Sep 23 2010
Project Euler Problem 52 Solution
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
My solution in Ruby:
i = 1; answer = nil; while answer == nil i += 1 x = i * 2 base = Array.new(x.to_s.scan(/./)) matchbase = true for t in 3..6 break if matchbase == false x = i * t digits = Array.new(x.to_s.scan(/./)) digits.each { |d| matchbase = false if base.index(d) == nil } end answer = i if matchbase == true end puts "Answer: #{answer}"
Apr 9 2009
Project Euler Problem 48 Solution
The series, 1^(1) + 2^(2) + 3^(3) + ... + 10^(10) = 10405071317.
Find the last ten digits of the series, 1^(1) + 2^(2) + 3^(3) + ... + 1000^(1000).
My solution in Ruby:
sum = 0 for i in 1..1000 do sum += i**i end str = sum.to_s puts str[str.length - 10,str.length]
UPDATE
s = 0 (1..1000).inject { |s, x| s + x ** x } % (10 ** 10) str = s.to_s puts str[str.length - 10,str.length]
Apr 9 2009
Project Euler Problem 28 Solution
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 13It 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?
My solution in Ruby:
corners = [1] num = 1 add = 2 500.times do 4.times do num += add corners << num end add += 2 end sum = 0 corners.each { |c| sum += c } puts sum
Apr 9 2009
Project Euler Problem 25 Solution
The Fibonacci sequence is defined by the recurrence relation:
F_(n) = F_(n−1) + F_(n−2), where F_(1) = 1 and F_(2) = 1.
Hence the first 12 terms will be:
F_(1) = 1
F_(2) = 1
F_(3) = 2
F_(4) = 3
F_(5) = 5
F_(6) = 8
F_(7) = 13
F_(8) = 21
F_(9) = 34
F_(10) = 55
F_(11) = 89
F_(12) = 144The 12th term, F_(12), is the first term to contain three digits.
What is the first term in the Fibonacci sequence to contain 1000 digits?
My solution in Ruby:
t1, t2, term = 1, 2, 3 loop do temp = t1 + t2 t1 = t2 t2 = temp term += 1 temp_s = temp.to_s break if temp_s.length >= 1000 end puts term
Apr 8 2009
Project Euler Problem 20 Solution
n! means n x (n - 1) x ... x 3 x 2 x 1
Find the sum of the digits in the number 100!
My solution in Ruby:
sum = 0 factorial = 1 for i in 1..100 factorial = factorial * i end str = factorial.to_s y = str.scan(/./) y.each do |c| sum += c.to_i end puts sum
Apr 8 2009
Project Euler Problem 17 Solution
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.
My solution in Ruby:
@@words = { 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",100 => "hundred", 1000 => "thousand",0 => "" } count = 0; def one_to_ninetynine(base) icount = 0 for i in 1..19 do icount += base + @@words[i].length end j = 10 until j == 90 j += 10 icount += base + @@words[j].length for k in 1..9 do icount += base + @@words[j].length + @@words[k].length end end icount end count += one_to_ninetynine(0) for l in 1..9 do count += @@words [l].length + (@@words[100].length) count += one_to_ninetynine(@@words[l].length + (@@words[100].length) + 3) end count += @@words[1].length + @@words[1000].length puts count
Apr 8 2009
Project Euler Problem 16 Solution
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)?
My solution in Ruby:
sum, number = 0, 2**1000 str = number.to_s y = str.scan(/./) y.each do |c| sum += c.to_i end puts sum
Apr 8 2009
Project Euler Problem 14 Solution
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.
My solution in Ruby:
highchain = {} high = 0 1000000.downto(1) do |i| links = 0 result = i loop do links += 1 if result.odd? result = (result * 3) + 1 else result = result / 2 end break if result == 1 end highchain = {i, links} if links > high high = links if links > high end highchain.each { |k,v| puts "#{k} has the longest chain of #{v} links!"}
Apr 8 2009
Project Euler Problem 13 Solution
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
(numbers omitted)
My solution in Ruby:
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 ] sum = 0 numbers.each do |i| sum += i end sum_s = sum.to_s puts sum_s[0,10]
Apr 8 2009
Project Euler Problem 12 Solution
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?
My solution in Ruby:
def calc_divisors(num) res=[1] 2.upto(Math.sqrt(num).floor) do |i| if num % i == 0 res << i end end res.reverse.each do |i| res << num / i end res.uniq end triangle = 0 count = 0 while calc_divisors(triangle).count < 500 count += 1 triangle += count end puts triangle
This code takes a little while to run, and so far, this is my favourite Project Euler Solution. I did originally try brute-forcing the divisors for each number, but as you can imagine, testing half of every possible candidate for numbers in the millions meant that the loops were getting too long. Interestingly, I never let that original method find the answer so instead I hit google, looking for a faster way. I discovered a lot about sieves and old math geniuses, but the the best example I found was using prime factorials. The function 'calc_divisors' in the above code seems to be the fastest way I could find to determine the factors of a specified Integer. If you think there is a faster way, I'd be interested to hear.