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Pythagorean triplets.

edothero

Got it. @Zacken could you close this please

Question was:

There is exactly one Pythagorean triplet for which

a + b + c = 1000

Find the product of

abc

(edited 8 years ago)

Oh ok nm

Original post by Bobjim12

Oh ok nm

Feel free to share your solution :biggrin:

Reply 3

Bobjim12

Original post by edothero

Feel free to share your solution :biggrin:

Oh no, i wasn't really on a solution, i wasn't too sure what you were looking for so i was trying to clarify it, but i guess it doesn't matter anymore XD

Reply 4

edothero

Original post by Bobjim12

Oh no, i wasn't really on a solution, i wasn't too sure what you were looking for so i was trying to clarify it, but i guess it doesn't matter anymore XD

Lmao yeah xD the solution is
a = 200
b = 375
c = 425

a^2 + b^2 = c^2
and
a+b+c = 1000
:biggrin:

Reply 5

Bobjim12

Original post by edothero

Lmao yeah xD the solution is
a = 200
b = 375
c = 425

a^2 + b^2 = c^2
and
a+b+c = 1000
:biggrin:

How did you get it? I like seeing code!!

Reply 6

Zacken

Original post by edothero

Got it. @Zacken could you close this please

I'll leave it open for the moment - looks like some people are interested in it. :biggrin:

Reply 7

Bobjim12

Original post by Zacken

I'll leave it open for the moment - looks like some people are interested in it. :biggrin:

heeeeeeeeey, i'm just curious :frown:

Reply 8

edothero

Original post by Zacken

I'll leave it open for the moment - looks like some people are interested in it. :biggrin:

That's fine, thank you anyway :biggrin:

Original post by Bobjim12

How did you get it? I like seeing code!!

Here's the code: I used Java.

public class Ptrips { 

 public static void main(String[] args) { 

 int a = 0; 
 int b = 0; 
 int c = 0; 
 int n = 1000; 
 boolean solution = false;

 for(a = 1; a < n; a++){ 
       for(b = a; b < n; b++) { 
             c = n - a - b; 


 if ((a*a) + (b*b) == (c*c)) { 
       solution = true; 
           break; 
      } 
 } 
        if(solution){ 
          System.out.println(a + ", " + b + " and " + c);	
          break; 
      } 
    } 
 }

(edited 8 years ago)

Reply 9

Bobjim12

Original post by edothero

Thanks fine, thank you anyway :biggrin:

.

Here's the code: I used Java.

public class Ptrips { 

 public static void main(String[] args) { 

 int a = 0; 
 int b = 0; 
 int c = 0; 
 int n = 1000; 
 solution = false;

 for(a = 1; a < n; a++){ 
       for(b = a; b < n; b++) { 
             c = n - a - b; 


 if ((a*a) + (b*b) == (c*c)) { 
       solution = true; 
           break; 
      } 
 } 
        if(found){ 
          System.out.println(a + ", " + b + ", " + c);	
          break; 
      } 
    } 
 }

Ok, i did misunderstand what you were asking,

I'm not sure if it would work but instead of having the c = n- a -b statement you could combine that with the if statement, if you want it to be slightly more efficient but if it works you probably don't want to bother!

how do indices work in Java? does 2**2 = 4?

(edited 8 years ago)

Reply 10

edothero

Original post by Bobjim12

I think if I combine it with the if statement then I would need two of them! One for

a^{2}+b^{2}=c^{2}

and one for

c = n - a - b

The only other thing I can think of atm is

if((c = n - a - b) && (a*a + b*b == c*c)) {
  solution = true;
  break;
}

Which would definitely be more efficient, good spot!

As for the indices thing, I'm not quite sure you know :rofl:

Give me a sec.

Aha, got it; unlike languages like Python you can't do a**b for

a^{b}

You need to use Math.pow(n, n1)Which would return

n^{n_{1}}

Though it would return it as a double (i.e)System.out.println(Math.pow(2, 2));would return 4.0

(edited 8 years ago)

Reply 11

Bobjim12

Original post by edothero

I think if I combine it with the if statement then I would need two of them! One for

a^{2}+b^{2}=c^{2}

and one for

c = n - a - b

The only other thing I can think of atm is

if((c = n - a - b) && (a*a + b*b == c*c)) {
     solution = true;
     break;
}

That.should work, if i am in the right frame of mind right now!

Reply 12

Zacken

Original post by Bobjim12

how do indices work in Java? does 2**2 = 4?

I think the usual thing is Math.Pow(a, b).

Reply 13

Bobjim12

Original post by Zacken

I think the usual thing is Math.Pow(a, b).

Hmm, that is interesting.

you could then just do Math.Pow(n-a-b, 2)

Assuming i am getting the args right.

Reply 14

edothero

Original post by Bobjim12

That.should work, if i am in the right frame of mind right now!

Check the post again, I've made a few edits :biggrin:

Zacken has beat me to it though, the guys unstoppable :biggrin:

Reply 15

Zacken

Original post by Bobjim12

Hmm, that is interesting.

you could then just do Math.Pow(n-a-b, 2)

Assuming i am getting the args right.

That would be (n-a-b)^2, yeah.

Reply 16

Student403

Original post by Zacken

That would be (n-a-b)^2, yeah.

Slightly off topic but do you know if you do much using programming in the Cambridge maths course?

Reply 17

Bobjim12

Splendid, well, this was fun haha!

Reply 18

Zacken

Original post by Student403

Slightly off topic but do you know if you do much using programming in the Cambridge maths course?

Very little in terms of using MatLab for a lengthy computation project in second and third years called "Catam", details should be available online, I think.

Reply 19

Student403

Original post by Zacken

Very little in terms of using MatLab for a lengthy computation project in second and third years called "Catam", details should be available online, I think.