Python Looping #2: Factorial, Perfect Numbers, and Pattern Printing Explained

Hey Folks! 👋

Good Day.

Welcome back to Day 2 of my Python Problem Solving journey.

A few days ago, our trainer assigned us 15 Python programming problems . Today, I solved another three problems from that list.

One thing I've started realizing is that programming is less about syntax and more about problem-solving. Once the logic becomes clear, writing the code becomes much easier.

In this article, I'll walk through three Python problems I solved and share the thinking process behind each solution.

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Problem 1: Factorial of a Number

Understanding the Problem

A factorial is the product of all positive integers from a given number down to 1.

For example:

5! = 5 × 4 × 3 × 2 × 1

Result:

120

Before writing any code, I asked myself:

If I were solving this manually on paper, what exact steps would I follow?

The answer became the algorithm itself.

1. Start with 1.
2. Multiply by the current number.
3. Move to the next smaller number.
4. Repeat until reaching 1.

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Python Solution Using While Loop

no = 5
factorial = 1

while no > 0:
    factorial = factorial * no
    no -= 1

print(factorial)

Output

120

---

Pseudocode

START

number = 5
factorial = 1

WHILE number > 0
    factorial = factorial × number
    number = number - 1
END WHILE

PRINT factorial

STOP

---

Flowchart

Start
  |
  v
number = 5
factorial = 1
  |
  v
number > 0 ?
  |
 Yes
  |
factorial = factorial * number
  |
number = number - 1
  |
  +------> Repeat
  |
 No
  |
Print factorial
  |
 End

---

Alternative Solution Using For Loop

no = 5
factorial = 1

for i in range(1, no + 1):
    factorial = factorial * i

print(factorial)

Output

120

---

Key Takeaway

When you already know how many iterations are required, a for loop is usually cleaner and easier to read.

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Problem 2: Print Odd Numbers First and Then Even Numbers

Expected Output

1 3 5 7 9 2 4 6 8 10

At first glance, most people would solve this using two separate loops:

• One loop for odd numbers
• One loop for even numbers

I wanted to see whether I could achieve the same result using a single loop.

---

My Thought Process

I imagined the output as two separate queues.

Odd numbers:

1 → 3 → 5 → 7 → 9

Even numbers:

2 → 4 → 6 → 8 → 10

The challenge wasn't generating the numbers.

The challenge was switching from the first sequence to the second without creating another loop.

---

Python Solution

no = 1

while no <= 10:
    print(no, end=" ")
    no += 2

    if no == 11:
        no = 2

Output

1 3 5 7 9 2 4 6 8 10

---

How the Logic Works

The loop starts at 1 and keeps incrementing by 2.

This generates:

1 3 5 7 9

After printing 9:

no += 2

changes the value to:

11

At that point:

if no == 11:
    no = 2

resets the sequence and begins printing even numbers.

A small condition completely changes the direction of the program.

---

Pseudocode

START

number = 1

WHILE number <= 10

    PRINT number

    number = number + 2

    IF number == 11
        number = 2

END WHILE

STOP

---

Flowchart

Start
  |
  v
number = 1
  |
  v
number <= 10 ?
  |
 Yes
  |
Print number
  |
number = number + 2
  |
number == 11 ?
 / \
Yes No
 |
number = 2
 |
 +------> Repeat
 |
 No
 |
 End

---

Key Takeaway

Sometimes a simple condition can eliminate the need for an additional loop.

Always ask yourself:

Can I simplify the solution without making it harder to understand?

---

Problem 3: Perfect Number Program in Python

What Is a Perfect Number?

A perfect number is a number whose proper divisors add up exactly to the original number.

Example:

6

Divisors:

1, 2, 3

Sum:

1 + 2 + 3 = 6

Since the sum equals the original number, 6 is called a Perfect Number.

---

Real-World Analogy

Imagine a group of friends contributing money to buy a gift.

If the total contribution exactly matches the target amount, everything balances perfectly.

That's exactly how a perfect number works.

The sum of all valid contributors (divisors) equals the target amount (the number itself).

---

Python Solution

no = 496
sum = 0
i = 1

while i < no:

    if no % i == 0:
        sum += i

    i += 1

if sum == no:
    print("Perfect")
else:
    print("Not Perfect")

---

How the Logic Works

The program checks every number from 1 to n - 1.

If a number divides the original number without leaving a remainder:

no % i == 0

it gets added to the running total.

After all checks are complete, we compare:

sum == no

If both values match, the number is perfect.

---

Dry Run Using 6

Divisor	Added?	Running Sum
1	Yes	1
2	Yes	3
3	Yes	6
4	No	6
5	No	6

Final Check:

6 == 6

Result:

Perfect Number

---

Pseudocode

START

number = n
sum = 0

FOR each value from 1 to n - 1

    IF value divides number
        sum = sum + value

END FOR

IF sum equals number
    PRINT Perfect Number
ELSE
    PRINT Not Perfect Number

STOP

---

Flowchart

Start
  |
  v
sum = 0
i = 1
  |
  v
i < number ?
  |
 Yes
  |
number % i == 0 ?
 / \
Yes No
 |
sum += i
 |
i += 1
 |
 +------> Repeat
 |
 No
 |
sum == number ?
 / \
Yes No
 |   |
Perfect
     |
Not Perfect

---

What I Learned from These Problems

Although these programs are small, they helped me strengthen some important programming fundamentals:

• Loops
• Conditional Statements
• Mathematical Logic
• Pattern Generation
• Problem Decomposition
• Dry Running Code

* Converting Logic into Programs

My Current Problem-Solving Workflow

Problem
    ↓
Understand the Requirement
    ↓
Break It Into Smaller Steps
    ↓
Write Pseudocode
    ↓
Code

This workflow has helped me reduce mistakes and understand programming concepts much more deeply.

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If you're new to this series, check out Part 1:

https://dev.to/buddingdeveloper/python-for-beginners-how-i-solved-two-classic-problems-in-multiple-ways-with-flowcharts-30ie