# Problem 5
# =========
# 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?
# Answer: 232792560 calculated in 1.09672546387e-05 seconds
from euler import runtime, lcm
from functools import reduce
def lcm_range(r):
return reduce(lcm, xrange(1, r+1))
runtime(lcm_range, 20)
# Answer: 1074 calculated in 3.38554382324e-05 seconds
from euler import runtime
def problem_18(p):
for x in xrange(len(p)-1, 0, -1):
for y in xrange(x):
p[x-1][y] += max(p[x][y], p[x][y+1])
return p[0]
pyramid = [
[75],
[95, 64],
[17, 47, 82],
[18, 35, 87, 10],
[20, 04, 82, 47, 65],
[19, 01, 23, 75, 03, 34],
[88, 02, 77, 73, 07, 63, 67],
[99, 65, 04, 28, 06, 16, 70, 92],
[41, 41, 26, 56, 83, 40, 80, 70, 33],
[41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
[53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
[70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
[91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
[63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
[04, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 04, 23]
]
runtime(problem_18, pyramid)
# Problem 29
# ==========
# Consider all integer combinations of a^b for 2 <= a <= 5 and 2 <= b <= 5:
# 2^2=4, 2^3=8, 2^4=16, 2^5=32
# 3^2=9, 3^3=27, 3^4=81, 3^5=243
# 4^2=16, 4^3=64, 4^4=256, 4^5=1024
# 5^2=25, 5^3=125, 5^4=625, 5^5=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 a^b for 2 <= a <=
# 100 and 2 <= b <= 100?
# Answer: 9183 calculated in 0.0198509693146 seconds
from euler import runtime
def distinct_terms(r):
return len({a**b for a in xrange(2, r+1) for b in xrange(2, r+1)})
runtime(distinct_terms, 100)
# ==========
# Pentagonal numbers are generated by the formula, P[n]=n(3n-1)/2. The first
# ten pentagonal numbers are:
# 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
# It can be seen that P[4] + P[7] = 22 + 70 = 92 = P[8]. However, their
# difference, 70 - 22 = 48, is not pentagonal.
# Find the pair of pentagonal numbers, P[j] and P[k], for which their sum
# and difference are pentagonal and D = |P[k] - P[j]| is minimised; what is
# the value of D?
# Answer: 5482660 calculated in 0.122126102448 seconds
from euler import runtime
def problem_44():
p = {x * (3 * x - 1) / 2 for x in xrange(1000, 3000)}
for x in p:
for y in p:
if x + y in p and abs(y - x) in p:
return abs(y - x)
runtime(problem_44)
# Problem 206
# ===========
# Find the unique positive integer whose square has the form
# 1_2_3_4_5_6_7_8_9_0,
# where each "_" is a single digit.
# Answer: 1389019170 calculated in 1.95045495033 seconds
from euler import runtime
def problem_206():
for x in xrange(150000001, 1, -2):
if str(x**2)[::2] == '123456789':
return x*10
runtime(problem_206)
# It is well known that if the square root of a natural number is not an
# integer, then it is irrational. The decimal expansion of such square roots
# is infinite without any repeating pattern at all.
# The square root of two is 1.41421356237309504880..., and the digital sum
# of the first one hundred decimal digits is 475.
# For the first one hundred natural numbers, find the total of the digital
# sums of the first one hundred decimal digits for all the irrational square
# roots.
# Answer: 40886 calculated in 0.00938415527344 seconds
from euler import runtime
from decimal import Decimal, getcontext
from math import sqrt
getcontext().prec = 102
def problem_80(s=0):
for x in xrange(2, 100):
if not sqrt(x) % 1:
continue
d = str(Decimal(x).sqrt()*10**99)
s += sum([int(x) for x in d[:100]])
return s
runtime(problem_80)
# Problem 113
# ===========
# Working from left-to-right if no digit is exceeded by the digit to its
# left it is called an increasing number; for example, 134468.
# Similarly if no digit is exceeded by the digit to its right it is called a
# decreasing number; for example, 66420.
# We shall call a positive integer that is neither increasing nor decreasing
# a "bouncy" number; for example, 155349.
# As n increases, the proportion of bouncy numbers below n increases such
# that there are only 12951 numbers below one-million that are not bouncy
# and only 277032 non-bouncy numbers below 10^10.
# How many numbers below a googol (10^100) are not bouncy?
# Answer: 51161058134250 calculated in 5.69820404053e-05 seconds
from euler import runtime, binomial
def problem_113():
return sum([binomial(110, 100), binomial(109, 100), -1002])
runtime(problem_113)
# Problem 3
# =========
# The prime factors of 13195 are 5, 7, 13 and 29.
# What is the largest prime factor of the number 600851475143 ?
# Answer: 6857 calculated in 0.0001540184021 seconds
from euler import runtime, prime_factorization
def largest_prime_factor(n):
return max(prime_factorization(n))
runtime(largest_prime_factor, 600851475143)
# The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4,
# and 5, giving the pandigital, 918273645, which is the concatenated product
# of 9 and (1,2,3,4,5).
# What is the largest 1 to 9 pandigital 9-digit number that can be formed as
# the concatenated product of an integer with (1,2, ... , n) where n > 1?
# Answer: 932718654 calculated in 0.00693202018738 seconds
from euler import runtime, pandigital
def concated_product(n, r):
s = ''
for x in xrange(r+1):
s += str(n*x)
return int(s)
def largest_concated_pandigital_product(largest=0):
for x in xrange(10000, 9000, -1):
for i in xrange(3):
c = concated_product(x, i)
if pandigital(c):
if c > largest:
largest = c
return largest
runtime(largest_concated_pandigital_product)
# * d[2]d[3]d[4]=406 is divisible by 2
# * d[3]d[4]d[5]=063 is divisible by 3
# * d[4]d[5]d[6]=635 is divisible by 5
# * d[5]d[6]d[7]=357 is divisible by 7
# * d[6]d[7]d[8]=572 is divisible by 11
# * d[7]d[8]d[9]=728 is divisible by 13
# * d[8]d[9]d[10]=289 is divisible by 17
# Find the sum of all 0 to 9 pandigital numbers with this property.
# Answer: 16695334890 calculated in 3.18974900246 seconds
from euler import runtime
from itertools import permutations
def problem_43(s=0):
p = permutations('1234567890')
for y in p:
y = ''.join(y)
if (not int(y[1:4]) % 2 and not int(y[2:5]) % 3 and
not int(y[3:6]) % 5 and not int(y[4:7]) % 7 and
not int(y[5:8]) % 11 and not int(y[6:9]) % 13 and
not int(y[7:]) % 17):
s += int(y)
return s
runtime(problem_43)
# F[3] = 2
# F[4] = 3
# F[5] = 5
# F[6] = 8
# F[7] = 13
# F[8] = 21
# F[9] = 34
# F[10] = 55
# F[11] = 89
# F[12] = 144
# The 12th term, F[12], is the first term to contain three digits.
# What is the first term in the Fibonacci sequence to contain 1000 digits?
# Answer: 4782 calculated in 0.0314209461212 seconds
from euler import runtime
from itertools import count
def fibonacci_digits(n):
x, y = 0, 1
for i in count():
if len(str(x)) == n:
return i
x, y = y, x+y
runtime(fibonacci_digits, 1000)
for x in xrange(1, len(n)-adj-1):
p = 1
for i in xrange(1, adj+1):
p *= int(n[x+i])
if p > m:
m = p
return m
num = ('73167176531330624919225119674426574742355349194934'
'96983520312774506326239578318016984801869478851843'
'85861560789112949495459501737958331952853208805511'
'12540698747158523863050715693290963295227443043557'
'66896648950445244523161731856403098711121722383113'
'62229893423380308135336276614282806444486645238749'
'30358907296290491560440772390713810515859307960866'
'70172427121883998797908792274921901699720888093776'
'65727333001053367881220235421809751254540594752243'
'52584907711670556013604839586446706324415722155397'
'53697817977846174064955149290862569321978468622482'
'83972241375657056057490261407972968652414535100474'
'82166370484403199890008895243450658541227588666881'
'16427171479924442928230863465674813919123162824586'
'17866458359124566529476545682848912883142607690042'
'24219022671055626321111109370544217506941658960408'
'07198403850962455444362981230987879927244284909188'
'84580156166097919133875499200524063689912560717606'
'05886116467109405077541002256983155200055935729725'
'71636269561882670428252483600823257530420752963450')
runtime(max_adjacent_product, num, 13)
# Problem 24
# ==========
# 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?
# Answer: 2783915460 calculated in 0.109308958054 seconds
from euler import runtime
from itertools import permutations
def nth_lexi_perm(r, n, c=0):
x = permutations(xrange(r), r)
for i in x:
if c == n-1:
return ''.join(str(x) for x in i)
c += 1
runtime(nth_lexi_perm, 10, 1000000)
# Problem 36
# ==========
# The decimal number, 585 = 1001001001[2] (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.)
# Answer: 872187 calculated in 0.553943872452 seconds
from euler import runtime, palindrome
def palindrome_base_2_and_10(r, s=0):
for x in xrange(r):
if palindrome(x) and palindrome(bin(x)[2:]):
s += x
return s
runtime(palindrome_base_2_and_10, 1000000)
# As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the
# smallest number that can be written as the sum of two abundant numbers is
# 24. By mathematical analysis, it can be shown that all integers greater
# than 28123 can be written as the sum of two abundant numbers. However,
# this upper limit cannot be reduced any further by analysis even though it
# is known that the greatest number that cannot be expressed as the sum of
# two abundant numbers is less than this limit.
# Find the sum of all the positive integers which cannot be written as the
# sum of two abundant numbers.
# Answer: 4179871 calculated in 0.436667919159 seconds
from euler import runtime, divisors
def problem_23(s=0):
n = set()
for x in xrange(1, 28123):
if sum(divisors(x)) > x:
n.add(x)
for y in n:
if x-y in n:
break
else:
s += x
return s
runtime(problem_23)
# Problem 21
# ==========
# 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.
# Evaluate the sum of all the amicable numbers under 10000.
# Answer: 31626 calculated in 0.08455991745 seconds
from euler import runtime, divisors
def d(n):
return sum(divisors(n))
def sum_amicable(m):
return sum([a for a in xrange(1, m) if d(d(a)) == a and a != d(a)])
runtime(sum_amicable, 10000)
# 15 = 3 x 5
# The first three consecutive numbers to have three distinct prime factors
# are:
# 644 = 2^2 x 7 x 23
# 645 = 3 x 5 x 43
# 646 = 2 x 17 x 19.
# Find the first four consecutive integers to have four distinct prime
# factors. What is the first of these numbers?
# Answer: 134043 calculated in 0.314765930176 seconds
from euler import runtime, prime_factorization
from itertools import count
def dpf(n):
return len(set(prime_factorization(n)))
def problem_47():
for x in count(100000):
if (dpf(x) == 4 and dpf(x+1) == 4 and
dpf(x+2) == 4 and dpf(x+3) == 4):
return x
runtime(problem_47)
# Problem 15 # ========== # Starting in the top left corner of a 2x2 grid, and only being able to move # to the right and down, there are exactly 6 routes to the bottom right # corner. # How many such routes are there through a 20x20 grid? # Answer: 137846528820 calculated in 5.96046447754e-06 seconds from euler import runtime, binomial runtime(binomial, 40, 20)
# Problem 9
# =========
# A Pythagorean triplet is a set of three natural numbers, a < b < c, for
# which,
# a^2 + b^2 = c^2
# For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
# There exists exactly one Pythagorean triplet for which a + b + c = 1000.
# Find the product abc.
# Answer: 31875000 calculated in 0.0185961723328 seconds
from euler import runtime
from math import sqrt
def prod_of_triplet(s):
for a in xrange(s/2):
for b in xrange(s/2):
if a**2 + b**2 == (s-a-b)**2:
return a*b*(s-a-b)
runtime(prod_of_triplet, 1000)
# ├────┼──────────────────┼──────┼───────────┤
# │ 7 │ 1,2,3,4,5,6 │ 6 │ 1.1666... │
# ├────┼──────────────────┼──────┼───────────┤
# │ 8 │ 1,3,5,7 │ 4 │ 2 │
# ├────┼──────────────────┼──────┼───────────┤
# │ 9 │ 1,2,4,5,7,8 │ 6 │ 1.5 │
# ├────┼──────────────────┼──────┼───────────┤
# │ 10 │ 1,3,7,9 │ 4 │ 2.5 │
# └────┴──────────────────┴──────┴───────────┘
# It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
# Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
# Answer: 510510 calculated in 2.19345092773e-05 seconds
from euler import runtime, prime_sieve
def problem_69(r, m=1):
# find the product of consecutive primes under r
primes = prime_sieve(50)
for x in primes:
if m * x > r:
return m
m *= x
runtime(problem_69, 1000000)
# 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 = t[10]. If the word
# value is a triangle number then we shall call the word a triangle word.
# Using [1]words.txt, a 16K text file containing nearly two-thousand common
# English words, how many are triangle words?
# Answer: 162 calculated in 0.00173592567444 seconds
from euler import runtime
from string import ascii_uppercase
def triangle_words(file, c=0):
triangles = {x*(x+1)/2 for x in xrange(1000)}
value = {c: ascii_uppercase.index(c)+1 for c in ascii_uppercase}
with open(file, 'r') as f:
words = list(f.read().replace('"', '').split(','))
for word in words:
s = 0
for char in word:
s += value[char]
if s in triangles:
c += 1
return c
runtime(triangle_words, 'txt/words.txt')
# Problem 10
# ==========
# The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
# Find the sum of all the primes below two million.
# Answer: 142913828922 calculated in 0.0908539295197 seconds
from euler import runtime, prime_sieve
def sum_of_primes(m):
return sum(prime_sieve(m))
runtime(sum_of_primes, 2000000)
# 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
# We can see that 28 is the first triangle number to have over five
# divisors.
# What is the value of the first triangle number to have over five hundred
# divisors?
# Answer: 76576500 calculated in 0.253170967102 seconds
from euler import runtime, prime_factorization, n_divisors
from itertools import count
def triangle_divisors(n, t=1):
for x in count(2):
if n_divisors(t) > n:
return t
t += x
runtime(triangle_divisors, 500)
# 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?
# Answer: 55 calculated in 3.2831389904 seconds
from euler import runtime, prime, prime_sieve
def circular_prime(n):
for i in xrange(len(str(n))):
if not prime(int(str(n)[i:] + str(n)[:i])):
return False
return True
def count_circular_primes(r, c=0):
for x in prime_sieve(r):
if circular_prime(x):
c += 1
return c
runtime(count_circular_primes, 1000000)
'97142617910342598647204516893989422179826088076852\n'
'87783646182799346313767754307809363333018982642090\n'
'10848802521674670883215120185883543223812876952786\n'
'71329612474782464538636993009049310363619763878039\n'
'62184073572399794223406235393808339651327408011116\n'
'66627891981488087797941876876144230030984490851411\n'
'60661826293682836764744779239180335110989069790714\n'
'85786944089552990653640447425576083659976645795096\n'
'66024396409905389607120198219976047599490197230297\n'
'64913982680032973156037120041377903785566085089252\n'
'16730939319872750275468906903707539413042652315011\n'
'94809377245048795150954100921645863754710598436791\n'
'78639167021187492431995700641917969777599028300699\n'
'15368713711936614952811305876380278410754449733078\n'
'40789923115535562561142322423255033685442488917353\n'
'44889911501440648020369068063960672322193204149535\n'
'41503128880339536053299340368006977710650566631954\n'
'81234880673210146739058568557934581403627822703280\n'
'82616570773948327592232845941706525094512325230608\n'
'22918802058777319719839450180888072429661980811197\n'
'77158542502016545090413245809786882778948721859617\n'
'72107838435069186155435662884062257473692284509516\n'
'20849603980134001723930671666823555245252804609722\n'
'53503534226472524250874054075591789781264330331690')
def problem_13(n):
return sum([int(i) for i in n.split('\n')])
runtime(sum_nums, num)
# Problem 1
# =========
# 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.
# Answer: 233168 calculated in 0.000159025192261 seconds
from euler import runtime
def sum_of_threes_and_fives(m):
return sum([x for x in xrange(m) if not x % 3 or not x % 5])
runtime(sum_of_threes_and_fives, 1000)
# 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
# In combinatorics, we use the notation, ^5C[3] = 10.
# In general,
# ^nC[r] = n! ,where r <= n, n! = n x(n-1) x...x3x2x1, and 0! = 1
# r!(n-r)!
# It is not until n = 23, that a value exceeds one-million: ^23C[10] =
# 1144066.
# How many, not necessarily distinct, values of ^nC[r], for 1 <= n <= 100,
# are greater than one-million?
# Answer: 4075 calculated in 0.0622220039368 seconds
from euler import runtime, binomial
def problem_53(r, m, c=0):
for x in xrange(r+1):
for y in xrange(r+1):
if binomial(x, y) > m:
c += 1
return c
runtime(problem_53, 100, 1000000)
# Problem 4
# =========
# A palindromic number reads the same both ways. The largest palindrome made
# from the product of two 2-digit numbers is 9009 = 91 x 99.
# Find the largest palindrome made from the product of two 3-digit numbers.
# Answer: 906609 calculated in 0.456642866135 seconds
from euler import runtime, palindrome
def largest_palindrome(m, largest=0):
for x in xrange(100, m):
for y in xrange(100, m):
p = x*y
if palindrome(p) and p > largest:
largest = p
return largest
runtime(largest_palindrome, 1000)
# 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.
# How many starting numbers below ten million will arrive at 89?
# Answer: 8581146 calculated in 37.9294171333 seconds
from euler import runtime, sum_digits
def problem_92(r, c89=0):
for x in xrange(1, r):
while x not in [1, 89]:
x = sum_digits(x, 2)
if x == 89:
c89 += 1
return c89
runtime(problem_92, 10000000)
# Problem 7
# =========
# By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
# that the 6th prime is 13.
# What is the 10 001st prime number?
# Answer: 104743 calculated in 0.00478291511536 seconds
from euler import runtime, prime_sieve
def nth_prime(i, m):
return prime_sieve(m)[i]
runtime(nth_prime, 10000, 110000)
