Posts Tagged ‘Project Euler’

Project Euler: Problem 11

Thursday, August 19th, 2010

This is problem 11 in Project Euler:

In the 20 x 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 x 20 grid?

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Project Euler: Problem 10

Thursday, August 12th, 2010

This is problem 10 on Project Euler:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

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Project Euler: Problem 9

Wednesday, August 11th, 2010

This is Problem 9 on Project Euler:

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.

Pythagorean triplets can be generated by using the following formulas, where m and n are integers and m is greater than n:

m^(2) – n^(2)
2mn
m^(2) + n^(2)

So for the purposes of this problem we are looking for values for m and n where these three formulas add up to 1000, that is:

(m^(2) – n^(2)) + (2mn) + (m^(2) + n^(2)) = 1000
This can be simplified to 2m^(2) + 2mn = 1000
Dividing by 2 and factoring by m we have m(m+n) = 500
Therefore m must be a factor of 500. Calculating m, m + n and n we get the following table:

m m + n n notes
1 500 499 Reject, because n is greater than m
2 250 248 Reject, because n is greater than m
5 100 95 Reject, because n is greater than m
10 50 40 Reject, because n is greater than m
20 25 5 A solution
25 20 -5 Reject, because n cannot be negative
50 10 -40 Reject, because n cannot be negative
100 5 -95 Reject, because n cannot be negative
250 2 -248 Reject, because n cannot be negative
500 1 -499 Reject, because n cannot be negative

Therefore the only solution is where m = 20 and n = 5. Plugging these values into our original expressions we have:

a = m^(2) – n^(2) = (20 * 20) – (5 * 5) = 375

b = 2mn = 2 * 20 * 5 = 200

c = m^(2) + n^(2) = (20 * 20) + (5 * 5) = 425

a + b + c = 1000 and ^(2) + b^(2) = c^(2), so the answer to the problem is a * b * c.

We have not had to use a computer to solve this problem, but a brute-force approach in Delphi could be this:

procedure TfrmMain.Problem9;
var
  a, b, c : integer;
begin
  for a := 1 to 998 do begin
    for b := a + 1 to 999 do begin
      c := 1000 - (a + b);
      if (a * a) + (b * b) = (c * c) then begin
        ShowMessage(IntToStr(a * b * c));
        exit;
      end;
    end;
  end;
end;

This code just generates sets of numbers where a + b + c = 1000 and then checks the set for being a Pythagorean triple. If it is, the product is displayed. There are no optimizations in this code, which could be improved considerably.

Project Euler: Problem 8

Monday, August 9th, 2010

Problem 8 in Project Euler is this

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

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