Crazy math problems.

via Project Euler

My Solution:

Code:

[SPOILER]            int result = 0;
            for (int i = 0; i < 1000; i++)
            {
                if (i%3 == 0 || i%5 == 0)
                {
                    Console.Write(i + ", ");
                    result += i;
                }
            }
            Console.WriteLine("");
            Console.WriteLine(result);
            Console.ReadLine();[/SPOILER]

Multiples of 3 or 5 below 1000:

Code:

[spoiler]3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42, 4
5, 48, 50, 51, 54, 55, 57, 60, 63, 65, 66, 69, 70, 72, 75, 78, 80, 81, 84, 85, 8
7, 90, 93, 95, 96, 99, 100, 102, 105, 108, 110, 111, 114, 115, 117, 120, 123, 12
5, 126, 129, 130, 132, 135, 138, 140, 141, 144, 145, 147, 150, 153, 155, 156, 15
9, 160, 162, 165, 168, 170, 171, 174, 175, 177, 180, 183, 185, 186, 189, 190, 19
2, 195, 198, 200, 201, 204, 205, 207, 210, 213, 215, 216, 219, 220, 222, 225, 22
8, 230, 231, 234, 235, 237, 240, 243, 245, 246, 249, 250, 252, 255, 258, 260, 26
1, 264, 265, 267, 270, 273, 275, 276, 279, 280, 282, 285, 288, 290, 291, 294, 29
5, 297, 300, 303, 305, 306, 309, 310, 312, 315, 318, 320, 321, 324, 325, 327, 33
0, 333, 335, 336, 339, 340, 342, 345, 348, 350, 351, 354, 355, 357, 360, 363, 36
5, 366, 369, 370, 372, 375, 378, 380, 381, 384, 385, 387, 390, 393, 395, 396, 39
9, 400, 402, 405, 408, 410, 411, 414, 415, 417, 420, 423, 425, 426, 429, 430, 43
2, 435, 438, 440, 441, 444, 445, 447, 450, 453, 455, 456, 459, 460, 462, 465, 46
8, 470, 471, 474, 475, 477, 480, 483, 485, 486, 489, 490, 492, 495, 498, 500, 50
1, 504, 505, 507, 510, 513, 515, 516, 519, 520, 522, 525, 528, 530, 531, 534, 53
5, 537, 540, 543, 545, 546, 549, 550, 552, 555, 558, 560, 561, 564, 565, 567, 57
0, 573, 575, 576, 579, 580, 582, 585, 588, 590, 591, 594, 595, 597, 600, 603, 60
5, 606, 609, 610, 612, 615, 618, 620, 621, 624, 625, 627, 630, 633, 635, 636, 63
9, 640, 642, 645, 648, 650, 651, 654, 655, 657, 660, 663, 665, 666, 669, 670, 67
2, 675, 678, 680, 681, 684, 685, 687, 690, 693, 695, 696, 699, 700, 702, 705, 70
8, 710, 711, 714, 715, 717, 720, 723, 725, 726, 729, 730, 732, 735, 738, 740, 74
1, 744, 745, 747, 750, 753, 755, 756, 759, 760, 762, 765, 768, 770, 771, 774, 77
5, 777, 780, 783, 785, 786, 789, 790, 792, 795, 798, 800, 801, 804, 805, 807, 81
0, 813, 815, 816, 819, 820, 822, 825, 828, 830, 831, 834, 835, 837, 840, 843, 84
5, 846, 849, 850, 852, 855, 858, 860, 861, 864, 865, 867, 870, 873, 875, 876, 87
9, 880, 882, 885, 888, 890, 891, 894, 895, 897, 900, 903, 905, 906, 909, 910, 91
2, 915, 918, 920, 921, 924, 925, 927, 930, 933, 935, 936, 939, 940, 942, 945, 94
8, 950, 951, 954, 955, 957, 960, 963, 965, 966, 969, 970, 972, 975, 978, 980, 98
1, 984, 985, 987, 990, 993, 995, 996, 999[/spoiler]

 

2300000?

 

My Solution:

Code:

[spoiler]            long current = 2;
            long previous = 1;
            long hold = 0;
            long result = 2;
            while (current <= 4000000)
            {
                hold = current;
                current = previous + current;
                previous = hold;
                if (current%2 == 0)
                {
                    Console.Write(current + ", ");
                    result += current;
                }

            }
            Console.WriteLine("");
            Console.WriteLine(result);
            Console.ReadLine();[/spoiler]

Even values in the Fibonacci sequence below 4 million:

Code:

[spoiler]8, 34, 144, 610, 2584, 10946, 46368, 196418, 832040, 3524578[/spoiler]

 

I opt to use a calculator or a Stephen Hawkings.

 

Nope try again.

 

The answer is

Spoiler

233168

calculated in 0.010s

In perl:

Code:

sub multi_under {
  my @pass = @_;
  my $sum = 0;
  for (my $counter = 1; $counter *  $pass[0] < $pass[1]; $counter++) {
    $sum += $pass[0] * $counter if $pass[0] * $counter % $pass[2] != 0;
  }
  return $sum;
}

my $total = multi_under(3, 1000, 5);
$total += multi_under(5, 1000, 1000);

print "$total\n";

While Arimil's would run faster with one pass I wrote mine out of habit by making the code fixable and reusable however I don't know why I ever would though.

 

Answer:

Spoiler

4613732

calculated in 0.009s

In Perl:

Code:

my $high = 1;
my $low = 1;
my $count = 0;
my $sum = 2;

while($sum < 4000000) {
  $count += $sum if $sum % 2 == 0;
  $low = $high;
  $high = $sum;
  $sum = $high + $low;
}

print "$count\n";

 

Alright lets try #3:

My solution in Perl took 0.012s

Code:

[SPOILER]sub factors {
  my @pass = @_;
  my @factors,
  $pass[1]++ while($pass[0] % $pass[1] != 0);
  push(@factors,$pass[1]);
  push(@factors, factors($pass[0]/$pass[1], $pass[1])) if $pass[0] > $pass[1];
  return @factors;
}

my @prime = factors(600851475143, 2);
print pop(@prime) . "\n";[/SPOILER]

I love recursion ^_^

 

I hate math, I hope someone figures this out.

 

You destroyed me on this one, mines been running for 2 days and still hasn't found it. '

EDIT: And... I just realized I don't know what a prime factor is and was trying to do it wrong.

My solution:

Code:

[spoiler]        public static bool isprime(long num)
        {
            for (int i = 2; i < num; i++)
            {
                if (num%i == 0)
                {
                    return false;
                }
            }
            return true;
        }
        static void Main(string[] args)
        {
            long val = 600851475143;
            List<long> factors = new List<long>();
            for (int i = 2; i < val; i++)
            {
                if (val%i == 0)
                {
                    factors.Add(i);
                    val = val / i;
                    i = 1;
                }
            }
            factors.Add(val);
            long result = 0;
            foreach (int factor in factors)
            {
                if (factor > result && isprime(factor))
                {
                    result = factor;
                }
            }
            Console.WriteLine(result);
            Console.ReadLine();
        }[/spoiler]

 

More of a coding question than a math question.

 

#4

My Solution:

Code:

[spoiler]        public static bool ispalindrome(string num)
        {
            char[] numbers = num.ToCharArray();
            for (int i = 1; i <= num.Length / 2; i++)
            {
                if (numbers[i - 1] != numbers[num.Length - i])
                {
                    return false;
                }
            }
            return true;
        }
        static void Main(string[] args)
        {
            int i2 = 100;
            int val = 0;
            int solution = 0;
            for (int i = 100; i < 1000; i++)
            {
                val = i * i2;
                if (val > solution && ispalindrome(val.ToString()))
                {
                    solution = val ;
                }
                if (i == 999 && i2 < 999)
                {
                    i = 100;
                    i2++;
                }
            }
            Console.WriteLine(solution);
            Console.ReadLine();
        }[/spoiler]

It asks you which language you use to solve the problems, there is an option for pencil and paper.

 

Coding scares the hell out of me, I could barely figure out grade ten programming class

 

Don't lie, don't care. <3

 

Consider yourself lucky, my school was technically retarded. Best tech class we had was HTML and they taught it all wrong.

 

Looks a lot like discreet structures which I'm taking this semester. My teach focused more on the mathmatics and algorithims instead of using that math to make programs.

 

I realized there was a site the second time around looking at the link lol. But the first paragraph is this.

No pencil is going to solve these problems in under a minute heh.

 

Dear Math,

Solve your own problems.

 

906609 by multiplying 993x913

Paper, pencil, calculator...figured for the highest palindrome, the last digit has to be a 9, since the first digit will most likely be a 9. Only three single digit integer products end in a 9 (1x9 or 3x3 or 7x7), so I decided to check the 3x3 possibilities first.

Fun fact: all (at least most) 6 digit palindromes are divisible by 11.

So at least one of the numbers has to be divisible by 11. The only number >900 that is divisible by 11 and ends in a 3 is 913. Then I started multiplying by numbers in the 900s ending in 3 so the last digit of the palindrome would be 9. Luckily the first one I tried (and the highest possible one), 993, created the product 906609, a palindrome.

Logic + guess and check ftw

 

Your better at math than me. I'll go with my solution of making a calculator to do it for me.