Problem 3

Find all positive integers

k<202

for which there exists a positive integer

n

such that

\left\{\frac{n}{202}\right\}+\left\{\frac{2n}{202}\right\}+\cdots+\left\{\frac{kn}{202}\right\}=\frac{k}{2},

where

\{x\}

denotes the fractional part of

x

. (Here

\{x\}

is the real number

r

with

0\le r<1

such that

x-r

is an integer.)

Step 3 of 5: Reduce modulo the prime 101

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101\mid k(k+1)\ \Longrightarrow\ k\in\{100,101,201\}\quad(1\le k<202)

Detailed analysis

Reducing $nk(k+1)-404z=202k$ modulo $101$ (noting $202=2\cdot101$ and $404=4\cdot101$ ) gives $101\mid nk(k+1)$ . Since $101$ is prime and $101\nmid n$ , this forces $101\mid k(k+1)$ , i.e. $101\mid k$ or $101\mid k+1$ . Together with $1\le k<202$ this leaves exactly $k\in\{100,101,201\}$ .