Find-the-number-of-positive-integers-that-are-factors-of-3-19-7-12-10-25-and-are-also-multiples-of-3-15-7-10-10-19-

Question Number 199311 by necx122 last updated on 01/Nov/23

$${Find}\:{the}\:{number}\:{of}\:{positive}\:{integers} \\ $$$${that}\:{are}\:{factors}\:{of}\:\mathrm{3}^{\mathrm{19}} .\mathrm{7}^{\mathrm{12}} .\mathrm{10}^{\mathrm{25}} \:{and}\:{are} \\ $$$${also}\:{multiples}\:{of}\:\mathrm{3}^{\mathrm{15}} .\mathrm{7}^{\mathrm{10}} .\mathrm{10}^{\mathrm{19}} \\ $$

Answered by AST last updated on 01/Nov/23

(3^(15) 7^(10) 2^(19) 5^(19) )(3^4 ×7^2 ×2^6 ×5^6 )⇒5×3×7^2 =735

$$\left(\mathrm{3}^{\mathrm{15}} \mathrm{7}^{\mathrm{10}} \mathrm{2}^{\mathrm{19}} \mathrm{5}^{\mathrm{19}} \right)\left(\mathrm{3}^{\mathrm{4}} ×\mathrm{7}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{6}} ×\mathrm{5}^{\mathrm{6}} \right)\Rightarrow\mathrm{5}×\mathrm{3}×\mathrm{7}^{\mathrm{2}} =\mathrm{735} \\ $$

Commented by necx122 last updated on 01/Nov/23

Please, what is the concept behind this?

Commented by AST last updated on 01/Nov/23

multiples of 3^(15) 7^(10) 2^(19) 5^(19) ⇒(3^(15) 7^(10) 2^(19) 5^(19) )k k=2^a 3^b 5^c ...;but k must divide 3^(19) 7^(12) 2^(25) 5^(25) ⇒k=2^a 3^b 5^c 7^d where 0≤a,d≤6;0≤b≤4;0≤c≤2 ⇒Total no. of choices=5×3×7^2

$${multiples}\:{of}\:\mathrm{3}^{\mathrm{15}} \mathrm{7}^{\mathrm{10}} \mathrm{2}^{\mathrm{19}} \mathrm{5}^{\mathrm{19}} \Rightarrow\left(\mathrm{3}^{\mathrm{15}} \mathrm{7}^{\mathrm{10}} \mathrm{2}^{\mathrm{19}} \mathrm{5}^{\mathrm{19}} \right){k} \\ $$$${k}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} …;{but}\:{k}\:{must}\:{divide}\:\mathrm{3}^{\mathrm{19}} \mathrm{7}^{\mathrm{12}} \mathrm{2}^{\mathrm{25}} \mathrm{5}^{\mathrm{25}} \\ $$$$\Rightarrow{k}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} \mathrm{7}^{{d}} \:{where}\:\mathrm{0}\leqslant{a},{d}\leqslant\mathrm{6};\mathrm{0}\leqslant{b}\leqslant\mathrm{4};\mathrm{0}\leqslant{c}\leqslant\mathrm{2} \\ $$$$\Rightarrow{Total}\:{no}.\:{of}\:{choices}=\mathrm{5}×\mathrm{3}×\mathrm{7}^{\mathrm{2}} \\ $$

Answered by BaliramKumar last updated on 01/Nov/23

((3^(19) ×7^(12) ×(2×5)^(25) )/(3^(15) ×7^(10) ×(2×5)^(19) )) = 3^4 ×7^2 ×(2×5)^6 3^4 ×7^2 ×2^6 ×5^6 = (4+1)(2+1)(6+1)(6+1) 5×3×7×7 = 105×7 = 735

$$\frac{\mathrm{3}^{\mathrm{19}} ×\mathrm{7}^{\mathrm{12}} ×\left(\mathrm{2}×\mathrm{5}\right)^{\mathrm{25}} }{\mathrm{3}^{\mathrm{15}} ×\mathrm{7}^{\mathrm{10}} ×\left(\mathrm{2}×\mathrm{5}\right)^{\mathrm{19}} }\:=\:\mathrm{3}^{\mathrm{4}} ×\mathrm{7}^{\mathrm{2}} ×\left(\mathrm{2}×\mathrm{5}\right)^{\mathrm{6}} \\ $$$$\:\mathrm{3}^{\mathrm{4}} ×\mathrm{7}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{6}} ×\mathrm{5}^{\mathrm{6}} \:=\:\left(\mathrm{4}+\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{6}+\mathrm{1}\right)\left(\mathrm{6}+\mathrm{1}\right) \\ $$$$\mathrm{5}×\mathrm{3}×\mathrm{7}×\mathrm{7}\:=\:\mathrm{105}×\mathrm{7}\:=\:\mathrm{735} \\ $$

Commented by necx122 last updated on 01/Nov/23

Wow! I love this too but why the addition of 1 at the end?

Commented by BaliramKumar last updated on 01/Nov/23

example How many factors of 4. 4 = 2^2 = (2^0 , 2^1 , 2^2 ) ⇒ (1, 2 & 4) = 3 factors 2^0 is extra so adding 1

$$\mathrm{example}\: \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{4}. \\ $$$$\mathrm{4}\:=\:\mathrm{2}^{\mathrm{2}} \:=\:\left(\mathrm{2}^{\mathrm{0}} ,\:\mathrm{2}^{\mathrm{1}} ,\:\mathrm{2}^{\mathrm{2}} \right)\:\Rightarrow\:\left(\mathrm{1},\:\mathrm{2}\:\&\:\mathrm{4}\right)\:=\:\mathrm{3}\:\mathrm{factors} \\ $$$$\mathrm{2}^{\mathrm{0}} \:\mathrm{is}\:\mathrm{extra}\:\mathrm{so}\:\mathrm{adding}\:\mathrm{1} \\ $$

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