Test 1

Algebra Difficulty 5.3 AIME, harder Prove it Hong Kong

Suppose $p$ and $q$ are positive integers such that $\frac{p}{q} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{1335}$ . Show that 2003 is a factor of $p$ .

Solution

$\begin{align*} & 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots - \frac{1}{1334} + \frac{1}{1335} \\ &= \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{1334} + \frac{1}{1335}\right) - 2\left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{1334}\right) \\ &= \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{1334} + \frac{1}{1335}\right) - \left(1 + \frac{1}{2} + \dots + \frac{1}{667}\right) \\ &= \frac{1}{668} + \frac{1}{669} + \dots + \frac{1}{1335} \\ &= \left(\frac{1}{668} + \frac{1}{1335}\right) + \left(\frac{1}{669} + \frac{1}{1334}\right) + \dots + \left(\frac{1}{1001} + \frac{1}{1002}\right) \\ &= 2003 \cdot \left(\frac{1}{668 \cdot 1335} + \frac{1}{669 \cdot 1334} + \dots + \frac{1}{1001 \cdot 1002}\right). \end{align*}$
One may check that 2003 is a prime. Thus, this factor cannot be cancelled out by the denominator. Thus, $p$ must be divisible by 2003.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.