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Miki Tebeka committed 97c594b

All under one roof

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+syntax: glob
+
+go/[0-9]+$
+haskell/*.hi
+csharp/*.exe
+
+; 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.
+;
+; Answer: 233168
+
+(defn pred? [n]
+ (or (zero? (mod n 3))
+     (zero? (mod n 5))))
+
+(println (apply + (filter pred? (range 1000))))
+; The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
+; 
+; Find the sum of all the primes below two million.
+;
+; Answer: 142913828922
+
+(defn primes []
+  (filter (fn [n] (.isProbablePrime (bigint n) 10)) (iterate inc 1)))
+
+(prn (apply + (take-while #(<= % 2000000) (primes))))
+; 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 any direction (up,
+; down, left, right, or diagonally) in the 20×20 grid?
+
+(def grid [
+  [ 8  2 22 97 38 15  0 40  0 75  4  5  7 78 52 12 50 77 91  8]
+  [49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48  4 56 62  0]
+  [81 49 31 73 55 79 14 29 93 71 40 67 53 88 30  3 49 13 36 65]
+  [52 70 95 23  4 60 11 42 69 24 68 56  1 32 56 71 37  2 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  3 45  2 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  2 62 12 20 95 63 94 39 63  8 40 91 66 49 94 21]
+  [24 55 58  5 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72]
+  [21 36 23  9 75  0 76 44 20 45 35 14  0 61 33 97 34 31 33 95]
+  [78 17 53 28 22 75 31 67 15 94  3 80  4 62 16 14  9 53 56 92]
+  [16 39  5 42 96 35 31 47 55 58 88 24  0 17 54 24 36 29 85 57]
+  [86 56  0 48 35 71 89  7  5 44 44 37 44 60 21 58 51 54 17 58]
+  [19 80 81 68  5 94 47 69 28 73 92 13 86 52 17 77  4 89 55 40]
+  [ 4 52  8 83 97 35 99 16  7 97 57 32 16 26 26 79 33 27 98 66]
+  [88 36 68 87 57 62 20 72  3 46 33 67 46 55 12 32 63 93 53 69]
+  [ 4 42 16 73 38 25 39 11 24 94 72 18  8 46 29 32 40 62 76 36]
+  [20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74  4 36 16]
+  [20 73 35 29 78 31 90  1 74 31 49 71 48 86 81 16 23 57  5 54]
+  [ 1 70 54 71 83 51 54 69 16 92 33 48 61 43 52  1 89 19 67 48]])
+
+
+(defn item [row col]
+  (nth (nth grid row) col))
+
+(defn horizontal []
+  (for [row (range 20)
+        col (range 16)]
+    [[row col] [row (+ col 1)] [row (+ col 2)] [row (+ col 3)]]))
+
+(defn vertical []
+  (for [col (range 20)
+        row (range 16)]
+    [[row col] [(+ row 1) col] [(+ row 2) col] [(+ row 3) col]]))
+
+(defn diagonal1 []
+  (for [row (range 16)
+        col (range 16)]
+    [[row col] [(+ row 1) (+ col 1)] 
+     [(+ row 2) (+ col 2)] [(+ row 3) (+ col 3)]]))
+
+(defn diagonal2 []
+  (for [row (range 16)
+        col (range 3 20)]
+    [[row col] [(+ row 1) (- col 1)] 
+     [(+ row 2) (- col 2)] [(+ row 3) (- col 3)]]))
+
+(defn product [v]
+  (apply * (map (fn [xy] (apply item xy)) v)))
+
+(let [vectors (concat (horizontal) (vertical) (diagonal1) (diagonal2))]
+  (prn (apply max (map product vectors))))
+; The sequence of triangle numbers is generated by adding the natural numbers.
+; So the 7^(th) 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?
+;
+; Answer: 76576500
+
+(defn num-factors [n]
+  (let [max (inc (int (Math/sqrt n)))]
+   (loop [fs [] i 1]
+     (if (= i max)
+       (count (set fs)) ; Unique items
+       (if (zero? (mod n i))
+         (recur (concat fs [i (/ n i)]) (inc i))
+         (recur fs (inc i)))))))
+
+(defn traignles-step [it]
+  (let [i (first it) t (last it)]
+    [(inc i) (+ i t)]))
+
+(defn triangles []
+  (map last (rest (iterate traignles-step [1 0]))))
+
+(defn find-n [n]
+  (let [fseq (pmap (fn [i] [i (num-factors i)]) (triangles))]
+    (ffirst (filter (fn [fs] (> (last fs) n)) fseq))))
+
+(prn (find-n 500))
+(shutdown-agents) ; So we'll exit nicely
+; Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
+;
+; Answer: 5537376230
+
+(def 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])
+
+(let [s (str (apply + numbers))]
+  (prn (apply str (take 10 s))))
+; 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.
+; 
+; Answer: 837799
+
+(defn- collatz-next [n]
+  (if (even? n)
+    (/ n 2)
+    (inc (* 3 n))))
+
+(defn collatz-length
+  ([n] (collatz-length n 1))
+  ([n len]
+   (if (= n 1)
+     len
+     (collatz-length (collatz-next n) (inc len)))))
+
+(def collatz-length (memoize collatz-length))
+
+(prn (apply max-key collatz-length (range 1 1000001)))
+; Starting in the top left corner of a 2×2 grid, there are 6 routes (without
+; backtracking) to the bottom right corner.
+; [15.gif]
+
+; How many routes are there through a 20×20 grid?
+;
+; Answer: 137846528820
+
+
+(defn num-paths [x y max-x max-y]
+  (if (or (= x max-x) (= y max-y))
+    1
+    (+ (num-paths (inc x) y max-x max-y) (num-paths x (inc y) max-x max-y))))
+
+(def num-paths (memoize num-paths))
+
+
+(prn (num-paths 0 0 20 20))
+; 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)?
+;
+; Answer: 1366
+
+(defn char->int [c]
+  (- (int c) (int \0)))
+
+(defn pow [a b]
+  (loop [acc a b b]
+    (if (= b 1)
+      acc
+      (recur (* a acc) (dec b)))))
+
+(let [s (str (pow 2 1000))]
+  (prn (apply + (map char->int s))))
+; 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.
+;
+; Answer: 21124
+
+(def numbers {
+    0 "" 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" })
+
+(def tens ["" "" "hundred" "thousand"])
+
+(defn div [n i]
+  (int (/ n i)))
+
+(defn num->eng [n]
+  (if (> n 999)
+    ; We know it can be only up to 1000
+    (concat ["one" "thousand"] (num->eng (rem n 1000)))
+    (if (<= n 20)
+      (let [v (numbers n)]
+        (if (empty? v) [] [v]))
+      (if (> n 99)
+        (if (= n (* 100 (div n 100)))
+          [(numbers (div n 100)) "hundred"]
+          (concat [(numbers (div n 100)) "hundred" "and"] (num->eng (rem n 100))))
+        (concat [(numbers (* (div n 10) 10))] (num->eng (rem n 10)))))))
+
+(defn num-length [n]
+  (apply + (map count (num->eng n))))
+
+(prn (apply + (map num-length (range 1 1001))))
+; 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 5
+; 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)
+; 
+; Answer: 1074
+
+(def triangle [
+  [75]
+  [95 64]
+  [17 47 82]
+  [18 35 87 10]
+  [20  4 82 47 65]
+  [19  1 23 75  3 34]
+  [88  2 77 73  7 63 67]
+  [99 65  4 28  6 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  4 68 89 53 67 30 73 16 69 87 40 31]
+  [ 4 62 98 27 23  9 70 98 73 93 38 53 60  4 23]])
+
+(defn max-sum-xy [x y triangle]
+  (if (or (< y 0) (>= y (count triangle))
+          (< x 0) (>= x (count (triangle y))))
+    0
+    (+ ((triangle y) x) (max (max-sum-xy x (inc y) triangle)
+                             (max-sum-xy (inc x) (inc y) triangle)))))
+
+(defn max-sum [triangle]
+  (max-sum-xy 0 0 triangle))
+
+(prn (max-sum triangle))
+; 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)?
+; 
+; Answer: 171
+
+(ns e19
+  (:import java.util.Calendar))
+
+(def cal (Calendar/getInstance))
+
+(def months 
+  (for [year (range 1901 2001) month (range 1 13)]
+    [year month]))
+
+(defn make-calendar [year month]
+  (doto cal
+    (.clear)
+    (.set year (dec month) 1)))
+
+(defn pred [n]
+  (let [c (apply make-calendar n)]
+    (= (.get c Calendar/DAY_OF_WEEK) Calendar/SUNDAY)))
+
+(prn (count (filter pred months)))
+; 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, ...
+
+; Find the sum of all the even-valued terms in the sequence which do not exceed
+; four million.
+;
+; Answer: 4613732
+
+(defn fibs []
+ (map first (iterate (fn [[a b]] [b (+ a b)]) [1 1])))
+
+(println (apply + (take-while #(< % 4000000) (filter even? (fibs)))))
+; n! means n × (n − 1) × ... × 3 × 2 × 1
+; 
+; Find the sum of the digits in the number 100!
+;
+; Answer: 648
+
+(defn fact [n]
+ (apply * (range 1 (inc n))))
+
+
+(defn char->int [c]
+  (- (int c) (int \0)))
+
+(let [s (str (fact 100))]
+  (prn (apply + (map char->int s))))
+; 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.
+;
+; Answer: 31626
+
+
+(defn factors [n]
+  (let [max (inc (int (Math/sqrt n)))]
+   (loop [fs [] i 2]
+     (if (= i max)
+       (set (cons 1 fs)) ; Unique items
+       (if (zero? (mod n i))
+         (recur (concat fs [i (/ n i)]) (inc i))
+         (recur fs (inc i)))))))
+
+(defn d [n]
+  (apply + (factors n)))
+
+(defn amicable? [n]
+  (let [i (d n)]
+    (and (not (= n i)) (= (d i) n))))
+
+(prn (apply + (filter amicable? (range 1 10001))))
+; 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?
+;
+; Answer: 871198282
+
+(defn char->int [c]
+  (- (int c) (dec (int \A))))
+
+(defn str->int [s]
+  (apply + (map char->int s)))
+
+(let [names (sort (re-seq #"[A-Z]+" (slurp "names.txt")))]
+ (prn (apply + (map #(* %1 (str->int %2))
+                    (iterate inc 1) names))))
+; 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 whose proper divisors are less than the number is called deficient
+; and a number whose proper divisors exceed the number is called abundant.
+; 
+; 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.
+;
+; Answer: 4179871
+ 
+(defn factors [n]
+  (let [max (inc (int (Math/sqrt n)))]
+   (loop [fs [] i 2]
+     (if (= i max)
+       (set (cons 1 fs)) ; Unique items
+       (if (zero? (mod n i))
+         (recur (concat fs [i (/ n i)]) (inc i))
+         (recur fs (inc i)))))))
+
+(def factors (memoize factors))
+
+(defn abundant? [n]
+  (> (apply + (factors n)) n))
+
+(defn sum-two-abundant [n]
+  (some (fn [i] (and (abundant? i) (abundant? (- n i))))
+        (range 1 (inc (Math/ceil (/ n 2))))))
+
+;(prn (apply + (filter (fn [i] (not (sum-two-abundant i))) (range 1 28123))))
+(dorun (map #(println % (sum-two-abundant %)) (range 1 28123)))
+; 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?
+;
+; Answer: 2783915460
+
+(defn last-permutation? [n]
+  (= n (reverse (sort n))))
+
+(defn next-head [n]
+  (first (filter #(> % (first n)) (sort (rest n)))))
+
+(defn permutation-step [n]
+  (if (last-permutation? n)
+    nil
+    (let [sub (permutation-step (rest n))]
+      (if sub
+        (cons (first n) sub)
+        (let [head (next-head n)]
+          (cons head (sort (cons (first n) (remove #(= % head) (rest n))))))))))
+
+(defn perm->str [n]
+  (apply str (map #(str (bigint %)) n)))
+
+(let [p (nth (iterate permutation-step (range 10)) 999999)]
+  (prn (perm->str p)))
+; For a positive integer n, let σ(n) be the sum of all divisors of n, so e.g.
+; σ(6) = 1 + 2 + 3 + 6 = 12.
+; 
+; A perfect number, as you probably know, is a number with σ(n) = 2n.
+; 
+; Let us define the perfection quotient of a positive integer as 
+; p(n)	= σ(n) / n
+; 
+; 
+; Find the sum of all positive integers n ≤ 10^(18) for which p(n) has the form
+; k + 1⁄2, where k is an integer.
+
+(defn factors [n]
+  (let [max (inc (int (Math/sqrt n)))]
+   (loop [fs [] i 2]
+     (if (= i max)
+       (set (cons n (cons 1 fs))) ; Unique items
+       (if (zero? (mod n i))
+         (recur (concat fs [i (/ n i)]) (inc i))
+         (recur fs (inc i)))))))
+
+(defn p [n]
+ (/ (apply + (factors n)) n))
+
+(defn k? [n]
+  (let [pn (p n)]
+    (and (= (class pn) clojure.lang.Ratio)
+         (= (.denominator pn) 2))))
+
+(defn sum-k [max-n]
+  (apply + (pmap #(if (k? %) % 0) 
+                 (take-while #(< % max-n) (iterate inc 1)))))
+
+(prn (sum-k (Math/pow 10 18)))
+(shutdown-agents)
+; 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) = 144
+; 
+; The 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?
+; 
+; Answer: 4782
+
+(defn fibs []
+ (map first (iterate (fn [[a b]] [b (+ a b)]) [1 1])))
+
+(defn find-fib [n]
+ (inc (count (take-while #(< (.length (str %)) n) (fibs)))))
+
+(prn (find-fib 1000))
+; 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.
+; 
+;
+
+
+; The idea is to create step of divison that will return the digit and the
+; leftover and run it with "iterate". Currently it seems like the best approach
+; is to have an embedded function where "n" is bound (or maybe use "bindings")
+(defn step [[digit a b]]
+  ; (divide 1 7) -> [1 3]
+  (let [a (* 10 a)
+        digit (int (/ a b))
+        leftover (- a (* b digit))]
+     [digit leftover b]))
+
+(defn digits [n]
+  (defn step [digit leftover]
+    
+  (map first (iterate divide [1 n])))
+; Euler published 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, 40^(2) + 40 + 41 = 40(40 + 1) + 41
+; is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly
+; divisible by 41.
+; 
+; Using computers, 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.
+; 
+; Answer: -59231
+
+(defn prime? [n]
+  (.isProbablePrime (bigint n) 10))
+
+(defn quad [a b n]
+  (+ (* n n) (* n a) b))
+
+(defn num-primes [a b]
+  (let [func (partial quad a b)]
+    (count (take-while prime? (map func (iterate inc 0))))))
+
+(let [abs (for [a (range -999 1001)  b (range -999 1001)] [a b])
+      results (pmap (fn [[a b]] [a b (num-primes a b)]) abs)
+      [a b _] (apply max-key #(% 2) results)]
+  (println (* a b)))
+(shutdown-agents)
+; 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 both diagonals is 101.
+; 
+; What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same
+; way?
+;
+; Answer: 669171001
+
+(defn corners [start size]
+  (take 4 (iterate #(+ % size) start)))
+
+(defn sum-diagonals [max-size]
+  (loop [start 3 size 2 sum-d 1]
+    (if (> (inc size) max-size)
+      sum-d
+      (let [cs (corners start size)
+            new-size (+ size 2)]
+        (recur (+ (last cs) new-size)
+               new-size
+               (apply + (cons sum-d cs)))))))
+
+(prn (sum-diagonals 1001))
+; Consider all integer combinations of a^(b) for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
+; 
+;     2^(2)=4, 2^(3)=8, 2^(4)=16, 2^(5)=32
+;     3^(2)=9, 3^(3)=27, 3^(4)=81, 3^(5)=243
+;     4^(2)=16, 4^(3)=64, 4^(4)=256, 4^(5)=1024
+;     5^(2)=25, 5^(3)=125, 5^(4)=625, 5^(5)=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 a^(b) for 2 ≤ a ≤
+; 100 and 2 ≤ b ≤ 100?
+; 
+; Answer: 9183
+
+(defn num-terms [min-a max-a min-b max-b]
+  (let [nums (for [b (range min-b (inc max-b))
+                   a (range min-a (inc max-a))]
+               (Math/pow a b))]
+    (count (set nums))))
+
+(prn (num-terms 2 100 2 100))
+; The prime factors of 13195 are 5, 7, 13 and 29.
+;
+; What is the largest prime factor of the number 600851475143 ?
+;
+; Answer: 6857
+
+(defn first-candidate [n]
+ (let [c (int (Math/sqrt n))]
+  (if (even? c)
+   (dec c)
+   c)))
+
+(defn dec2 [n]
+ (- n 2))
+
+(defn biggest-factor [n]
+ (first (filter (fn [x] (and (zero? (mod n x)) 
+                             (.isProbablePrime (bigint x) 10)))
+                (iterate dec2 (first-candidate n)))))
+
+(println (biggest-factor 600851475143))
+; Surprisingly there are only three numbers that can be written as the sum of
+; fourth powers of their digits:
+; 
+;     1634 = 1^(4) + 6^(4) + 3^(4) + 4^(4)
+;     8208 = 8^(4) + 2^(4) + 0^(4) + 8^(4)
+;     9474 = 9^(4) + 4^(4) + 7^(4) + 4^(4)
+; 
+; As 1 = 1^(4) 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.
+;
+; Answer: 443839
+
+(defn char->num [n]
+  (- (int n) (int \0)))
+
+(defn sum5 [n]
+  (apply + (map #(Math/pow % 5) (map char->num (str n)))))
+
+; FIXME: I just chose 1000000 as arbitrary large
+(let [ints (take-while #(< % 1000000) (iterate inc 2))
+      nums (filter #(= (sum5 %) %) ints)]
+  (prn (apply + nums)))
+
+; In England the currency is made up of pound, £, and pence, p, and there are
+; eight coins in general circulation:
+; 
+;     1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
+; 
+; It is possible to make £2 in the following way:
+; 
+;     1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
+; 
+; How many different ways can £2 be made using any number of coins?
+; 
+; Answer:
+
+(defn num-ways [sum coins path]
+  (if (zero? sum)
+    path
+    (let [possible-coins (filter #(<= % sum) coins)]
+      (for [path paths
+      (apply + (map #(num-ways (- sum %) coins) possible-coins)))))
+
+(def num-ways (memoize num-ways))
+
+;(println (num-ways 200 [1 2 5 10 20 50 100 200]))
+; We shall say that an n-digit number is pandigital if it makes use of all the
+; digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1
+; through 5 pandigital.
+; 
+; The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
+; multiplicand, multiplier, and product is 1 through 9 pandigital.
+; 
+; Find the sum of all products whose multiplicand/multiplier/product identity
+; can be written as a 1 through 9 pandigital.
+;
+; HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
+;
+; Answer:
+
+; FIXME: This is wrong!
+;
+(use 'clojure.contrib.combinatorics)
+(use '[clojure.set :only (difference)])
+
+(def *digits* #{\1 \2 \3 \4 \5 \6 \7 \8 \9})
+(def *sizes* '(1 2 3 4))
+
+(defn pandigital? [lhs rhs product]
+  (= (sort *digits*) (sort (str lhs rhs product))))
+
+(defn all-perms [nums sizes]
+  (mapcat identity 
+          (for [size sizes] (mapcat permutations (combinations nums size)))))
+
+(defn all-pairs [digits sizes]
+  (for [lhs (all-perms digits sizes)
+        rhs (all-perms (difference digits lhs) sizes)]
+    [lhs rhs]))
+
+(defn chars->num 
+  "(\1 \2 \3) -> 123"
+  [cs]
+  (Integer/valueOf (apply str cs)))
+
+(defn map-fn [[lhs rhs]]
+  (let [lhs (chars->num lhs) rhs (chars->num rhs) product (* lhs rhs)]
+    (if (pandigital? lhs rhs product)
+      product)))
+
+(let [products (pmap map-fn (all-pairs *digits* *sizes*))]
+  (println (reduce + (filter (complement nil?) products))))
+(shutdown-agents)
+; The number, 197, is called a circular prime because all rotations of the
+; digits: 197, 971, and 719, are themselves prime.
+; 
+; There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71,
+; 73, 79, and 97.
+; 
+; How many circular primes are there below one million?
+;
+; Answer: 55
+
+(defn rotations [n]
+  (let [n (str n) size (count n)]
+    (for [i (range size)]
+      (Integer/parseInt (apply str (.substring n i) (.substring n 0 i))))))
+
+(defn primes-seq []
+  (filter (fn [n] (.isProbablePrime (bigint n) 10)) (iterate inc 1)))
+
+(defn primes-upto [n]
+  (take-while #(< % n) (primes-seq)))
+
+(defn circular? [n primes]
+  (every? #(contains? primes %) (rotations n)))
+
+(defn num-circular [n]
+  (let [primes (set (primes-upto n))]
+    (count (filter #(circular? % primes) primes))))
+
+(prn (num-circular 1000000))
+; If p is the perimeter of a right angle triangle with integral length sides,
+; {a,b,c}, there are exactly three solutions for p = 120.
+; 
+; {20,48,52}, {24,45,51}, {30,40,50}
+; 
+; For which value of p ≤ 1000, is the number of solutions maximised?
+;
+; Answer: 840
+
+(defn triangle? [a b c]
+  (= (* c c) (+ (* a a) (* b b))))
+
+(defn triangles [n]
+  (for [a (range 1 (- n 1)) b (range a (- n a 1)) 
+        :when (triangle? a b (- n a b))]
+    [a b (- n a b)]))
+
+(defn max-triangles [max-p]
+  (loop [p 1 best 1 best-n 0]
+    (if (> p max-p)
+      best
+      (let [n (count (triangles p))]
+        (if (> n best-n)
+          (recur (inc p) p n)
+          (recur (inc p) best best-n))))))
+
+(prn (max-triangles 1000))
+; 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.
+;
+; Answer: 906609
+
+(defn palindrom? [s]
+  (let [size (count s)
+        middle (int (/ size 2))
+        pre (.substring s 0 middle)
+        i (if (even? size) middle (inc middle))]
+    (= (list* pre) (reverse (.substring s i size)))))
+
+
+(defn palindrom-numbers []
+  (for [a (range 100 999)
+        b (range 100 999)
+        :when (palindrom? (format "%d" (* a b)))]
+    (* a b)))
+
+(prn (apply max (palindrom-numbers)))
+; We shall say that an n-digit number is pandigital if it makes use of all the
+; digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is
+; also prime.
+; 
+; What is the largest n-digit pandigital prime that exists?
+;
+; Answer:
+
+(defn str->digits [s]
+  (sort (map #(- (int %) (int \0)) s)))
+
+(defn pandigital? [n]
+  (let [n (str n)]
+    (= (range 1 (inc (count n))) (str->digits n))))
+
+(defn primes-from [n]
+  (filter (fn [n] (.isProbablePrime (bigint n) 10)) 
+          (take-while #(> % 1) (iterate dec n))))
+
+(prn (max (filter pandigital? (primes-from 987654321))))
+; The n^(th) term of the sequence of triangle numbers is given by, t_(n) =
+; ½n(n+1); so the first ten triangle numbers are:
+; 
+; 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+; 
+; By converting each letter in a word to a number corresponding to its
+; alphabetical position and adding these values we form a word value. For
+; example, the word value for SKY is 19 + 11 + 25 = 55 = t_(10). If the word
+; value is a triangle number then we shall call the word a triangle word.
+; 
+; Using words.txt (right click and 'Save Link/Target As...'), a 16K text file
+; containing nearly two-thousand common English words, how many are triangle
+; words?
+; 
+; Answer: 162
+
+(defn word->int [w]
+  (apply + (map #(inc (- (int %) (int \A))) w)))
+
+(defn triangles [n]
+  (map #(* (/ % 2) (inc %)) (range 1 (inc n))))
+
+(let [words (re-seq #"[A-Z]+" (slurp "words.txt"))
+      values (map word->int words)
+      ts (set (triangles (apply max values)))]
+  (prn (count (filter #(contains? ts %) values))))
+; 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).
+;
+; Answer: 9110846700 
+
+(defn power-seq [n]
+  (map #(.pow % %) (map bigint (range 1 (inc n)))))
+
+(let [n (str (apply + (power-seq 1000)))]
+  (prn (.substring n (- (count n) 10))))
+
+; 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?
+;
+; Answer: 232792560
+
+(defn divisible? [n]
+  (every? #(zero? (mod n %)) (range 1 20)))
+
+(prn (first (filter divisible? (iterate inc 19))))
+; If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
+; 
+; Not all numbers produce palindromes so quickly. For example,
+; 
+; 349 + 943 = 1292,
+; 1292 + 2921 = 4213
+; 4213 + 3124 = 7337
+; 
+; That is, 349 took three iterations to arrive at a palindrome.
+; 
+; Although no one has proved it yet, it is thought that some numbers, like 196,
+; never produce a palindrome. A number that never forms a palindrome through
+; the reverse and add process is called a Lychrel number. Due to the
+; theoretical nature of these numbers, and for the purpose of this problem, we
+; shall assume that a number is Lychrel until proven otherwise. In addition you
+; are given that for every number below ten-thousand, it will either (i) become
+; a palindrome in less than fifty iterations, or, (ii) no one, with all the
+; computing power that exists, has managed so far to map it to a palindrome. In
+; fact, 10677 is the first number to be shown to require over fifty iterations
+; before producing a palindrome: 4668731596684224866951378664 (53 iterations,
+; 28-digits).
+; 
+; Surprisingly, there are palindromic numbers that are themselves Lychrel
+; numbers; the first example is 4994.
+; 
+; How many Lychrel numbers are there below ten-thousand?
+;
+; Answer: 249
+
+(defn palindrom? [s]
+  (let [size (count s)
+        middle (int (/ size 2))
+        pre (.substring s 0 middle)
+        i (if (even? size) middle (inc middle))]
+    (= (list* pre) (reverse (.substring s i size)))))
+
+(defn strrev [s]
+  (apply str (reverse s)))
+
+(defn step [s]
+  (str (+ (bigint s) (bigint (strrev s)))))
+
+(defn lychrel? [n]
+  (let [n (step (str n)) ; Don't count the number itself
+        lseq (take-while #(not (palindrom? %)) (iterate step n))]
+    (> (count (take 51 lseq)) 50)))
+
+(prn (count (filter lychrel? (range 1 10000))))
+; A googol (10^(100)) is a massive number: one followed by one-hundred zeros;
+; 100^(100) is almost unimaginably large: one followed by two-hundred zeros.
+; Despite their size, the sum of the digits in each number is only 1.
+; 
+; Considering natural numbers of the form, a^(b), where a, b < 100, what is the
+; maximum digital sum?
+;
+; Answer: 972
+
+(defn char->num [n]
+  (- (int n) (int \0)))
+
+(defn sum-digits [n]
+  (apply + (map char->num (str (bigint n)))))
+
+(defn max-digits [n]
+  (let [ab (for [a (range 1 n) b (range 1 n)] [(bigint a) (bigint b)])]
+    (apply max (map #(sum-digits (.pow (first %) (last %))) ab))))
+
+(prn (max-digits 100))
+; The sum of the squares of the first ten natural numbers is,
+; 1^(2) + 2^(2) + ... + 10^(2) = 385
+; 
+; The square of the sum of the first ten natural numbers is,
+; (1 + 2 + ... + 10)^(2) = 55^(2) = 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.
+;
+; Answer: 25164150
+
+(defn square [n]
+  (* n n))
+
+(defn sum [items] 
+  (apply + items))
+
+(let [nums (range 1 101)]
+  (prn (Math/abs (- (sum (map square nums)) (square (sum nums))))))
+; 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?
+;
+; Answer: 104743
+
+(defn primes []
+  (filter #(.isProbablePrime (bigint %) 10) (iterate inc 1)))
+
+(prn (nth (primes) 10000))
+; 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
+;
+; Answer: 40824
+
+
+(def N (str
+  "73167176531330624919225119674426574742355349194934"
+   "96983520312774506326239578318016984801869478851843"
+   "85861560789112949495459501737958331952853208805511"
+   "12540698747158523863050715693290963295227443043557"
+   "66896648950445244523161731856403098711121722383113"
+   "62229893423380308135336276614282806444486645238749"
+   "30358907296290491560440772390713810515859307960866"
+   "70172427121883998797908792274921901699720888093776"
+   "65727333001053367881220235421809751254540594752243"
+   "52584907711670556013604839586446706324415722155397"
+   "53697817977846174064955149290862569321978468622482"
+   "83972241375657056057490261407972968652414535100474"
+   "82166370484403199890008895243450658541227588666881"
+   "16427171479924442928230863465674813919123162824586"
+   "17866458359124566529476545682848912883142607690042"
+   "24219022671055626321111109370544217506941658960408"
+   "07198403850962455444362981230987879927244284909188"
+   "84580156166097919133875499200524063689912560717606"
+   "05886116467109405077541002256983155200055935729725"
+   "71636269561882670428252483600823257530420752963450"))
+       
+
+(defn prod [start]
+  (apply * (map (fn [i] 
+                   (- (int (nth N i)) (int \0)))
+                 (range start (+ start 5)))))
+
+(prn (apply max (map prod (range 0 (- (count N) 5)))))
+; A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+; a^(2) + b^(2) = c^(2)
+; 
+; For example, 3^(2) + 4^(2) = 9 + 16 = 25 = 5^(2).
+; 
+; There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+; Find the product abc.
+;
+; Answer: 31875000
+ 
+(defn triplets []
+  (for [a (range 1 999)
+        b (range 1 (- 999 a))]
+    [a b (- 1000 (+ a b))]))
+
+(defn pythagorean? [a b c]
+  (= (+ (* a a) (* b b)) (* c c)))
+
+(let [abc (first (filter #(apply pythagorean? %) (triplets)))]
+  (prn (apply * abc)))
+; See http://projecteuler.net/index.php?section=problems&id=96 :)
+;
+; Answer: 
+
+(refer 'clojure.set :only '(difference))
+
+(def problem '(
+  "003020600"
+  "900305001"
+  "001806400"
+  "008102900"
+  "700000008"
+  "006708200"
+  "002609500"
+  "800203009"
+  "005010300"))
+
+(defn char->num [c]
+  (- (int c) (int \0)))
+
+(defstruct cell :row :col)
+
+(defn parse-row [grid row s]
+  (loop [grid grid i 0]
+    (if (= i (count s))
+      grid
+      (let [v (char->num (nth s i))]
+        (if (zero? v)
+          (recur grid (inc i))
+          (recur (assoc grid (struct cell row i) v) (inc i)))))))
+
+(defn parse-rows [rows]
+  (loop [grid {} row 0]
+    (if (= row (count rows))
+      grid
+      (recur (parse-row grid row (nth rows row)) (inc row)))))
+
+(defn cell-row [c]
+  (for [col (range 9)]
+    (struct cell (:row c) col)))
+
+(defn cell-col [c]
+  (for [row (range 9)]
+    (struct cell row (:col c))))
+
+(defn box-first [n]
+  (int (Math/floor (/ n 3))))
+
+(defn cell-box [c]
+  (let [row (box-first (:row c))
+        col (box-first (:col c))]
+    (for [r (range 3) c (range 3)]
+      (struct cell (+ row r) (+ col c)))))
+
+(defn cell-peers [c]
+  (let [peers (set (concat (cell-row c) (cell-col c) (cell-box c)))]
+    (disj peers c)))
+

clojure/README.txt

+Solving "Project Euler" (http://projecteuler.net) using Clojure (http://clojure.org/)
+
+After finishing the first 25 in order, I just jump around and try to solve what
+seems interesting.

clojure/names.txt

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