This notebook contains my problem-solving code used to answer the various challenges posted on the website https://projecteuler.net. It is often cited as good coding practice, to get used to things like basic algorithms, data structures, optimization, and so on. My main focus in coding practice is undoubtedly about data, statistics, inference and knowledge discovery. On the other hand, I am starting more and more to think that a strong base in algorithms and use of Python's built-ins and will be more beneficial than just starting in the deep end with numpy, pandas, sk-learn, tensorflow and whatnot -- right away. Code complexity and performance when submitting to production might just we worth more in the long run.
Sideways inspiration for this practice is the Google whitepaper Machine Learning: The High-Interest Credit Card of Technical Debt.
In [2]:
print(sum([n for n in range(1000) if n % 3 == 0 or n % 5 == 0]))
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.
In [23]:
"""I end up using this decorator often to add 'dynamic programming'
style solving capabilities to problems. It can't be used with success
for most function calls, but it can sometimes speed some things up
dramatically with the addition of just an extra line of code.
Shamelessly copied from:
https://wiki.python.org/moin/PythonDecoratorLibrary
"""
import collections
import functools
class memoized(object):
'''Decorator. Caches a function's return value each time it is called.
If called later with the same arguments, the cached value is returned
(not reevaluated).
'''
def __init__(self, func):
self.func = func
self.cache = {}
def __call__(self, *args):
if not isinstance(args, collections.Hashable):
# uncacheable. a list, for instance.
# better to not cache than blow up.
return self.func(*args)
if args in self.cache:
return self.cache[args]
else:
value = self.func(*args)
self.cache[args] = value
return value
def __repr__(self):
'''Return the function's docstring.'''
return self.func.__doc__
def __get__(self, obj, objtype):
'''Support instance methods.'''
return functools.partial(self.__call__, obj)
In [3]:
@memoized
def fib(n):
a = b = 1
for i in range(n):
yield a
a, b = b, a + b
def even_fib():
sum_ = 0
for i in list(fib(100)):
if i < 4000000:
if i % 2 == 0:
sum_ += i
else:
pass
else:
pass
return sum_
print(even_fib())
In [24]:
import math
def primes_sieve(limit):
"""Sieve of Erastostenes up to $limit.
TODO: Set up as generator
int -> [int,]"""
limitn = limit+1
not_prime = set()
primes = []
for i in range(2, limitn):
if i in not_prime:
continue
for f in range(i*2, limitn, i):
not_prime.add(f)
primes.append(i)
return primes
def problem_3():
"""Generate list of candidate primes, return first one
that matches. Get list of primes by starting at sqrt($big_num)
and going downwards"""
big_num = 600851475143
start_n = int(math.sqrt(big_num) + 1)
return max([n for n in primes_sieve(start_n) if big_num % n == 0])
print(problem_3())
In [5]:
def is_palindrome(num):
"""Takes list of three integer strings, outputs True or False"""
return str(num) == str(num)[::-1]
def problem_4():
"""Goes through every pair of 3-digit numbers, sees if
product is a palindrome."""
palindromes = []
for i in range(100, 1000):
for j in range(100, 1000):
if is_palindrome(i * j) == True:
palindromes.append(i * j)
else:
pass
return palindromes
print(max(problem_4()))
In [6]:
# Python has a Greatest Common Divisor (GCD) method we can
# use to get to the Least Common Denominator (LCD).
# Instead of looping over numbers, simply multiply the terms.
from math import gcd
def problem_5(n=20):
"""The LCD is the product of the greatest common divisors
between all numbers from 1 to n, default n = 20.
(int) -> int"""
num = 1
for i in range(1, n + 1):
num *= i // gcd(i, num)
return num
print(problem_5())
In [7]:
%matplotlib inline
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
first_500_sm = [float(problem_5(n)) for n in range(1, 501)]
fig = plt.figure()
ax = fig.add_subplot(111)
ax.set_yscale('log')
ax.set_title('Smallest Multiple of the x First Numbers, up to 500\n Axes: log-normal\n')
ax.plot(first_500_sm, color='#348ABD', lw=1.4, alpha=0.6)
plt.show()
Now that's really interesting! The fact that it plots as an almost straight line in a log-normal plot means it can be approximated well with a best-fit exponential curve. Let's take a look at the parameters that come out:
In [8]:
x = np.array(range(1, 501))
y = np.array(first_500_sm)
# Fit for y = Ae^(Bx)
B, A = np.polyfit(-x, -np.log(y), 1)
print('\nCurve of best fit: y = {0} * exp({1} x)'.format(round(A, 8), round(B, 8)))
z = A*(np.exp(B*x))
fig = plt.figure()
ax = fig.add_subplot(111)
ax.set_yscale('log')
ax.set_title('Best Fit for Smallest Multiple of the x First Numbers\n Axes: log-normal\n')
ax.plot(x, z, lw=2.1, alpha=0.3, c="#A60628", label='best fit, ~1.806e^x')
ax.plot(x, y, color='#348ABD', lw=1.4, alpha=0.8, label='smallest multiple of x first numbers')
plt.legend(loc=2, prop={'size':9})
plt.show()
Now as the B parameter is suspiciously close to 1, and -- at least according to Wolfram Alpha -- the A parameter is close to
$$2 \frac{(1 + \pi)}{(\pi - 1)^2},$$I'm guessing the whole equation boils down to some number-theoretical equation proven a long time ago. Maybe something along the lines of:
$$ SM_x \approx 2 \frac{(1 + \pi)}{(\pi - 1)^2} \exp{(x)}?$$Now granted, I might be over-reaching a little in my extrapolation, but it is always so fascinating to see e and pi magically appear when working with natural numbers. In any case, if you have any leads, please reach out! Shedding more light on this would be appreciated. I can't say I ever took a number theory course either, so, win-win.
</addendum>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.
In [9]:
def problem_6():
sum_of_squares = sum([n**2 for n in range(1, 101)])
square_of_sums = sum([n for n in range(1, 101)])**2
return square_of_sums - sum_of_squares
print(problem_6())
In [10]:
"""We already have a primes_sieve function set up, all we need is an upper bound
so that we get the 10001th prime.
"""
print(primes_sieve(500000)[10001 - 1])
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
In [11]:
def prod(iterable):
from functools import reduce
"""There is no built-in method for product similar to sum().
This funciton is inelegant and a little cryptic, but I'm
compensating with this long-winded explanation instead.
The lambda expression creates an accumulator y, and multiplies
it with every element of the iterable.
Note: itertools.accumulate(iterable, func=operator.mul) would
work, but not on bare ints.
iterable_containing_int -> int"""
return reduce(lambda x, y: int(x)*int(y), iterable)
def problem_8():
big_num = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
slices = [num for i in range(0, 1000 - 13) for num in [big_num[i:i + 13]]]
return max([prod(s) for s in slices])
print(problem_8())
In [12]:
# Get the set of int. triplets for which a + b + c = 1000
def problem_9():
for i in range(1, 1001):
for j in range(1 + i, 1001):
k = 1000 - i - j
if i**2 + j**2 == k**2:
return i * j * k
print(problem_9())
In [13]:
print(sum(primes_sieve(1999999)))
In the 20×20 grid below, four numbers along a diagonal line have been marked in red pointed with arrows a-d.
a b c d
v v v v
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
a> 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
b> 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
c> 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
d> 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
In [14]:
import numpy as np
def problem_11():
"""There are 4 distinct ways with which the array can be sliced,
including two diagonals. We create a big list of 4-number lists,
and take the maximum product of these. The inverse dataframe is
used to take the 'anti-clockwise' diagonals.
The indexes will go out of bounds for a few cases, but their results
are simply discarded as we are only looking for the maximum.
Using numpy, as diagonal slicing and products are directly handled.
/lazy"""
df = np.array([
[ 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8],
[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0],
[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65],
[52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91],
[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],
[24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50],
[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],
[67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21],
[24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],
[21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95],
[78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92],
[16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57],
[86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58],
[19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40],
[ 4,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66],
[88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69],
[ 4,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36],
[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16],
[20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54],
[ 1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48],
], dtype=np.uint8)
df_inv = df[::-1]
max_product = 0
for i in range(20):
for j in range(20):
horiz = np.prod( df[i][j:j+4] )
verti = np.prod( df[:,j][i:i+4] )
d = min(i, j)
diag_1 = np.prod( np.diagonal(df, j-i)[ d:d+4 ] )
diag_2 = np.prod( np.diagonal(df_inv, j-i)[ d:d+4 ] )
max_product = max([max_product, horiz, verti, diag_1, diag_2])
return max_product
print(problem_11())
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
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?
In [15]:
import itertools
import math
@memoized
def triangle_number(n):
"""Returns nth Pascal's triangular number
int -> int"""
return sum(range(n + 1))
def num_divisors(num):
"""Returns the number of distinct divisors a given integers has.
End number given as sqrt(num) as we get no bigger. Special case
when num is perfect square, divs -= 1.
int -> int"""
divs = 0
end = int(math.sqrt(num))
for i in range(1, end + 1):
if num % i == 0:
divs += 2
if end**2 == num:
divs -= 1
return divs
def problem_12(n=1):
for i in itertools.count(n): # count() is open-ended range()
t = triangle_number(i)
divs = num_divisors(t)
if divs > 500:
return t
else:
pass
print(problem_12(10000)) # Refined search after the fact for faster load times
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
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
In [16]:
nums = [
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,
]
print(str(sum(nums))[0:10])
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.
In [17]:
import time
from tqdm import *
@memoized
def collatz(value):
"""Generator that spits out next in Collatz series given 'value'.
int -> int"""
n = value
yield n
while n != 1:
if n % 2 == 0:
n = n // 2
yield n
else:
n = 3 * n +1
yield n
def problem_14():
max_seq = 0
max_num = 1
for i in tqdm(range(1, 1000001)):
seq = len(list(collatz(i)))
if seq > max_seq:
max_seq = seq
max_num = i
return max_num
print(problem_14())
In [18]:
# Assume "paths of length 40" was meant? Otherwise this becomes
# a whole different problem altogether. Seems like a '40 choose 20'
# -type problem from combinatorics.
# Anyhow, naively brute-forcing worked, so I'm not touching it.
@memoized
def lattice(node=(0, 0)):
n = 0
k = 20
if node[0] == k and node[1] == k:
return 1
if node[0] < k + 1 and node[1] < k + 1:
n += lattice((node[0] + 1, node[1]))
n += lattice((node[0], node[1] + 1))
return n
print(lattice())
In [19]:
print(sum([int(n) for n in str(2**1000)]))
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?
NOTE: 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.
In [20]:
def number_name(n):
"""Returns the written-out string of the integer in the argument.
Range: (1, 1000).
int -> str"""
digits = ['', 'one', 'two', 'three', 'four', 'five', 'six', 'seven',
'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen',
'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen',
'nineteen']
decades = ['-', '-', 'twenty', 'thirty', 'forty', 'fifty', 'sixty',
'seventy', 'eighty', 'ninety']
if 0 < n < 20:
return digits[n]
elif 20 <= n < 100:
return decades[n // 10] + ('-' + digits[n % 10]
if (n % 10 != 0) else "")
elif 100 <= n < 1000:
return digits[n // 100] + " hundred " + (("and " +
number_name(n % 100)) if (n % 100 != 0) else "")
elif 1000:
return 'one thousand'
else:
raise ValueError('Please use numbers 1 to 1000.')
# I know, it's odd to add spaces and dashes in the function,
# only to strip them away in the compute phase. But I figure
# the number_name(n) function might come in handy down the line.
print(len([char for n in range(1, 1001) for char in number_name(n)
if char in 'abcdefghijklmnopqrstuvwxyz']))
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.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
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 09 70 98 73 93 38 53 60 04 23
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)
In [21]:
"""Given the note, well, I guess I'll cross that bridge if and when
I get to problem 67. For now, I'll just brute force and memoize. ;)
From the data 'triangle' below, we have to bound our path from
one level to the next to either the same index or the following one."""
problem_18_data = [ [75],
[95,64],
[17,47,82],
[18,35,87,10],
[20, 4,82,47,65],
[19, 1,23,75, 3,34],
[88, 2,77,73, 7,63,67],
[99,65, 4,28, 6,16,70,92],
[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, 4,68,89,53,67,30,73,16,69,87,40,31],
[ 4,62,98,27,23, 9,70,98,73,93,38,53,60, 4,23],
]
@memoized
def all_paths(r=0, c=0, df=problem_18_data):
"""Takes path with stride 0 or +1, returns list of all paths
possible in the data triangle.
(int, int) -> [[int,],]"""
current = df[r][c]
if r < len(df) - 1:
down_paths = all_paths(r + 1, c) + all_paths(r + 1, c + 1)
return [[current] + path for path in down_paths]
else:
return [[current]]
print(max([sum(path) for path in all_paths()]))
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)?
In [22]:
def problem_19():
days = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun']
reg_days_months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
leap_days_months = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
years = [n for n in range(1901, 2001)]
leap_years = [n for n in range(1901, 2001) if
(n % 4 == 0 and n % 100 != 0) or n % 400 == 0]
reg_calendar = [i for n in reg_days_months for i in range(1, n+1)]
leap_calendar =[i for n in leap_days_months for i in range(1, n+1)]
millenium = []
for y in years:
if y in leap_years:
for d in leap_calendar:
millenium.append(d)
else:
for d in reg_calendar:
millenium.append(d)
# Cycle the days of the week, attach it every day number in
# the millenium calendar.
millenium = list(zip(millenium, itertools.cycle(days)))
return len([day for day in millenium if day == (1, 'Mon')])
print(problem_19())
In [23]:
print(sum([int(n) for n in str(math.factorial(100))]))
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.
In [24]:
from tqdm import *
@memoized
def divisors_sum(i):
"""Returns the sum of all proper divisors for integer i
int -> int."""
return sum([d for d in range(1, (i+1 // 2)) if i % d == 0])
def amicable_numbers(n=10000):
"""Returns the list of amicable number pairs up to n;
Default: n = 10000.
Amicable numbers are defined as pairs of numbers whose
sum of proper divisors is equal. e.g.: 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.
int -> [int,]"""
divisors_sum_tuple = [(n, divisors_sum(n)) for n in tqdm(range(1, n+1))]
return [key for (key, value) in divisors_sum_tuple if (
(value, key) in divisors_sum_tuple and value != key)]
assert divisors_sum(220) == 284
assert divisors_sum(284) == 220
print(sum(amicable_numbers()))
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.
What is the total of all the name scores in the file?
In [25]:
with open('names.txt') as f:
names = f.read().strip('"').split('","')
print(sum([(i + 1) * (ord(char) - ord('A') + 1) for (i, name)
in enumerate(sorted(names)) for char in name]))
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.
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.
In [26]:
# Re-use the divisors_sum function from problem 21.
# It's a little slow, but still solves the problem
# in about a minute.
assert divisors_sum(28) == 28
def abundant_number(n):
"""Return a boolean value, whether or not the supplied number is
abundant. e.g. A number n is called abundant if the sum of its
proper divisors exceeds n.
int -> bool."""
return divisors_sum(n) > n
assert abundant_number(12) == True
def problem_23():
limit = 28123
ab_numbers = [n for n in tqdm(range(1, limit + 1))
if abundant_number(n) == True]
# Cut down on for-loop by using list index as data (with enumerate)
non_sum = (limit + 1) * [False]
for i in ab_numbers:
for j in ab_numbers:
if i + j < limit + 1:
non_sum[i + j] = True
else:
break
return sum([i for (i, x) in enumerate(non_sum) if not x])
print(problem_23())
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?
In [27]:
def nth_lexicographic_perm(n=1000000):
"""Returns the nth lexicographic permutation of numbers.
Default: n = 1 000 000
(int) -> int"""
perm = itertools.islice(itertools.permutations('0123456789'), n - 1, None)
return str("".join(next(perm)))
print(nth_lexicographic_perm())
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
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
In [28]:
# We use the generator we setup in Problem 2.
# 10 000 seems like a good place to start our search.
def problem_25():
for i, n in enumerate(fib(10000)):
if len(str(n)) >= 1000:
return i + 1 # Because zero-based index
print(problem_25())
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.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
In [29]:
def recip_cycle_len(d):
"""Returns the reciprocal's recurring cycle
with the longest length.
e.g. 1/6 = 0.1(6) -> rcl = 1;
1/7 = 0.(142857) -> rcl = 6;
1/8 = 0.125 -> rcl = 0;
int -> int."""
# Strategy: By multiplying n by multiples of 10,
# we can get rcl by str(n)[i] == 10 * str(n)[i].
# Cycle starts after p numbers
cycle = []
i = 10 % d
while i not in (cycle):
cycle.append(i)
i = (i * 10) % d
return len(cycle)
assert recip_cycle_len(6) == 1
assert recip_cycle_len(7) == 6
print(max(range(1, 1000), key=recip_cycle_len))
Euler discovered the remarkable quadratic formula:
n2 + n + 41
It turns out that the formula will produce 40 primes for the consecutive integer values 0 ≤ n ≤39. However, when n=40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and when n = 41, 412 + 41 + 41 is clearly divisible by 41.
The incredible formula
n2 − 79n + 1601
was discovered, which produces 80 primes for the consecutive values 0 ≤ n ≤ 79. The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form: n2 + an + b , where |a| <1000 and |b| ≤1000
where |n| is the modulus/absolute value of n, e.g. |11| = 11 and |−4| = 4.
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
In [30]:
def max_cons_primes(lim=1000):
"""Returns the a and b coefficients of the quadratic
expression that produces the maximum number of primes.
e.g. n^2 - 79n + 1601 produces primes for n in range(0, 79).
Search space for a, b is range(-max, max), def. max=1000.
(int, int) -> int, int"""
# First, set up list of the 10000 first primes. My intuition is
# that it will be way easier to catch the primes closest to zero
# than farther away, especially given that 'starting with n=0'
primes = primes_sieve(10000)
def num_consecutive_primes(tup):
a, b = tup
for n in itertools.count():
y = n**2 + a * n + b
if y not in primes:
return n
assert num_consecutive_primes((1, 41)) == 40
assert num_consecutive_primes((-79, 1601)) == 80
# For n = 0, y = 0 + 0 + b so b needs to be prime;
# We can filter them out right away.
search_space = [(a, b) for a in range(-lim, lim) for b
in range(1, lim) if b in primes]
result = max(search_space, key=num_consecutive_primes)
return result[0] * result[1]
print(max_cons_primes())
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?
In [63]:
# Non-bruteforce attempt. For n*n spiral, sum is:
# 1 + 4*(n + 2) + 4*(n + 4) + 4*(n + 6) + ... + 4*(n + n-1)
def spiral_sum(n=1001):
"""Returns sum of diagonals of number spiral
of size n * n. n must be uneven.
Default: n = 1001.
(int) -> int."""
if n % 2 == 0:
raise ValueError('n must be uneven.')
agg, cur = 1, 1
for i in range(2, n, 2):
for _ in range(4):
cur += i
agg += cur
return agg
assert spiral_sum(5) == 101
print(spiral_sum())
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?
In [32]:
# Gotta love those one-liners!
print(len(set(a**b for a in range(2, 101) for b in range(2, 101))))
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.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
In [33]:
# So far in Project Euler, this is my favorite one-liner.
# Don't forget to take out 1 = 1^5 at the end, though!
print(sum([n for n in range(1, 500000) if
sum([int(i)**5 for i in str(n)]) == n]) - 1)
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?
In [34]:
from tqdm import *
def problem_31():
"""Picks permutations of 100 elements from coins, returns
number of different ways they can add up to 200.
None -> int"""
# TODO: Come back to this one once I understand dynamic
# programming a little better. For now, it's a really
# ugly bruteforce thing, and takes about ten minutes.
# p = [1, 2, 5, 10, 20, 50, 100, 200]
# coins = [p for p in p for l in range(0, 201, p) if l != 0]
one = [i for i in range(0, 201)]
two = [i for i in range(0, 201) if i % 2 == 0]
five = [i for i in range(0, 201) if i % 5 == 0]
ten = [i for i in range(0, 201) if i % 10 == 0]
twenty = [i for i in range(0, 201) if i % 20 == 0]
fifty = [i for i in range(0, 201) if i % 50 == 0]
pound = [i for i in range(0, 201) if i % 100 == 0]
count = 1
for i in one:
for j in two:
for k in five:
for l in ten:
for m in twenty:
for n in fifty:
for o in pound:
if i + j + k + l + m + n + o == 200:
count += 1
return count
# print(problem_31())
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. HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
In [35]:
def is_pandigital(tup):
"""A number is pandigital iif the numbers 1-9
appear in i, j, k for the product i * j = k.
Expects a group of two integers.
(int, int) -> bool."""
i, j = tup
chars = [i for i in (str(i) + str(j) + str(i * j))]
return '123456789' == ''.join(sorted(chars))
def pandigitals(lim=10000):
"""Returns a list of tuples who, when their digits
and their products' digits are """
# Product has to be < 10k, as no product between any two numbers
# whose sum of digits is 4 can give that product.
tuples = [tup for tup in itertools.combinations(range(1, lim // 5), 2)
if prod(tup) < lim]
return [tup for tup in tuples if is_pandigital(tup) == True]
assert is_pandigital((39, 186)) == True
assert is_pandigital((39, 231)) == False
assert (39, 186) in pandigitals()
assert (234, 186) not in pandigitals()
print(sum(set(i * j for i, j in pandigitals())))
The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
In [36]:
def digit_cancelling(tup):
"""Takes a tuple containing two sorted integers < 100,
returns True if they are digit-cancelling, else None.
(int, int) -> bool"""
i, j = tup
different_digits = set(str(i)) ^ set(str(j)) # set difference
if len(different_digits) == 2:
# Crunch these only if necessary
common_digit = set(str(i)) & set(str(j)) # set intersection
d = [n for n in str(i) + str(j) if n not in common_digit]
if (int(d[0]) / int(d[1])) == (i / j):
return True
def d_c_fractions():
"""Gives the list of all digit cancelling fractions,
for pairs of two-digit integers.
None -> [(int, int),]"""
# We get rid of all numbers of the form nn, and those
# that contain the number 0. In other words, those divisible
# by 10 and 11.
nums = (i for i in range(12, 99) if i % 11 != 0 and i % 10 != 0)
pairs = itertools.combinations(nums, 2)
return [pair for pair in pairs if digit_cancelling(pair)]
def problem_33():
product_of_fractions = [1, 1] # [numerator, denominator]
for frac in d_c_fractions():
product_of_fractions = [i*j for i, j in zip(product_of_fractions, frac)]
return product_of_fractions[1] // product_of_fractions[0]
assert digit_cancelling((49, 98)) == True
assert digit_cancelling((43, 93)) == None
assert len(d_c_fractions()) == 4
print(problem_33())
In [37]:
def curious_numbers(lim=100000):
"""Returns list of integers whose sum of
the factorial of each of its digits equals the
integer itself.
Argument: (optional) limit of search.
Default: 100'000.
(int) -> int."""
return [n for n in range(lim) if n ==
sum([math.factorial(int(i)) for i in str(n)])]
assert 145 in curious_numbers()
print(sum(curious_numbers()) - 3) # - 3 as per problem note
In [38]:
# I totally misunderstood this at first. 'Rotations' really means
# 'string rotations'. The smart thing to rotate a string would be
# to use a double-ended queue, as it already has the rotate method
# built-in.
from tqdm import *
from collections import deque
def rotations(num):
"""Returns the set of string rotations for the
digits in num. e.g. 197 -> {197, 971, 719}.
int -> set(int,)"""
d = deque(digit for digit in str(num))
candidates = []
for i in range(len(d)):
d.rotate()
candidates.append(int(''.join(d)))
return set(candidates)
def circular_primes(lim=1000000):
"""Returns the list of circular primes under n.
Default: n = 1'000'000.
(int) -> [int,]"""
# Note: numbers containing '0' are automatically out, as
# one member of that rotation will be divisible by 10.
primes = set(primes_sieve(lim))
nums = [False] * lim
for num, boolean in enumerate(tqdm(nums)):
# TODO: Tighten that loop to go through primes_sieve
if boolean == True or num < 2 or '0' in str(num):
pass
else:
candidates = rotations(num)
if candidates < primes: # c subset of p
for n in candidates:
nums[n] = True
return sum(nums) # True = 1, False = 0.
assert rotations(197) == {971, 197, 719}
assert circular_primes(100) == 13
print(circular_primes())
In [64]:
def is_doubly_palindromic(num):
"""Returns True is number has palindromic digits in
both base 10 and base 2.
E.g. 585 = 2b(1001001001), both of which palindromic.
int -> bool."""
b = str(bin(num))[2:]
num = str(num)
return b == b[::-1] and num == num[::-1]
assert is_doubly_palindromic(585) == True
assert is_doubly_palindromic(586) == False
print(sum([n for n in tqdm(range(1, 1000000)) if is_doubly_palindromic(n)]))
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.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
In [34]:
# So the best strategy is probably to keep appending single
# digit primes and keep only the trees that survive. But in
# this case, the upper bound of 11 primes also nicely becomes
# a stopping condition for 'directed bruteforce'. Note that
# a truncatable number can't be 'built up' from existing ones.
from tqdm import *
def truncate_left_right(num):
"""Returns True if the integer n is a truncatable
prime, that is, a prime number such that removing
digits one at a time from both sides, every truncated
number is also prime.
E.g. 3797, (797, 97, 7), (379, 37, 3) are all prime.
int -> bool."""
s = str(num)
candidates = set()
for i in range(1, len(s)):
candidates.add(int(s[i:]))
candidates.add(int(s[:i]))
return candidates
def problem_37(lim=1000000):
primes = set(primes_sieve(lim))
trunc_primes = set()
for prime in tqdm(primes):
if truncate_left_right(prime) < primes:
trunc_primes.add(prime)
return [p for p in trunc_primes if p > 7]
print(sum(problem_37()))
Take the number 192 and multiply it by each of 1, 2, and 3:
192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
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?
In [72]:
def problem_38():
"""For numbers n, m where concat(n * range(m)),
we can tighten the solution space by searching
for m up to only the order of magnitude that
yields 9-digit numbers."""
max_num = ""
for n in range(2, 10):
for i in range(1, 10**(9 // n)):
s = "".join(str(i * j) for j in range(1, n + 1))
if "".join(sorted(s)) == "123456789":
max_num = max(s, max_num)
return max_num
print(problem_38())
In [89]:
def right_angle_solutions(p):
"""Returns the number of solutions that give
an integral length side p for a right triangle.
For example:
p = 120: 20,48,52; 24,45,51; 30,40,50;
i.e. we have 3 right angle solutions.
int -> int"""
solutions = 0
for i in range(1, p + 1):
for j in range(i, (p - i) // 2 + 1):
k = p - i - j
if i ** 2 + j ** 2 == k ** 2:
solutions += 1
return solutions
assert right_angle_solutions(120) == 3
print(max([p for p in range(1, 1001)], key=right_angle_solutions))
An irrational decimal fraction is created by concatenating the positive integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If dn represents the nth digit of the fractional part, find the value of the following expression.
d1 × d10 × d100 × d1000 × d10'000 × d100'000 × d1'000'000
In [103]:
champ_c = [d for n in range(10 ** 6) for d in str(n)]
print(prod([int(i) for j in range(7) for i in champ_c[10 ** j]]))
In [146]:
# If we start from 9-digit number permutations and move
# downwards, that's our stopping condition. Since we get
# pandigital numbers only scarcely scattered through the
# primes, it's much faster to just generate the pandigitals
# and check for primacy.
# Now this implementaion is about two orders of magnitude faster than
# the naive code I had came up with for problem 3. Next steps for
# faster implementation would probably need numpy, and two-stage
# probabilistic -> deterministic checking. Prime number algorithms
# are an unsurprisingly very mature field, and even though fascinating,
# most of that complexity does not suit our needs.
# http://stackoverflow.com/questions/1801391/what-is-the-best-algorithm-for-checking-if-a-number-is-prime
def is_prime(n):
"""Returns True if n is prime using the AKS algorithm.
int -> bool"""
if n == 2:
return True
if n == 3:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
def problem_41():
"""Returns the largest pandigital prime.
None -> int"""
for i in range(10, 2, -1):
digits = [str(n) for n in range(1, i)][::-1]
a = 0
for num in (int(''.join(s)) for s in itertools.permutations(digits)):
if is_prime(num):
return num
print(problem_41())
Also of note, the following paragraph found when researching prime-searching algorithms, from the following link: http://archive.oreilly.com/pub/a/python/excerpt/pythonckbk_chap1/index1.html?page=last
Be aware that one liners, even clever ones, are generally anything but speed demons! primes_oneliner takes 2.9 seconds to complete the same small task (computing primes up to 8,192) which, eratosthenes does in 22 milliseconds, and primes_less_than in 9.7 -- so you're slowing things down by 130 to 300 times just for the fun of using a clever, opaque one liner, which is clearly not a sensible tradeoff. Clever one liners can be instructive but should almost never be used in production code, not just because they're terse and make maintenance harder than straightforward coding (which is by far the main reason), but also because of the speed penalties they may entail.
Interesting to say the least! I guess experience and knowing your co-workers is still the best gauge of what is to be considered terse and unmaintainable. A balance must be met set between readability and code unfolding -- Python's style must always stay in the gray zone between "too blown-up" à la C, versus "too terse", à la Haskell.
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); 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.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
In [21]:
import csv
def problem_42(f='words.txt'):
"""Returns the number of coded triangle words for the
given text file, containing words.
Default: /words.txt.
'filename' -> int"""
t_numbers = [(n * n + n) // 2 for n in range(1, 100000)]
letters = [c for c in '-ABCDEFGHIJKLMNOPQRSTUVWXYZ']
words = open('words.txt').read().replace('"', '').split(',')
sum_coded = 0
for word in words:
chars = [c for c in word]
nums = [letters.index(c) for c in chars]
if sum(nums) in t_numbers:
# print(word, nums)
sum_coded += 1
return sum_coded
print(problem_42())
The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
In [56]:
import itertools
def is_substring_d(num):
"""Returns True if given pandigital number
is substring divisible by its 'primacy'.
int -> bool."""
primes = [2, 3, 5, 7, 11, 13, 17]
num = str(num)
for i, p in enumerate(primes):
if int(num[i + 1: i + 4]) % p != 0:
return False
return True
def problem_43():
pan = (int(''.join(n)) for n in itertools.permutations('1234567890'))
substringable = []
for num in pan:
if num > 1234567889:
if is_substring_d(num):
substringable.append(num)
return substringable
assert is_substring_d(1406357289) == True
assert is_substring_d(6574892951) == False
print(sum(problem_43()))
Pentagonal numbers are generated by the formula, Pn=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 P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
In [6]:
# After many an hour spent on this problem, this is where
# I learn the hard way that dictionaries are really fast
# when you want to check iterable membership.
from tqdm import *
def problem_44(lim=2000):
"""Returns the smallest difference of D such that
D = abs(Pk - Pj) is minimised for pairs of nums
Pk and Pj being pentagonal numbers, where their
sum and difference are both also pentagonal.
Argument: lim(int), the nth pentagonal number to
search to. Default: 1000.
(int) -> int"""
pent_nums = [(3 * n * n - n) // 2 for n in range(1, 2 * lim)]
min_dict = dict.fromkeys(pent_nums)
for i in trange(lim):
for j in range(i + 1, 2 * lim - 1):
if (pent_nums[i] + pent_nums[j]) in min_dict:
diff = pent_nums[j] - pent_nums[i]
if diff in min_dict:
# As we're going up a quadratic series,
# the first diff should be the smallest.
return diff
print(problem_44())
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ...
Hexagonal Hn=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.
In [22]:
def problem_45(lower=40755, lim=100000):
"""Returns second number that is triangle, pentagonal,
and hexagonal.
Args: lower, the lowest integer after which to start.
lim, the maximum number to search to.
Default: lower=40755, lim=1'000'000.
(int, int) -> int"""
triangle = [n * (n + 1) // 2 for n in range(0, lim)]
pentagon = dict.fromkeys([n * (3 * n - 1) // 2 for n in range(0, lim)], 1)
hexagon = dict.fromkeys([n * (2 * n - 1) for n in range(0, lim)], 1)
for i in tqdm(triangle[286:]):
if i in pentagon and i in hexagon:
return i
print(problem_45())
It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
9 = 7 + 2×12
15 = 7 + 2×22
21 = 3 + 2×32
25 = 7 + 2×32
27 = 19 + 2×22
33 = 31 + 2×12
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
In [49]:
def problem_46(lim=100000):
"""Returns the smallest odd composite that cannot
be written as the sum of a prime and twice a square.
Args: lim(int), the limit of the search space.
Default: lim=100'000.
(int) -> int"""
squares = dict.fromkeys((n * n for n in range(1, lim)), 1)
primes = primes_sieve(lim)
prime_dict = dict.fromkeys(primes, 1)
for i in trange(33, lim, 2):
if i not in prime_dict:
goldbach = False
for p in primes[1:]:
if p >= i:
break
if ((i - p) // 2) in squares:
# You win this one, Goldbach.
goldbach = True
break
if not goldbach:
# One conjecture is enough, sir.
return i
print(problem_46())
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 each. What is the first of these numbers?
In [55]:
def prime_factors(n):
"""For given n, returns set of prime factors.
int -> set(int,)"""
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return set(factors)
def problem_47(lim=1000000):
"""Returns the first of four consecutive numbers for
which they share four distinct prime factors each.
Args: lim(int), limit of the search space.
Default: 1'000'000
(int) -> int"""
primes = primes_sieve(lim)
prime_dict = dict.fromkeys(primes, 1)
consecutive = 0
for i in trange(647, lim):
if len(prime_factors(i)) == 4:
consecutive += 1
if consecutive == 4:
return i - 3
else:
consecutive = 0
print(problem_47())
In [58]:
# Gets the last 10 digits of the sum of n^n for n in (1, 1000)
print(str(sum([n ** n for n in range(1, 1001)]))[-10:])
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
What 12-digit number do you form by concatenating the three terms in this sequence?
In [85]:
import itertools
def problem_49():
"""Returns the concatenated digits of the second 4-digit
arithmetic sequence of primes which are digit permutations
of each other, the first being 1487, 4817 and 8147. In
this case, arithmetic means that the difference between
the terms is the same; it being 3330 in our example.
None -> int"""
lim = 9999
primes = [n for n in primes_sieve(lim) if n > 1487 and '0' not in str(n)]
primes.remove(4817)
primes.remove(8147)
for p in tqdm(primes):
prime_candidates = [int(''.join(i)) for i in itertools.permutations(str(p))]
sequence_candidates = [n for n in prime_candidates if n in primes]
# Get the unique members, and re-order list.
sequence_candidates = sorted(list(set(sequence_candidates)))
for i, a in enumerate(sequence_candidates):
for b in sequence_candidates[i + 1:]:
c = (a + b) // 2
if c in sequence_candidates:
return ''.join(str(a) + str(c) + str(b))
print(problem_49())
The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one million, can be written as the sum of the most consecutive primes?
In [101]:
def problem_50(lim=1000000):
"""Returns the prime number under a million which can be
written as the sum of the most consecutive prime numbers.
Args: lim(int); upper bound. Default: 1'000'000.
None -> int"""
primes = primes_sieve(lim)
prime_dict = dict.fromkeys(primes, 1)
def sum_candidate(n, bound):
candidate = 0
for i in range(0, bound):
# Thank God they need to be consecutive
total = sum(primes[i:i + n])
if total > bound:
break
if total in prime_dict:
candidate = total
return candidate
# This seems like a good heuristic to start bruteforce.
for n in trange(1000, 21, -1):
candidate = sum_candidate(n, lim)
if candidate:
return candidate
print(problem_50())
Congratulations, the answer you gave to problem 50 is correct.
You are the 45560th person to have solved this problem.
Nice work, leblancfg, you've just advanced to Level 2. 40494 members (5.9%) have made it this far.
You have earned 1 new award:
Flawless Fifty: Solve fifty consecutive problems