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"Peter Norvig
April 2026
\n",
"\n",
"# Project Euler Problem 3\n",
"\n",
"Project Euler's [Problem #3](https://projecteuler.net/problem=3) states:\n",
"\n",
"\n",
" The prime factors of 13195 are 5, 7, 13 and 29.\n",
" What is the largest prime factor of the number 600851475143?\n",
"\n",
"I'll describe here how I think about this problem, at alevel suitable for a novice programmer. There are two things to get right: the general plan of how to attack this problem, and the details of how to say it in Python.\n",
"\n",
"## The General Plan\n",
"\n",
"I would like to be able to find the largest prime factor of *any* integer, not just 600851475143; I want `largest_prime_factor(n)`. Here's how I think about it:\n",
"\n",
"- I don't know how to immediately find the *largest* prime factor of *n*. \n",
"- I do know how to find the *smallest* prime factor: just go through the integers from 2 to *n*, in order. The first integer that evenly divides *n* must be the smallest prime factor. It is a factor because it evenly divides, it is smallest because we didn't find a smaller one first, and it is prime because if it were a composite number (like 4 or 9), then we would have found one of its factors (like 2 or 3) first.\n",
"- Once I have the smallest prime factor ***p*** then I know: **the largest prime factor of *n* is the maximum of *p* and the largest prime factor of *n*/*p*.**\n",
"- If *n* is prime, then the largest prime factor is *n*.\n",
"- If *n* is 1, then [by convention](https://oeis.org/wiki/Greatest_prime_factor_of_n) we say the largest prime factor is 1, even though 1 is usually not considered a prime."
]
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"## How to Say It in Python\n",
"\n",
"There are a few things to remember (or look up):\n",
"- `def` is the keyword for [defining a function](https://k21academy.com/ai-ml/functions/) in Python.\n",
" - `def` is followed by the function name, the function parameter(s), and the function body.\n",
" - It is good practice to include a [documentation string](https://www.educative.io/answers/what-is-a-python-docstring) as the first thing in the function body, for clarification. \n",
"- `range(2, n)` means all the integers starting at 2 and stopping just before *n*. E.g., `range(2, 6)` is [2, 3, 4, 5].\n",
"- `n % p` means \"the remainder of *n* divided by *p*\"; e.g., `13 % 10` is `3`.\n",
"- `n % p == 0` means \"is the remainder of *n* divided by *p* equal to zero?\" or \"does *p* evenly divide *n*?\" or \"is *p* a factor of *n*\"?\n",
"- `n // p` means \"integer division\"; `30 // 2` is the integer 15, while `30 / 2` is the decimal number 15.0. We want integers.\n",
"- `return` means to immediately return a value from a function; don't do any more statements in the function.\n",
"\n",
"\n",
"## The Implementation and the Solution\n",
"\n",
"The **implementation**:"
]
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"def largest_prime_factor(n):\n",
" \"\"\"The largest prime that evenly divides n.\n",
" Find the smallest prime p that evenly divides n, \n",
" and return the maximum of p and the largest prime factor of n/p.\n",
" If n is prime or if n = 1, this will return n.\"\"\"\n",
" for p in range(2, n):\n",
" if n % p == 0: # n is composite\n",
" return max(p, largest_prime_factor(n // p))\n",
" return n # n is prime or 1"
]
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"source": [
"\n",
"\n",
"The **solution**:"
]
},
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"execution_count": 11,
"id": "734de0e3-aafc-438a-8332-09741eef39aa",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"6857"
]
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"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
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],
"source": [
"largest_prime_factor(600851475143)"
]
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"source": [
"## Example Factor Tree\n",
"\n",
"[Here](https://www.cuemath.com/numbers/factor-tree/) is the \"factor tree\" of *n* = 36:\n",
"\n",
"![](\"factor36.png\")
\n",
"\n",
"\n",
"We could also represent this as a series of equations, one per line:\n",
"\n",
" 36 = 2 × 18\n",
" 18 = 2 × 9\n",
" 9 = 3 × 3\n",
" 3 = 3\n",
" 36 = 2 × 2 × 3 × 3\n",
"\n",
"\n",
"Or as equations for how `largest_prime_factor(36)` is broken down, with `lpf` as an abbreviation for `largest_prime_factor`:\n",
"\n",
" lpf(36) = max(2, lpf(18))\n",
" lpf(18) = max(2, lpf(9))\n",
" lpf(9) = max(3, lpf(3))\n",
" lpf(3) = 3\n",
" lpf(36) = max(2, max(2, max(3, 3))) = 3"
]
},
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"id": "893a4079-d73d-47c9-970f-f886f4761580",
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"source": [
"## Some Tests\n",
"\n",
"Are we sure our function is correct? It is good idea to define some **tests** to gain confidence. "
]
},
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"cell_type": "code",
"execution_count": 12,
"id": "9bedb1c9-4138-4250-90e6-77b10ba01586",
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{
"data": {
"text/plain": [
"'all tests pass'"
]
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"source": [
"def tests():\n",
" assert largest_prime_factor(1) == 1 # by convention, 1\n",
" assert largest_prime_factor(2) == 2 # even prime\n",
" assert largest_prime_factor(3) == 3 # odd prime\n",
" assert largest_prime_factor(6) == 3 # composite\n",
" assert largest_prime_factor(8) == 2 # power of 2\n",
" assert largest_prime_factor(36) == 3 # example from the diagram\n",
" assert largest_prime_factor(49) == 7 # square of a prime\n",
" assert largest_prime_factor(97) == 97 # bigger prime\n",
" assert largest_prime_factor(97 ** 9) == 97 # even bigger prime\n",
" assert largest_prime_factor(600851475143) == 6857 # really big number\n",
" return 'all tests pass'\n",
"\n",
"tests()"
]
},
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"cell_type": "markdown",
"id": "c51ab097-5ba7-4fdf-9725-944e31aa2336",
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"source": [
"## Efficiency\n",
"\n",
"How long does it take to get our answer? We can use `%time` to see it is just a few hundred microseconds (μs):"
]
},
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"execution_count": 13,
"id": "0e4a1dc6-5969-49fd-9d65-11d9801cbea2",
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"name": "stdout",
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"text": [
"CPU times: user 633 μs, sys: 71 μs, total: 704 μs\n",
"Wall time: 706 μs\n"
]
},
{
"data": {
"text/plain": [
"6857"
]
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"execution_count": 13,
"metadata": {},
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"source": [
"%time largest_prime_factor(600851475143)"
]
},
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"source": [
"The algorithm is slowest when *n* is prime, because the `for` loop has to go all the way up to *n*. How long would it take for the largest 8-digit prime, 99,999,989?"
]
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"id": "7323d528-96d7-4c05-a05d-125e99605443",
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"text": [
"CPU times: user 1.77 s, sys: 21.7 ms, total: 1.79 s\n",
"Wall time: 1.79 s\n"
]
},
{
"data": {
"text/plain": [
"99999989"
]
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"source": [
"n8 = 99999989 # An 8-digit number that happens to be prime \n",
"%time largest_prime_factor(n8)"
]
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"About 2 seconds. Maybe that's good enough. But could we speed things up?\n",
"\n",
"In trying to find a *p* that evenly divides *n*, the algorithm tests all the integers from 2 to *n*. But do we really have to test all those potential factors? No! Either *n* is prime, or it has two factors *p* and *q* such that *p* × *q* = *n*. Of those two factors, one must be less than or equal to the square root of *n*. So to determine if *n* has a prime factor other than itself, we only have to check integers up to √*n*, not all the way up to *n*. That's a big difference! for an 8-digit prime it is the difference between roughly 100 million steps versus a mere 10 thousand steps.\n",
"\n",
"Let's change the definition of `largest_prime_factor` to incorporate this new trick. (We will `import` the square root function, `sqrt`, from the `math` module.)"
]
},
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"execution_count": 15,
"id": "b90b5407-4666-4925-99d2-6a3a6b192fac",
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"source": [
"from math import sqrt\n",
"\n",
"def largest_prime_factor(n):\n",
" \"\"\"The largest prime that evenly divides n.\n",
" Find the smallest prime p that evenly divides n, \n",
" and return the maximum of p and the largest prime factor of n/p.\n",
" If n is prime or if n = 1, this will return n.\"\"\"\n",
" for p in range(2, int(sqrt(n) + 1)): # <<<< only need to go up to √n\n",
" if n % p == 0: # n is composite\n",
" return max(p, largest_prime_factor(n // p))\n",
" return n # n is prime or 1"
]
},
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"id": "c39b08b9-06ed-49c0-a277-db248de909fd",
"metadata": {},
"source": [
"Any time you change a function, you should re-run the tests to give you some confidence that you didn't introduce a bug:"
]
},
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"cell_type": "code",
"execution_count": 16,
"id": "4026dc87-a0aa-4c75-b92a-96ec24cda1b7",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"'all tests pass'"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
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"source": [
"tests()"
]
},
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"id": "6cddf393-a0fd-4cc2-9eef-3341339106bd",
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"source": [
"Now we can see how fast the new function is on the 8-digit prime:"
]
},
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"cell_type": "code",
"execution_count": 17,
"id": "d3d0c13e-5c01-4112-b372-60f7eb302d25",
"metadata": {},
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{
"name": "stdout",
"output_type": "stream",
"text": [
"CPU times: user 209 μs, sys: 0 ns, total: 209 μs\n",
"Wall time: 210 μs\n"
]
},
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"data": {
"text/plain": [
"99999989"
]
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"execution_count": 17,
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"source": [
"%time largest_prime_factor(n8)"
]
},
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"source": [
"As predicted, this is about 10,000 times faster.\n",
"\n",
"We should be able to handle a 16-digit prime in about 2 seconds:"
]
},
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"cell_type": "code",
"execution_count": 18,
"id": "9b6030ef-626a-4195-aedb-ef2edac65da4",
"metadata": {},
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{
"name": "stdout",
"output_type": "stream",
"text": [
"CPU times: user 2.15 s, sys: 22 ms, total: 2.17 s\n",
"Wall time: 2.17 s\n"
]
},
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"data": {
"text/plain": [
"9927935178558959"
]
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"execution_count": 18,
"metadata": {},
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"source": [
"n16 = 9927935178558959 # Largest 16-digit prime\n",
"%time largest_prime_factor(n16)"
]
},
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"source": [
"## Imperative versus Declarative (or Functional) Style\n",
"\n",
"Our definition of largest_prime_factor mixed paradigms, using some [imperative](https://en.wikipedia.org/wiki/Imperative_programming) features (a for loop with a return in the middle) and some [functional](https://en.wikipedia.org/wiki/Functional_programming) features (an implementation of the equation `largest_prime_factor(n) = max(p, largest_prime_factor(n // p))`).\n",
"\n",
"Here's what it might look like if we leaned into the functional style more, making the equation more explicit:"
]
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"execution_count": 19,
"id": "0308612c-6860-49b0-bcd6-856fa08133b1",
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"'all tests pass'"
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"execution_count": 19,
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"source": [
"def largest_prime_factor(n):\n",
" \"\"\"The largest prime that evenly divides n.\n",
" Find the smallest prime p that evenly divides n, \n",
" and return the maximum of p and the largest prime factor of n/p.\n",
" If n is prime or if n = 1, this will return n.\"\"\"\n",
" p = smallest_prime_factor(n)\n",
" return 1 if n == 1 else max(p, largest_prime_factor(n // p))\n",
"\n",
"def smallest_prime_factor(n):\n",
" \"\"\"The smallest prime that evenly divides n (or n itself if no prime divisors).\"\"\"\n",
" return next((p for p in range(2, int(sqrt(n) + 1)) if n % p == 0), n)\n",
"\n",
"tests()"
]
},
{
"cell_type": "markdown",
"id": "00508ec5-85db-4329-a50d-7c3176bff203",
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"source": [
"And here's what it looks like in an imperative style, updating `n`, `p`, and `largest` within a `while` loop that contains another `while` loop to take care of the case where *n* has *p* as a repeated factor."
]
},
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"cell_type": "code",
"execution_count": 20,
"id": "b8907a8f-872f-4531-825c-fae9c70211c1",
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"data": {
"text/plain": [
"'all tests pass'"
]
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"execution_count": 20,
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"source": [
"def largest_prime_factor(n):\n",
" \"\"\"The largest prime that evenly divides n.\n",
" Find the smallest prime p that evenly divides n, \n",
" and return the maximum of p and the largest prime factor of n/p.\n",
" If n is prime or if n = 1, this will return n.\"\"\"\n",
" largest = 1\n",
" p = 2\n",
" while p * p <= n:\n",
" while n % p == 0:\n",
" largest = p\n",
" n = n // p\n",
" p = p + 1\n",
" return max(n, largest)\n",
" \n",
"tests()"
]
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