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 2 of 5: Clear denominators

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nk(k+1)-404z=202k,\qquad z=\sum_{i=1}^{k}\left\lfloor\frac{in}{202}\right\rfloor

Detailed analysis

Writing $\{in/202\}=in/202-\lfloor in/202\rfloor$ and summing for $i=1,\ldots,k$ , the equation becomes $\frac{n}{202}\cdot\frac{k(k+1)}{2}-z=\frac{k}{2}$ where $z$ is the sum of the floors above. Multiplying by $404$ gives the integer identity shown.