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Question Number 160092 by abdullah_ff last updated on 24/Nov/21

x and y are positive integers and   x × x − 8y = 4x   .  If x is not a multiple of 8, then what is  the minimum possible value for y?

$${x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{and}\: \\ $$$${x}\:×\:{x}\:−\:\mathrm{8}{y}\:=\:\mathrm{4}{x}\:\:\:. \\ $$$$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{8},\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{for}\:{y}? \\ $$

Answered by FongXD last updated on 24/Nov/21

x^2 −8y=4x  ⇒ y=((x(x−4))/8)∈N  ⇒ x−4 is a multiple of 8, ⇒ x=8t+4, t∈N  minimum possible value for y=((12(12−4))/8)=12

$$\mathrm{x}^{\mathrm{2}} −\mathrm{8y}=\mathrm{4x} \\ $$$$\Rightarrow\:\mathrm{y}=\frac{\mathrm{x}\left(\mathrm{x}−\mathrm{4}\right)}{\mathrm{8}}\in\mathbb{N} \\ $$$$\Rightarrow\:\mathrm{x}−\mathrm{4}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{8},\:\Rightarrow\:\mathrm{x}=\mathrm{8t}+\mathrm{4},\:\mathrm{t}\in\mathbb{N} \\ $$$$\mathrm{minimum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{for}\:\mathrm{y}=\frac{\mathrm{12}\left(\mathrm{12}−\mathrm{4}\right)}{\mathrm{8}}=\mathrm{12} \\ $$

Commented by abdullah_ff last updated on 24/Nov/21

thank you very much sir

$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 24/Nov/21

Compact & Efficient!

$${Compact}\:\&\:\mathcal{E}{fficient}! \\ $$

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