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Question Number 19239 by Tinkutara last updated on 07/Aug/17

Assume that a, b, c and d are positive integers such that a^5 = b^4 , c^3 = d^2 and c − a = 19. Determine d − b.

$$\mathrm{Assume}\:\mathrm{that}\:{a},\:{b},\:{c}\:\mathrm{and}\:{d}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{such}\:\mathrm{that}\:{a}^{\mathrm{5}} \:=\:{b}^{\mathrm{4}} ,\:{c}^{\mathrm{3}} \:=\:{d}^{\mathrm{2}} \:\mathrm{and} \\ $$$${c}\:−\:{a}\:=\:\mathrm{19}.\:\mathrm{Determine}\:{d}\:−\:{b}. \\ $$

Answered by ajfour last updated on 07/Aug/17

let a^5 =b^4 =n^(20) ⇒ a=n^4 and b=n^5 ....(i) let c^3 =d^2 =m^6 ⇒ c=m^2 and d=m^3 ...(ii) Since c−a=19 m^2 −n^4 =19 or (m−n^2 )(m+n^2 )=19 ⇒ m−n^2 =1 and m+n^2 =19 ⇒ 2m=20 and 2n^2 =18 m=10 ; n=3 So from (i) and (ii) a=n^4 =81 b=n^5 =243 c=m^2 =100 d=m^3 =1000 . d−b=757 .

$$\mathrm{let}\:\mathrm{a}^{\mathrm{5}} =\mathrm{b}^{\mathrm{4}} =\mathrm{n}^{\mathrm{20}} \\ $$$$\:\:\:\Rightarrow\:\mathrm{a}=\mathrm{n}^{\mathrm{4}} \:\:\mathrm{and}\:\:\mathrm{b}=\mathrm{n}^{\mathrm{5}} \:\:\:....\left(\mathrm{i}\right) \\ $$$$\mathrm{let}\:\:\mathrm{c}^{\mathrm{3}} =\mathrm{d}^{\mathrm{2}} =\mathrm{m}^{\mathrm{6}} \\ $$$$\:\:\:\Rightarrow\:\mathrm{c}=\mathrm{m}^{\mathrm{2}} \:\:\mathrm{and}\:\mathrm{d}=\mathrm{m}^{\mathrm{3}} \:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{Since}\:\:\:\mathrm{c}−\mathrm{a}=\mathrm{19} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{4}} =\mathrm{19} \\ $$$$\:\:\:\:\:\:\:\mathrm{or}\:\:\:\left(\mathrm{m}−\mathrm{n}^{\mathrm{2}} \right)\left(\mathrm{m}+\mathrm{n}^{\mathrm{2}} \right)=\mathrm{19} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\mathrm{m}−\mathrm{n}^{\mathrm{2}} =\mathrm{1}\:\mathrm{and}\:\mathrm{m}+\mathrm{n}^{\mathrm{2}} =\mathrm{19} \\ $$$$\:\:\:\:\Rightarrow\:\:\mathrm{2m}=\mathrm{20}\:\:\:\mathrm{and}\:\:\mathrm{2n}^{\mathrm{2}} =\mathrm{18} \\ $$$$\:\:\:\:\mathrm{m}=\mathrm{10}\:\:;\:\:\mathrm{n}=\mathrm{3}\: \\ $$$$\mathrm{So}\:\:\mathrm{from}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\mathrm{a}=\mathrm{n}^{\mathrm{4}} =\mathrm{81} \\ $$$$\:\:\:\:\mathrm{b}=\mathrm{n}^{\mathrm{5}} =\mathrm{243} \\ $$$$\:\:\:\:\mathrm{c}=\mathrm{m}^{\mathrm{2}} =\mathrm{100} \\ $$$$\:\:\:\:\mathrm{d}=\mathrm{m}^{\mathrm{3}} =\mathrm{1000}\:. \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{d}}−\boldsymbol{\mathrm{b}}=\mathrm{757}\:.\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by Tinkutara last updated on 08/Aug/17

Thank you very much Sir!

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

Commented by malwaan last updated on 08/Aug/17

can you solve 19204 please ?

$$\mathrm{can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{19204}\:\mathrm{please}\:? \\ $$

Commented by ajfour last updated on 08/Aug/17

I know not.

$$\mathrm{I}\:\mathrm{know}\:\mathrm{not}. \\ $$

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