Python Program for Smallest K digit number divisible by X

In this article, we will learn how to find the smallest K-digit number that is divisible by a given number X. This is a common mathematical programming problem that involves understanding number ranges and divisibility concepts.

Problem Statement

Given two integers K and X, we need to find the smallest K-digit number that is divisible by X.

For example:

• K=3, X=7: Find smallest 3-digit number divisible by 7
• K=4, X=13: Find smallest 4-digit number divisible by 13

Approach

The solution follows these steps:

1. Calculate MIN: The smallest K-digit number is 10^(K-1). For K=3, MIN = 100; for K=4, MIN = 1000.
2. Check divisibility: If MIN is divisible by X, then MIN is our answer.
3. Find next divisible number: If not, find the next number after MIN that is divisible by X using the formula: MIN + (X - MIN % X).

Implementation

def smallest_k_digit_divisible(k, x):
    # Calculate smallest k-digit number
    min_num = pow(10, k - 1)
    
    # If already divisible by x
    if min_num % x == 0:
        return min_num
    else:
        # Find next number divisible by x
        return min_num + (x - min_num % x)

# Test cases
k = 3
x = 7
result = smallest_k_digit_divisible(k, x)
print(f"Smallest {k}-digit number divisible by {x}: {result}")

k = 4
x = 13
result = smallest_k_digit_divisible(k, x)
print(f"Smallest {k}-digit number divisible by {x}: {result}")

Smallest 3-digit number divisible by 7: 105
Smallest 4-digit number divisible by 13: 1001

How It Works

Let's trace through an example where K=3 and X=7:

k, x = 3, 7

# Step 1: Find smallest 3-digit number
min_num = pow(10, k - 1)  # 10^(3-1) = 100
print(f"Smallest {k}-digit number: {min_num}")

# Step 2: Check if divisible by x
remainder = min_num % x
print(f"{min_num} % {x} = {remainder}")

if remainder == 0:
    result = min_num
else:
    # Step 3: Find next divisible number
    result = min_num + (x - remainder)
    print(f"Next divisible number: {min_num} + ({x} - {remainder}) = {result}")

print(f"Answer: {result}")

Smallest 3-digit number: 100
100 % 7 = 2
Next divisible number: 100 + (7 - 2) = 105
Answer: 105

Edge Cases

The algorithm handles various edge cases correctly:

# Edge case 1: When MIN is already divisible
print("K=2, X=5:")
result = smallest_k_digit_divisible(2, 5)
print(f"Result: {result}")  # 10 is divisible by 5

# Edge case 2: Large divisor
print("\nK=3, X=999:")
result = smallest_k_digit_divisible(3, 999)
print(f"Result: {result}")

# Edge case 3: X = 1 (every number is divisible by 1)
print("\nK=4, X=1:")
result = smallest_k_digit_divisible(4, 1)
print(f"Result: {result}")

K=2, X=5:
Result: 10

K=3, X=999:
Result: 999

K=4, X=1:
Result: 1000

Conclusion

This algorithm efficiently finds the smallest K-digit number divisible by X using mathematical properties of divisibility. The time complexity is O(1) as it uses direct mathematical calculations rather than iterating through numbers.