Problem: Find the greastest product of 4 adjacent numbers along a horizontal, vertical, or diagonal line in the following grid:
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
This problem is small enough to solve by brute force. We can use the sliding window technique described in problem 8, but I'll avoid it here. One thing to remember is there are 2 diagonal directions. Here is the code:
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]]
def product(nums):
ret = 1
for n in nums:
ret *= n
return ret
L = 4 # product length
R,C = len(grid),len(grid[0]) # grid size
largest = 0
sol_rc = (-1,-1)
sol_nums = []
sol_dir = ''
for r in range(len(grid)):
for c in range(len(grid[0])):
if r + L <= R: # vertical
nums = [grid[r+i][c] for i in range(L)]
prod = product(nums)
if prod > largest:
largest = prod
sol_rc = (r,c)
sol_nums = nums
sol_dir = 'down'
if c + L <= C: # horizontal
nums = [grid[r][c+i] for i in range(L)]
prod = product(nums)
if prod > largest:
largest = prod
sol_rc = (r,c)
sol_nums = nums
sol_dir = 'right'
if r + L <= R and c + L <= C: # diagonal 1
nums = [grid[r+i][c+i] for i in range(L)]
prod = product(nums)
if prod > largest:
largest = prod
sol_rc = (r,c)
sol_nums = nums
sol_dir = 'down-right'
if r + L <= R and c >= L-1: # diagonal 2
nums = [grid[r+i][c-i] for i in range(L)]
prod = product(nums)
if prod > largest:
largest = prod
sol_rc = (r,c)
sol_nums = nums
sol_dir = 'down-left'
print(largest)
print(f'start at {sol_rc} going {sol_dir} multiply {sol_nums}')
Problem: Find the smallest triangle number with over 500 divisors. The \(n\)th triangle number is \(1+2+\ldots+n\).
This could be done by brute force once we have a way to calculate how many prime factors a number has, but we'll use a faster method. First we will look at 2 ways to count divisors of \(n\). We could have a loop for all numbers \(1,2,\ldots,n\), but if \(p\leq\lfloor\sqrt{n}\rfloor\) then \(n=pq\) with \(q\geq\lfloor\sqrt{n}\rfloor\) so for each divisor below the square root, there is a corresponding divisor. This allows us to loop over \(1,2,\ldots,\lfloor\sqrt{n}\rfloor\) which is much faster. The only special case to handle is for perfect squares.
The other method we'll look at is by factoring the number. If we have a prime factorization \(n=p_1^{a_1}p_2^{a_2}\ldots\), then we can choose exponents for each prime from 0 to \(a_i\) to form a divisor. This tells us that the number of divisors is \((a_1+1)(a_2+1)\ldots\). This may be just as slow as the other method, such as when \(n\) is a semiprime, but it helps significantly when \(n\) has some small prime factors, so this is the method we will use. This divisor counting function is similar to the solution for problem 3.
Now, we will look at a method of generating triangle nummbers. Every triangle number is \(\sum_{i=1}^ni=n(n+1)/2\) for some \(n\). So we can start a loop at \(n=1\) going up. Now we compute divisors of \(n(n+1)/2\). There is a more efficient way to do this by observing that \(n\) and \(n+1\) are coprime. One of those 2 will be even. If \(n\) is even, then \(n/2\) and \(n+1\) are coprime. If \(n+1\) is even, then \(n\) and \((n+1)/2\) are coprime. Now if \(\gcd(p,q)=1\), then \(D(pq)=D(p)D(q)\) where \(D(x)\) is the number of divisors of \(x\), because \(p\) and \(q\) do not share prime factors. It is much faster this way because we can find the number of divisors by factoring 2 much smaller numbers instead of 1 big number.
def divcount(n):
factors = 1
while n % 2 == 0: # divide out 2s
factors += 1
n //= 2
d = 3
while d*d <= n: # find multiplicity of prime factors
multiplicity = 0
while n % d == 0:
multiplicity += 1
n //= d
factors *= multiplicity+1
d += 2
if n != 1: # last prime factor
factors *= 2
return factors
D = 500
n = 1
while True: # try n(n+1)/2
if n % 2 == 0: # n/2,n+1
d1 = divcount(n//2)
d2 = divcount(n+1)
else: # n,(n+1)/2
d1 = divcount(n)
d2 = divcount((n+1)//2)
if d1*d2 > D:
print(n*(n+1)//2)
print(f'{d1*d2} divisors, n = {n}')
break
n += 1
Problem: Find the first 10 digits of the sum of these 100 numbers (50 digits each):
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
In Python, we can just sum the numbers easily since large integers are supported:
s = sum([
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
])
print(str(s)[:10])
If we want to challenge ourselves a little more or do this in a language without large integer support, we can perform the add and carry operations:
nums = [
'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'
]
total = [0]*50 # digits starting from most significant
for num_str in nums:
carry = 0
i = len(total)-1 # i,j are which index to add
j = len(num_str)-1
while i >= 0 and j >= 0: # add num_str to the total
sum_ = total[i] + int(num_str[j]) + carry
total[i] = sum_ % 10
carry = sum_ // 10
i -= 1
j -= 1
while i >= 0 and carry > 0: # propogate carry
sum_ = total[i] + carry
total[i] = sum_ % 10
carry = sum_ // 10
i -= 1
if carry > 0 and i < 0: # need to extend by a digit
total.insert(0,carry)
print(''.join(map(str,total[:10])))
Problem: Find the starting number under 1000000 produces the longest Collatz sequence chain. Given a starting number \(n\), we generate the Collatz sequence by updating it as shown below, terminating at 1 (it is conjectured that every sequence ends at 1, but not proven).
\[n\rightarrow\begin{cases} n/2 & \text{if $n$ is even}\\ 3n+1 & \text{if $n$ is odd}\\ \end{cases}\]The simplest way to solve this is by iterating Collatz sequences for all numbers below 1000000. We can speed this up by caching sequence lengths so when a larger starting number reaches a number we already had, we do not have to repeat the work.
L = 1000000
cache_len = [0]*L
cache_max = [0]*L # track max in sequence (extra, unneeded)
cache_len[1] = 1
cache_max[1] = 1
for n_start in range(2,L):
len_ = 0
max_ = 0
n = n_start
while n >= n_start:
len_ += 1
max_ = max(max_,n)
if n % 2 == 0:
n //= 2
else:
n = 3*n + 1
# use cache to finish
cache_len[n_start] = len_ + cache_len[n]
cache_max[n_start] = max(max_,cache_max[n])
# number with longest chain
max_len = max(cache_len)
max_len_nums = [i for i in range(L) if cache_len[i] == max_len]
# number with highest peak value in its chain
max_max = max(cache_max)
max_max_nums = [i for i in range(L) if cache_max[i] == max_max]
print(f'highest number iterated {max_max}')
print(f'reached from starting number(s) {max_max_nums}')
print(f'maximum chain length {max_len}')
for n in max_len_nums:
print(f'starting number {n} reaches max of {cache_max[n]}')
print(n)
Note: the longest chain found has length 525. The highest number ever reached during chain iteration is 56991483520, which is too big for 32 bit integers.
Problem: Starting from the top left corner of a \(20\times20\) grid, find how many possible paths there are to the bottom right corner by moving only right and down. For example, in a \(2\times2\) grid, the answer is 6, as shown in the image below, an SVG based on the image from Project Euler.
One way is to recursively find all paths, but that is too slow for this problem. We could use dynamic programming to solve this, because a solution for the number of paths starting at a position depends only on that of positions right and down from it. Along the bottom and right edge, there is only 1 path. Everything else in the middle can be filled in by adding the solution for the points directly adjacent on the right and down.
R = 20
C = 20
# dp[r][c] is the solution for a r*c grid
dp = [[0]*(C+1) for _ in range(R+1)]
# 1 path along the edges
for r in range(R+1):
dp[r][0] = 1
for c in range(C+1):
dp[0][c] = 1
# fill in the rest of the sub problems
for r in range(1,R+1):
for c in range(1,C+1):
dp[r][c] = dp[r-1][c] + dp[r][c-1]
# the solution is dp[20][20]
print(dp[R][C])
Another way is to notice how each path will contain 20 steps right and 20 steps down, for 40 steps total. If we choose which of these 40 positions are right steps, and make the rest down steps, there are \({40}\choose{20}\) ways to do this. This can be calculated by hand but is tedious.
Note: the solution does not fit in a 32 bit integer.
Problem: Find the sum of the digits in the number \(2^{1000}\).
Since Python supports large integers, we can solve it with this single line:
print(sum(map(int,str(2**1000))))
In a language without large integers, we have to implement some large integer arithmetic. Here is an example that keeps track of the decimal digits (least significant first) and doubles it repeatedly.
P = 1000
number = [1] # start at 1
for _ in range(P):
carry = 0
for i in range(len(number)): # multiply by 2
tmp = 2*number[i] + carry
number[i] = tmp % 10
carry = tmp // 10
if carry > 0: # number gets a digit longer
number.append(carry)
print(sum(number))
Problem: Find how many letters are used to write the word form of all numbers from 1 to 1000. (Note: the word "and" is used in the hundreds to follow British English, for example "three hundred and forty two" or "one hundred and fifteen")
We will look at a programming way to do this by writing a function that generates the word form of numbers. For small numbers, we use a lookup table. Then in the range 20-99, we spit words into 2 parts. For example, 22 would be "twenty" and "two" so we take the "twenty" from a lookup table for the tens place and the ones part with the other lookup table. The ones part is nothing when the ones place is 0, so 20 would just by "twenty". For hundreds, we need the words for the hundreds digit followed by "and" and words for the other 2 digits when it is nonzero, which can be done with the work already done. For example, 100 is "one hundred" and 122 is "one hundred and twenty two". For simplicity, we handle the special case of 1000 separately.
def word_form(n):
if n < 0:
return 'negative' + word_form(-n)
if n < 20: # use lookup table for small numbers
return [
'zero', 'one', 'two', 'three', 'four', 'five',
'six', 'seven', 'eight', 'nine', 'ten',
'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen',
'sixteen', 'seventeen', 'eighteen', 'nineteen'
][n]
elif n < 100: # 20-99
tens,ones = divmod(n,10)
ret = [
'', '', 'twenty', 'thirty', 'forty', 'fifty',
'sixty', 'seventy', 'eighty', 'ninety'
][tens]
if ones != 0:
ret += word_form(ones)
return ret
elif n < 1000: # 100-999
hundreds,remainder = divmod(n,100)
ret = word_form(hundreds) + 'hundred'
if remainder != 0:
ret += 'and' + word_form(remainder)
return ret
else:
raise Exception('too big')
print(sum(len(word_form(n)) for n in range(1,1000)) + len('onethousand'))
We can also solve this by counting. It is a bit tedious, but doable by hand. First, we count the lengths of the numbers 1-19. Below is a table for 1-9 and 10-19 separately:
| number | word form | length |
|---|---|---|
| 1 | one | 3 |
| 2 | two | 3 |
| 3 | three | 5 |
| 4 | four | 4 |
| 5 | five | 4 |
| 6 | six | 3 |
| 7 | seven | 5 |
| 8 | eight | 5 |
| 9 | nine | 4 |
| total | 36 |
| number | word form | length |
|---|---|---|
| 10 | ten | 3 |
| 11 | eleven | 6 |
| 12 | twelve | 6 |
| 13 | thirteen | 8 |
| 14 | fourteen | 8 |
| 15 | fifteen | 7 |
| 16 | sixteen | 7 |
| 17 | seventeen | 9 |
| 18 | eighteen | 8 |
| 19 | nineteen | 8 |
| total | 70 |
Next, for each range of 10 numbers (20-29, 30-39, ...), we count the tens word like "twenty" which is used 10 times and each number 1-9.
| tens word | length | letters used |
|---|---|---|
| twenty | 6 | 10×6 + 36 = 96 |
| thirty | 6 | 10×6 + 36 = 96 |
| forty | 5 | 10×5 + 36 = 86 |
| fifty | 5 | 10×5 + 36 = 86 |
| sixty | 5 | 10×5 + 36 = 86 |
| seventy | 7 | 10×7 + 36 = 106 |
| eighty | 6 | 10×6 + 36 = 96 |
| ninety | 6 | 10×6 + 36 = 96 |
| total | 748 |
So far we have used 36 + 70 + 748 = 854 letters for numbers 1-99. Now to count the ranges 100-199, 200-299, and so on, we have a few parts. First are the words like "one hundred". We use each single digit (1-9) word 100 times and the word "hundred" 900 times so that is 100×36 + 900×7 = 9900. The word "and" is used 99 times in each of the 9 ranges so that is another 9×99×3 = 2673 letters. Finally, in each of the 9 ranges we use the words for the numbers 1-99 once each so that is another 9×854 = 7686 letters. Lastly is "one thousand" with 11 letters. By adding all these together, we get the answer.
Problem: Find the maximum total sum following a path from top to bottom in the triangle below. (For each step down, there are 2 numbers to choose from)
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
The most obvious solution to this is to try all paths from top to bottom. There is a nicer solution, but we will save that for problem 67 which asks the same thing on a much larger triangle where brute force is too slow. We can do this by recursion. The triangle will be a 2D array with varying row lengths. If we are at position \((r,c)\) in the array, then going down a row we can move to either \((r+1,c)\) or \((r+1,c+1)\). The following code uses a recursive function to try all paths and updates a global variable at the end of each path.
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]]
largestsum = 0
largestpath = []
path = [] # path during recursion (stack)
def recur(r,c):
global largestsum,largestpath,path,triangle
path.append(triangle[r][c]) # add to path
if r == len(triangle)-1: # bottom row
if sum(path) > largestsum:
largestsum = sum(path)
largestpath = path[:]
else:
recur(r+1,c)
recur(r+1,c+1)
path.pop() # backtrack
recur(0,0)
print(f'the path is {largestpath}')
print(largestsum)
Problem: Count how many Sundays were on the first of a month during the 20th century (1 Jan 1901 to 31 Dec 2000). The following information is provided:
To solve this, we can begin at 1 Jan 1900 and go up a month at a time. The day of the week can be kept track of with an integer modulo 7. The number of days to go up at each step is calculated by finding the month length.
weekday = 1 # 0 = sunday, 1 = monday, ...
count = 0
for y in range(1900,2001):
months = [31,28,31,30,31,30,31,31,30,31,30,31]
# set february to 29 days if leap year
if y % 4 == 0 and not (y % 100 == 0 and y % 400 != 0):
months[1] = 29
for m in months:
if weekday == 0 and 1901 <= y <= 2000:
count += 1
weekday = (weekday + m) % 7
print(count)
Problem: Find the sum of digits in the number \(100!\).
This is straight forward in Python with large integer support and using the factorial function.
from math import factorial
print(sum(map(int,str(factorial(100)))))
For a bit more of a challenge, we can represent a decimal number and repeatedly multiply it which lets us manage large integer arithmetic code. Here is an example where the number is stored as an array of base 10 digits starting from least significant. This method works similar to writing out a row of the grade school multiplication method.
num = [1] # start at 1
for n in range(1,1+100): # multiply 1,2,..,100
carry = 0
for i in range(len(num)):
tmp = n*num[i] + carry
num[i] = tmp % 10
carry = tmp // 10
# at the end, we may need to append extra digits for carry
while carry > 0:
num.append(carry % 10)
carry //= 10
print(sum(num))