Problem 5

The equation

(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots(x-2016)

is written on the board, with

2016

linear factors on each side. What is the least possible value of

k

for which it is possible to erase exactly

k

of these

4032

linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?

Step 4 of 6: Case 3: far outside, or strictly between two blocks

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In plain words

Rewriting the equation as a product of factors each strictly between $0$ and $1$ makes it impossible for the product to equal $1$ .

\text{pile}:=\{\text{coins with denominator }\ge 2k+1\}

Detailed analysis

All coins with denominator at least $2k+1=201$ have value at most $1/201$ ; collect them in one pile. This includes any surviving coins of denominator $201$ , which the previous boxes did not use.