Unique Factor.

Level 2

$A_{1} = \{1\}$ and $1|1$

$A_{2} = \{2,3\}$ and $2|2$

$A_{3} = \{4,5,6\}$ and $3|6$

$A_{4} = \{7,8,9,10\}$ and $4|8$

$A_{5} = \{11,12,13,14,15\}$ and $5|15$

$A_{6} = \{16,17,18,19,20,21\}$ and $6|18$

In general:

$\exists! \alpha \in A_{n} \ni n|\alpha.$

False True

1 solution

Rocco Dalto
Dec 20, 2017

Let $n$ be a fixed positive integer.

In general the set $A_{n} = \{(n + \dfrac{(n - 2) (n - 1)}{2} + j)|(0 <= j <= n - 1\}$ , and we can write the elements of $A_{n}$ backwards so that $A_{n} = \{\dfrac{(n)(n + 1)}{2} - j|(0 <= j <= n - 1\}$ .

For $n$ even $j = \dfrac{n}{2} \implies p_{(\dfrac{n}{2})} = \dfrac{n^2}{2} = n * (\dfrac{n}{2}) = n * k^{*}$ and for $n$ odd $j = 0 \implies p_{0} = \dfrac{n(n + 1)}{2} = n * (\dfrac{n + 1}{2}) = n * k$ .

$\therefore \exists \alpha \in A_{n} \ni n|\alpha$ and since set $A_{n}$ contains $n$ consecutive positive integers $\implies \exists! \alpha \in A_{n} \ni n|\alpha.$

Unique Factor.

1 solution

0 pending reports