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Question Number 35425 by Rio Mike last updated on 18/May/18

Given that a number is a factor of 144 and the square of the number added to five times the number is ≥ −6 find the number

$$\:{Given}\:{that}\:{a}\:{number}\:{is}\:{a}\:{factor}\: \\ $$$${of}\:\mathrm{144}\:{and}\:{the}\:{square}\:{of}\:{the}\:{number} \\ $$$${added}\:{to}\:{five}\:{times}\:{the}\:{number} \\ $$$${is}\:\geqslant\:−\mathrm{6}\:{find}\:{the}\:{number} \\ $$

Answered by Rasheed.Sindhi last updated on 19/May/18

n∣144 ∧ n^2 +5n≥−6 n^2 +5n+6≥0 (n+2)(n+3)≥0 (n+2)(n+3)=0 ∣ ^∗ (n+2)(n+3)>0 n=−2 ∣ n=−3.........I ^∗ (n+2)(n+3)>0 n+2>0 ∧ n+3>0 ∣ n+2<0 ∧ n+3<0 n>−2 ∧ n>−3 ∣ n<−2 ∧ n<−3 n>−2 ∣ n<−3...........II I & II: n≥−2 or n≤−3 Hence Factors of 144 ≥−2 or ≤−3 {,..−8−6 ,−4,−3,−2,−1,1,2,3,4,6,8....} ={±1,±2,±3,±4,±6,±8,±9,....} Or All the factors of 144

$$\mathrm{n}\mid\mathrm{144}\:\wedge\:\mathrm{n}^{\mathrm{2}} +\mathrm{5n}\geqslant−\mathrm{6} \\ $$$$\mathrm{n}^{\mathrm{2}} +\mathrm{5n}+\mathrm{6}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)=\mathrm{0}\:\:\mid\:\:^{\ast} \left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)>\mathrm{0} \\ $$$$\:\mathrm{n}=−\mathrm{2}\:\mid\:\mathrm{n}=−\mathrm{3}.........\mathrm{I} \\ $$$$\:^{\ast} \left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)>\mathrm{0} \\ $$$$\:\:\mathrm{n}+\mathrm{2}>\mathrm{0}\:\wedge\:\mathrm{n}+\mathrm{3}>\mathrm{0}\:\mid\:\mathrm{n}+\mathrm{2}<\mathrm{0}\:\wedge\:\mathrm{n}+\mathrm{3}<\mathrm{0} \\ $$$$\:\mathrm{n}>−\mathrm{2}\:\wedge\:\mathrm{n}>−\mathrm{3}\:\:\mid\:\mathrm{n}<−\mathrm{2}\:\wedge\:\mathrm{n}<−\mathrm{3} \\ $$$$\mathrm{n}>−\mathrm{2}\:\:\mid\:\:\mathrm{n}<−\mathrm{3}...........\mathrm{II} \\ $$$$\mathrm{I}\:\&\:\mathrm{II}:\:\:\mathrm{n}\geqslant−\mathrm{2}\:\:\mathrm{or}\:\:\mathrm{n}\leqslant−\mathrm{3} \\ $$$$\mathrm{Hence} \\ $$$$\:\mathrm{Factors}\:\mathrm{of}\:\mathrm{144}\:\geqslant−\mathrm{2}\:\mathrm{or}\:\leqslant−\mathrm{3} \\ $$$$\:\left\{,..−\mathrm{8}−\mathrm{6}\:,−\mathrm{4},−\mathrm{3},−\mathrm{2},−\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{8}....\right\} \\ $$$$=\left\{\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},\pm\mathrm{4},\pm\mathrm{6},\pm\mathrm{8},\pm\mathrm{9},....\right\} \\ $$$$\mathrm{Or}\:\mathrm{All}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{144} \\ $$

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