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Question Number 77459 by jagoll last updated on 06/Jan/20
    the greatest value of n so that   only one value of k satisfies   (8/(15))<(n/(n+k))<(7/(13)) is
$$ \\ $$$$ \\ $$$${the}\:{greatest}\:{value}\:{of}\:{n}\:{so}\:{that}\: \\ $$$${only}\:{one}\:{value}\:{of}\:{k}\:{satisfies}\: \\ $$$$\frac{\mathrm{8}}{\mathrm{15}}<\frac{{n}}{{n}+{k}}<\frac{\mathrm{7}}{\mathrm{13}}\:{is}\: \\ $$
Answered by MJS last updated on 06/Jan/20
(8/(15))<(n/(n+k)) ⇒ k<(7/8)n  (n/(n+k))<(7/(13)) ⇒ k>(6/7)n  ⇒  (6/7)n<k<(7/8)n  to get at least m possible values for k  n≥7×8×m+1  ⇒ answer is n=7×8×2=112
$$\frac{\mathrm{8}}{\mathrm{15}}<\frac{{n}}{{n}+{k}}\:\Rightarrow\:{k}<\frac{\mathrm{7}}{\mathrm{8}}{n} \\ $$$$\frac{{n}}{{n}+{k}}<\frac{\mathrm{7}}{\mathrm{13}}\:\Rightarrow\:{k}>\frac{\mathrm{6}}{\mathrm{7}}{n} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{6}}{\mathrm{7}}{n}<{k}<\frac{\mathrm{7}}{\mathrm{8}}{n} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{at}\:\mathrm{least}\:{m}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{for}\:{k} \\ $$$${n}\geqslant\mathrm{7}×\mathrm{8}×{m}+\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:{n}=\mathrm{7}×\mathrm{8}×\mathrm{2}=\mathrm{112} \\ $$

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