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 1 of 5: Dispose of the residues n=0 and n=101

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n=0\Rightarrow\text{sum}=0\ne\tfrac{k}{2};\qquad n=101\Rightarrow\text{condition holds}\iff k=1

Detailed analysis

Since $\{in/202\}$ only depends on $n$ modulo $202$ , it suffices to take $0\le n<202$ . If $n=0$ every term is $0$ , so the equation fails for every $k$ . If $n=101$ , one checks directly that the equation holds exactly when $k=1$ . From now on assume $101\nmid n$ .