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

q+

$${q}+\: \\ $$

Commented by FilupSmith last updated on 27/May/16

∃x∈Z^+ :x=pq, p,q∈Z^+ \{0, 1} n,i∈P p=Π_(n∣p) n q=Π_(i∣q) i ∴ x = product of factors of p and q if p=a_1 ×a_2 ×...×a_n q=b_1 ×b_2 ×...×b_i x=(a_1 ×...×a_n )(b_1 ×...×b_i ) x=(a_1 ×...×a_n ×b_1 )(b_2 ×...×b_i ) x=(a_2 ×...×a_n )(a_1 ×b_1 ×...×b_i ) etc number of pq that make x=#number of ways to arrange RHS =(n+i)! e.g. x=12 =2×6 =2×2×3 #=3!=6 x_1 =2×2×3=2×6 x_2 =2×2×3=4×6 x_3 =2×2×3=6×2 (later half is reverse ab=ba) x_4 =6×2 x_5 =6×4 x_6 =2×6 ∴number of ways to multiply two numbers to get any +ve integer x is equal to the factorial of the number of prime factors x has. (excluding a=1a=a1) i.e. p,q≠1 ^� ^� \( °⌣°)/ ^� ^�

$$\exists{x}\in\mathbb{Z}^{+} :{x}={pq},\:{p},{q}\in\mathbb{Z}^{+} \backslash\left\{\mathrm{0},\:\mathrm{1}\right\} \\ $$$${n},{i}\in\mathbb{P} \\ $$$${p}=\underset{{n}\mid{p}} {\prod}{n} \\ $$$${q}=\underset{{i}\mid{q}} {\prod}{i} \\ $$$$\therefore\:{x}\:=\:\mathrm{product}\:\mathrm{of}\:\mathrm{factors}\:\mathrm{of}\:{p}\:\mathrm{and}\:{q} \\ $$$$\mathrm{if}\:{p}={a}_{\mathrm{1}} ×{a}_{\mathrm{2}} ×...×{a}_{{n}} \\ $$$$\:\:\:\:{q}={b}_{\mathrm{1}} ×{b}_{\mathrm{2}} ×...×{b}_{{i}} \\ $$$${x}=\left({a}_{\mathrm{1}} ×...×{a}_{{n}} \right)\left({b}_{\mathrm{1}} ×...×{b}_{{i}} \right) \\ $$$${x}=\left({a}_{\mathrm{1}} ×...×{a}_{{n}} ×{b}_{\mathrm{1}} \right)\left({b}_{\mathrm{2}} ×...×{b}_{{i}} \right) \\ $$$${x}=\left({a}_{\mathrm{2}} ×...×{a}_{{n}} \right)\left({a}_{\mathrm{1}} ×{b}_{\mathrm{1}} ×...×{b}_{{i}} \right) \\ $$$${etc} \\ $$$$\mathrm{number}\:\mathrm{of}\:{pq}\:\mathrm{that}\:\mathrm{make}\:{x}=#{number}\:{of}\:{ways}\:{to}\:{arrange}\:\boldsymbol{{RHS}} \\ $$$$=\left({n}+{i}\right)! \\ $$$$ \\ $$$${e}.{g}. \\ $$$${x}=\mathrm{12} \\ $$$$=\mathrm{2}×\mathrm{6} \\ $$$$=\mathrm{2}×\mathrm{2}×\mathrm{3} \\ $$$$#=\mathrm{3}!=\mathrm{6} \\ $$$${x}_{\mathrm{1}} =\mathrm{2}×\mathrm{2}×\mathrm{3}=\mathrm{2}×\mathrm{6} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}×\mathrm{2}×\mathrm{3}=\mathrm{4}×\mathrm{6} \\ $$$${x}_{\mathrm{3}} =\mathrm{2}×\mathrm{2}×\mathrm{3}=\mathrm{6}×\mathrm{2} \\ $$$$\left(\mathrm{later}\:\mathrm{half}\:\mathrm{is}\:\mathrm{reverse}\:{ab}={ba}\right) \\ $$$${x}_{\mathrm{4}} =\mathrm{6}×\mathrm{2} \\ $$$${x}_{\mathrm{5}} =\mathrm{6}×\mathrm{4} \\ $$$${x}_{\mathrm{6}} =\mathrm{2}×\mathrm{6} \\ $$$$ \\ $$$$\therefore\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{multiply}\:\mathrm{two} \\ $$$$\mathrm{numbers}\:\mathrm{to}\:\mathrm{get}\:\mathrm{any}\:+\mathrm{ve}\:\mathrm{integer}\:{x} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{factorial}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{prime}\:\mathrm{factors}\:{x}\:\mathrm{has}. \\ $$$$\left({excluding}\:{a}=\mathrm{1}{a}={a}\mathrm{1}\right) \\ $$$${i}.{e}.\:\:{p},{q}\neq\mathrm{1} \\ $$$$ \\ $$$$\bar {\:}\bar {\:}\backslash\left(\:\:°\smile°\right)/\bar {\:}\bar {\:} \\ $$

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