64Digits

MATH CHALLENGE

Posted by Glen on Sept. 1, 2010, 8:08 p.m.

So uh.

a^2 + b^2 + c^2 = 2009

What's a,b,c. There's more than one possibility. My professor asked this question today. No one got it. (Including me. idk the answer).

Comments

Killpill28 14 years, 2 months ago

sqrt(2009/3) = 25.87791851

a, b, and c are these.

I couldn't make the fraction.

As long as a, b and c add up to 2009/3, it works. Right?

DesertFox 14 years, 2 months ago

One answer is:

8, 24, 37

and so is:

22, 25, 30

Edit: Meh, I was shh-ing these to Glen but since _Player_ spoiled…

All possible integer answers (excluding answers with 0 as a number):

(2, 18, 41)

(2, 22, 39)

(3, 8, 44)

(3, 20, 40)

(4, 12, 43)

(6, 23, 38)

(7, 14, 42)

(8, 24, 37)

(9, 22, 38)

(12, 29, 32)

(16, 27, 32)

(18, 23, 34)

(21, 28, 28)

(22, 25, 30)

noshenim 14 years, 2 months ago

Here's my solution xD:

Quote:

#include <cstdlib>

#include <iostream>

#include <fstream>

#include <math.h>

using namespace std;

#define PREC 256

#define EPSILON 0.0000000001

#define rt2009 44.821869662029940805417523722353

int main(int argc, char *argv[])

{

ofstream thefile;

thefile.open("result.txt");

double a,b,c;

for(a=0;a<=rt2009;a+=rt2009/PREC){

for(b=0;b<=sqrt(2009-a*a);b+=sqrt(2009-a*a)/PREC){

if(a*a+b*b<=2009){

c = sqrt( 2009-a*a-b*b );

if( fabs(a*a+b*b+c*c-2009.0f) <= EPSILON ){

printf("A: %0.0f, B: %0.0f, C: %0.0f; \n",a,b,c);

thefile << "A: " << a << ",B: " << b << ",C: " << c << "\n";}}}}

thefile.close();

while(1){ /* lololol */ }

return EXIT_SUCCESS;

}

(warning – 2mb) http://64digits.com/users/_Player_/result.txt

I don't know the algebraic solution though, although there is probably none…

Integer solutions:

A: 0,B: 28,C: 35

A: 0,B: 35,C: 28

A: 2,B: 18,C: 41

A: 2,B: 22,C: 39

A: 2,B: 39,C: 22

A: 2,B: 41,C: 18

A: 3,B: 8,C: 44

A: 3,B: 20,C: 40

A: 3,B: 40,C: 20

A: 3,B: 44,C: 8

A: 4,B: 12,C: 43

A: 4,B: 43,C: 12

A: 6,B: 23,C: 38

A: 6,B: 38,C: 23

A: 7,B: 14,C: 42

A: 7,B: 42,C: 14

A: 8,B: 3,C: 44

A: 8,B: 24,C: 37

A: 8,B: 37,C: 24

A: 8,B: 44,C: 3

A: 9,B: 22,C: 38

A: 9,B: 38,C: 22

A: 12,B: 4,C: 43

A: 12,B: 29,C: 32

A: 12,B: 32,C: 29

A: 12,B: 43,C: 4

A: 14,B: 7,C: 42

A: 14,B: 42,C: 7

A: 16,B: 27,C: 32

A: 16,B: 32,C: 27

A: 18,B: 2,C: 41

A: 18,B: 23,C: 34

A: 18,B: 34,C: 23

A: 18,B: 41,C: 2

A: 20,B: 3,C: 40

A: 20,B: 40,C: 3

A: 21,B: 28,C: 28

A: 22,B: 2,C: 39

A: 22,B: 9,C: 38

A: 22,B: 25,C: 30

A: 22,B: 30,C: 25

A: 22,B: 38,C: 9

A: 22,B: 39,C: 2

A: 23,B: 6,C: 38

A: 23,B: 18,C: 34

A: 23,B: 34,C: 18

A: 23,B: 38,C: 6

A: 24,B: 8,C: 37

A: 24,B: 37,C: 8

A: 25,B: 22,C: 30

A: 25,B: 30,C: 22

A: 27,B: 16,C: 32

A: 27,B: 32,C: 16

A: 28,B: 0,C: 35

A: 28,B: 21,C: 28

A: 28,B: 28,C: 21

A: 28,B: 35,C: 0

A: 29,B: 12,C: 32

A: 29,B: 32,C: 12

A: 30,B: 22,C: 25

A: 30,B: 25,C: 22

A: 32,B: 12,C: 29

A: 32,B: 16,C: 27

A: 32,B: 27,C: 16

A: 32,B: 29,C: 12

A: 34,B: 18,C: 23

A: 34,B: 23,C: 18

A: 35,B: 0,C: 28

A: 35,B: 28,C: 0

A: 37,B: 8,C: 24

A: 37,B: 24,C: 8

A: 38,B: 6,C: 23

A: 38,B: 9,C: 22

A: 38,B: 22,C: 9

A: 38,B: 23,C: 6

A: 39,B: 2,C: 22

A: 39,B: 22,C: 2

A: 40,B: 3,C: 20

A: 40,B: 20,C: 3

A: 41,B: 2,C: 18

A: 41,B: 18,C: 2

A: 42,B: 7,C: 14

A: 42,B: 14,C: 7

A: 43,B: 4,C: 12

A: 43,B: 12,C: 4

A: 44,B: 3,C: 8

A: 44,B: 8,C: 3

DesertFox 14 years, 2 months ago

@Player - I'm assuming that he means integers only. Also, there's a much simpler way to do it:

int max_val=ceil(sqrt(2009));//maximum viable number (0^2 + 0^2 + max_val^2)

for(int x = 1; x <= max_val; x++)
{
[tab]for(int y = x; y <= max_val; y++)
[tab]{
[tab][tab]for(int z = y; z <= max_val; z++)
[tab][tab]{
[tab][tab][tab]if(x*x + y*y + z*z == 2009)
[tab][tab][tab][tab]cout << x << "," << y << "," << z << "\n";
[tab][tab]}
[tab]}
}

Not only will this code run faster, it also removes duplicate entries like (A: 44,B: 8,C: 3) and (A: 3,B: 8,C: 44)

Josea 14 years, 2 months ago

Also, you can consider that as the sphere centered on (0,0,0) with radius sqrt(2009), and then you would have infinite real solutions.

sirxemic 14 years, 2 months ago

Yeah, that is the answer: the sphere with center point (0,0,0) with radius sqrt(2009). The perfect representation of the infinite number of answers :D

Scott_AW 14 years, 2 months ago

a=3

b=20

c=40

Mordi 14 years, 2 months ago

Wouldn't be too hard to write a small program that could "brute force" it, but that's probably considered cheating.

noshenim 14 years, 2 months ago

@Mordi

Desert fox and I have each done that and posted the codes >_>

Glen 14 years, 2 months ago

Thanks :D