamos el cambio a un
    // valor divisible por €0,05.
    int ajustado = (int)(cambio * 20M);
    cambio = (decimal)ajustado / 20M;

    decimal[,] devolucion =
        new decimal[monedas.Length, 2];

    // Pasar por cada valor de moneda y agregar
    // a la devolución las cantidades necesarias.
    for (int i = 0; i < monedas.Length; i++)
    {
        decimal m = monedas[i];
        devolucion[i, 0] = m;
        devolucion[i, 1] = 0;
        for (; cambio >= m; cambio -= m)
            devolucion[i, 1]++;
        if (cambio == 0M) break;
    }

    return devolucion;
}

This routine could be called like this:

decimal[] monedas = new decimal[]
    { 2M, 1M, 0.5M, 0.2M, 0.1M, 0.05M };
decimal[,] devolucion =
    ProducirCambio(cambio, monedas);
for (int i = 0; i < monedas.Length; i++)
    Console.WriteLine("{1} monedas de {0}",
                      devolucion[i, 0],
                      devolucion[i, 1]);