## Posts Tagged ‘Programming’

### 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,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

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:

m2 – n2
2mn
m2 + n2

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:

(m2 – n2) + (2mn) + (m2 + n2) = 1000
This can be simplified to 2m2 + 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 = m2 – n2 = (20 * 20) – (5 * 5) = 375

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

c = m2 + n2 = (20 * 20) + (5 * 5) = 425

a + b + c = 1000 and 2 + b2 = c2, 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.

### Battles With the Android

Tuesday, August 10th, 2010

As I have posted previously I am converting my PC application SecureIT to the Android smartphone platform. (more…)

### 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

### Project Euler: Problem 7

Sunday, August 8th, 2010

This is problem 7 on Project Euler:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?