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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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