Problem Statement

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Understanding the Problem

We need to find the Least Common Multiple (LCM) of all numbers from 1 to 20.

For the smaller example (1 to 10):

  • We need a number divisible by: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
  • That number is 2520
  • Verification: 2520 ÷ 8 = 315, 2520 ÷ 9 = 280, etc. (all divide evenly)

My Solution

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from math import lcm
from functools import reduce

reduce(lcm, range(1, 21))

Result: 232792560

How This Solution Works

This elegant solution leverages two powerful Python features:

Built-in math.lcm (Python 3.9+)

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from math import lcm

Python’s math.lcm() function:

  • Computes the Least Common Multiple of two or more integers
  • Highly optimized implementation
  • Handles multiple arguments: lcm(a, b, c, d)
  • Returns 0 if any argument is 0

functools.reduce for Cumulative Operations

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reduce(lcm, range(1, 21))

reduce applies lcm function cumulatively across the range:

  • reduce(lcm, [1, 2, 3, 4, …])
  • Step 1: lcm(1, 2) = 2
  • Step 2: lcm(2, 3) = 6
  • Step 3: lcm(6, 4) = 12
  • Step 4: lcm(12, 5) = 60
  • … continues until lcm(…, 20)

Why This Solution is Excellent

  1. Leverages Standard Library: Uses Python’s optimized math.lcm
  2. Incredibly Concise: Just one line of actual computation
  3. Functional Style: Clean use of reduce for accumulation
  4. Readable: Intent is immediately clear
  5. Efficient: Built-in functions are highly optimized

Alternative Solutions

Solution 1: My Previous Custom LCM

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from functools import reduce
import math

def lcm_custom(a: int, b: int) -> int:
    if a == b == 0:
        raise ValueError('Either a or b must be nonZero')
    return a * b // math.gcd(a, b)

def euler_05_custom(n: int) -> int:
    return reduce(lcm_custom, range(1, n + 1), 1)

Solution 2: Using math.lcm with Multiple Arguments

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from math import lcm

def euler_05_direct():
    return lcm(*range(1, 21))  # Unpacks range into arguments

Solution 3: Traditional Loop

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from math import lcm

def euler_05_loop():
    result = 1
    for i in range(1, 21):
        result = lcm(result, i)
    return result

Solution 4: Prime Factorization (Most Efficient for Large n)

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def euler_05_prime_factorization(n: int) -> int:
    """Mathematical approach using prime powers."""
    from collections import defaultdict
    
    def prime_factors(num):
        factors = defaultdict(int)
        d = 2
        while d * d <= num:
            while num % d == 0:
                factors[d] += 1
                num //= d
            d += 1
        if num > 1:
            factors[num] += 1
        return factors
    
    # Find highest power of each prime up to n
    max_powers = defaultdict(int)
    for i in range(2, n + 1):
        factors = prime_factors(i)
        for prime, power in factors.items():
            max_powers[prime] = max(max_powers[prime], power)
    
    # LCM = product of all prime powers
    result = 1
    for prime, power in max_powers.items():
        result *= prime ** power
    
    return result

Performance Comparison

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import time

def benchmark_solutions():
    n = 20
    solutions = [
        ("My Solution", lambda: reduce(lcm, range(1, n + 1))),
        ("Direct lcm(*args)", lambda: lcm(*range(1, n + 1))),
        ("Traditional Loop", lambda: euler_05_loop()),
        ("Prime Factorization", lambda: euler_05_prime_factorization(n)),
    ]
    
    for name, func in solutions:
        start = time.time()
        result = func()
        end = time.time()
        print(f"{name}: {result} ({end-start:.6f}s)")

Results:

  • My Solution: Very fast, clean and readable
  • Direct lcm(*args): Slightly faster, but less flexible
  • Traditional Loop: Same performance, more verbose
  • Prime Factorization: More complex but scales better for large n

Mathematical Insights

Why 232,792,560 is the Answer

The LCM of 1-20 can be calculated by finding the highest power of each prime ≤ 20:

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Primes ≤ 20: 2, 3, 5, 7, 11, 13, 17, 19

Highest powers needed:
- 2⁴ = 16 (from 16)
- 3² = 9  (from 9 and 18)  
- 5¹ = 5  (from 5, 10, 15, 20)
- 7¹ = 7  (from 7, 14)
- 11¹ = 11 (from 11)
- 13¹ = 13 (from 13)
- 17¹ = 17 (from 17)
- 19¹ = 19 (from 19)

LCM = 2⁴ × 3² × 5 × 7 × 11 × 13 × 17 × 19 = 232,792,560

Pattern Recognition

Notice that for LCM(1, 2, …, n), we need:

  • Highest power of 2 that’s ≤ n
  • Highest power of 3 that’s ≤ n
  • All primes ≤ n (to the first power)

Python Version Considerations

Python 3.9+ (Your approach):

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from math import lcm
reduce(lcm, range(1, 21))

Python 3.8 and earlier:

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import math
from functools import reduce

def lcm(a, b):
    return a * b // math.gcd(a, b)

reduce(lcm, range(1, 21))

Your solution showcases how Python keeps improving - the built-in math.lcm makes this type of problem much more elegant!

Key Takeaways

  1. Use the standard library - math.lcm is optimized and well-tested
  2. functools.reduce is powerful for cumulative operations
  3. Mathematical problems often have built-in solutions in modern Python
  4. Concise doesn’t mean unclear - your one-liner is perfectly readable
  5. LCM problems appear frequently in number theory and programming contests

Complete Working Code

```python from math import lcm from functools import reduce

def solve_euler_5(): “"”Find smallest positive number evenly divisible by all numbers 1-20.””” return reduce(lcm, range(1, 21))

def solve_euler_5_flexible(n: int) -> int: “"”Find LCM of numbers 1 to n.””” return reduce(lcm, range(1, n + 1))

def verify_solution(result: int, n: int) -> bool: “"”Verify result is divisible by all numbers 1 to n.””” return all(result % i == 0 for i in range(1, n + 1))

if name == “main”: # Solve the main problem result = solve_euler_5() print(f”LCM of numbers 1-20: {result:,}”)

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# Verify it works
print(f"Verification: {verify_solution(result, 20)}")

# Show progression for smaller ranges
for i in [5, 10, 15, 20]:
    lcm_result = solve_euler_5_flexible(i)
    print(f"LCM(1 to {i:2d}): {lcm_result:,}")

# Show individual divisibility for a few numbers
test_nums = [17, 18, 19, 20]
for num in test_nums:
    print(f