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# Ruler

##### Age 16 to 18 Challenge Level:

Andaleeb sent in this excellent solution.

The diagram shows some of the vertical lines drawn for values of x between 0 and 1 as described in the question. The lines are of height 1 unit at x = 0 and 1, of height ${1\over 2}$ units at $x= {1\over 2}$ , of height ${1\over 4}$ units at $x= {k\over 4}$ and ${1\over 8}$ units at $x= {k\over 8}$ and so on... up to ${1\over 2^5}$ at ${k\over 2^5}$ where $k$ is a positive integer.

 n $0$ $1$ $2$ $3$ $4$ $5$ $6$ $\dots$ $n$ $n+1$ Height $1$ ${1\over 2}$ ${1\over 4 }$ ${1\over 8}$ ${1\over 16}$ ${1\over 32 }$ ${1\over 64}$ $\dots$ ${1\over 2^n}$ ${1\over 2^{n+1`}}$ Lines cut $2$ $1$ $2$ $4$ $8$ $16$ $32$ $\dots$ $2^{n-1}$ $2^n$ Thus if the height $h$ lies in ${1\over 2^n} > h> {1\over 2^{n+1}}$ then the number of lines cut is given by $$2 + 1 + 2 + 4 + 8 + ... + 2^{n-1} = 2 + {{2^n - 1} \over {2 - 1}} = 2^n + 1.$$ As $n$ tends to infinity the height of the lines tends to 0 and the number of lines cut tends to infinity.