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 5 of 5: Conclude

Previous step Next step

k\in\{1,100,101,201\}

Detailed analysis

Combining the trivial case $k=1$ (from $n=101$ ) with the three candidates confirmed by explicit constructions, the full answer is $k\in\{1,100,101,201\}$ .