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Question Number 22042 by Tinkutara last updated on 10/Oct/17

$$\mathrm{Determine}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ordered}$$$$\mathrm{pairs}\mathrm{of}\mathrm{positive}\mathrm{integers}(a,b)\mathrm{such}$$$$\mathrm{that}\mathrm{the}\mathrm{least}\mathrm{common}\mathrm{multiple}\mathrm{of}a$$$$\mathrm{and}b\mathrm{is}{2}^3\cdot 5^7\cdot {11}^{13}.$$

Answered by Rasheed.Sindhi last updated on 13/Oct/17

$$\mathrm{Case}-1:a=2^{<3}.5^{<7}.{11}^{<13}$$$$a=2^{0,1,2}.5^{0,..,6}.{11}^{0,..,12}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}3.7.13=273\mathrm{ways}$$$$b=2^3.5^7.{11}^{13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.1.1=1\mathrm{way}.$$$$\mathrm{Numb}.\mathrm{of}\mathrm{ways}\mathrm{of}(a,b)=273.1=273$$$$\mathrm{Case}-2:a=2^{<3}.5^{<7}.{11}^{13}$$$$a=2^{0,1,2}.5^{0,..,6}.{11}^{13}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}3.7.1=21\mathrm{ways}.$$$$b=2^3.5^7.{11}^{0,1...13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}1\times 1\times 14=14\mathrm{ways}.$$$$\mathrm{No}\mathrm{of}(a,b)=21.14=294$$$$\mathrm{Case}-3:a=2^{<3}.5^7.{11}^{<13}$$$$a=2^{0,1,2}.5^7.{11}^{0,..,12}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}3.1.13=39$$$$b=2^3.5^{0,..7}.{11}^{13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.8.1=8\mathrm{ways}$$$$\mathrm{No}\mathrm{of}(a,b)=39.8=312$$$$\mathrm{Case}-4:a=2^{<3}.5^7.{11}^{13}$$$$a=2^{0,1,2}.5^7.{11}^{13}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}3.1.1=3\mathrm{ways}.$$$$b=2^3.5^{0,..7}.{11}^{0,...13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}1.8.14=112\mathrm{ways}$$$$\mathrm{No}\mathrm{of}\mathrm{ways}\mathrm{for}(a,b)=3\times 112=336$$$$\mathrm{Case}-5:a=2^3.5^{<7}.{11}^{<13}$$$$a=2^3.5^{0,..,6}.{11}^{0,..,12}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.7.13=91$$$$b=2^{0,..3}.5^7.{11}^{13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}4.1.1=4\mathrm{ways}.$$$$\mathrm{No}\mathrm{of}\mathrm{ways}\mathrm{for}(a,b)=91\times 4=364$$$$\mathrm{Case}-6:a=2^3.5^{<7}.{11}^{13}$$$$a=2^3.5^{0,..,6}.{11}^{13}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.7.1=7\mathrm{ways}$$$$b=2^{0,..3}.5^7.{11}^{0,..13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}4.1.14=56\mathrm{ways}.$$$$\mathrm{No}\mathrm{of}\mathrm{ways}\mathrm{for}(a,b)=7\times 56=392$$$$\mathrm{Case}-7:a=2^3.5^7.{11}^{<13}$$$$a=2^3.5^7.{11}^{0,..,12}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.1.13=13\mathrm{ways}.$$$$b=2^{0,..,3}.5^{0,..,7}.{11}^{13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}4.8.1=32\mathrm{ways}$$$$\mathrm{No}.\mathrm{of}\mathrm{ways}\mathrm{for}(a,b)=13\times 32=416$$$$\mathrm{Case}-8:a=2^3.5^7.{11}^{13}$$$$a\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}1.1.1=1\mathrm{way}.$$$$b=2^{⩽3}.5^{⩽7}.{11}^{⩽13}$$$$b\mathrm{can}\mathrm{be}\mathrm{chosen}\mathrm{in}4.8.14=448\mathrm{ways}.$$$$\mathrm{No}.\mathrm{of}\mathrm{ways}\mathrm{for}(a,b)=1\times 448=448\mathrm{ways}.$$$$$$$$\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{ways}\mathrm{for}(a,b):$$$$273+294+312+336+364+392+416+448$$$$=2835$$

Commented by Tinkutara last updated on 13/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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