The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143?
The Fundamental Theorem of Arithmetic guarantees that every integer greater than 1 has a unique prime factorisation. Our strategy is to repeatedly extract the smallest prime factor of n, yielding each one, until nothing remains. The last factor yielded will be the largest.
The key efficiency insight: if n has no divisor in the range [2, √n], then n is prime. This bounds our search at ceil(sqrt(n)), giving O(√n) trial divisions per call — fast even for numbers in the hundreds of billions.
The script decomposes the problem into three clean, single-responsibility functions:
"""
Project Euler Problem 3 - Largest Prime Factor.
This module contains functions to find the largest prime factor of a number.
It includes utilities for finding the smallest divisor and generating prime factors.
"""
from typing import Iterator
from math import (ceil, sqrt)
def smallest_divisor(number: int) -> int:
"""
Returns the smallest divisor of number bigger than 1.
For example
smallest_divisor(100) = 2,
smallest_divisor(15) = 3,
smallest_divisor(125) = 5.
"""
assert (number > 1), f"Not a valid input number = {number} must be > 1"
upper_bound = ceil(sqrt(number))
if number % 2 == 0:
return 2
if number % 3 == 0:
return 3
divisors = (d for d in range(5, upper_bound + 1, 2) if number % d == 0)
try:
return next(divisors)
except StopIteration:
return number
def prime_factors(number: int) -> Iterator[int]:
"""
:param number: Integer > 1.
:return: Returns a generator of the prime factors of number.
"""
if number > 1:
prime = smallest_divisor(number)
yield prime
yield from prime_factors(number // prime)
def answer(number: int) -> int:
"""
Return the largest prime factor of the given number.
This function solves Project Euler problem 3 by finding all prime factors
of the number and returning the largest one.
Parameters
----------
number : int
The number to find the largest prime factor of.
Returns
-------
int
The largest prime factor of the input number.
Examples
--------
>>> answer(13195)
29
"""
return list(prime_factors(number))[-1]
print(answer(600_851_475_143))
smallest_divisor WorksThe function finds the smallest prime factor of number in three steps:
range(5, ub + 1, 2) steps by 2, skipping all even numbers. Since 2 has already been ruled out as a factor, no even number can divide number, so this halves the remaining search space.next(). The generator g is lazy — it produces candidates on demand. Calling next(g) retrieves only the first divisor found and stops immediately. If no divisor exists, the StopIteration exception is caught and number itself is returned, indicating number is prime.prime_factors Works — Recursion and yield fromprime_factors is a recursive generator. At each call it:
p of number and yields it.yield from prime_factors(number // p) to recursively yield all prime factors of the reduced number.yield from is Python 3’s way of delegating to a sub-generator — it forwards every value the recursive call produces directly to the caller, without needing an explicit loop. The recursion bottoms out when number == 1, at which point the if number > 1 guard is False and the generator returns.
For example, factorising 12:
prime_factors(12) → yields 2, then delegates to
prime_factors(6) → yields 2, then delegates to
prime_factors(3) → yields 3, then delegates to
prime_factors(1) → base case, stops
Result: [2, 2, 3]
IteratorThe signature def prime_factors(n: int) -> Iterator[int] uses Python’s typing module to declare that this function returns an iterator of integers. This is the correct annotation for a generator function — it signals to type checkers (and human readers) that the return value is not a list but something to be consumed lazily. In Python 3.9+, Iterator can be imported directly from collections.abc instead.
def answer(number: int) -> int:
return list(prime_factors(number))[-1]
prime_factors yields factors in ascending order because smallest_divisor always returns the smallest factor first. Collecting them into a list and taking the last element [-1] gives the largest prime factor directly.
For 600_851_475_143, the full factorisation is 71 × 839 × 1471 × 6857, so the answer is the final element: 6857.
Answer: 6857
]]>If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.
We want every integer x in the range [1, 999] such that 3 | x or 5 | x (read: “3 divides x” or “5 divides x”).
The elegant closed-form approach uses the inclusion–exclusion principle. Define S(k, n) as the sum of all multiples of k strictly below n. Since these form an arithmetic series k, 2k, 3k, … with p = ⌊(n−1)/k⌋ terms:
# S(k, n) = k · p · (p + 1) / 2
def S(k, n):
p = (n-1) // k
return k * p * (p + 1) / 2
S(3, 1_000) + S(5, 1_000) - S(15, 1_000)
> 233168
The answer is then S(3, 1,000) + S(5, 1,000) − S(15, 1,000) — we subtract multiples of 15 because they were counted twice (once as multiples of 3, once as multiples of 5). This runs in O(1) time regardless of n.
The one-liner below takes a more direct, Pythonic approach using a generator expression inside sum():
def answer(n: int) -> int:
return sum(
x for x in range(n)
if (x % 3) * (x % 5) == 0
)
# "below 1_000" means range(1_000)
print(answer(1_000)) # 233168
The condition (x % 3) * (x % 5) == 0 exploits the zero-product property: a product of integers is zero if and only if at least one factor is zero. So the expression is True whenever x % 3 == 0 or x % 5 == 0 — exactly our condition — without needing the or keyword.
Compared to the O(1) formula this is O(n), but for n = 1,000 the difference is imperceptible. Python’s generator expression also avoids materialising a list in memory; it streams values one by one into sum().
Answer (n = 1000): 233168
]]>Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
The Fibonacci sequence is defined by the recurrence $F(n) = F(n−1) + F(n−2)$ with seeds $F(0) = 0, F(1) = 1$.
A beautiful observation: every third Fibonacci number is even. This follows from the parity pattern of the sequence — even, odd, odd, even, odd, odd, even, … — which repeats with period 3. So rather than checking each term for evenness, we could jump three steps at a time using the identity $F(n+3) = 4·F(n) + F(n−3)$. Our solution below takes the simpler filtering approach, which is perfectly readable and fast enough for this limit.
from typing import Iterator
from itertools import takewhile
def fibonacci(a: int = 0, b: int = 1) -> Iterator[int]:
"""
Generate an infinite Fibonacci sequence.
This function returns a generator that yields Fibonacci numbers indefinitely,
starting from the given initial values a and b.
Parameters
----------
a : int, optional
The first number in the sequence, by default 0
b : int, optional
The second number in the sequence, by default 1
Yields
------
Iterator[int]
An infinite generator of Fibonacci numbers.
Examples
--------
>>> gen = fibonacci()
>>> next(gen)
0
>>> next(gen)
1
>>> next(gen)
1
>>> next(gen)
2
"""
while True:
yield a
a, b = b, a + b
def answer(limit: int) -> int:
"""
Return the sum of even-valued Fibonacci numbers not exceeding n.
This function solves Project Euler problem 2 by generating Fibonacci numbers
and summing only the even-valued ones that are less than or equal to 4,000,000.
Returns
-------
int
The sum of even Fibonacci numbers not exceeding 4 million.
Examples
--------
>>> answer(4_000_000)
4613732
"""
return sum(
x for x in takewhile(lambda x: x <= limit, fibonacci())
if x % 2 == 0
)
print(answer(4_000_000)) # 4613732
The yield keyword transforms fibonacci into a generator function. Instead of computing the entire sequence and returning a list, it produces one value at a time — resuming where it left off each time the caller asks for the next item. Memory usage is O(1) regardless of how large limit gets.
The tuple-swap a, b = b, a + b is idiomatic Python: the right-hand side is fully evaluated before any assignment occurs, so no temporary variable is needed — the Fibonacci step is atomic.
Answer (limit = 4,000,000): 4613732
]]>Inline: $x^2$ or \\(\frac{y}{z}\\).
Display: \[\frac{y}{z}\]
Display: \\[\frac{y}{z}\\]
_posts directory. Go ahead and edit it and re-build the site to see your changes. You can rebuild the site in many different ways, but the most common way is to run jekyll serve, which launches a web server and auto-regenerates your site when a file is updated.
To add new posts, simply add a file in the _posts directory that follows the convention YYYY-MM-DD-name-of-post.ext and includes the necessary front matter. Take a look at the source for this post to get an idea about how it works.
Jekyll also offers powerful support for code snippets:
def print_hi(name)
puts "Hi, #{name}"
end
print_hi('Tom')
#=> prints 'Hi, Tom' to STDOUT.
Check out the Jekyll docs for more info on how to get the most out of Jekyll. File all bugs/feature requests at Jekyll’s GitHub repo. If you have questions, you can ask them on Jekyll Talk.
]]>link: http://url-you-want-linked to the post’s YAML front matter and you’re done.
And this is how a quote looks.
Some link can also be shown.
]]>When using Kramdown {: .notice} can be added after a sentence to assign the .notice to the <p></p> element.
Changes in Service: We just updated our privacy policy here to better service our customers. We recommend reviewing the changes.
Primary Notice: Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero. Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet.
Info Notice: Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero. Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet.
Warning Notice: Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero. Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet.
Danger Notice: Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero. Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet.
Success Notice: Lorem ipsum dolor sit amet, consectetur adipiscing elit. Integer nec odio. Praesent libero. Sed cursus ante dapibus diam. Sed nisi. Nulla quis sem at nibh elementum imperdiet.
Want to wrap several paragraphs or other elements in a notice? Using Liquid to capture the content and then filter it with markdownify is a good way to go.
{% capture notice-2 %}
#### New Site Features
* You can now have cover images on blog pages
* Drafts will now auto-save while writing
{% endcapture %}
<div class="notice">{{ notice-2 | markdownify }}</div>
Or you could skip the capture and stick with straight HTML.
<div class="notice">
<h4>Message</h4>
<p>A basic message.</p>
</div>
A basic message.
Only one thing is impossible for God: To find any sense in any copyright law on the planet.
]]>
Costello: Funny names?
Abbott: Nicknames, nicknames. Now, on the St. Louis team we have Who’s on first, What’s on second, I Don’t Know is on third–
Costello: That’s what I want to find out. I want you to tell me the names of the fellows on the St. Louis team.
Abbott: I’m telling you. Who’s on first, What’s on second, I Don’t Know is on third–
Costello: You know the fellows’ names?
Abbott: Yes.
Costello: Well, then who’s playing first?
Abbott: Yes.
Costello: I mean the fellow’s name on first base.
Abbott: Who.
Costello: The fellow playin’ first base.
Abbott: Who.
Costello: The guy on first base.
Abbott: Who is on first.
Costello: Well, what are you askin’ me for?
Abbott: I’m not asking you–I’m telling you. Who is on first.
Costello: I’m asking you–who’s on first?
Abbott: That’s the man’s name.
Costello: That’s who’s name?
Abbott: Yes.
Costello: When you pay off the first baseman every month, who gets the money?
Abbott: Every dollar of it. And why not, the man’s entitled to it.
Costello: Who is?
Abbott: Yes.
Costello: So who gets it?
Abbott: Why shouldn’t he? Sometimes his wife comes down and collects it.
Costello: Who’s wife?
Abbott: Yes. After all, the man earns it.
Costello: Who does?
Abbott: Absolutely.
Costello: Well, all I’m trying to find out is what’s the guy’s name on first base?
Abbott: Oh, no, no. What is on second base.
Costello: I’m not asking you who’s on second.
Abbott: Who’s on first!
Costello: St. Louis has a good outfield?
Abbott: Oh, absolutely.
Costello: The left fielder’s name?
Abbott: Why.
Costello: I don’t know, I just thought I’d ask.
Abbott: Well, I just thought I’d tell you.
Costello: Then tell me who’s playing left field?
Abbott: Who’s playing first.
Costello: Stay out of the infield! The left fielder’s name?
Abbott: Why.
Costello: Because.
Abbott: Oh, he’s center field.
Costello: Wait a minute. You got a pitcher on this team?
Abbott: Wouldn’t this be a fine team without a pitcher?
Costello: Tell me the pitcher’s name.
Abbott: Tomorrow.
Costello: Now, when the guy at bat bunts the ball–me being a good catcher–I want to throw the guy out at first base, so I pick up the ball and throw it to who?
Abbott: Now, that’s he first thing you’ve said right.
Costello: I DON’T EVEN KNOW WHAT I’M TALKING ABOUT!
Abbott: Don’t get excited. Take it easy.
Costello: I throw the ball to first base, whoever it is grabs the ball, so the guy runs to second. Who picks up the ball and throws it to what. What throws it to I don’t know. I don’t know throws it back to tomorrow–a triple play.
Abbott: Yeah, it could be.
Costello: Another guy gets up and it’s a long ball to center.
Abbott: Because.
Costello: Why? I don’t know. And I don’t care.
Abbott: What was that?
Costello: I said, I DON’T CARE!
Abbott: Oh, that’s our shortstop!
]]>All children, except one, grow up. They soon know that they will grow up, and the way Wendy knew was this. One day when she was two years old she was playing in a garden, and she plucked another flower and ran with it to her mother. I suppose she must have looked rather delightful, for Mrs. Darling put her hand to her heart and cried, “Oh, why can’t you remain like this for ever!” This was all that passed between them on the subject, but henceforth Wendy knew that she must grow up. You always know after you are two. Two is the beginning of the end.
]]>