Program to find last digit of the given sequence for given n in Python

We need to find the last digit of a sequence S for a given value n. The sequence S is defined by a mathematical formula involving powers of 2.

Understanding the Problem

The sequence formula is complex, but we can simplify it to find only the last digit. For n = 2, we calculate specific terms and sum them to get 126, where the last digit is 6.

Algorithm Steps

To solve this problem efficiently ?

• Initialize total = 0 and temp = 1
• While temp ? n, add (2^temp mod 10) to total
• Multiply temp by 2 for next iteration
• Apply final multiplication factor based on whether n is odd or even
• Return the last digit

Example

Here's the implementation to find the last digit of the sequence ?

def solve(n):
    total = 0
    temp = 1
    while temp <= n:
        total += pow(2, temp, 10)  # 2^temp mod 10
        temp *= 2
    total = total * (1 + (4 if n % 2 == 1 else 0)) % 10
    return total

# Test with n = 2
n = 2
result = solve(n)
print(f"For n = {n}, last digit is: {result}")

# Test with additional values
test_cases = [1, 3, 4, 5]
for n in test_cases:
    print(f"For n = {n}, last digit is: {solve(n)}")

For n = 2, last digit is: 6
For n = 1, last digit is: 2
For n = 3, last digit is: 2
For n = 4, last digit is: 6
For n = 5, last digit is: 6

How It Works

The algorithm uses modular arithmetic to find only the last digit instead of computing huge numbers. The pow(2, temp, 10) function efficiently calculates 2^temp mod 10. The final multiplication factor depends on whether n is odd (multiply by 5) or even (multiply by 1).

Key Points

• We use pow(a, b, m) for efficient modular exponentiation
• Only the last digit matters, so we work with mod 10 throughout
• The algorithm handles the complex mathematical sequence efficiently

Conclusion

This solution efficiently finds the last digit of a complex sequence using modular arithmetic. The key insight is avoiding large number calculations by working with remainders throughout the process.