[SOLVED] Show that p is divisible by 641?

• May 21st 2009, 10:20 AM
fardeen_gen
[SOLVED] Show that p is divisible by 641?
If $p$ and $q$ are positive integers such that $\frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}$, show that $p$ is divisible by 641.
• May 21st 2009, 10:02 PM
CaptainBlack
Quote:

Originally Posted by fardeen_gen
If $p$ and $q$ are positive integers such that $\frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}$, show that $p$ is divisible by 641.

The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

CB
• May 21st 2009, 10:21 PM
fardeen_gen
Even I couldn't find the general term. Thats why I posted.
• May 21st 2009, 11:50 PM
Swlabr
Quote:

Originally Posted by CaptainBlack
The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

CB

Yes they are as 3 divides 480 properly, 3.160 = 480. I believe the series is,

$\displaystyle\sum_{i=1}^{160} 1/(3i-2) +
\displaystyle\sum_{i=1}^{160} 1/(3i-1) - \displaystyle\sum_{i=1}^{160} 2/3i$
$
= \displaystyle\sum_{i=1}^{160} (1/(3i-2) + 1/(3i-1) - 2/3i) =
\displaystyle\sum_{i=1}^{160} (9i-4)/(3i-2)(3i-1)3i$
.

I'm sure that there is a neater what though! I hope that helps.
• May 21st 2009, 11:54 PM
fardeen_gen
Quote:

Originally Posted by CaptainBlack
The last three terms above are not consistent with those before the ellipsis, can you give the expression for the general term in this summation.

CB

Quote:

Originally Posted by Swlabr
Yes they are as 3 divides 480 properly, 3.160 = 480. I believe the series is,

$\displaystyle\sum_{i=1}^{160} 1/(3i-2) +
\displaystyle\sum_{i=1}^{160} 1/(3i-1) - \displaystyle\sum_{i=1}^{160} 2/3i$
$
= \displaystyle\sum_{i=1}^{160} (1/(3i-2) + 1/(3i-1) - 2/3i) =
\displaystyle\sum_{i=1}^{160} (9i-4)/(3i-2)(3i-1)3i$
.

I'm sure that there is a neater what though! I hope that helps.

We have the series then. Anyone with any ideas on proving the required divisibility?
• May 22nd 2009, 12:04 AM
Swlabr
Quote:

Originally Posted by fardeen_gen
We have the series then. Anyone with any ideas on proving the required divisibility?

• May 22nd 2009, 12:13 AM
fardeen_gen
641 seems to be completely random to me. I do not see any pattern.
• May 22nd 2009, 12:23 AM
Moo
Hi,

I found the same sum as above.

And as for 641, notice that it is prime. So I guess the result of the summation will have a factor 641 in the numerator.
That's all I can guess about.

I tried to compute $\int_0^1 \frac{1-x^{480}}{1-x^3} ~dx=\int_0^1 \sum_{k=1}^{160} x^{3k-3} ~dx=\sum_{k=1}^{160}\frac{1}{3k-2}$

But didn't give it enough time (it was late :p)...
• May 22nd 2009, 12:35 AM
Swlabr
Quote:

Originally Posted by fardeen_gen
641 seems to be completely random to me. I do not see any pattern.

641 is prime, as Moo mentioned, so it does not divide the denominator of our fraction, as the denominator is just 1.2.3. ... .480. Thus, if 641 divides the numerator of our product it will divide every such numerator (this is not clear as 3 divides the numerator but also the denominator in our sum).
• May 22nd 2009, 01:44 AM
Laurent
Quote:

Originally Posted by fardeen_gen
641 seems to be completely random to me. I do not see any pattern.

I have no idea how it would come up in this problem, but 641 doesn't seem random to me: it reminds me immediately the proof by Euler that the Fermat number $F_5=2^{2^5}+1$ is not prime (thus breaking down Fermat's conjecture). Euler showed that 641 divides $F_5$. I guess this is not a coincidence, but I may be totally wrong. If it may help anyone...
• May 22nd 2009, 05:10 AM
Isomorphism
Quote:

Originally Posted by fardeen_gen
[SIZE=3]If $p$ and $q$ are positive integers such that $\frac{p}{q} = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \frac{1}{7} + \frac{1}{8} - \frac{2}{9} + \mbox{...} + \frac{1}{478} + \frac{1}{479} - \frac{2}{480}$, show that $p$ is divisible by 641.

Lets say $S = 1 + \frac{1}{2} - \frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} + \mbox{...} +$ $\frac{1}{478} + \frac{1}{479} - \frac{2}{480}$

$S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \mbox{...} + \frac{1}{478} +$ $\frac{1}{479} + \frac{1}{480} - 3\left(\frac13 + \frac{1}{6} + \frac{1}{9} + \mbox{...} +\frac1{480}\right)$

$S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \mbox{...} + \frac{1}{478}$ $+ \frac{1}{479} + \frac{1}{480} - \left(1 + \frac{1}{2} + \frac{1}{3} + \mbox{...} +\frac1{160}\right)$

$S = \frac1{161} + \frac{1}{162} + \frac{1}{163} + \frac{1}{164} + \mbox{...} +\frac{1}{480}$

Now group terms:

$S = \left(\frac1{161} + \frac{1}{480}\right) +\left(\frac1{162} + \frac{1}{479}\right) \mbox{...} +$ $\left(\frac1{320} + \frac{1}{321}\right)$

$S = \frac{641}{161 \times 480} +\frac{641}{162 \times 479} + \mbox{...} +\frac{641}{320 \times 321}$

Finally note that 641 will remain on the top always because its prime and all the denominator fellows are less than 641. So done (Happy)

Nice problem!
• May 22nd 2009, 05:17 AM
Isomorphism
Quote:

Originally Posted by Laurent
I have no idea how it would come up in this problem, but 641 doesn't seem random to me: it reminds me immediately the proof by Euler that the Fermat number $F_5=2^{2^5}+1$ is not prime (thus breaking down Fermat's conjecture). Euler showed that 641 divides $F_5$

Thats exactly what I thought when he said it looks random. Its a very famous Number Theory anecdote.
• May 22nd 2009, 06:47 AM
TiRune
Brilliant proof isomorphism o_O