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 4 of 5: Exhibit n for each candidate k

Previous step Next step

k=201,n=1;\quad k=100,n=2;\quad k=101,n=51

Detailed analysis

For $k=201$ , take $n=1$ : the remainders of $n,2n,\ldots,kn$ modulo $202$ are $1,2,\ldots,201$ , averaging $101$ . For $k=100$ , take $n=2$ : the remainders are $2,4,\ldots,200$ , again averaging $101$ . For $k=101$ , take $n=51$ ; splitting the $101$ remainders into $25$ blocks of four plus one extra term, each block's average rises steadily and the overall average works out to exactly $101$ , as a direct computation confirms.