3 and 7 in Python

Sometimes we need to determine if a positive number can be expressed as a sum of non-negative multiples of 3 and 7. This problem involves checking if n = a×3 + b×7 where a and b are non-negative integers.

For example, 13 can be written as 1×7 + 2×3 = 13, so the answer is True.

Algorithm

The approach is to iterate through all possible multiples of 7 up to n and check if the remainder is divisible by 3 ?

• For each multiple of 7 from 0 to n (step by 7)

• Check if (n - multiple_of_7) % 3 == 0

• If yes, return True (we found a valid combination)

• If no valid combination found, return False

Implementation

class Solution:
    def solve(self, n):
        for i in range(0, n+1, 7):
            if (n - i) % 3 == 0:
                return True
        return False

# Test the solution
ob = Solution()
print(ob.solve(13))
print(ob.solve(1))
print(ob.solve(10))

True
False
True

How It Works

Let's trace through the example with n = 13 ?

• i = 0: (13 - 0) % 3 = 13 % 3 = 1 ? 0

• i = 7: (13 - 7) % 3 = 6 % 3 = 0 ? (Found: 2×3 + 1×7 = 13)

Alternative Direct Function

def can_sum_with_3_and_7(n):
    """Check if n can be expressed as a×3 + b×7"""
    for multiples_of_7 in range(0, n+1, 7):
        remainder = n - multiples_of_7
        if remainder % 3 == 0:
            return True
    return False

# Test cases
test_numbers = [13, 1, 6, 7, 9, 10, 14]
for num in test_numbers:
    result = can_sum_with_3_and_7(num)
    print(f"{num}: {result}")

13: True
1: False
6: True
7: True
9: True
10: True
14: True

Time Complexity

The time complexity is O(n/7) since we iterate through multiples of 7 up to n. The space complexity is O(1) as we use constant extra space.

Conclusion

This algorithm efficiently determines if a number can be expressed as a sum of non-negative multiples of 3 and 7. The key insight is iterating through multiples of 7 and checking divisibility by 3 for the remainder.