rhys.codes

Problem 31

Fri, 25 Sep 2015 18:01:09 -0400

Solution to Problem 31, written in Python.

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p How many different ways can £2 be made using any number of coins?

Getting Functional

Mon, 23 Feb 2015 22:21:16 -0500

This is my first Haskell program, I plan to solve more Project Euler problems using Haskell as I get better at it.

Problem 1 - Project Euler

main = print $ sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]

Problem 55

Tue, 20 Jan 2015 09:51:26 -0500

Solution to Problem 55. Written in Python.

Problem: If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Problem 79

Tue, 20 Jan 2015 08:45:58 -0500

Solution to Problem 79. Solved using pencil/paper.

Problem: A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.

The text file, keylog.txt, contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.

Problem 41

Mon, 19 Jan 2015 12:08:36 -0500

Solutions to Problem 41. Written in Python.

Problem: We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?

Method 1: This is my slowest (and original) solution without using any tricks or analysis of the problem. I also used a fast prime number generator from this StackOverflow answer. I first generate all primes up to 987654321 (the largest pandigital integer) and then parallelize the pandigital check and print the maximum of the returned values.

Problem 39

Sat, 17 Jan 2015 13:28:10 -0500

Solution to Problem 39. Written in Python.

Problem: If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

import math
import itertools
from problems.euler_lib import euler_lib as lib

c = lambda a, b: math.sqrt(a ** 2 + b ** 2)
test = lambda a, b, p: (a + b + c(a, b)) == p


def count_permutations(p):
 permutes = []
 a_b_range = range(1, p + 1)

 for a in a_b_range:
 for b in a_b_range:
 c_val = c(a, b)
 test = a + b + c_val

 if test == p:
 permutes.append(sorted([a, b, int(c_val)]))

 permutes.sort()
 return len(list(k for k, _ in itertools.groupby(permutes)))

max_permutes = 0
perimeter = 0

for p in range(1, 1001):
 value = count_permutations(p)

 if value > max_permutes:
 perimeter = p
 max_permutes = value

print perimeter

Problem 179

Fri, 16 Jan 2015 15:22:50 -0500

Solution to Problem 179. Written in Python.

Problem: Find the number of integers 1 < n < 10⁷, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.

from problems.euler_lib import euler_lib as lib
from multiprocessing import Pool

p = Pool(8)
divisors = p.map(lib.get_divisor_count, range(2, 10 ** 7))
p.close()

ctr = 0
for n in range(0, len(divisors) - 1):
 if divisors[n] == divisors[n + 1]:
 ctr = ctr + 1

print ctr

Problem 32

Fri, 16 Jan 2015 14:18:42 -0500

Solution to Problem 32. Written in Python.

Problem: We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

Problem 27

Fri, 16 Jan 2015 13:00:14 -0500

Solution to Problem 27. Written in Python.

Problem: Euler discovered the remarkable quadratic formula: n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

Problem 52

Thu, 15 Jan 2015 11:03:38 -0500

Solution to Problem 52. Written in Python.

Problem: It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.

def same_digits(numbers):
 digits_list = [sorted([int(digit) for digit in str(number)]) for number in numbers]
 return all(cmp(x, digits_list[0]) == 0 for x in digits_list)

x = 2
while True:
 array = [x, 2 * x, 3 * x, 4 * x, 5 * x, 6 * x]

 if same_digits(array):
 print x
 break

 x = x + 1

Problem 42

Thu, 15 Jan 2015 10:05:28 -0500

Solution to Problem 42. Written in Python.

Problem: The n^th term of the sequence of triangle numbers is given by, tn = n(n+1)/2; so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t₁₀. If the word value is a triangle number then we shall call the word a triangle word.

Problem 53

Thu, 15 Jan 2015 09:07:28 -0500

Solution to Problem 53. Written in Python.

Problem: There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, ⁵C₃ = 10.

In general, ⁿC_r = n!/r!(n−r)!, where r ≤ n, n! = n * (n−1) * … * 3 * 2 * 1, and 0! = 1. It is not until n = 23, that a value exceeds one-million: ²³C₁₀ = 1144066.

Problem 45

Thu, 15 Jan 2015 09:07:23 -0500

Solution to Problem 45. Written in Python.

Problem: Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle T(n) = n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal P(n) = n(3n−1)/2 1, 5, 12, 22, 35, ...
Hexagonal H(n) = n(2n−1) 1, 6, 15, 28, 45, ...

It can be verified that T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

Problem 40

Thu, 15 Jan 2015 09:03:08 -0500

Solution to Problem 40. Written in Python.

Problem: An irrational decimal fraction is created by concatenating the positive integers:

0.123456789101112131415161718192021...

It can be seen that the 12^th digit of the fractional part is 1.

If d_n represents the n^th digit of the fractional part, find the value of the following expression.

d₁ * d₁₀ * d₁₀₀ * d₁₀₀₀ * d₁₀₀₀₀ * d₁₀₀₀₀₀ * d_1000000

d = "-" + "".join([str(i) for i in range(1, 1000000)])

print int(d[1]) * int(d[10]) * int(d[100]) * int(d[1000]) * int(d[10000]) * int(d[100000]) * int(d[1000000])

Problem 23

Thu, 16 Oct 2014 13:00:27 -0400

Solution to Problem 23. Written in Python.

Problem: A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

Problem 25

Tue, 23 Sep 2014 14:39:20 -0400

Solution to Problem 25. Written in Python.

Problem: The Fibonacci sequence is defined by the recurrence relation:

Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1. Hence the first 12 terms will be:

F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144

Problem 47

Tue, 23 Sep 2014 14:26:45 -0400

Solution to Problem 47. Written in Python.

Problem: The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19

Find the first four consecutive integers to have four distinct prime factors. What is the first of these numbers?

Problem 37

Mon, 22 Sep 2014 21:26:13 -0400

Solution to Problem 37. Written in Python.

Problem: The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

Problem 36

Mon, 22 Sep 2014 21:26:11 -0400

Solution to Problem 36. Written in Python.

Problem: The decimal number, 585 = 1001001001 (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

from problems.euler_lib import euler_lib as lib

print sum([num for num in xrange(1000000) if lib.is_palindrome(num) and lib.is_palindrome(bin(num)[2:])])

Problem 35

Mon, 22 Sep 2014 21:26:08 -0400

Solution to Problem 35. Written in Python.

Problem: The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

import collections
from problems.euler_lib import euler_lib as lib

def get_rotations(number):
 rotations = [number]
 d = collections.deque([int(n) for n in str(number)])

 for i in range(len(d) - 1):
 d.rotate(1)
 rotations.append(int(''.join(str(digit) for digit in d)))

 return rotations

answer = 0
sieve = lib.eratosthenes_sieve(1000000)
prime_numbers = filter(lambda x: x < 1000000, sieve)

for n in prime_numbers:
 rotations = get_rotations(n)
 if all(p in prime_numbers for p in rotations):
 answer += 1

print answer

Problem 92

Sat, 20 Sep 2014 10:34:24 -0400

Solution to Problem 92. Written in Python.

Problem: A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,

44 → 32 → 13 → 10 → 1 → 1
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

Problem 34

Wed, 17 Sep 2014 20:19:29 -0400

Solution to Problem 34. Written in Python.

Problem: 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

import math

answer = []
for n in range(3, 10000000):
 sum_of_factorials = sum(map(lambda x: math.factorial(int(x)), str(n)))
 if(sum_of_factorials == n):
 answer.append(sum_of_factorials)

print sum(answer)

Problem 29

Wed, 17 Sep 2014 19:42:13 -0400

Solution to Problem 29. Written in Python.

Problem: Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

2²=4, 2³=8, 2⁴=16, 2⁵=32
3²=9, 3³=27, 3⁴=81, 3⁵=243
4²=16, 4³=64, 4⁴=256, 4⁵=1024
5²=25, 5³=125, 5⁴=625, 5⁵=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Problem 24

Mon, 25 Aug 2014 09:30:37 -0400

Solution to Problem 24. Written in Python.

Problem: A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Khal Drobo

Sun, 11 May 2014 23:42:36 -0400

I just recently bought a Drobo 5N, and stuffed it with 5x 3TB WD Red drives.

Overall, it’s been a pretty decent experience. Transfer speeds have been really good, but I did notice that the Plex Media Server performance could use a lot of work.

The point of this post was to make a note of the mounting instructions for Ubuntu. Install CIFS with sudo apt-get install cifs-utils. Then add these lines to /etc/fstab:

Problem 30

Sun, 23 Feb 2014 13:17:05 -0500

Solution to Problem 30. Written in Python.

Problem: Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1⁴ + 6⁴ + 3⁴ + 4⁴
8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴
9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴

As 1 = 1⁴ is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Sublime Text 2 Workflow

Sun, 23 Feb 2014 13:11:32 -0500

Prerequisites Link to heading

node.js Link to heading

First we’ll install node.js, as SublimeLinter depends on it.

sudo apt-get install build-essential
wget http://nodejs.org/dist/v0.10.26/node-v0.10.26.tar.gz
tar xzf node-v0.10.26.tar.gz
cd node-v0.10.26/
./configure && make && sudo make install
sudo npm install -g jshint

PHP5 Link to heading

This is required for all PHP related linting.

sudo apt-get install php5 php5-cli

Problem 17

Mon, 22 Jul 2013 22:42:00 +0000

Solution to Problem 17. Solved in Python.

Problem: If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used? Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.

Problem 26

Mon, 22 Jul 2013 22:40:00 +0000

Solution to Problem 26. Solved in Python.

Problem: A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1

Where 0.1(6) means 0.166666…, and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

Problem 22

Mon, 22 Jul 2013 22:39:00 +0000

Solution to Problem 22. Written in Python.

Problem: Using names.txt (right click and ‘Save Link/Target As…’), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 53 = 49714.

Problem 21

Mon, 22 Jul 2013 22:38:00 +0000

Solution to Problem 21. Written in Python.

Problem: Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Problem 19

Mon, 22 Jul 2013 22:36:00 +0000

Solution to Problem 19. Written in Python.

Problem: You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September, April, June and November. All the rest have thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Problem 18/67

Mon, 22 Jul 2013 22:33:00 +0000

Solution to Problem 18 and 67. Written in Python.

Problem: By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

Problem 28

Mon, 22 Jul 2013 22:16:00 +0000

Solution to Problem 28. Written in Python.

Problem: Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101. What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

Problem 20

Mon, 22 Jul 2013 22:15:00 +0000

Solution to Problem 20. Written in Python.

Problem: n! means n x (n - 1) x … x 3 x 2 x 1. For example, 10! = 10 x 9 x … x 3 x 2 x 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the number 100!

Problem 48

Mon, 22 Jul 2013 22:12:00 +0000

Solution to Problem 48. Written in Python.

Problem: The series, 1¹ + 2² + 3³ + … + 10¹⁰ = 10405071317. Find the last ten digits of the series, 1¹ + 2² + 3³ + … + 1000¹⁰⁰⁰.

print str(sum([pow(x, x) for x in xrange(1, 1001)]))[-10:]

Problem 16

Mon, 22 Jul 2013 22:11:00 +0000

Solution to Problem 16. Written in Python.

Problem: 2¹⁵ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 2¹⁰⁰⁰?

print str(sum(map(int, str(pow(2, 1000)))))

Problem 15

Mon, 22 Jul 2013 22:07:00 +0000

Solution to Problem 15. Solved using combinatorics.

Problem: Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20 x 20 grid?

The answer is the binomial coefficient: C(40,20)

Problem 14

Mon, 22 Jul 2013 22:06:00 +0000

Solution to Problem 14. Written in Python.

Problem: The following iterative sequence is defined for the set of positive integers: n -> n/2 (n is even); n-> 3n+1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence: 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? Note: Once the chain starts the terms are allowed to go above one million.

Problem 13

Mon, 22 Jul 2013 22:04:00 +0000

Solution to Problem 13. Written in Python.

Problem: Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

I have a text file named “Problem0013.txt” with the numbers in it:

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

import os

file_path = os.path.join(os.path.dirname(__file__), '..', 'euler_lib/problem13.txt')

file = open(file_path, 'r')
x = file.readlines()
file.close()

print str(sum(map(int, x)))[:10]

Problem 12

Mon, 22 Jul 2013 21:59:00 +0000

Solution to Problem 12. Written in Python.

Problem: The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,…

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

Problem 11

Mon, 22 Jul 2013 21:58:00 +0000

Solution to Problem 11. Written in Python.

Problem: In the 20 x 20 grid below, four numbers along a diagonal line have been marked in red. The product of these numbers is 26 x 63 x 78 x 14 = 1788696. What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20 x 20 grid? You can find the 20 x 20 grid on Project Euler.

Problem 10

Mon, 22 Jul 2013 21:57:00 +0000

Solution to Problem 10. Written in Python.

Problem: The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.

from problems.euler_lib import euler_lib as lib

print sum(lib.eratosthenes_sieve(2000000))

# For reference:
def eratosthenes_sieve(n):
 candidates = list(range(n + 1))
 fin = int(n ** 0.5)

 for i in xrange(2, fin + 1):
 if candidates[i]:
 candidates[2 * i::i] = [None] * (n // i - 1)

 return [i for i in candidates[2:] if i]

Problem 9

Mon, 22 Jul 2013 21:55:00 +0000

Solution to Problem 9. Written in Python.

Problem: A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a² + b² = c². For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

my_range = xrange(1, 1001)
cross_product = [[x, y] for x in my_range for y in my_range]

for [x, y] in cross_product:
 a = pow(x, 2) - pow(y, 2)
 b = 2 * x * y
 c = pow(x, 2) + pow(y, 2)

 if(a + b + c == 1000):
 product = a * b * c
 if product > 0:
 print str(product)
 break

Problem 8

Mon, 22 Jul 2013 21:53:00 +0000

Solution to Problem 8. Written in Python.

Problem: Find the greatest product of five consecutive digits in the 1000-digit number:

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

from operator import mul

my_input = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'

window = 13
max_product = 0
start = 0
stop = start + window
input_length = len(my_input)

while stop != input_length + 1:
 my_input_ints = map(int, my_input[start:stop])
 product = reduce(mul, my_input_ints, 1)

 if product > max_product:
 max_product = product

 start += 1
 stop = start + window

print max_product

Problem 6

Mon, 22 Jul 2013 21:46:00 +0000

Solution to Problem 6. Written in Python.

Problem: The sum of the squares of the first ten natural numbers is, 12 + 22 + … + 102 = 385. The square of the sum of the first ten natural numbers is, (1 + 2 + … + 10)2 = 552 = 3025. Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Problem 7

Mon, 22 Jul 2013 21:46:00 +0000

Solution to Problem 7. Written in Python.

Problem: By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^th prime is 13. What is the 10001^st prime number?

from problems.euler_lib import euler_lib as lib

print lib.eratosthenes_sieve(1000000)[10000]

# For reference:
def eratosthenes_sieve(n):
 candidates = list(range(n + 1))
 fin = int(n ** 0.5)

 for i in xrange(2, fin + 1):
 if candidates[i]:
 candidates[2 * i::i] = [None] * (n // i - 1)

 return [i for i in candidates[2:] if i]

Problem 5

Mon, 22 Jul 2013 21:45:00 +0000

Solution to Problem 5. Written in Python.

Problem: 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?

my_range = xrange(1, 21)
my_range_len = len(my_range)
is_divisible_by_all = lambda number: len([ctr for ctr in my_range if number % ctr == 0]) == my_range_len

answer = 1
while not is_divisible_by_all(answer):
 answer += 1

print(str(answer))

Problem 4

Mon, 22 Jul 2013 21:43:00 +0000

Solution to Problem 4. Written in Python.

Problem: A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers.

from problems.euler_lib import euler_lib as lib

print max([i * j for i in range(100, 1000) for j in range(100, 1000) if lib.is_palindrome(i * j)])

Problem 3

Mon, 22 Jul 2013 21:40:00 +0000

Solution to Problem 3. Written in Python.

Problem: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143?

from problems.euler_lib import euler_lib as lib

sieve = lib.eratosthenes_sieve(10000000)
print max(lib.prime_factorization(600851475143, sieve))

# For reference:
def eratosthenes_sieve(n):
 candidates = list(range(n + 1))
 fin = int(n ** 0.5)

 for i in xrange(2, fin + 1):
 if candidates[i]:
 candidates[2 * i::i] = [None] * (n // i - 1)

 return [i for i in candidates[2:] if i]

def prime_factorization(n, primes):
 return [prime for prime in primes if n % prime == 0]

Problem 2

Mon, 22 Jul 2013 21:39:00 +0000

Solution to Problem 2. Written in Python.

Problem: 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.

num_sum = 0
ctr = 0
temp = 0
found = False


def fib(x):
 if (x == 0):
 return 0
 elif (x == 1):
 return 1

 return (fib(x - 1) + fib(x - 2))

while (found == False):
 temp = fib(ctr)
 if (temp <= 4000000):
 if (temp % 2 == 0):
 num_sum += temp
 ctr += 1
 else:
 found = True

print num_sum

Problem 1

Mon, 22 Jul 2013 21:29:00 +0000

Solution to Problem 1. Written in Python.

Problem: 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.

sum_num = 0
for x in range(1, 1000):
 if (x % 3 == 0) or (x % 5 == 0):
 sum_num += x

print sum_num

About

Mon, 01 Jan 0001 00:00:00 +0000

I’m a Software Engineer based in Atlanta, GA. Currently at Pardot, a salesforce.com company.

All opinions on this site are my own.