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]
s = 0 (1..1000).inject { |s, x| s + x ** x } % (10 ** 10) str = s.to_s puts str[str.length - 10,str.length]
The Fibonacci sequence is defined by the recurrence relation:
F_(n) = F_(nā1) + F_(nā2), 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
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
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
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]
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
My solution in Ruby:
def is_prime ( p ) if p == 2 return true elsif p <= 1 || p % 2 == 0 return false else (3 .. Math.sqrt(p)).step(2) do |i| if p % i == 0 return false end end return true end end sum = 0 for i in 2..2000000 if is_prime(i) sum += i end end puts sum
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
My solution in Ruby:
var = %& 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450& arr = var.split( // ) arr.delete "\n" big = 0; for i in 0..arr.length - 5 next if Integer(arr[i]) == 0 tmp = Integer(arr[i]) 1.upto(4) { |j| tmp = tmp * Integer(arr[i + j]) } big = tmp if tmp > big end puts big
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^(th) prime is 13.
What is the 10001^(st) prime number?
My solution in Ruby:
def is_prime ( p ) if p == 2 return true elsif p <= 1 || p % 2 == 0 return false else (3 .. Math.sqrt(p)).step(2) do |i| if p % i == 0 return false end end return true end end prime_count = 6 prime_number = 13 number = 13 while prime_count < 10001 do number += 2 if is_prime(number) prime_count += 1 prime_number = number end end puts '***********' puts "#{prime_count}: #{prime_number}"
The sum of the squares of the first ten natural numbers is,
1^(2) + 2^(2) + ... + 10^(2) = 385The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^(2) = 55^(2) = 3025Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 ā 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
My solution in Ruby:
def sum_and_square(j, k) tmp = 0 for i in 1..100 tmp += i**j end tmp**k end puts sum_and_square(1,2) - sum_and_square(2,1)
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 number that is evenly divisible by all of the numbers from 1 to 20?
My solution in Ruby:
def has_remainder?(var) 1.upto(20) { |i| return true if var % i != 0 } false end number = 0 loop do number += 1 break if !has_remainder?(number) end puts numbe