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Question Number 19516 by Tinkutara last updated on 12/Aug/17

Let Akbar and Birbal together have n marbles, where n > 0. Akbar says to Birbal, ♮If I give you some marbles then you will have twice as many marbles as I will have.ε Birbal says to Akbar, ♮If I give you some marbles then you will have thrice as many marbles as I will have.ε What is the minimum possible value of n for which the above statements are true?

$$\mathrm{Let}\:\mathrm{Akbar}\:\mathrm{and}\:\mathrm{Birbal}\:\mathrm{together}\:\mathrm{have}\:{n} \\ $$ $$\mathrm{marbles},\:\mathrm{where}\:{n}\:>\:\mathrm{0}. \\ $$ $$\mathrm{Akbar}\:\mathrm{says}\:\mathrm{to}\:\mathrm{Birbal},\:\natural\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$ $$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{twice}\:\mathrm{as} \\ $$ $$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.\varepsilon\:\mathrm{Birbal} \\ $$ $$\mathrm{says}\:\mathrm{to}\:\mathrm{Akbar},\:\natural\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$ $$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{thrice}\:\mathrm{as} \\ $$ $$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.\varepsilon \\ $$ $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$ $${n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{above}\:\mathrm{statements}\:\mathrm{are} \\ $$ $$\mathrm{true}? \\ $$

Answered by dioph last updated on 12/Aug/17

As n ∈ N, if there is the possibility B = 2x_1 and A = x_1 ⇒ n = 3x_1 . Similarly, when A = 3x_2 and B = x_2 ⇒ n = 4x_2 . Hence 3 ∣ n and 4 ∣ n ⇒ n ≥ 12. If n = 12, 4<A<9, and B = 12−A, then we have a solution where A can give marbles so that A = 4 and B = 8, and B can give marbles so that A = 9 and B = 3.

$$\mathrm{As}\:{n}\:\in\:\mathbb{N},\:\mathrm{if}\:\mathrm{there}\:\mathrm{is}\:\mathrm{the}\:\mathrm{possibility}\: \\ $$ $$\mathrm{B}\:=\:\mathrm{2}{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{A}\:=\:{x}_{\mathrm{1}} \:\Rightarrow\:{n}\:=\:\mathrm{3}{x}_{\mathrm{1}} . \\ $$ $$\mathrm{Similarly},\:\mathrm{when}\:{A}\:=\:\mathrm{3}{x}_{\mathrm{2}} \:\mathrm{and}\:{B}\:=\:{x}_{\mathrm{2}} \\ $$ $$\Rightarrow\:{n}\:=\:\mathrm{4}{x}_{\mathrm{2}} .\:\mathrm{Hence}\:\mathrm{3}\:\mid\:{n}\:\mathrm{and}\:\mathrm{4}\:\mid\:{n} \\ $$ $$\Rightarrow\:{n}\:\geqslant\:\mathrm{12}.\:\mathrm{If}\:{n}\:=\:\mathrm{12},\:\mathrm{4}<{A}<\mathrm{9}, \\ $$ $$\mathrm{and}\:{B}\:=\:\mathrm{12}−{A},\:\mathrm{then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{solution} \\ $$ $$\mathrm{where}\:{A}\:\mathrm{can}\:\mathrm{give}\:\mathrm{marbles}\:\mathrm{so}\:\mathrm{that}\:{A}\:=\:\mathrm{4} \\ $$ $$\mathrm{and}\:{B}\:=\:\mathrm{8},\:\mathrm{and}\:{B}\:\mathrm{can}\:\mathrm{give}\:\mathrm{marbles} \\ $$ $$\mathrm{so}\:\mathrm{that}\:{A}\:=\:\mathrm{9}\:\mathrm{and}\:{B}\:=\:\mathrm{3}. \\ $$

Commented byTinkutara last updated on 12/Aug/17

Thank you very much Sir!

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

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