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Determine-the-number-of-ordered-pairs-of-positive-integers-a-b-such-that-the-least-common-multiple-of-a-and-b-is-2-3-5-7-11-13-

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Question Number 22042 by Tinkutara last updated on 10/Oct/17
Determine the number of ordered pairs of positive integers (a, b) such that the least common multiple of a and b is 2^3 ∙5^7 ∙11^(13) .
Determinethenumberoforderedpairsofpositiveintegers(a,b)suchthattheleastcommonmultipleofaandbis23⋅57⋅1113.
Answered by Rasheed.Sindhi last updated on 13/Oct/17
Case-1: a=2^(<3) .5^(<7) .11^(<13) a=2^(0,1,2) .5^(0,..,6) .11^(0,..,12) a can be chosen in 3.7.13=273 ways b=2^3 .5^7 .11^(13) b can be chosen in 1.1.1=1 way. Numb. of ways of (a,b)=273.1=273 Case-2: a=2^(<3) .5^(<7) .11^(13) a=2^(0,1,2) .5^(0,..,6) .11^(13) a can be chosen in 3.7.1=21 ways. b=2^3 .5^7 .11^(0,1...13) b can be chosen 1×1×14=14 ways. No of (a,b)=21.14=294 Case-3: a=2^(<3) .5^7 .11^(<13) a=2^(0,1,2) .5^7 .11^(0,..,12) a can be chosen in 3.1.13=39 b=2^3 .5^(0,..7) .11^(13) b can be chosen in 1.8.1=8 ways No of (a,b)=39.8=312 Case-4: a=2^(<3) .5^7 .11^(13) a=2^(0,1,2) .5^7 .11^(13) a can be chosen in 3.1.1=3 ways. b=2^3 .5^(0,..7) .11^(0,...13) b can be chosen 1.8.14=112 ways No of ways for (a,b)=3×112=336 Case-5: a=2^3 .5^(<7) .11^(<13) a=2^3 .5^(0,..,6) .11^(0,..,12) a can be chosen in 1.7.13=91 b=2^(0,..3) .5^7 .11^(13) b can be chosen in 4.1.1=4 ways. No of ways for (a,b)=91×4=364 Case-6: a=2^3 .5^(<7) .11^(13) a=2^3 .5^(0,..,6) .11^(13) a can be chosen in 1.7.1=7 ways b=2^(0,..3) .5^7 .11^(0,..13) b can be chosen in 4.1.14=56 ways. No of ways for (a,b)=7×56=392 Case-7: a=2^3 .5^7 .11^(<13) a=2^3 .5^7 .11^(0,..,12) a can be chosen in 1.1.13=13 ways. b=2^(0,..,3) .5^(0,..,7) .11^(13) b can be chosen in 4.8.1=32 ways No. of ways for (a,b)=13×32=416 Case-8:a=2^3 .5^7 .11^(13) a can be chosen in 1.1.1=1 way. b=2^(≤3) .5^(≤7) .11^(≤13) b can be chosen in 4.8.14=448 ways. No. of ways for (a,b)=1×448=448 ways. Total no. of ways for (a,b): 273+294+312+336+364+392+416+448 =2835
Case−1:a=2<3.5<7.11<13a=20,1,2.50,..,6.110,..,12acanbechosenin3.7.13=273waysb=23.57.1113bcanbechosenin1.1.1=1way.Numb.ofwaysof(a,b)=273.1=273Case−2:a=2<3.5<7.1113a=20,1,2.50,..,6.1113acanbechosenin3.7.1=21ways.b=23.57.110,1…13bcanbechosen1×1×14=14ways.Noof(a,b)=21.14=294Case−3:a=2<3.57.11<13a=20,1,2.57.110,..,12acanbechosenin3.1.13=39b=23.50,..7.1113bcanbechosenin1.8.1=8waysNoof(a,b)=39.8=312Case−4:a=2<3.57.1113a=20,1,2.57.1113acanbechosenin3.1.1=3ways.b=23.50,..7.110,…13bcanbechosen1.8.14=112waysNoofwaysfor(a,b)=3×112=336Case−5:a=23.5<7.11<13a=23.50,..,6.110,..,12acanbechosenin1.7.13=91b=20,..3.57.1113bcanbechosenin4.1.1=4ways.Noofwaysfor(a,b)=91×4=364Case−6:a=23.5<7.1113a=23.50,..,6.1113acanbechosenin1.7.1=7waysb=20,..3.57.110,..13bcanbechosenin4.1.14=56ways.Noofwaysfor(a,b)=7×56=392Case−7:a=23.57.11<13a=23.57.110,..,12acanbechosenin1.1.13=13ways.b=20,..,3.50,..,7.1113bcanbechosenin4.8.1=32waysNo.ofwaysfor(a,b)=13×32=416Case−8:a=23.57.1113acanbechosenin1.1.1=1way.b=2⩽3.5⩽7.11⩽13bcanbechosenin4.8.14=448ways.No.ofwaysfor(a,b)=1×448=448ways.Totalno.ofwaysfor(a,b):273+294+312+336+364+392+416+448=2835
Commented by Tinkutara last updated on 13/Oct/17
Thank you very much Sir!
ThankyouverymuchSir!

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