Approximating π

I found this written in the back of an old Math book today:
[math] \displaystyle \pi \approx \frac{208341}{66371}= \frac{A}{17}+ \frac{B}{47}+ \frac{C}{83}[/math]
and it appears to be within one-half part per trillion of the actual value. Find the values of A, B, and C, if you're hard enough.

Do your own homework.

>and it appears to be within one-half part per trillion of the actual value.
You sure about that, m8?

OP is a faggot and can't type 66317 properly.

probably why he can't do his homework

hard to decompose into partial fractions when you start with the wrong factors

oops

You'd think that with modern super computers, the last digit of pie would be found by now.

Pi actually ends with 51413, it's a palindromic number.

How am I suppose to solve this?
>A B C

>66371 = 17 * 47 *83
try harder baguette

To fulfill this, you'd need, for example, this equality to be true as well:
6 = 8a (mod 17)

Up to you to figure out why.

you doing chemistry mate ?
what uni you at ?

The actual number es 66317, not 66371...

>homework
No, when I wrote that I found this written in the back of an old book (Mathematical Handbook by Murray R.Spiegel) that was true, unlike what you wrote.

Correct, I transposed the 1 and 7 because fatigue.

Stop talking like a stuck-up faggot. Just talk like a normal person and say "I fucked up".

stop talking like an inbred faggot and talk like a normal person

Find out how to solve linear Diophantine equations in three variables.
I found this solution:
A = -64169028
B = 176464827
C = 1666728

>it appears to be within one-half part per trillion of the actual value
it's only ~39 ppb by my calculation.

I found a other approximation which is slightly more accurate:
[math]\frac{312689}{99532} = \frac{A}{4} + \frac{B}{149} + \frac{C}{167}[/math]

positive integers A, B, C are all
smaller than your value for C

lol Americans...

your inferiority is showing

too lazy to find the particular solution :^)

[math]
\frac{5}{17} + \frac{37}/{47} + \frac{171}/{83}
[/math]

baka

[math]
\frac{5}{17} + \frac{37}{47} + \frac{171}{83}
[/math]

How did you arrive?

it's pretty trivial after re-expressing it as
[math]
799\,C+1411\,B+3901\,A=208341
[/math]
so trivial, in fact, that i used a CAS to give the general solution

I plugged it into Mathematica, but I don't know how to specify that I want only natural number solutions.

Does anybody know how to do that?

Reduce[3901 A + 1411 B + 799 C == 208341, {A, B, C}, Integers]
:^)

Cool, thx
(I see ", Integers" also works with Solve)

found this solution as well

[eqn]\frac{312689}{99532} = \frac{3}{4} + \frac{144}{149} + \frac{238}{167}[/eqn]

>within one-half part per trillion
>it's only ~39 ppb by my calculation
We were both wrong, it's 39 per trillion.

I'm a rebel
[eqn]\frac{22}{17} + \frac{37}{47} + \frac{87}{83}[/eqn]

>french
>page in the background literally says croissante
Really makes you think.

here's the solution I found with some modular algebra:
[math]208341=(47)(83)A+(17)(83)B+(17)(47)C[/math]
remainders on division by 17:
[math]6 \equiv 8A(mod~17)[/math] so A = 5,22,39...
remainders on division by 47:
[math]37 \equiv B(mod~47)[/math] so B = 37,84,131...
remainders on division by 83:
[math]11 \equiv 52C(mod~83)[/math] so C = 5,88,171...
picking B = 37 (for example) gives the choices
(A, C) = (5, 171), (22, 88), (39, 5)

>3.1295444607
Pi as fuck.

>Joke's on you I was only pretending to be retarded