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 6 of 6: Conclusion

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In plain words

All four cases together cover every real number, so the candidate erasure genuinely has no real root, matching the lower bound.

k=2016

Detailed analysis

Every real number $x$ falls into exactly one of Case 1 (an integer $1,\dots,2016$ ), Case 2 (near a block boundary), Case 3 (far outside or strictly between two blocks), or Case 4 (in the middle of a block), and in each case the equation of Step 2 fails. So the erasure of Step 2 (exactly $2016$ factors) produces an equation with no real solution. Combined with the necessity bound of Step 1, the least possible value is $k=2016$ .