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.
function Mult3or5_below(num)
sum_all = 0
for i = 1:num
if ((i%5 == 0) || (i%3 == 0))
sum_all += i
end
end
return sum_all
end
Mult3or5_below (generic function with 1 method)
Mult3or5_below(999)
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.
function SumFibEvens_below(num)
fib_num_prev = 1
fib_num = 1
fib_sum = 0
answer = 0
while fib_sum < num
fib_sum = fib_num_prev + fib_num
fib_num_prev = fib_num
fib_num = fib_sum
if fib_sum%2 == 0
answer += fib_sum
end
end
return answer
end
SumFibEvens_below (generic function with 1 method)
SumFibEvens_below(4000000)
4613732
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
function LargestPrimeFactor(num)
factor_dict = factor(num)
return maximum(factor_dict)[1]
end
LargestPrimeFactor (generic function with 1 method)
LargestPrimeFactor(600851475143)
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.
function IsPalindrome(num1, num2)
product = string(num1 * num2)
return product == reverse(product)
end
IsPalindrome (generic function with 1 method)
IsPalindrome(91,99)
true
function MaxPalindrome_digits(num)
min_num = 10^(num-1)
max_num = int(repeat("9",num))
product = 0
answer = 0
for x = min_num:max_num
for y = min_num:max_num
product = x * y
if IsPalindrome(x,y) && product > answer
answer = product
end
end
end
return answer
end
MaxPalindrome_digits (generic function with 1 method)
MaxPalindrome_digits(3)
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?
function SmallestMult(num)
range_list = Int64[]
for i = 1:num
push!(range_list,i)
end
answer = lcm(range_list...)
return answer
end
SmallestMult (generic function with 1 method)
SmallestMult(20)
232792560
lcm(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
232792560
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.
function SumSqDif(num)
sumsquares = 0
runningsum = 0
for i = 1:num
sumsquares += i^2
runningsum += i
end
squaresum = runningsum^2
answer = squaresum - sumsquares
return answer
end
SumSqDif (generic function with 1 method)
SumSqDif(100)
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 10001st prime number?
function Prime(n)
x = 2
i = 0
answer = 0
while i < n
if isprime(x)
answer = x
i = i + 1
end
x = x + 1
end
return answer
end
Prime (generic function with 1 method)
Prime(10001)
104743
Find the greatest product of five consecutive digits in the 1000-digit number.
x = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
"7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
function GrtstProdofFive(input_string)
answer = 0
for i = 1:length(input_string)-4
product = input_string[i] - '0' # subtract '0' to convert char into int
for k = 1:4
product = product * (input_string[i+k] - '0')
end
if product > answer
answer = product
end
end
return answer
end
GrtstProdofFive (generic function with 1 method)
GrtstProdofFive(x)
40824
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
function PythagTriple()
answer = 0
for a = 1:1000
for b = a:1000
c = sqrt(a^2 + b^2)
if (a+b+c) == 1000
answer = a*b*c
return int(answer)
end
end
end
end
PythagTriple (generic function with 1 method)
PythagTriple()
31875000
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
function Prime2(n)
x = 1
answer = 2
while x < n
if isprime(x)
answer += x
end
x = x + 2
end
return answer
end
Prime2 (generic function with 1 method)
Prime2(2000000)
142913828922
In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
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?
My2DArray = [[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]]
20x20 Array{Int64,2}: 8 2 22 97 38 15 0 40 0 … 5 7 78 52 12 50 77 91 8 49 49 99 40 17 81 18 57 60 40 98 43 69 48 4 56 62 0 81 49 31 73 55 79 14 29 93 67 53 88 30 3 49 13 36 65 52 70 95 23 4 60 11 42 69 56 1 32 56 71 37 2 36 91 22 31 16 71 51 67 63 89 41 54 22 40 40 28 66 33 13 80 24 47 32 60 99 3 45 2 44 … 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 67 59 54 70 66 18 38 64 70 67 26 20 68 2 62 12 20 95 39 63 8 40 91 66 49 94 21 24 55 58 5 66 73 99 26 97 78 96 83 14 88 34 89 63 72 21 36 23 9 75 0 76 44 20 14 0 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 … 80 4 62 16 14 9 53 56 92 16 39 5 42 96 35 31 47 55 24 0 17 54 24 36 29 85 57 86 56 0 48 35 71 89 7 5 37 44 60 21 58 51 54 17 58 19 80 81 68 5 94 47 69 28 13 86 52 17 77 4 89 55 40 4 52 8 83 97 35 99 16 7 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 3 … 67 46 55 12 32 63 93 53 69 4 42 16 73 38 25 39 11 24 18 8 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 69 82 67 59 85 74 4 36 16 20 73 35 29 78 31 90 1 74 71 48 86 81 16 23 57 5 54 1 70 54 71 83 51 54 69 16 48 61 43 52 1 89 19 67 48
function FourProd(array_2d)
enum = size(array_2d)
transpose_array = array_2d'
rotate_array = rotl90(array_2d)
answer = 0
for x = 1:enum[1]-4
for y = 1:enum[2]-4
product1 = array_2d[x,y]
product2 = transpose_array[x,y]
product3 = array_2d[x,y]
product4 = rotate_array[x,y]
for z = 1:3
h = x + z
v = y + z
product1 *= array_2d[h,y] # search for up/down product
product2 *= transpose_array[h,y] # search for left/right product
product3 *= array_2d[h,v] # search for diagonal \ product
product4 *= rotate_array[h,v] # search for diagonal / product
end
max_product = max(product1, product2, product3, product4)
if max_product > answer
answer = max_product
end
end
end
return answer
end
FourProd (generic function with 1 method)
FourProd(My2DArray)
70600674
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:
Let us list the factors of the first seven triangle numbers:
What is the value of the first triangle number to have over five hundred divisors?
function NumDivisors(n)
answer = 0
x = floor(sqrt(n))
for i=1:x
if n%i == 0
answer += 2
end
end
if n/x == x
answer -= 1
end
return answer
end
function TriangleDivisors(n)
count_div = 0
counter = 1
answer = 0
while count_div < n
answer += counter
count_div = NumDivisors(answer)
counter += 1
end
return answer
end
TriangleDivisors (generic function with 1 method)
TriangleDivisors(500)
76576500
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
LargeNumString = "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"
"37107287533902102798797998220837590246510135740250\n46376937677490009712648124896970078050417018260538\n74324986199524741059474233309513058123726617309629\n91942213363574161572522430563301811072406154908250\n23067588207539346171171980310421047513778063246676\n89261670696623633820136378418383684178734361726757\n28112879812849979408065481931592621691275889832738\n44274228917432520321923589422876796487670272189318\n47451445736001306439091167216856844588711603153276\n70386486105843025439939619828917593665686757934951\n62176457141856560629502157223196586755079324193331\n64906352462741904929101432445813822663347944758178\n92575867718337217661963751590579239728245598838407\n58203565325359399008402633568948830189458628227828\n80181199384826282014278194139940567587151170094390\n35398664372827112653829987240784473053190104293586\n86515506006295864861532075273371959191420517255829\n71693888707715466499115593487603532921714970056938\n54370070576826684624621495650076471787294438377604\n53282654108756828443191190634694037855217779295145\n36123272525000296071075082563815656710885258350721\n45876576172410976447339110607218265236877223636045\n17423706905851860660448207621209813287860733969412\n81142660418086830619328460811191061556940512689692\n51934325451728388641918047049293215058642563049483\n62467221648435076201727918039944693004732956340691\n15732444386908125794514089057706229429197107928209\n55037687525678773091862540744969844508330393682126\n18336384825330154686196124348767681297534375946515\n80386287592878490201521685554828717201219257766954\n78182833757993103614740356856449095527097864797581\n16726320100436897842553539920931837441497806860984\n48403098129077791799088218795327364475675590848030\n87086987551392711854517078544161852424320693150332\n59959406895756536782107074926966537676326235447210\n69793950679652694742597709739166693763042633987085\n41052684708299085211399427365734116182760315001271\n65378607361501080857009149939512557028198746004375\n35829035317434717326932123578154982629742552737307\n94953759765105305946966067683156574377167401875275\n88902802571733229619176668713819931811048770190271\n25267680276078003013678680992525463401061632866526\n36270218540497705585629946580636237993140746255962\n24074486908231174977792365466257246923322810917141\n91430288197103288597806669760892938638285025333403\n34413065578016127815921815005561868836468420090470\n23053081172816430487623791969842487255036638784583\n11487696932154902810424020138335124462181441773470\n63783299490636259666498587618221225225512486764533\n67720186971698544312419572409913959008952310058822\n95548255300263520781532296796249481641953868218774\n76085327132285723110424803456124867697064507995236\n37774242535411291684276865538926205024910326572967\n23701913275725675285653248258265463092207058596522\n29798860272258331913126375147341994889534765745501\n18495701454879288984856827726077713721403798879715\n38298203783031473527721580348144513491373226651381\n34829543829199918180278916522431027392251122869539\n40957953066405232632538044100059654939159879593635\n29746152185502371307642255121183693803580388584903\n41698116222072977186158236678424689157993532961922\n62467957194401269043877107275048102390895523597457\n23189706772547915061505504953922979530901129967519\n86188088225875314529584099251203829009407770775672\n11306739708304724483816533873502340845647058077308\n82959174767140363198008187129011875491310547126581\n97623331044818386269515456334926366572897563400500\n42846280183517070527831839425882145521227251250327\n55121603546981200581762165212827652751691296897789\n32238195734329339946437501907836945765883352399886\n75506164965184775180738168837861091527357929701337\n62177842752192623401942399639168044983993173312731\n32924185707147349566916674687634660915035914677504\n99518671430235219628894890102423325116913619626622\n73267460800591547471830798392868535206946944540724\n76841822524674417161514036427982273348055556214818\n97142617910342598647204516893989422179826088076852\n87783646182799346313767754307809363333018982642090\n10848802521674670883215120185883543223812876952786\n71329612474782464538636993009049310363619763878039\n62184073572399794223406235393808339651327408011116\n66627891981488087797941876876144230030984490851411\n60661826293682836764744779239180335110989069790714\n85786944089552990653640447425576083659976645795096\n66024396409905389607120198219976047599490197230297\n64913982680032973156037120041377903785566085089252\n16730939319872750275468906903707539413042652315011\n94809377245048795150954100921645863754710598436791\n78639167021187492431995700641917969777599028300699\n15368713711936614952811305876380278410754449733078\n40789923115535562561142322423255033685442488917353\n44889911501440648020369068063960672322193204149535\n41503128880339536053299340368006977710650566631954\n81234880673210146739058568557934581403627822703280\n82616570773948327592232845941706525094512325230608\n22918802058777319719839450180888072429661980811197\n77158542502016545090413245809786882778948721859617\n72107838435069186155435662884062257473692284509516\n20849603980134001723930671666823555245252804609722\n53503534226472524250874054075591789781264330331690"
num_array = split(LargeNumString, "\n")
A = [ float(num_array[i]) for i=1:length(num_array) ]
answer = sum(A)
5.537376230390877e51
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:
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
function CollatzSeq(n)
A = [n]
while n > 1
n = iseven(n)? int(n/2) : 3*n + 1
push!(A, n)
end
return A
end
CollatzSeq (generic function with 1 method)
begin_time = time()
answer = []
seq = []
for i = 700000:1000000
seq = CollatzSeq(i)
length(seq) >= length(answer)? answer = seq : pass
end
print("runtime: ", time() - begin_time)
return answer[1]
runtime: 6.568037033081055
837799
begin_time = time()
seqs = map((x) -> length(CollatzSeq(x)), 700000:1000000)
print("runtime: ",time() - begin_time)
return 700000+indmax(seqs)-1
runtime: 3.7657041549682617
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?
function LatticeRecursive(x, y)
if x == 1.0 && y == 1.0
return 2.0
elseif x == 1.0
return 1.0 + LatticeRecursive(x, y-1.0)
elseif y == 1.0
return 1.0 + LatticeRecursive(x-1.0, y)
else
return LatticeRecursive(x-1.0, y) + LatticeRecursive(x, y-1.0)
end
end
LatticeRecursive (generic function with 1 method)
function Lattice(x, y)
answer = zeros(x+1, y+1)
for i = 1:y+1
for j = 1:x+1
if i == 1.0 || j == 1.0
answer[i, j] = 1.0
else
answer[i, j] = answer[i-1, j] + answer[i, j-1]
end
end
end
return answer[x+1, y+1]
end
Lattice (generic function with 1 method)
big(int((Lattice(20,20))))
137846528820
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?
function PowerDigit(x)
n = string(big(2)^x)
answer = 0
for i= 1:length(n)
answer += int(n[i] - '0') # subtract '0' to convert char into int
end
return answer
end
PowerDigit (generic function with 1 method)
PowerDigit(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.
function LetterCounts(n)
l = {
1 => "one",
2 => "two",
3 => "three",
4 => "four",
5 => "five",
6 => "six",
7 => "seven",
8 => "eight",
9 => "nine",
10 => "ten",
11 => "eleven",
12 => "twelve",
13 => "thirteen",
14 => "fourteen",
15 => "fifteen",
16 => "sixteen",
17 => "seventeen",
18 => "eighteen",
19 => "nineteen",
20 => "twenty",
30 => "thirty",
40 => "forty",
50 => "fifty",
60 => "sixty",
70 => "seventy",
80 => "eighty",
90 => "ninety",
1000 => "onethousand"
}
# adds:
# 100, 200, 300...
# 101, 102, 103...
# 201, 202, 203...
# 110, 120, 130...
for h=1:9
l[(h*100)] = l[h] * "hundred"
for ones=1:9
l[(h*100) + ones] = l[h*100] * "and" * l[ones]
l[(h*100) + (ones*10)] = l[h*100] * "and" * l[ones*10]
end
end
# adds:
# 21, 22, 23...
# 31, 32, 33...
# 111, 112, 113...
# 121, 122, 123...
# 231, 232, 233...
for ones=1:9
for tens=2:9
l[(tens*10) + ones] = l[tens*10] * l[ones]
for h=1:9
l[(h*100) + (tens*10) + ones] = l[h*100] * "and" * l[tens*10] * l[ones]
l[(h*100) + (10) + ones] = l[h*100] * "and" * l[10 + ones]
end
end
end
return length(l[n])
end
answer = 0
for x=1:1000
answer += LetterCounts(x)
end
return answer
21124
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)
function MakeTriangle()
triangle = zeros(Int64, 15, 15)
triangle[1,1] = 75
triangle[2,1:2] = [95 64]
triangle[3,1:3] = [17 47 82]
triangle[4,1:4] = [18 35 87 10]
triangle[5,1:5] = [20 04 82 47 65]
triangle[6,1:6] = [19 01 23 75 03 34]
triangle[7,1:7] = [88 02 77 73 07 63 67]
triangle[8,1:8] = [99 65 04 28 06 16 70 92]
triangle[9,1:9] = [41 41 26 56 83 40 80 70 33]
triangle[10,1:10] = [41 48 72 33 47 32 37 16 94 29]
triangle[11,1:11] = [53 71 44 65 25 43 91 52 97 51 14]
triangle[12,1:12] = [70 11 33 28 77 73 17 78 39 68 17 57]
triangle[13,1:13] = [91 71 52 38 17 14 91 43 58 50 27 29 48]
triangle[14,1:14] = [63 66 04 68 89 53 67 30 73 16 69 87 40 31]
triangle[15,1:15] = [04 62 98 27 23 09 70 98 73 93 38 53 60 04 23]
return triangle
end
function MaxPathSum(triangle)
sums = zeros(Int64, size(triangle))
sums[1,1] = triangle[1,1]
rows, cols = size(triangle)
for row = 2:rows
for col = 1:row
val = triangle[row,col]
if col == 1
sums[row,col] = val + sums[row-1,col]
else
sums[row,col] = val + max(sums[row-1,col], sums[row-1,col-1])
end
end
end
return maximum(sums[end,1:end])
end
MaxPathSum (generic function with 2 methods)
MaxPathSum(MakeTriangle())
1074
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 in triangle.txt, a 15K text file containing a triangle with one-hundred rows.
NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
function MakeBigTriangle()
big_triangle = open("triangle.txt")
lines = readlines(big_triangle)
rows = length(lines)
triangle = zeros(Int64, rows, rows)
for x=1:rows
triangle[x,1:x] = [int(split(lines[x], ' '))]
end
return triangle
end
MakeBigTriangle (generic function with 1 method)
MaxPathSum(MakeBigTriangle())
7273
You are given the following information, but you may prefer to do some research for yourself.
1 Jan 1900 was a Monday.
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)?
first_sundays = 0
months = [
31, #jan
28, #feb
31, #mar
30, #apr
31, #may
30, #jun
31, #jul
31, #aug
30, #sep
31, #oct
30, #nov
31 #dec
]
day = 3 # jan 1, 1901 is a tue (3)
m = 1 # m = month in year
for m_range = 1:1200 # m_range = total months elapsed
yr = floor((m_range - 1) / 12) + 1901
if day == 1
# print("$m/$yr: $day\n")
first_sundays += 1
end
if (m == 2) && (yr % 4 == 0) && ((yr % 100 != 0) || (yr % 400 == 0))
day += 1 # leap year
end
drift = months[m] - 28
day = ((day + drift) % 7 == 0) ? 7 : (day + drift) % 7
m = (m_range % 12 == 0) ? 1 : m + 1
end
return first_sundays
171
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!
sum(digits(factorial(big(100)))) # digits does not work on BigInt
no method convert(Type{Unsigned},BigInt) at In[45]:1 in digits at intfuncs.jl:289 in digits at intfuncs.jl:288
function FactDigitSum(n)
fact = string(reduce(*, 1:big(n)))
answer = 0
for i= 1:length(fact)
answer += int(fact[i] - '0') # subtract '0' to convert char into int
end
return answer
end
FactDigitSum (generic function with 1 method)
FactDigitSum(100)
648
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.
function ProperDivisor(n)
divisors = [1]
n_sqrt = floor(sqrt(n))
for i = 2:n_sqrt
if n % i == 0
if n/i == i
push!(divisors, i)
else
push!(divisors, i)
push!(divisors, n/i)
end
end
end
return divisors
end
function Amicable(n)
sum1 = sum(ProperDivisor(n))
sum2 = sum(ProperDivisor(sum1))
return sum2 == n && sum1 != sum2
end
amicables = Int64[]
for num = 2:10000
if Amicable(num)
print(num, " ")
push!(amicables, num)
end
end
return sum(amicables)
220 284 1184 1210 2620 2924 5020 5564 6232 6368
31626
Using names.txt, 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?
alpha_score = Dict{Char,Int64}()
for character = 1:26
alpha_score[(char(character + 64))] = character
end
name_list = open(readall, "names.txt")
names = sort(split(replace(name_list, '"',""), ","))
score = 0
for name = 1:length(names)
name_score = 0
for letter = 1:length(names[name])
name_score += alpha_score[names[name][letter]]
end
name_score *= name
score += name_score
end
return score
871198282
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.
begin_time = time()
abundant_nums = [12]
for n = 13:20161
if sum(ProperDivisor(n)) > n
push!(abundant_nums, n)
end
end
answer = [1:23]
for i = 24:20161
found = false
for num in abundant_nums
if (i - num) in abundant_nums
found = true
break
end
end
if found == false
answer = push!(answer, i)
end
end
print("runtime: ", time() - begin_time)
return sum(answer)
runtime: 129.338219165802
4179871
# scratchpad
ab_num_sums = Array(Int64, 1)
for num in abundant_nums
sums_array = pmap((x) -> x+num, abundant_nums)
ab_num_sums = cat(1, ab_num_sums, sums_array)
end
answer = [1]
for i = 2:28123
answer = sum(push!(i in ab_num_sums? i : pass))
end
return answer
abundant_nums not defined at In[31]:6 in anonymous at no file
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:
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
sum(map(x -> 10^(x[1]-1) * x[2], enumerate(nthperm([0:9], 1000000))))
645193872
The Fibonacci sequence is defined by the recurrence relation:
The 12th term, F12, is the first term to contain three digits.
What is the first term in the Fibonacci sequence to contain 1000 digits?
function Fib_num_digits(n)
fib_seq = ["1","1"]
while length(fib_seq[end]) < n
push!(fib_seq, string(BigInt(fib_seq[end-1]) + BigInt(fib_seq[end])))
# print(fib_seq[end], " ")
end
return length(fib_seq)
end
Fib_num_digits (generic function with 1 method)
begin_time = time()
answer = Fib_num_digits(1000)
print("runtime: ", time() - begin_time)
return answer
runtime: 0.20959210395812988
4782
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.
function ReciprocalCycles(n)
seq_len = 0
answer = 0
for i = 2:n
remainders = zeros(Int64,i)
value = 1
len = 0
while value < i
value *= 10
end
while value != 0
if value in remainders
break
end
push!(remainders, value)
while (value < i)
value *= 10
len += 1
end
value %= i
end
if value != 0 && len > seq_len
seq_len = len
answer = i
end
end
return answer, seq_len
end
ReciprocalCycles (generic function with 1 method)
begin_time = time()
answer = ReciprocalCycles(1000)
print("runtime: ", time() - begin_time)
return answer
runtime: 0.3948509693145752
(983,982)
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 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.
function QuadraticPrimes()
max_len = 0
best_a = 0
best_b = 0
for a = -999:999
for b = -999:999
if isprime(b)
n = 0
while isprime(abs((n^2) + (a*n) + b))
n += 1
end
if n > max_len
max_len = n
best_a = a
best_b = b
end
end
end
end
return best_a * best_b
end
QuadraticPrimes (generic function with 1 method)
begin_time = time()
answer = QuadraticPrimes()
print("runtime: ", time() - begin_time)
return answer
runtime: 6.396510124206543
-59231
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?
function NumSpiralDiag(n)
i = 1
result = 1
while i < n
i += 2
result = result + (i^2 * 4) - ((i-1) * 6)
end
return result
end
NumSpiralDiag (generic function with 2 methods)
begin_time = time()
answer = NumSpiralDiag(1001)
print("runtime: ", time() - begin_time)
return answer
runtime: 0.005847930908203125
669171001
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?
function GeneratePrimes(n)
primes = Int[]
for i = 1:n
if isprime(i)
push!(primes, i)
end
end
return primes
end
GeneratePrimes (generic function with 1 method)
function ConsecutivePrimes(n)
primes_list = GeneratePrimes(n/250)
l = length(primes_list)
count = 0
prime = 0
count_result = 0
prime_result = 0
for i = 1:l
for j = 1:l
x = l-j
count = x-i
if count > count_result
num = sum({primes_list[y] for y=i:x})
if num < n
if isprime(num)
count_result = count
prime_result = num
print("Result: $prime_result, $count_result\n")
end
end
end
end
end
return prime_result, count_result
end
ConsecutivePrimes (generic function with 1 method)
begin_time = time()
answer = ConsecutivePrimes(1000000)
print("runtime: ", time() - begin_time)
return answer
Result: 958577, 535 Result: 978037, 538 Result: 997651, 542 runtime: 0.0333561897277832
(997651,542)
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?
function DistinctPowers(a, b)
matrix = zeros(BigFloat, a, b)
matrix = [ BigFloat(row)^col for row=2:a, col=2:b ]
return length(unique(matrix))
end
DistinctPowers (generic function with 1 method)
begin_time = time()
answer = DistinctPowers(100, 100)
print("runtime: ", time() - begin_time)
return answer
runtime: 0.6840958595275879
9183