RugvAI Labs

Mini-Project: FizzBuzz

You'll build FizzBuzz in 20 minutes -- the most famous interview problem in software history. By the end, you'll have written a single program that practices for-loops, modulo, and the if/elif/else chain that catches countless beginners.

#What you'll build

A program that prints the numbers 1 to 100, but with a twist:

For multiples of 3, print "Fizz" instead of the number.
For multiples of 5, print "Buzz" instead.
For multiples of both 3 and 5 (i.e., multiples of 15), print "FizzBuzz".

Sample interaction

text

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
...

#What you'll learn

How to iterate over a range of numbers with for ... in range(...)
How the modulo operator % reveals divisibility
How to order if/elif/else branches so the most specific case wins

#Step 1: Print 1 to 15

Start by printing the first 15 numbers, one per line, with a for loop and range(). Remember that range(1, 16) produces 1, 2, 3, ... 15 -- the upper bound is excluded.

pythonrunnable cell

for n in range(1, 16):
    print(n)

range(1, 16) and not range(1, 15) -- the second argument is exclusive. Off-by-one errors here are the most common beginner mistake in Python.

What Do You Think?

What does 10 % 3 evaluate to?

The answer is 1. The % operator (modulo) returns the REMAINDER after division, not the quotient. 10 ÷ 3 = 3 with 1 left over — that 1 is what modulo returns. This is the foundation of FizzBuzz: n % 3 == 0 is True only when there's no remainder, meaning n is exactly divisible by 3. Modulo is also how you check if a number is even (n % 2 == 0) or get the last digit of a number (n % 10).

Quick check

How many numbers does `range(1, 16)` produce?

#Step 2: Detect multiples of 15

Modulo (%) returns the remainder after division. If n % 15 == 0, then 15 divides n evenly -- meaning n is a multiple of 15.

pythonrunnable cell

for n in range(1, 31):
    if n % 15 == 0:
        print("FizzBuzz")
    else:
        print(n)

15 and 30 should print FizzBuzz; everything else prints the number. Verify before moving on.

What Do You Think?

What does this print for n = 15? if n % 3 == 0: print("Fizz") elif n % 5 == 0: print("Buzz") else: print(n)

The answer is Fizz. Even though 15 is divisible by BOTH 3 and 5, elif is short-circuit: as soon as n % 3 == 0 matches, Python runs that branch and skips the rest. The % 5 check never even happens. This is exactly why FizzBuzz needs the % 15 check FIRST — without it, every multiple of 15 silently becomes "Fizz" instead of "FizzBuzz". One missing branch ruins the whole program in a way no error catches.

#Step 3: Add Fizz (3) and Buzz (5)

Now extend with two more branches -- multiples of 3 print "Fizz", multiples of 5 print "Buzz". Order matters here.

pythonrunnable cell

for n in range(1, 16):
    if n % 15 == 0:
        print("FizzBuzz")
    elif n % 3 == 0:
        print("Fizz")
    elif n % 5 == 0:
        print("Buzz")
    else:
        print(n)

The FizzBuzz check has to come FIRST. If you check n % 3 first, then 15 would print "Fizz" because 15 is also a multiple of 3 -- and the elif short-circuits before the "FizzBuzz" branch ever runs.

Quick check

What happens if you check `n % 3 == 0` before `n % 15 == 0`?

#Step 4: Scale to 100

The pattern is the same -- just change the upper bound on range().

pythonrunnable cell

for n in range(1, 101):
    if n % 15 == 0:
        print("FizzBuzz")
    elif n % 3 == 0:
        print("Fizz")
    elif n % 5 == 0:
        print("Buzz")
    else:
        print(n)

That's it -- the entire FizzBuzz problem in 9 lines. The output is 100 lines long, with "FizzBuzz" at 15, 30, 45, 60, 75, 90.

What Do You Think?

What does this print for n = 30? print(n % 3 == 0 and n % 5 == 0)

The answer is True. 30 % 3 is 0 (30 divides evenly by 3), and 30 % 5 is also 0. Both comparisons 0 == 0 are True, and True and True is True. This expression is the LOGICAL equivalent of n % 15 == 0 — both detect multiples of 15. The and version is sometimes clearer because it explicitly shows both rules; the % 15 version is faster because it's one operation. Both are correct.

#Now build it yourself

Mini project

FizzBuzz from 1 to 100

~20 min

Print each number from 1 to 100 on its own line. Replace multiples of 3 with 'Fizz', multiples of 5 with 'Buzz', and multiples of both (i.e., 15) with 'FizzBuzz'. Use only one loop and a single if/elif/else chain.

Expected output

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
... (continues to 100)

Your code

pythonrunnable cell

# FizzBuzz 1 to 100

for n in range(1, 101):
  # TODO: if n is a multiple of 15 -> print 'FizzBuzz'
  # TODO: elif n is a multiple of 3 -> print 'Fizz'
  # TODO: elif n is a multiple of 5 -> print 'Buzz'
  # TODO: else -> print n
  pass

#Now make it your own

Try one of these variations:

FizzBuzz string concat -- solve it without the FizzBuzz branch. Build the output by concatenating "Fizz" if divisible by 3 and "Buzz" if divisible by 5; if the result is empty, print the number. (Pure FizzBuzz pattern -- no % 15 check.)
Custom divisors -- ask the user for two divisors and two words at runtime. "Give me a number and a word -- I'll print that word for every multiple." Then run FizzBuzz with their inputs.
Collect, don't print -- build a LIST of the results instead of printing immediately. Print the whole list at the end with print(", ".join(results)).

#What's next

FizzBuzz is the smallest complete loop+branch program. Once you can write it cold, you're ready for functions -- the unit Python organizes everything around. From here on, instead of writing one-shot scripts, you'll start packaging logic into named, reusable blocks that take inputs and return outputs.