[Monte314]

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[TheLastMohican]

Yet another that looks so simple...

[tp6626]

I remember seeing this one in New Scientist when I was 17 (I was lucky enough to have a subscription back then). There was a £15 prize. I never looked out for the solution though, I think it was a bit of a stumper!

I'm not going to attempt it, mathematics like this just makes my head hurt nowadays. (I feel too old for it!).

[Monte314]

	Originally Posted by tp6626 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I remember seeing this one in New Scientist when I was 17 (I was lucky enough to have a subscription back then). There was a £15 prize. I never looked out for the solution though, I think it was a bit of a stumper! I'm not going to attempt it, mathematics like this just makes my head hurt nowadays. (I feel too old for it!).

This particular problem was formulated by yours truly this very afternoon. But I have no doubt that there are many like it out there.

[Rudy]

Solution:

( Show Spoiler )

24965, 24964. Typing up how I got the answer now...

Okay, I started by trying to factor 1869678781. Surprise! Turns out that it's a prime number.

So, let us note that N = M + C, for some positive number C.

N^3 = (M+C)^3 = M^3 + 3*M^2*C + 3*M*C^2 + C^3 = M^3 + 1869678781.

SO:

3*M^2*C + 3*M*C^2 + C^3 = 1869678781

HOWEVER, since 1869678781 is prime, and 1869678781 is divisible by C, C must equal 1.

SO:

3*M^2 + 3*M + 1 = 1869678781

3*M^2 + 3*M - 1869678780 = 0

*Quadratic Formula Magic*

M = 24964. N = M+C = M+1 = 24965.

That was fun. Thanks, Monte!

[Monte314]

OK, rudyhenkel has provided a correct answer. Anybody else?

[tp6626]

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	This particular problem was formulated by yours truly this very afternoon. But I have no doubt that there are many like it out there.

Oh right, hehe.

I just remember someone writing in saying it was a diophantine problem (as they didn't even mention that, they just showed the original problem - and most likely different integers used than yours).


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[Doppelbock]

Here's my solution. I took the lazy man's approach:

( Show Spoiler )

#include <stdlib.h>
#include <stdio.h>

int main( int argc, char* argv[] )
{
long TheNumber = 1869678781;
long n, m, n3, m3;
for (n=1; n<50000; n++) {
for (m=0; m<n; m++) {
n3 = n*n*n;
m3 = m*m*m;
if (n3 - m3 == TheNumber) {
printf( "n: %ld, m: %ld\n", n, m );
exit(0);
}
}
}
printf( "oops, didn't find n or m\n" );
exit(-1);
}

* * *

output:

n: 24965, m: 24964

[Rudy]

Mmm... math plus delicious computer programs. Oh, C++, I remember you fondly...

[Monte314]

Originally Posted by Doppelbock To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Here's my solution. I took the lazy man's approach: ( Show Spoiler ) #include <stdlib.h> #include <stdio.h> int main( int argc, char* argv[] ) { long TheNumber = 1869678781; long n, m, n3, m3; for (n=1; n<50000; n++) { for (m=0; m<n; m++) { n3 = n*n*n; m3 = m*m*m; if (n3 - m3 == TheNumber) { printf( "n: %ld, m: %ld\n", n, m ); exit(0); } } } printf( "oops, didn't find n or m\n" ); exit(-1); } * * * output: n: 24965, m: 24964 DB ;-)

Lazy, perhaps. Correct? Yes.

[Doppelbock]

And extra lazy because I didn't bother to correct the nice code indentation that the forum editor was kind enough to erase for me.

[Monte314]

	Originally Posted by Doppelbock To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	And extra lazy because I didn't bother to correct the nice code indentation that the forum editor was kind enough to erase for me.

I hate it when that happens! You go to all the trouble of completely negating the intellectual value of a worthy pursuit by means of some cheap dodge, and then a scrap of poorly crafted software takes the polish off your gloating. Sheesh.

But, you still have our admiration... for whatever that's worth.

[Doppelbock]

Yeah, your admiration and $4 will buy me a latte. ;-)

[Jonathan Brewer]

Would I be getting it correct if I merely copied and pasted the solutions previously provided?

[nacht]

I solved it by using a search engine, does that matter? Making use of the internet to research a problem is to be encouraged as there could be hidden treasures of mathematics to be discovered beneath the surface of many of these problems. However, there is a fine line between researching ideas and using the answer you found on another website. If you photocopy a crossword solution then what have you achieved?

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[Rudy]

	Originally Posted by Jonathan Brewer To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Would I be getting it correct if I merely copied and pasted the solutions previously provided?

Well, you've highlighted well exactly why there are no prizes offered for getting the problems correct =D

[Monte314]

	Originally Posted by RudyHenkel To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Well, you've highlighted well exactly why there are no prizes offered for getting the problems correct =D

On the Forum, copied answers are fine, as long as you change the font size or something. As we all know, it's appearance that matters, not substance.

I used to teach a numerical analysis class at Penn State, and kids would hand in computer programs they copied from someone else; they'd change the order of the subroutines or something, thinking that would obscure their duplicity. This is how I got my start in automated pattern recognition....

BWA HA HA HA HA HA HA!!!

[sagewolf]

	Would I be getting it correct if I merely copied and pasted the solutions previously provided?

That's no fun. Do like me and just bang your head against your desk until it magically makes sense.

And the answer is clearly marshmallow.

(I will get this...
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)

[Monte314]

I agree with sagewolf. There are some problems I struggled with (off and on) for over 20 years... and there are some I'm still trying to wrestle to the ground.

Of course, if I can solve it in 20 years, it's fair game as homework for my students....

( Show Spoiler )

That big, long number is prime, and the difference of two cubes can be factored....

[AnotherNormal]

My lazy man solution ...

( Show Spoiler )

Did it in Excel vba


Option Explicit

Sub solveformonte()
Dim Done As Boolean
Dim N As Long, M As Long
Const Monte = 1869678781

Done = False
N = 1
M = 1
Do
If N ^ 3 < M ^ 3 + Monte Then
N = N + 1
ElseIf N ^ 3 > M ^ 3 + Monte Then
M = M + 1
Else
Done = True
Range("A1").Value = "N ="
Range("B1").Value = N
Range("A2").Value = "M ="
Range("B2").Value = M
End If
Loop Until Done
End Sub

[SirJac]

Lazy man in python:

( Show Spoiler )

import math
a = False
n = 1231
monte = 18697871
while a == False:
n = n + 1
mrange = int(math.pow((n**3 - monte),0.33333333333))
for m in range(mrange,mrange + 2):
if monte == n**3 - m**3:
a = True
break
print n,m

results: N = 24965 M = 24964

[nacht]

Those of you using code on this one should
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~_^

[Monte314]

	Originally Posted by nacht To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Those of you using code on this one should To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. roblem ~_^

Egyptian fractions! I posted a problem in this area awhile back... very interesting subject.

Elementary number theory has always been interesting to me, as have some more advanced topics (e.g., factorization in extension fields, fourier transforms over finite fields)... but there is this broad area in between involving congruences, residues, and the like that has always left me cold. I think it's the notation; I just don't like the way it looks or "feels".

[Jonathan Brewer]

Almost like it never truly resolves anything?

[Gwargh]

I think I beat everyone in lazyness, since I went with a gut feeling which I guess worked. woot!

( Show Spoiler )

Ok, So I'm going along with a gut feeling (yay for Ni!)
I'm not sure about this particular problem, but I'm getting a strong hunch that: N=M+1

So I'm just going to go ahead with the assumption that this is so and solve as follows:

(M+1)^3 = M^3 + 1869678781
M^3 + 3M^2 + 3M + 1 = M^3 +1869678781
3M^2 + 3M - 1869678780 = 0
Solve for M
M= 24964
N = 24965
Checked it and it fits. Hooray!