Problem 30

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.

continue →

Problem 29

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?

continue →

Problem 28

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?

continue →

Problem 27

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 certainly 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.

continue →

Problem 26

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.

continue →

Problem 25

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?

continue →

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?
continue →

Problem 23

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.

Solution

def sum_divisors(n, divisor):
sqrtN = n**0.5
i = divisor
t = 0
while i <= sqrtN:
if not n%i:
divisor = i
if i == n//i or i == 1:
t+= i
break
t += sum([i,n//i])
break
i += 1
if i < sqrtN:
divisor += 1
t += sum_divisors(n, divisor)
return t

abund = []
abundSum = []
for i in range(1, 28124):
if sum_divisors(i, 1) > i:
abund += [i]
for j in abund:
if i+j > 28123:
break
abundSum += [i+j]

print(sum(range(1,28124)) - sum(set(abundSum)))
###################
##  END
###################
from time import time
st = time()

n = 2200000000000

def sum_divisors(n, divisor):
sqrtN = n**0.5
i = divisor
t = 0
while i <= sqrtN:
if not n%i:
divisor = i
if i == n//i or i == 1:
#print(i, n//i)
t+= i
break
t += sum([i,n//i])
#print(i, n//i)
break
i += 1
if i < sqrtN:
divisor += 1
t += sum_divisors(n, divisor)
return t

def amic(n):
t =0
for i in range(n):
if sum_divisors(sum_divisors(i,1), 1)== i:
print(i)
t+=i
print(t)

def primeFactorize(n):
li = []
p = []
i = 1
while n > 1:
i += 1
if not n%i:
n = n//i
p += [i]
li += [n]
#print(li)

i -= 1
print(p)
return p

def prod(t):
t = list(t)
#print(t)
p = 1
for i in t:
p *= i
return p

abundant= 
ans = []
lab = 1
n= 37
for i in range(12,n):
p = sum_divisors(i, 1)
if p>i:
abundant += [i]
lab += 1
for j in range(lab):
if abundant[j] + i<n:
#print(abundant[j], ans)
ans += [abundant[j] + i]
else:
break

#print(lab, abundant)
ans = list(set(ans))
print((ans))

print(sum(list(range(28124))),sum(ans))

print('___________________%s__________________'%(time() - st))
st = time()
abundant = []
n= 5000
for i in range(12,n):
p = sum_divisors(i, 1)
if p>i:
abundant += [i]

#print(len(abundant), abundant)
ab, na = [],0
l = len(abundant)
for n in range(1,n):
for i in range(l):
ai = abundant[i]
if n > ai and n-ai >= ai:
try:
abundant.index(n-ai,i,n)
#print("sum of two abundant", n, i)
ab += [n]
break
except:
""
#else:
#print("not sum of 2 abundant", n, i)
#na += n
#break

print(len(ab))
#print(na)

print('___________________%s__________________'%(time() - st))
continue →

Problem 22

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?

Solution

s = s[1:len(s)-1].split('","')
s.sort()
s = '","'.join(s)
for i in range(26):
s = s.replace(chr(ord('A')+i),str(i+1)+"+")

s = s.split('","')
ans = 0
for i in range(len(s)):
ans += (i+1)*eval(s[i]+"0")

print(ans)
continue →

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.

Solution

def sum_divisors(n, divisor):
sqrtN = n**0.5
i = divisor
t = 0
while i <= sqrtN:
if not n%i:
divisor = i
if i == n//i or i == 1:
t+= i
break
t += sum([i,n//i])
break
i += 1
if i < sqrtN:
divisor += 1
t += sum_divisors(n, divisor)
return t

n = 100000
li = list(range(2,n))
for i in range(2,n):
di = sum_divisors(i,1)
if di == i:
try:
li.remove(i)
except:
pass
continue
ddi = sum_divisors(di,1)
if ddi!=i:
try:
li.remove(i)
except:
pass

print(sum(li))
continue →

Problem 20

n! means n × (n − 1) × ... × 3 × 2 × 1

For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

Solution

def fact(n):
if n > 1:
return n*fact(n-1)
else:
return 1

print(sum([int(i) for i in str(fact(100))]))
continue →

Problem 19

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)?

Solution 1

def isleap(year):
if not year%100:
if not year%400:
return 29
return 28
if not year%4:
return 29
return 28

ans = 0
days = sum([31,isleap(1900),31,30,31,30,31,31,30,31,30,31])

for y in range(1, 101):

month = 0
for m in [0,31,isleap(1900+y),31,30,31,30,31,31,30,31,30]:
month += m
if not (days+1+month)%7:
ans += 1

days += sum([31,isleap(1900+y),31,30,31,30,31,31,30,31,30,31])

print(ans)
continue →

Problem 18

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)

Solution 1: Harsh code

s='''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'''

lroute = list(map(int, s.split("\n")[len(s.split("\n"))-1].split()))
for i in s.split("\n")[len(s.split("\n"))-2::-1]:
route = list(map(int, i.split()))
at = 0
temp = []
for j in route:
if lroute[at] > lroute[at+1]:
temp += [j+lroute[at]]
#print(temp)
else:
temp += [j+lroute[at+1]]
#print(temp)
at += 1
lroute = temp

print(lroute)

Though run fast, it takes time to understand.
The output: 1074.

continue →

Problem 17

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.

Solution 1

tens = [6,6,5,5,5,7,6,6]
AND = 3
ten = 3
hundred = 7
thousand = 8

_1_9 = [3,3,5,4,4,3,5,5,4]
_11_19 = [6,6,8,8,7,7,9,8,8]
_20_99 = []
_100_999 = []

#21_99
for i in tens:
_20_99 += [i]
for j in _1_9:
_20_99 += [i+j]

#100_999
for i in _1_9:
_100_999 += [i+hundred]
for j in _1_9:
_100_999 += [i+hundred+AND+j]

_100_999 += [i+hundred+AND+ten]
for j in _11_19:
_100_999 += [i+hundred+AND+j]
for j in _20_99:
_100_999 += [i+hundred+AND+j]

for i in _1_9:

for i in _11_19:
for i in _20_99:
for i in _100_999:

print(answer + 3 + thousand)

The output is 21124.
This run:~0.

continue →

Problem 16

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

Solution 1:

So pythonic code!

print(sum(int(i) for i in str(2**1000)))

The output: 1366.
This run who cares it just a line, bro! I mean sis!

continue →

Problem 15

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?

Solution 1:

Look at the picture! How many lines do pass through specific box with number inside. The result would be the sum of all boxes and 1 which did not pass through box but upper sides.      Here is the code.

n = 20
li = []
temp = []
for i in range(n):
temp += 
li += [temp]

for j in range(n-1):
temp = []
for i in range(n, 0, -1):
temp += [sum(li[j][:i])]
temp.reverse()
li += [temp]

tot = 0
for i in range(n):
tot += sum(li[i])
print(tot + 1)

The output is: 137846528820.
This run:~ 0.017461299896240234.

continue →

Problem 14

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.

Solution: 1 Brute Forcing

Iterating through all whole numbers to 1000,000 and counting number of chain(not counting number itself).

longest = 0
l = 0

for n in range(1000000):
if l > longest:
longest = l

l = 0
while n > 1:
if n%2:
n = 3*n+1
else:
n = n//2
l += 1

The output: 837799
This run:~ 29.196775913238525. Oh alot of work is waiting to improve.

continue →

Problem 13

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

Solution 1:

Just added all 50-digits numbers. Then print the first ten. I am doing it literally. Why not?

s = '''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'''
s = s.split('\n')

for i in s:

The output is: 5537376230
This run:~ 0.00961923599243164. Fair enough.

continue →

Problem 12

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?

Solution 1: Rough answer

'''
#This method num_divisors(Num, start_divisor) only checks half of divisors
#and with out counting 1 and itself if start_divisor is > 1
#The number of full divisors is 2*ret of num_divisors
'''
def num_divisors(n, divisor):
sqrtN = n**0.5
i = divisor
c = 0
while i < sqrtN:
if not n%i:
divisor = i
c += 1
break
i += 1
if i < sqrtN:
divisor += 1
c += num_divisors(n, divisor)
return c

" Brute forcing which takes only 9.84299993515 sec"
i = 1
while True:
n = (i**2 + i)//2
num = num_divisors(n, 1)

if 2*num > 500:
break
i += 1

This run:~ 6.115420579910278 in good time. The output: 76576500

Solution 2:

I am gonna use this theorem for good!

Theorem:

The number of factors of N is equal to the product of the power of prime factorization of N + 1.

def num_factors(n):
num = n
i = 2

temp = 0
while not num%i:
num //= i
temp += 1

i = 3
while i < num**0.5:

# Reducing num to not multiple of i
temp = 0
while not num%i:
num //= i
temp += 1

i += 2

if not num**0.5%1:

" Brute forcing which takes only 9.84299993515 sec"
i = 1
while True:
n = (i**2 + i)//2
num = num_factors(n)

if num > 500:
break
i += 1

This run:~ 0.5420579910278. It is open for improvement.
The output: 76576500.

continue →

Problem 11

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?

Solution 1:

#I used formatter so that I can use 1 function for all up or right and diagonal

s = '''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'''

#This will return list of list formatted
#cod 0 in right or
#cod 1 down or
#cod 2 and cod 3 right diagonal and left diagonal
#Only this function is bulky and unorganized

def formatter(s, cod):
li = s.split("\n")
lis = []

for i in range(len(li)):
lis.insert(len(lis), list(map(int, li[i].split())))

lis1 = lis[:]
if cod:
lis2 = []
for i in range(len(lis)):
temp = []
for j in range(len(lis)):
temp += [lis[j][i]]
lis2.insert(len(lis2), temp)
lis = lis2
if cod == 2:
liss2 = []
lis1.reverse()
lis.reverse()
for i in range(len(lis)):
temp = []
temp2 = []
for j in range((i + 1)):
temp += [lis[j][i-j]]
temp2 += [lis1[j][i-j]]
liss2.insert(len(liss2), temp)
liss2.insert(len(liss2), temp2)
lis = liss2
elif cod == 3:
lis1.reverse()
for i in range(len(lis1)):
lis1[i].reverse()
liss2 = []
for i in range(len(lis)):
temp = []
temp2 = []
for j in range((i + 1)):
temp += [lis[j][i-j]]
temp2 += [lis1[j][i-j]]
liss2.insert(len(liss2), temp)
liss2.insert(len(liss2), temp2)
lis = liss2

#print("[+] Out of formatter func\n[+] Returning formatted list of len")
#print(lis)
return lis

def prod(li):
li = list(map(int, li))
return reduce(mul, li, 1)

def max_adj(s, k, i):
li = formatter(s, i)
ma = 0
trace = 0
l1 = len(li)
for i in range(l1):
li2 = li[i]
l2 = len(li2)

#if li contains list
for j in range(l2 - k + 1):
if ma < prod(li2[j:j+k]):
ma = prod(li2[j:j+k])
trace = li2[j:j+k]

#print("[+] Out of max_adj() func\n[+] Returning max product of "
#+ str(k) + " adj\n[+]found at " + str(trace)
#+ "\n" + str(ma))
return ma

cod = [0,1,2,3]
ma = []
for i in cod:
ma += [max_adj(s, 4, i)]
print("The answer is " + str(max(ma)))

This python code run :~ 0.010417938232421875
The output: 70600674

continue →

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.

Solution 1

Let’s look at the answer:

n = 3
primes = []
total = 2

# no multiples of 2
while n < 2000000:
isPrime = True
sqrt_n = n**0.5

for p in primes:

if p > sqrt_n:
break

if not n%p:
isPrime = False
break

if isPrime:
primes += [n]
total += n

n += 2

print(total)

The output: 142913828922

This run: ~5.3197691440582275

continue →

Problem 9

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Solution 1: Stupid Brute Forcing

import math

def tri(n): for i in range(1, n): for j in range(i+1, n): if i + j + math.sqrt(i**2 + j**2) == float(n): return i*j*int(math.sqrt(i**2 + j**2))

print(tri(1000))

This run: ~0.3749995231628418. It is much less stupid than I imagined at first.
The output: 31875000

Just making n to (n - i)//2 in the second for loop for making hypotanous bigger.

initial = time()

def tri(n): for i in range(1, n): for j in range(i, (n - i)//2): # changed from n if i + j + math.sqrt(i**2 + j**2) == n: return i*j*int(math.sqrt(i**2 + j**2))

print(tri(1000))

This run: ~0.1249995231628418. Halved.
The output: 31875000

Solution 2: Smartest way

[The m,n formula for generating Pythagorean Triples. Read here!](https://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#section2.3)

def tri(num): for m in range(num): for n in range(1,m): a = m**2-n**2 b = 2*m*n c = m**2+n**2 tot = a + b + c
# We got primitive which can # be multiplied to make answer if not num%tot: return (num // tot)**3 * (a * b * c)

print(tri(1000))

This run: ~0.0. That is it.
The output: 31875000

continue →

Problem 8

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?

Solution 1: Brute Forcing

Iterating through all 13 adjacents and saving maximum.

from functools import reduce from operator import mul

s = '''73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450'''
s = "".join(s.split()) s = list(map(int, s))

def prod(l): return reduce(mul, l, 1)

def max_adj(s): ma = 0
for i in range(987): if prod(s[i:i+13]) > ma: ma = prod(s[i:i+13])
return ma

This run: ~0.0. I dont know the best or the worst way to solve.
The output: 23514624000

continue →

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?

Solution 1: Brute forcing

I just stored 10001 primes in the way. I used the primes to check for the next primes.

n = 2
counter = 0
primes = []

while counter < 10001: isPrime = True
for p in primes: if not n%p: isPrime = False break
if isPrime: primes += [n] counter += 1
n += 1

print(primes[-1:])

This run: ~15.046829223632812. I feel like I am stupid. May b I am.
The output: 104743.

I assumed jumping over multiples of 2, 3 and 5 would reduce the time significantly. Guess what? It just reduce readablity. Of course, it reduced iteration one third of 1 million iteration, while the iteration is much bigger than 1 million.
Here is my try.

n = 29
counter = 0
primes = [2,3,5,7,11,13,17,19,23,29]

while counter < 10001: for i in [2,6,4,2,4,2,4,6]: n += i isPrime = True
for p in primes: if not n%p: isPrime = False break
if isPrime: primes += [n] counter += 1
print(primes)

This run: ~14.499957799911499. I feel like I am stupid. Sure I am.
The output: 104743.

Solution 2: Improved Time

I checked only half of primes to judge the number as prime. This one saved half of the time and iteration not readablity. Feeling smart.

n = 2
counter = 0
primes = []

while counter < 10001: isPrime = True
for p in primes[:counter//2]: if not n%p: isPrime = False break
if isPrime: primes += [n] counter += 1
n += 1

print(n-1)

This run: ~7.953100919723511. This is me.
The output: 104743.

We can use jumping of multiples of 2, 3 and 5 for better. right?

n = 29
counter = 0
primes = [7,11,13,17,19,23,29]

# no multiples of 2, 3, 5
while counter < 9998:
for i in [2,6,4,2,4,2,4,6]:
n += i
isPrime = True

for p in primes[:counter//5]: if not n%p: isPrime = False break
if isPrime: primes += [n] counter += 1

print(primes)

This run: ~2.5937414169311523. Challange is opportunity.
The output: 104743.

Reducing checking of prime to primes less than or equal to square root of n.

n = 29
counter = 0
primes = [7,11,13,17,19,23,29]

# no multiples of 2, 3, 5 while counter < 9998: for i in [2,6,4,2,4,2,4,6]: n += i isPrime = True sqrt_n = n**0.5
for p in primes:
# This is game changer if p > sqrt_n: break
if not n%p: isPrime = False break
if isPrime: primes += [n] counter += 1

print(primes)

This run: ~0.2968735694885254. Wow. I did not dare to think to get answer such time without using Sieve or Miller algorithm.
The output: 104743.

continue →

Problem 6

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

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Solution 1: Maths

To get sum of squares we just need to calculate for primes below 100. For (2, 4, 8, 16, ...), (3, 9, 24, ...), ... are progression with pattern. right? so we use mathematical equation.

Here is the result of progression of all primes.

To get square of sum, we just find arithmetic series and square it.

Solution 2: Brute Forcing

Brute forcing is enough here.

def square_sum(n):
return sum(map(lambda x: x**2, list(range(1,n+1))))

def sum_square(n): return sum(list(range(1,n+1)))**2

print(sum_square(100)- square_sum(100))

The code run: ~0.0s.
The output: 25164150

.

continue →

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?

Solution 1: Maths

1*2*3*4*5*6*7*8*9*10 = 3628800 can be divided by each of the numbers from 1 to 10 but not the smallest nor the greatest of this kind. 10*9 which implies it is divisible by 10, 9, 6, 5, 3, 2, 1. We only need to multiply it by 4 to make it divisible by 8. So 10*9*4*7 is divisible for number between 1 to 11.
Similarly: 20*19*9*17*4*7*13*11 = 232792560 would be divisible by number between 1 and 21. That is the answer.

Solution 2: Brute Forcing

This is when you dont want to use maths.

n = 20
found = 0

while not found: for i in range(1,21): if n%i: break found = i//20
n += 1
print(n-1)

This code run: ~484.1459057331085s.
The output: 232792560. Still correct.

Note: We can reduce redundancy of checking for 2, 4, 8 while we know it is divisible by 16. We may iterate for multiples of 20 cause the number we want is also multiples of 20.

continue →

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 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

Note: It takes some trial, maths equation & observation to do it manually. You do the maths.

Solution 1: Brute Forcing

Let's iterate the variable a & b through all three digit numbers(100-999) and check if it is palindrome and largest.

def isPalindrome(n):
return n == int(str(n)[::-1])

largest = 0
for a in range(100, 1000): for b in range(100, 1000): if isPalindrome(a*b) and largest < a*b: largest = a*b

print(largest)

This code run: ~1.453120231628418.
The output: 906609

Now improving is so interesting. Did you see the redundancy! we checked when (a = 991 & b = 100) and (a = 100 & b = 991). C'mon multiplication is commutative. This will reduce the time by half.

def isPalindrome(n):
return n == int(str(n)[::-1])

largest = 0 for a in range(100, 1000):
# We reduce half of redundancy by starting from a for b in range(a, 1000): if isPalindrome(a*b) and largest < a*b: largest = a*b

print(largest)

This code run: ~0.7499971389770508. Reduced by half.
The output: 906609

Solution 2: Improved

Why just we start from greatest three digit number?

def isPalindrome(n):
return n == int(str(n)[::-1])

ans = 0
for a in range(999, 99, -1): for b in range(a, 99, -1): if isPalindrome(a*b): if ans < a*b: ans = a*b
break print(ans)

This code run: ~0.5156266689300537. Not much.
The output: 906609

You know when previous a is greater than a b must greater than previous b to be in the business. So let's make b greater than the previous b

def isPalindrome(n):
return n == int(str(n)[::-1])

ans = 0 low = 99
for a in range(999, 99, -1): for b in range(a, low, -1): if isPalindrome(a*b): if ans < a*b: ans = a*b
low = b # Min a and max b may exceed break # max a and min b.

print(ans)

This code run: ~0.015625476837158203. This is my best record.
The output: 906609

continue →

Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

Solution 1: Brute forcing

It is wasting to iterate through all natural numbers to 600851475143 and check if divisible by iter and iter is prime. So we just iterate through all natuaral numbers to square root of 600851475143 since two primes greater than the square root is also greater than the number.
So,

num = 600851475143
i = 2

while i <= num**0.5:

Now we need to check if the number is divisible by iteration. If we get a number that divedes, reduce our question to lowest number and our iteration continue from there.
Here is the code

num = 600851475143
i = 2

while i <= num**0.5:

# Python would optimize this!
# Removing would make the code
# slower but readable.
while num%i:
i += 1

# Reducing num to not multiple of i
while num%i == 0:
num //= i

i += 1

print(num)

The output: 6857.
This run ~0.0s.

Solution 2: Improved

The above took 1473 iterations before getting the answer. The number iterations depends on number of divisors that number have. In the worst case, when the number is prime number of iterations is equal to the number itself(not even its square root?). We can improve the worst case(at least to square root) and number of iterations by jumping over divisible of 2 which will reduce the number of iterations by half.

n = 600851475143
num = n

# To jump over multiples of 2.
# Since num would be not multiple of 2,
# multiples of 2 cant be factor
i = 2
while not num%i:
num //= i

i = 3
while i <= num**0.5:
# Reducing num to not multiple of i
while not num%i:
num //= i

i += 2

if num ==  n:
print("No Factor!")
else:
print(num)

continue →

Problem 2

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.

Solution 1: Brute forcing

By just iterating through all fibonnacci sequence below 4,000,001 (31 iteration) and checking wether even-valued and adding to the variable, we can find the answer.

The python code:

total = 0
pre_f = 1
next_f = 2

while next_f < 4000001: if next_f % 2 == 0: total += next_f
#swaping next_f to pre_f #and making next_f the sum of #pre_f & next_f temp_f2 = next_f next_f = next_f + pre_f pre_f = temp_f2
print(total)

The output: 4613732.
This code run ~0.0.

Solution 2: Improved Iteration

We may not improve the time it took to run but we can reduce the number of iteration from 31 to 11. It takes sum of 2 even numbers or 2 odd numbers to be even. We can see that fibonacci sequence term is even @ terms: 2, 5, 8,11, 14. After two odd valued term we get even valued term.

total = 0
f = 1
even_f = 2

while even_f < 4000001: total += even_f
# finding next even term next_f = f + even_f f = next_f + even_f even_f = next_f + f
print(total)

The output: 4613732.
This code run ~0.0.

continue →

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.

Solution 1: Mathematical View.

Multiples of 3: 3, 6, 9, 12, 15, ... Multiples of 5: 5, 10, 15, 20, 25, ... Multiples of 15(3 & 5): 15, 30, 45, 60, 75, ...

So we were asked the sum of arithmetic progression - arithmetic series. This can be solved by the equation of arithmetic series. You can read it here. Arithmetic progression.

Using the equation of arithmetic series: The arithmetic series of multiples of 3 below 1000 is 333 * (3 + 999) / 2 = 166833.0. The arithmetic series of multiples of 5 below 1000 is 199 * (5 + 995) / 2 = 99500. The arithmetic series of multiples of 15 below 1000 is 66 * (15 + 990) / 2 = 33165. We are not asked to add multiples of 15 but we need to subtruct multiples of 15 since it was counted on both side.

So the answer is summing the sum of multiples of 3 and 5 and subtructing sum multiples of 15.

The output: 233168

This only takes pencil and paper. May be calculator.

Solution 2: Brute Forcing.

Let's go through all natural numbers below 1000 and add to our variable if it is divisible by 3 or 5.

Here is my code in python:

total = 0

for n in range(1, 1000): if n%3 == 0 or n%5 == 0: total += n
print(total)

The output: 233168

This code run ~0.0. So may be there is not the room for improvement!

Solution 3: Improved Brute Forcing.

We can improve this by making the iteration to the candidates. By iterating through multiples of 15 and adding the pattern of multiples of 3 and 5, we will reduce the iteration from 999 to 66.

Improved code:

total = 0

for n in range(0, 990, 15): total += 30 + 4*n total += 30 + 3*n
total += 993 + 996 + 999 total += 995
print(total)

The output: 233168

This code run ~0.0

continue →

Project Euler It is a website dedicated to a series of computational problems intended to be solved with computer programs. The project attracts adults and students interested in mathematics and computer programming. Since its creation in 2001 by Colin Hughes, Project Euler has gained notability and popularity worldwide. It includes over 600 problems, with a new one added once every two weeks. I have compiled my try for sharing.

continue →