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Question Number 5796 by FilupSmith last updated on 28/May/16
How many ways can you express  30,030 as the product of 4 positive numbers?  (excluding 1)
$$\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{you}\:\mathrm{express} \\ $$$$\mathrm{30},\mathrm{030}\:\mathrm{as}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{4}\:\mathrm{positive}\:\mathrm{numbers}? \\ $$$$\left(\mathrm{excluding}\:\mathrm{1}\right) \\ $$
Commented by Rasheed Soomro last updated on 28/May/16
I didn′t think  k,l,m,n to be necessarily prime!
$${I}\:{didn}'{t}\:{think}\:\:{k},{l},{m},{n}\:{to}\:{be}\:{necessarily}\:{prime}! \\ $$
Commented by FilupSmith last updated on 28/May/16
ok so  x as an integer can be split into a maximum  amount of products as its prime factors.  eg. 100=2^2 5^2 =2×2×5×5  so 100 can be split to a max of 4 numbers  the number of ways ylu can express 100  as 4 numbers is 4!=24 ways  100=2×2×5×5=5×5×2×2=...etc    to look at combinatioms of 3 it is^4 C_3   PLUS the number of ways you can turn  2×2×5×5 into 3 numbers.  i.e. 2×2×5×5:  a_1 =4×5×5      a_2 =5×4×5    a_3 =...  a_4 =2×10×2    ...etc  a_n =...  the number of ways you can do that = n  ∴ combinations of 3=^4 C_3 +n    thoughts or simplifications?
$${ok}\:{so} \\ $$$${x}\:\mathrm{as}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{can}\:\mathrm{be}\:\mathrm{split}\:\mathrm{into}\:\mathrm{a}\:\mathrm{maximum} \\ $$$$\mathrm{amount}\:\mathrm{of}\:\mathrm{products}\:\mathrm{as}\:\mathrm{its}\:\mathrm{prime}\:\mathrm{factors}. \\ $$$$\mathrm{eg}.\:\mathrm{100}=\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{2}} =\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5} \\ $$$$\mathrm{so}\:\mathrm{100}\:\mathrm{can}\:\mathrm{be}\:\mathrm{split}\:\mathrm{to}\:\mathrm{a}\:\mathrm{max}\:\mathrm{of}\:\mathrm{4}\:\mathrm{numbers} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{ylu}\:\mathrm{can}\:\mathrm{express}\:\mathrm{100} \\ $$$$\mathrm{as}\:\mathrm{4}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{4}!=\mathrm{24}\:\mathrm{ways} \\ $$$$\mathrm{100}=\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5}=\mathrm{5}×\mathrm{5}×\mathrm{2}×\mathrm{2}=…\mathrm{etc} \\ $$$$ \\ $$$$\mathrm{to}\:\mathrm{look}\:\mathrm{at}\:\mathrm{combinatioms}\:\mathrm{of}\:\mathrm{3}\:\mathrm{it}\:\mathrm{is}\:^{\mathrm{4}} \mathrm{C}_{\mathrm{3}} \\ $$$$\boldsymbol{{PLUS}}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{you}\:\mathrm{can}\:\mathrm{turn} \\ $$$$\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5}\:\mathrm{into}\:\mathrm{3}\:\mathrm{numbers}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{5}: \\ $$$${a}_{\mathrm{1}} =\mathrm{4}×\mathrm{5}×\mathrm{5}\:\:\:\:\:\:{a}_{\mathrm{2}} =\mathrm{5}×\mathrm{4}×\mathrm{5}\:\:\:\:{a}_{\mathrm{3}} =… \\ $$$${a}_{\mathrm{4}} =\mathrm{2}×\mathrm{10}×\mathrm{2}\:\:\:\:…{etc}\:\:{a}_{{n}} =… \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{you}\:\mathrm{can}\:\mathrm{do}\:\mathrm{that}\:=\:{n} \\ $$$$\therefore\:\mathrm{combinations}\:\mathrm{of}\:\mathrm{3}=^{\mathrm{4}} {C}_{\mathrm{3}} +{n} \\ $$$$ \\ $$$${thoughts}\:{or}\:{simplifications}? \\ $$
Commented by FilupSmith last updated on 28/May/16
note:  30,030=2×3×5×7×11×13
$$\mathrm{note}: \\ $$$$\mathrm{30},\mathrm{030}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$
Commented by Rasheed Soomro last updated on 29/May/16
30,030=2×3×5×7×11×13 (total factors 6)  Actually we have to  divide above 6 primes  into four  groups whose products  will be required four positive numbers.  Two cases are possible.    Case-I : When groups contain 3,1,1,1 primes    ^6 C_3 =((6×5×4)/(1×2×3))=20 ways  Case-II : When groups contain 2,2,1,1 primes       ^6 C_2 ×^4 C_2 =((6×5)/(1×2))×((4×3)/(1×2))=15×6=90 ways  Total ways in which the number can be   expressed as product of four positive numbers                =20+90=110  Remember that order of number  is not important.  And  if  order is important then number of ways  =110×4!=2640  I am not confident.  prakashjain, Yozzi and  others can help me.
$$\mathrm{30},\mathrm{030}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13}\:\left({total}\:{factors}\:\mathrm{6}\right) \\ $$$$\mathcal{A}{ctually}\:{we}\:{have}\:{to}\:\:{divide}\:{above}\:\mathrm{6}\:{primes} \\ $$$${into}\:{four}\:\:{groups}\:{whose}\:{products} \\ $$$${will}\:{be}\:{required}\:{four}\:{positive}\:{numbers}. \\ $$$${Two}\:{cases}\:{are}\:{possible}. \\ $$$$ \\ $$$$\mathcal{C}{ase}-{I}\::\:\mathcal{W}{hen}\:{groups}\:{contain}\:\mathrm{3},\mathrm{1},\mathrm{1},\mathrm{1}\:{primes} \\ $$$$\:\:\:^{\mathrm{6}} \mathrm{C}_{\mathrm{3}} =\frac{\mathrm{6}×\mathrm{5}×\mathrm{4}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}=\mathrm{20}\:\mathrm{ways} \\ $$$$\mathcal{C}{ase}-{II}\::\:\mathcal{W}{hen}\:{groups}\:{contain}\:\mathrm{2},\mathrm{2},\mathrm{1},\mathrm{1}\:{primes} \\ $$$$\:\:\:\:\:\:^{\mathrm{6}} \mathrm{C}_{\mathrm{2}} ×^{\mathrm{4}} \mathrm{C}_{\mathrm{2}} =\frac{\mathrm{6}×\mathrm{5}}{\mathrm{1}×\mathrm{2}}×\frac{\mathrm{4}×\mathrm{3}}{\mathrm{1}×\mathrm{2}}=\mathrm{15}×\mathrm{6}=\mathrm{90}\:\mathrm{ways} \\ $$$$\mathrm{Total}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{number}\:\mathrm{can}\:\mathrm{be}\: \\ $$$$\mathrm{expressed}\:\mathrm{as}\:\mathrm{product}\:\mathrm{of}\:\mathrm{four}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{20}+\mathrm{90}=\mathrm{110} \\ $$$${Remember}\:{that}\:{order}\:{of}\:{number}\:\:{is}\:{not}\:{important}. \\ $$$${And}\:\:{if}\:\:{order}\:{is}\:{important}\:{then}\:{number}\:{of}\:{ways} \\ $$$$=\mathrm{110}×\mathrm{4}!=\mathrm{2640} \\ $$$${I}\:{am}\:{not}\:{confident}. \\ $$$${prakashjain},\:{Yozzi}\:{and}\:\:{others}\:{can}\:{help}\:{me}. \\ $$
Commented by FilupSmith last updated on 28/May/16
One thing to consider is that  30030=2×3×5×7×11×13               =6×5×7×11×13               =2×15×7×11×13  etc.               =30×7×11×13               =2×35×11×13  etc.    i.e. k, l, m, n   dont have to be prime  the trick, i think, is in the combinations  of products, and find the combination  out of each one.
$$\mathrm{One}\:\mathrm{thing}\:\mathrm{to}\:\mathrm{consider}\:\mathrm{is}\:\mathrm{that} \\ $$$$\mathrm{30030}=\mathrm{2}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{6}×\mathrm{5}×\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}×\mathrm{15}×\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$$$\mathrm{etc}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30}×\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}×\mathrm{35}×\mathrm{11}×\mathrm{13} \\ $$$$\mathrm{etc}. \\ $$$$ \\ $$$${i}.{e}.\:{k},\:{l},\:{m},\:{n}\:\:\:\mathrm{dont}\:\mathrm{have}\:\mathrm{to}\:\mathrm{be}\:\mathrm{prime} \\ $$$$\mathrm{the}\:\mathrm{trick},\:\mathrm{i}\:\mathrm{think},\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{combinations} \\ $$$$\mathrm{of}\:\mathrm{products},\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{combination} \\ $$$$\mathrm{out}\:\mathrm{of}\:\mathrm{each}\:\mathrm{one}. \\ $$
Commented by Rasheed Soomro last updated on 29/May/16
As the order doesn′t matter in multiplication  so I considered 6×5×77×13  and 13×5×6×77  same.And this is what I have understood from  the question. Anyway if we had an answer  in which order is not considered important then  change it in answer in which order is importanyt  is very simple!
$${As}\:{the}\:{order}\:{doesn}'{t}\:{matter}\:{in}\:{multiplication} \\ $$$${so}\:{I}\:{considered}\:\mathrm{6}×\mathrm{5}×\mathrm{77}×\mathrm{13}\:\:{and}\:\mathrm{13}×\mathrm{5}×\mathrm{6}×\mathrm{77} \\ $$$${same}.{And}\:{this}\:{is}\:{what}\:{I}\:{have}\:{understood}\:{from} \\ $$$${the}\:{question}.\:{Anyway}\:{if}\:{we}\:{had}\:{an}\:{answer} \\ $$$${in}\:{which}\:{order}\:{is}\:{not}\:{considered}\:{important}\:{then} \\ $$$${change}\:{it}\:{in}\:{answer}\:{in}\:{which}\:{order}\:{is}\:{importanyt} \\ $$$${is}\:{very}\:{simple}! \\ $$

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