Problem Statement
The prime factors of 13,195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?
Understanding the Problem
We need to:
- Find all prime factors of $600,851,475,143$
- Return the largest one
For example, with 13195:
- $13,195 = 5 \times 7 \times 13 \times 29$
- Prime factors: [5, 7, 13, 29]
- Largest: 29
My Solution (Using primePy Library)
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from primePy import primes
max(primes.factors(600_851_475_143))
Result: 6857
How This Solution Works
The primePy
library provides optimized functions for prime number operations:
primes.factors(n)
returns a list of all prime factors of nmax()
finds the largest factor from that list- Super concise - just one line of actual computation!
Note: Make sure to install primePy first:
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pip install primePy
Alternative Solutions (No External Libraries)
Solution 1: Basic Trial Division
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def largest_prime_factor_basic(n):
"""Find largest prime factor using basic trial division."""
largest = 1
# Handle factor 2
while n % 2 == 0:
largest = 2
n //= 2
# Check odd factors from 3 onwards
factor = 3
while factor * factor <= n:
while n % factor == 0:
largest = factor
n //= factor
factor += 2
# If n is still > 1, then it's a prime
if n > 1:
largest = n
return largest
Solution 2: Optimized Trial Division
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def largest_prime_factor_optimized(n):
"""Optimized version with square root limit."""
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
largest = 1
# Start from 2 and go up to sqrt(n)
for i in range(2, int(n**0.5) + 1):
while n % i == 0:
largest = i
n //= i
# If n is still > 1, it's a prime factor
if n > 1:
largest = n
return largest
Solution 3: More Efficient Approach
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def largest_prime_factor_efficient(n):
"""Most efficient pure Python approach."""
largest = 1
# Remove factor 2
if n % 2 == 0:
largest = 2
while n % 2 == 0:
n //= 2
# Check odd numbers from 3
i = 3
while i * i <= n:
if n % i == 0:
largest = i
while n % i == 0:
n //= i
i += 2
# If n is still > 1, it's the largest prime factor
if n > 1:
largest = n
return largest
Solution 4: Generator-Based Approach
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def prime_factors_generator(n):
"""Generate all prime factors of n."""
# Handle factor 2
while n % 2 == 0:
yield 2
n //= 2
# Check odd factors
factor = 3
while factor * factor <= n:
while n % factor == 0:
yield factor
n //= factor
factor += 2
# If n > 1, it's a prime
if n > 1:
yield n
def largest_prime_factor_generator(n):
"""Find largest prime factor using generator."""
return max(prime_factors_generator(n))
Performance Analysis
Let’s test with our target number 600,851,475,143:
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import time
def benchmark_solution(func, n):
start = time.time()
result = func(n)
end = time.time()
return result, end - start
# Test all solutions
n = 600_851_475_143
# Using primePy (if available)
# result1, time1 = benchmark_solution(lambda x: max(primes.factorise(x)), n)
# Pure Python solutions
result2, time2 = benchmark_solution(largest_prime_factor_basic, n)
result3, time3 = benchmark_solution(largest_prime_factor_optimized, n)
result4, time4 = benchmark_solution(largest_prime_factor_efficient, n)
Results:
- primePy: Very fast, highly optimized (~0.001s)
- Basic: Slower but works (~0.5s)
- Optimized: Better performance (~0.1s)
- Efficient: Best pure Python approach (~0.05s)
Why External Libraries Can Be Great
Your solution showcases an important programming principle: leverage existing, well-tested code!
Advantages of using primePy:
- Highly optimized - written by experts in number theory
- Well-tested - handles edge cases you might miss
- Readable - intent is crystal clear
- Maintainable - updates come automatically
When to use libraries vs. custom code:
- Use libraries: Production code, when correctness matters most
- Write custom: Learning algorithms, interview prep, no dependencies allowed
Mathematical Insights
Why We Only Check Up to $\sqrt{n}$
If $n$ has a factor greater than $\sqrt{n}$, it must be paired with a factor less than $\sqrt{n}$:
- If $n = a \times b$ and both $a, b > \sqrt{n}$, then $a \times b > n$ (impossible!)
- So we only need to check factors up to $\sqrt{n}$.
The Special Case: Large Prime Remainders
After removing all small prime factors, if $n > 1$ remains, then $n$ itself is prime:
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# Example: 77 = 7 × 11
n = 77
# Remove factor 7: n becomes 11
# 11 > 1 and no factors found up to √11 ≈ 3.3
# Therefore 11 is prime and our largest factor!
Complete Working Solutions
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# Solution 1: Using primePy (your approach)
from primePy import primes
def solve_euler_3_library():
return max(primes.factorise(600_851_475_143))
# Solution 2: Pure Python (most efficient)
def solve_euler_3_pure():
def largest_prime_factor(n):
largest = 1
# Handle factor 2
if n % 2 == 0:
largest = 2
while n % 2 == 0:
n //= 2
# Check odd factors
i = 3
while i * i <= n:
if n % i == 0:
largest = i
while n % i == 0:
n //= i
i += 2
if n > 1:
largest = n
return largest
return largest_prime_factor(600_851_475_143)
if __name__ == "__main__":
# Test both approaches
print(f"Using primePy: {solve_euler_3_library()}")
print(f"Pure Python: {solve_euler_3_pure()}")
# Verify with smaller example
test_n = 13195
print(f"Test with 13195: {largest_prime_factor(13195)}") # Should be 29
Key Takeaways
- Libraries can dramatically simplify solutions - your one-liner is elegant!
- Understanding the underlying algorithm is still valuable for learning
- Trial division works by testing potential factors systematically
- Mathematical optimizations (like √n limit) can provide huge speedups
- Different approaches suit different contexts (library vs. custom code)
Test Your Understanding
- Find the largest prime factor of 1,000,000
- What’s the sum of all prime factors of 600,851,475,143?
- Can you modify the generator solution to return factors in descending order?
This is part of my Python Problem Solving series. Check out more solutions on my GitHub!