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Question Number 5772 by sagarvijay444@gmail.com last updated on 27/May/16

$$q+$$

Commented by FilupSmith last updated on 27/May/16

$$\mathrm{\exists }x\in {\mathbb{Z}}^{+}:x=pq,p,q\in {\mathbb{Z}}^{+}\mathrm{\setminus }\{0,1\}$$$$n,i\in \mathbb{P}$$$$p=\prod _{n\mid p}n$$$$q=\prod _{i\mid q}i$$$$\therefore x=\mathrm{product}\mathrm{of}\mathrm{factors}\mathrm{of}p\mathrm{and}q$$$$\mathrm{if}p=a_1\times a_2\times ...\times a_n$$$$q=b_1\times b_2\times ...\times b_i$$$$x=(a_1\times ...\times a_n)(b_1\times ...\times b_i)$$$$x=(a_1\times ...\times a_n\times b_1)(b_2\times ...\times b_i)$$$$x=(a_2\times ...\times a_n)(a_1\times b_1\times ...\times b_i)$$$$etc$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$=(n+i)!$$$$$$$$e.g.$$$$x=12$$$$=2\times 6$$$$=2\times 2\times 3$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$x_1=2\times 2\times 3=2\times 6$$$$x_2=2\times 2\times 3=4\times 6$$$$x_3=2\times 2\times 3=6\times 2$$$$(\mathrm{later}\mathrm{half}\mathrm{is}\mathrm{reverse}ab=ba)$$$$x_4=6\times 2$$$$x_5=6\times 4$$$$x_6=2\times 6$$$$$$$$\therefore \mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{to}\mathrm{multiply}\mathrm{two}$$$$\mathrm{numbers}\mathrm{to}\mathrm{get}\mathrm{any}+\mathrm{ve}\mathrm{integer}x$$$$\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{factorial}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{of}$$$$\mathrm{prime}\mathrm{factors}x\mathrm{has}.$$$$(excludinga=1a=a1)$$$$i.e.p,q\ne 1$$$$$$$$\overline{}\overline{}\mathrm{\setminus }(°⌣°)/\overline{}\overline{}$$

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