the solution is hère.

I thi nk it is a problem inChina

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

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]

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 $.

Old imo lol

Paragdey12 wrote: Old imo lol It is from IMO $1979$