p is divisible by 2003

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Source: Pan African 2000

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3498 posts

#1 h Oct 3, 2005, 2:45 PM • 2 Y

Y by Adventure10, Mango247

Let $p$ and $q$ be coprime positive integers such that:
$\dfrac{p}{q}=1-\frac12+\frac13-\frac14 \cdots -\dfrac{1}{1334}+\dfrac{1}{1335}$
Prove $p$ is divisible by 2003.

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#2 h Oct 3, 2005, 10:58 PM • 2 Y

Y by Adventure10, Mango247

the solution is hère.

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#3 h May 17, 2006, 11:09 AM • 2 Y

Y by Adventure10, Mango247

I thi nk it is a problem inChina

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#4 h May 17, 2006, 6:12 PM • 2 Y

Y by Adventure10, Mango247

There is a more generalized problem
If p is grater than 3, q=[2p/3] and let m and n be positive integers such that

m/n = 1- 1/2 +1/3 -1/4 + ... +(-1)exp(q-1) 1/q

then m is divisible by p

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#5 h May 17, 2006, 6:24 PM • 2 Y

Y by Adventure10, Mango247

You just use Catalan Identity And then group since 1/([(q+1)/2]+k) + 1/(q-k)=p/(..)
(where k is 0,1,2...,(q-1)/2 )
And then p divides m

1335 = [2*2003/3]

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#6 h Mar 22, 2017, 12:02 PM • 3 Y

Y by Dattier, RockmanEX3, Adventure10

Given,
$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1334}+\frac{1}{1335}$

We notice that the negative terms have even denominators. Now we write each fraction of the form $\frac{1}{2k}$ as ,

$\frac{1}{2k} - \frac{1}{k}$

So our given quantity can be written as,

$\frac{p}{q}=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1334}+\frac{1}{1335}) - 2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1334})$

$= (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1334}+\frac{1}{1335}) - (\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{667})$

$= (\frac{1}{668}+\frac{1}{669}+\frac{1}{670}+...+\frac{1}{1334}+\frac{1}{1335})$

Now, we notice

$\frac{1}{668 + k} + \frac {1}{1335-k} = \frac {2003}{(668+k)(1335-k)}$ (Partial Fractions)

Therefore,

we can write , $\frac{p}{q} = ( \frac{1}{668} + \frac{1}{1335}) + \cdots + (\frac{1}{1001} + \frac{1}{1002})$

$= \frac {2003}{668.1335} + \frac {2003}{669.1334} + \cdots + \frac{2003}{1001.1002}$

$= 2003\frac{p'}{q'}$ , where $\frac{p'}{q'}$ is the sum of the fractions $\frac {1}{668.1335} , \frac {1}{669.1334} , \cdots , \frac{1}{1001.1002}$ .

$q'$ is the product of the denominators of those fractions, and it is relatively prime to 2003, so the fraction $2003\frac{p'}{q'}$ is in its lowest terms, so $2003p' = p$ and therefore $p$ is divisible by $2003$ .

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#7 h Jan 6, 2018, 7:46 PM • 3 Y

Y by Adventure10, ericxyzhu, Mango247

Old imo lol

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#8 h Jun 20, 2019, 12:45 AM • 3 Y

Y by ericxyzhu, Adventure10, Mango247

Paragdey12 wrote:

Old imo lol :rotfl:

:rotfl:

It is from IMO $1979$

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#9 h Apr 24, 2025, 4:40 PM • 1 Y

Y by ATM_

Just IMO 1979 P1 idea

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