最近在巩固自己的 python，查找资料的时候看到有博主推荐 欧拉计划，记得之前阿良也推荐过，就尝试做了十几个题目。
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.
def msum(x1=3, x2=5, limit=1000):
lx1 = {i for i in range(1,limit) if i%x1 == 0}
lx2 = {i for i in range(1,limit) if i%x2 == 0}
return sum(lx1 | lx2)
msum()
233168
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
解题思路：简单递归方法速度贼慢，参考 stackoverflow: How to write the Fibonacci Sequence in Python
#f(n+2)=f(n+1)+f(n)
def fib(n, first=1, second=2):
if n == first:
return 1
elif n == second:
return 2
else:
return fib(n-1) + fib(n-2)
fib(9)
55
def fib2(n, first=1, second=2):
a,b = first,second
m = 0
while m < n:
yield a
a,b = b,a+b
m+=1
for i in fib2(10):print(i)
1
2
3
5
8
13
21
34
55
89
def fib3(n, first=1, second=2):
a,b = first,second
k=1
while k < n:
a,b = b,a+b
k+=1
return a
fib3(5)
8
n = 1
totalsum = 0
while fib(n) <= 4000000:
if fib(n)%2 == 0:
totalsum+=fib(n)
n+=1
totalsum
4613732
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
注意：质数算法是很多题目的关键，对速度要求很高。具体实现可以参考 编程随想的博客：求质数算法的 N 种境界[1] - 试除法和初级筛法
import math
def factor(n):
l = []
k = 2
limit = math.sqrt(n)
while k < limit:
if n%k == 0:
l.append(k)
n=n/k
else:
k+=1
return l
factor(600851475143)
[71, 839, 1471, 6857]
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.
def isreverseNum(number):
if str(number)[::-1] == str(number):
return True
def palin(start=100, end=1000):
l =[]
for x in range(start, end):
for y in range(start, end):
z = x*y
if isreverseNum(z):
l.append(z)
return max(l)
palin()
906609
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?
解题思路：先对 1—20 中的每个数进行因式分解，并将结果变为“因子-个数”，如 20 就是两个2，一个5，然后对这20组数中的因子个数取并集。
def factor2(n):
l=[]
k=2
n0=n
while k <= n0:
if n%k == 0:
l.append(k)
n=n/k
else:
k+=1
return l
factor2(20)
[2, 2, 5]
from collections import Counter
def findmul(n):
cnt = Counter()
for i in range(3,n):
cnt = cnt | Counter(factor2(i))
return cnt
findmul(20)
Counter({2: 4, 3: 2, 5: 1, 7: 1, 11: 1, 13: 1, 17: 1, 19: 1})
total = 1
for k,v in findmul(20).items():
print(k,v)
total*=k**v
total
3 2
2 4
5 1
7 1
11 1
13 1
17 1
19 1
232792560
The sum of the squares of the first ten natural numbers is,
The square of the sum of the first ten natural numbers is,
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.
import numpy as np
np.square(np.sum(np.arange(1,101))) - np.sum(np.square(np.arange(1,101)))
25164150
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?
n = 3
i = 1
while n < 10002:
k = 2*i+1
if k%3 != 0 and k%5 != 0:
if len(factor(k)) == 0:
n+=1
i+=1
if n < 5 or n%2000 == 0 or n == 10001:
print('第 {} 个质数是 {}'.format(n,k))
else:
i+=1
else:
i+=1
第 4 个质数是 7
第 2000 个质数是 17107
第 4000 个质数是 37379
第 6000 个质数是 58889
第 8000 个质数是 81131
第 10000 个质数是 104033
第 10001 个质数是 104047
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?
seed = """
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450"""
seed2 = ''.join(seed.split())
j = 0
large = 1
while j+13 <= len(seed2):
text = list(seed2[j:(j+13)])
digits = [int(i) for i in text]
product = 1
for i in digits:
product*=int(i)
if product >= large:
large = product
j+=1
large
23514624000
A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,
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 $abc$.
for a in range(1,500):
for b in range(1,500):
if a < b and a**2 + b**2 == (1000-a-b)**2:
print(a, b, a*b*(1000-a-b))
200 375 31875000
The sum of the primes below 10 is $2 + 3 + 5 + 7 = 17$.
Find the sum of all the primes below two million.
prime_sum = 10
i = 1
n = 3
for i in range(1,1000000):
k = 2*i+1
if k%3!=0 and k%5 != 0 and k < 2000000:
if len(factor(k)) == 0:
n+=1
prime_sum+=k
if n < 6 or n%20000 == 0:
print('第 {} 个质数是 {}，他们的和为 {}'.format(n,k,prime_sum))
prime_sum
第 4 个质数是 7，他们的和为 17
第 5 个质数是 11，他们的和为 28
第 20000 个质数是 223633，他们的和为 2124103173
第 40000 个质数是 478399，他们的和为 9120139073
第 60000 个质数是 744661，他们的和为 21334048815
第 80000 个质数是 1018123，他们的和为 38948745271
第 100000 个质数是 1297019，他们的和为 62092494303
第 120000 个质数是 1580573，他们的和为 90861601085
第 140000 个质数是 1867477，他们的和为 125342898663
143042032078
In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
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 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 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?
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?
def triangle(n):
return int(n*(n+1)/2)
def triangle2(n):
if n == 1:
return 1
else:
return int(triangle(n-1)*(n+1)/(n-1))
triangle2(100)
5050
def triNum(n):
if n == 1:
return 1
tri=triangle(n)
limit = math.sqrt(tri)
k = 2
count = 2
while k <= limit:
if tri%k == 0:
count+=2
k+=1
return count
triNum(7)
6
count = 2
i = 1
while count <= 500:
i+=1
count = triNum(i)
print(i, triangle(i),count)
12375 76576500 576
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
The following iterative sequence is defined for the set of positive integers:
(n is even)
(n is odd)
Using the rule above and starting with 13, we generate the following sequence:
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.
def Collatz(n):
count=0
while n != 1:
if n%2 == 0:
n = int(n/2)
else:
n = 3*n+1
count+=1
return count
Collatz(13)
9
Collatz_dict = {i:Collatz(i) for i in range(1, 1000001)}
max(Collatz_dict, key=Collatz_dict.get)
837799
Starting in the top left corner of a 2×2 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 20×20 grid?
$ 2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$.
What is the sum of the digits of the number $2^{1000}$?
sum([int(i) for i in list(str(2**1000))])
1366
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.
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.
def div_fac(n):
limit = math.sqrt(n)
k = 2
divi = set([1])
while k <= limit:
if n%k == 0:
divi.add(k)
divi.add(int(n/k))
k=k+1
return sum(divi)
div_fac(220)
284
total=0
for i in range(1,10000):
if div_fac(div_fac(i)) == i and i != div_fac(i):
total+=i
print(i,div_fac(i))
total
220 284
284 220
1184 1210
1210 1184
2620 2924
2924 2620
5020 5564
5564 5020
6232 6368
6368 6232
31626
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.
abundant=set()
for i in range(12,28123):
if div_fac(i) > i:
abundant.add(i)
l = set()
for i in range(1,28124):
m = [j for j in abundant if j < i]
for t in m:
if (i-t) in abundant:
l.add(i)
cannot = {i for i in range(1,28124)} - l
sum(cannot)
4179871
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?
i = 1
a,b=1,1
while len(str(a)) < 1000:
i+=1
a,b = b, a+b
print(i,len(str(a)),a)
4782 1000 1070066266382758936764980584457396885083683896632151665013235203375314520604694040621889147582489792657804694888177591957484336466672569959512996030461262748092482186144069433051234774442750273781753087579391666192149259186759553966422837148943113074699503439547001985432609723067290192870526447243726117715821825548491120525013201478612965931381792235559657452039506137551467837543229119602129934048260706175397706847068202895486902666185435124521900369480641357447470911707619766945691070098024393439617474103736912503231365532164773697023167755051595173518460579954919410967778373229665796581646513903488154256310184224190259846088000110186255550245493937113651657039447629584714548523425950428582425306083544435428212611008992863795048006894330309773217834864543113205765659868456288616808718693835297350643986297640660000723562917905207051164077614812491885830945940566688339109350944456576357666151619317753792891661581327159616877487983821820492520348473874384736771934512787029218636250627816
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
As $1 = 1^4$ is not a sum it is not included.
The sum of these numbers is .
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
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:
How many different ways can £2 be made using any number of coins?
coins = [1,2,5,10,20,50,100,200]
def findcoin(target, n):
if n <= 1:
return 1
res = 0
while target >= 0:
res+=findcoin(target,n-1)
target = target - coins[n-1]
return res
findcoin(200,8)
73682
findcoin(231,8)
139462