🚀 Day 35/150 – Count Digits in a Number in Python

Counting digits means finding how many digits are present in a number.

Examples:
12345 → 5 digits
900 → 3 digits
0 → 1 digit

Let’s explore different ways to count digits in Python 👇

🔹 Method 1 – Using while Loop

count = 0 while n > 0: n //= 10 count += 1  

print("Digits:", count)

 digit at a time using integer division.

🔹 Method 2 – Taking User Input

n = int(input("Enter a number: "))
count = 0

temp = abs(n)

while temp > 0:
    temp //= 10
    count += 1

print("Digits:", count)









✅ Works with negative numbers too.

🔹 Method 3 – Using String Method

n = 12345

count = len(str(abs(n)))

print("Digits:", count)

✅ Easiest and most beginner-friendly method.

🔹 Method 4 – Using Recursion

def count_digits(n): n = abs(n) if n < 10: return 1 return 1 + count_digits(n // 10) print(count_digits(12345))

✅ Great for learning recursive logic.

 🎯 Output
    Digits: 5

🔑 Key Takeaways 
Use // 10 to remove the last digit step by step.

len(str(n)) is the easiest way to count digits.

Use abs(n) to handle negative numbers.

Special case: 0 has 1 digit.

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Solve Any Quadratic Equation in Python Using User Input (Step-by-Step Guide)

 

Mathematics meets programming in one of the most practical ways—solving equations using code.

In this guide, you’ll learn how to build a Python program that takes user input and solves any quadratic equation instantly.

Let’s turn a classic math formula into real-world code 👇

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What is a Quadratic Equation?

A quadratic equation looks like this:

$$a{x}^2+bx+c=0$$

Where:

• a, b, c are constants
• x is the variable we want to find

To solve it, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Understanding the Discriminant

The part inside the square root is called the discriminant:

$$D = b^2 - 4ac$$

It determines the type of roots:

• D > 0 → Two real and distinct roots
• D = 0 → One real root
• D < 0 → Complex (imaginary) roots

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 Python Implementation

Now let’s convert this logic into Python code that takes input from the user. 

import math # taking input a = float(input("Enter a: ")) b = float(input("Enter b: ")) c = float(input("Enter c: ")) # discriminant d = b**2 - 4*a*c # solving if d > 0: x1 = (-b + math.sqrt(d)) / (2*a) x2 = (-b - math.sqrt(d)) / (2*a) print("Two real roots:", x1, x2) elif d == 0: x = -b / (2*a) print("One real root:", x) else: real = -b / (2*a) imag = math.sqrt(-d) / (2*a) print("Complex roots:", real, "+", imag, "i and", real, "-", imag, "i")

Example Run

Enter a: 1
Enter b: -3
Enter c: 2

Output:

Two real roots: 2.0 1.0

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Key Concepts You Learned

• Taking user input in Python
• Using the math module
• Applying mathematical formulas in code
• Handling different cases (real & complex roots)

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 Pro Tip

Always make sure:

• a ≠ 0, otherwise it's not a quadratic equation
• Use float() to handle decimal values

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Conclusion

With just a few lines of Python, you can solve any quadratic equation automatically. This is a perfect beginner project that combines math + programming logic.

Once you understand this, you can extend it further:

• Build a GUI calculator 🖥️
• Plot graphs of equations 📊
• Turn it into a web app 🌐

Mathematics with Python Solving Problems and Visualizing Concepts

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Python Coding Challenge - Question with Answer (ID -300426)

 

Explanation:

🔹 Step 1: Understand Boolean Values in Python

In Python, booleans are treated like integers:

True = 1

False = 0

🔹 Step 2: Replace Boolean with Integer Values

print(1 + 0 * 5)

🔹 Step 3: Follow Operator Precedence

Python follows BODMAS/PEMDAS

Multiplication (*) happens before addition (+)

So:

0 * 5 = 0

🔹 Step 4: Perform Addition

1 + 0 = 1

🔹 Step 5: Final Output

print(1)

👉 Output:

1

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🚀 Day 34/150 – Armstrong Number in Python

An Armstrong number is a number that is equal to the sum of its own digits raised to the power of total digits.
Example: 153 = 1³ + 5³ + 3³ = 153

Let’s explore different ways to check Armstrong number in Python 👇

🔹 Method 1 – Using while Loop

n = 153
temp = n
digits = len(str(n))
total = 0

while n > 0:
    digit = n % 10
    total += digit ** digits
    n //= 10

if temp == total:
    print("Armstrong Number")
else:
    print("Not Armstrong Number")

✅ Best numeric method.

🔹 Method 2 – Taking User Input

n = int(input("Enter a number: ")) temp = n digits = len(str(n)) total = 0 while n > 0: digit = n % 10 total += digit ** digits n //= 10 print("Armstrong Number" if temp == total else "Not Armstrong Number")

✅ Useful for dynamic programs.

🔹 Method 3 – Using for Loop + String

n = 153
digits = len(str(n))

total = sum(int(i) ** digits for i in str(n))

if n == total:
    print("Armstrong Number")
else:
    print("Not Armstrong Number")

✅ Short and clean method.

🔹 Method 4 – Using Function

def is_armstrong(n):
    digits = len(str(n))
    total = sum(int(i) ** digits for i in str(n))
    return n == total

print(is_armstrong(153))

✅ Reusable for projects.

📌 Example Output

For 153

Armstrong Number

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🎯 Best Method?

✔ while loop → best for logic building
✔ for loop + string → shortest method
✔ function → reusable and clean

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