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given-that-432-n-413-n-11-10-find-the-value-of-n-




Question Number 68280 by mezihloic last updated on 08/Sep/19
given that 432_n −413_n =11_(10) .find the value of n
$${given}\:{that}\:\mathrm{432}_{{n}} −\mathrm{413}_{{n}} =\mathrm{11}_{\mathrm{10}} .{find}\:{the}\:{value}\:{of}\:{n} \\ $$
Commented by TawaTawa last updated on 08/Sep/19
4 × n^2  + 3 × n^1  + 2 × n^0  − (4 × n^2  + 1 × n^1  + 3 × n^0 )  =  11  4n^2  + 3n + 2 − (4n^2  + n + 3)  =  11  4n^2  + 3n + 2 − 4n^2  − n − 3  =  11  3n + 2 − n − 3  =  11  2n − 1  =  11  2n  =  11 + 1  2n  =  12    n  =  ((12)/2)    n  =  6
$$\mathrm{4}\:×\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{3}\:×\:\mathrm{n}^{\mathrm{1}} \:+\:\mathrm{2}\:×\:\mathrm{n}^{\mathrm{0}} \:−\:\left(\mathrm{4}\:×\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}\:×\:\mathrm{n}^{\mathrm{1}} \:+\:\mathrm{3}\:×\:\mathrm{n}^{\mathrm{0}} \right)\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{3n}\:+\:\mathrm{2}\:−\:\left(\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{n}\:+\:\mathrm{3}\right)\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{3n}\:+\:\mathrm{2}\:−\:\mathrm{4n}^{\mathrm{2}} \:−\:\mathrm{n}\:−\:\mathrm{3}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{3n}\:+\:\mathrm{2}\:−\:\mathrm{n}\:−\:\mathrm{3}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{2n}\:−\:\mathrm{1}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{2n}\:\:=\:\:\mathrm{11}\:+\:\mathrm{1} \\ $$$$\mathrm{2n}\:\:=\:\:\mathrm{12} \\ $$$$\:\:\mathrm{n}\:\:=\:\:\frac{\mathrm{12}}{\mathrm{2}} \\ $$$$\:\:\mathrm{n}\:\:=\:\:\mathrm{6} \\ $$
Answered by mr W last updated on 08/Sep/19
4n^2 +3n+2−(4n^2 +n+3)=11  2n−1=11  ⇒n=6
$$\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}−\left(\mathrm{4}{n}^{\mathrm{2}} +{n}+\mathrm{3}\right)=\mathrm{11} \\ $$$$\mathrm{2}{n}−\mathrm{1}=\mathrm{11} \\ $$$$\Rightarrow{n}=\mathrm{6} \\ $$

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