for (Int32 i = 1; i <= 100; i++)
{
if (i % 3 == 0 && i % 5 == 0)
Console.WriteLine(i + " is a multiple of both 3 AND 5!");
if (i % 3 == 0)
Console.WriteLine(i + " is a multiple of 3");
if (i % 5 == 0)
Console.WriteLine(i + " is a multiple of 5");

}

I'd use else ifs instead of ifs otherwise you'd get unnecessary repitition.

for (Int32 i = 1; i <= 100; i++)
{
if (i % 3 == 0 && i % 5 == 0)
Console.WriteLine(i + " is a multiple of both 3 AND 5!");
else if (i % 3 == 0)
Console.WriteLine(i + " is a multiple of 3");
else if (i % 5 == 0)
Console.WriteLine(i + " is a multiple of 5");

}

i.e you get:

12 is a multiple of 3
15 is a multiple of both 3 AND 5!
18 is a multiple of 3

instead of this:

12 is a multiple of 3
15 is a multiple of both 3 AND 5!
15 is a multiple of 3
15 is a multiple of 5
18 is a multiple of 3

(defun quick-sort (a-list)
(let ((pivot nil)
(lesser nil)
(greater nil))
(when a-list
(cond ((= (length a-list) 1)
(list (first a-list)))
(t
(setf pivot (first a-list))
(dolist (item (rest a-list))
(if (<= item pivot)
(push item lesser)
(push item greater)))
(append (quick-sort lesser) (list pivot) (quick-sort greater))
)))))

Which produces:

1 is not a multiple of 3 (remainder 1) or a multiple of 5 (remainder 1)
2 is not a multiple of 3 (remainder 2) or a multiple of 5 (remainder 2)
3 is a multiple of 3
4 is not a multiple of 3 (remainder 1) or a multiple of 5 (remainder 4)
5 is a multiple of 5
6 is a multiple of 3
7 is not a multiple of 3 (remainder 1) or a multiple of 5 (remainder 2)
8 is not a multiple of 3 (remainder 2) or a multiple of 5 (remainder 3)
9 is a multiple of 3
10 is a multiple of 5
11 is not a multiple of 3 (remainder 2) or a multiple of 5 (remainder 1)
12 is a multiple of 3
13 is not a multiple of 3 (remainder 1) or a multiple of 5 (remainder 3)
14 is not a multiple of 3 (remainder 2) or a multiple of 5 (remainder 4)
15 is a multiple of both 3 AND 5!
...

for (int i=0;i<=100;System.out.print(++i%3 == 0 && i%5 ==0 ? (i+" is a multiple of 3 and 5\n") : i%5 == 0 ? (i+" is a multiple of 5\n") : i%3 == 0 ? (i+" is a multiple of 3\n") : ""));

import java.util.Random;

public class MontyHall {

private enum Door { CAR, GOAT }

public static void main(String[] args) {

int switchWins = 0;
int iterations = 100000;


for (int i=1;i<=iterations;i++) {
if (doMontyHall())
switchWins++;
}

System.out.println("Switch wins = " + switchWins +
", No Switch wins = " + (iterations - switchWins) +
", Percentage wins from switching = " + ((float)(switchWins * 100) / iterations));


}

public static boolean doMontyHall() {
Door[] doors = new Door[3];
doors[0] = Door.GOAT;
doors[1] = Door.GOAT;
doors[2] = Door.GOAT;

Random r = new Random();
int carPosition = r.nextInt(3);
doors[carPosition] = Door.CAR;

int contestantSelection = r.nextInt(3);
boolean switchToWin = false;

if (doors[contestantSelection] == Door.GOAT)
switchToWin = true;

// This next step is not at all necessary to establish whether or not switching is beneficial
// and illustrates the point that once a door has been selected, the problem becomes a
// simple issue of only picking the correct door 1/3 times and the whole Host opening
// a door and asking is actually just a charade that serves to confuse people into thinking
// they now have a 50/50 chance of picking the correct door, which, of course they do, but
// 2/3 is always better than 50/50.

int hostSelection = r.nextInt(3);
while (hostSelection == contestantSelection || doors[hostSelection] == Door.CAR)
hostSelection = r.nextInt(3);

System.out.println("Contestant picked door " + contestantSelection +
", Host opened door " + hostSelection +
", Car was in door " + carPosition + ", Switch to win = " + switchToWin);

return switchToWin;

}
}

(defun monty-hall (times)
(let ((change-hits 0)
(change-misses 0)
(no-change-hits 0)
(no-change-misses 0)
(n-change-trials 0)
(doors '(nil nil nil)))
(dotimes (n times)
(dotimes (x 3)
(setf (nth x doors) 0))
(setf (nth (random 3) doors) 1)
(let* ((choice (random 3))
(choice-set (remove choice '(0 1 2)))
(alternative nil)
(change-trial (random 2)))
(if (equal (nth choice doors) 1)
(setf alternative (nth (random 2) choice-set))
(progn
(if (equal (nth (first choice-set) doors) 0)
(setf alternative (second choice-set))
(setf alternative (first choice-set)))))
(if (equal change-trial 1)
(progn
(incf n-change-trials)
(if (equal (nth alternative doors) 1)
(incf change-hits)
(incf change-misses)))
(if (equal (nth choice doors) 1)
(incf no-change-hits)
(incf no-change-misses)))))
(format t "~% change-hits ~A ~A" change-hits (float (/ change-hits n-change-trials)))
(format t "~% no-change-hits ~A ~A" no-change-hits (float (/ no-change-hits (- times n-change-trials))))
))

Result:

Contestant swapping? False


Ferraris: 330 33%
Goats: 670 67%

Contestant swapping? True


Ferraris: 671 67.1%
Goats: 329 32.9%

The For/Next Loops step through each multiple
of 3*5, 5 and 3 and adds the multiples to a Sorted List
if they haven't already been added.

A Sorted List is a list where each item in the list
consists of two parts: a Key part and a Value part.
The list is automatically sorted on the Key part.

We store each iterated multiple in the the Key part
and text explaining the type of multiple is
stored in the corresponding Value part.

Then we just display the contents of the Sorted List.

We choose one of the 3 doors at random.
Monty removes one of the remaining 2 doors
that has a goat behind it.
We don't need to know which door, just that a goat
has been removed and now we only have 2 doors left
- A goat behind one and a Car behind the other, and
that our chosen door is one of them.

The If/THEN/ELSE line of the code works as follows:

If we stick with our choice and win, then
increase the count of wins for the Chosen Door.
(in the "Swapping Doors Scenario" we would have lost)

If we stick with our choice and lose, then in the
"Swapping Doors Scenario" we would have won, so
increase the count of wins for the Alternate Door.

I'm sure there's a way to do it without a loop, but can't think of it atm.

Assuming £100,000 with a 4.75% rate and paying off £700/month
Result:

Repayment duration: 211 months - 17 Years and 7 months.

(defun mortgage (balance payment rate months)
(if
(< balance 0)
(print months)
(mortgage (- (+ (* (/ rate 12) balance) balance) payment) payment rate (+ months 1))))

Result:

Repayment duration: 211 months - 17 Years and 7 months.