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Question Number 68280 by mezihloic last updated on 08/Sep/19

$$giventhat{432}_n-{413}_n={11}_{10}.findthevalueofn$$

Commented by TawaTawa last updated on 08/Sep/19

$$4\times {\mathrm{n}}^2+3\times {\mathrm{n}}^1+2\times {\mathrm{n}}^0-(4\times {\mathrm{n}}^2+1\times {\mathrm{n}}^1+3\times {\mathrm{n}}^0)=11$$$${4\mathrm{n}}^2+3\mathrm{n}+2-({4\mathrm{n}}^2+\mathrm{n}+3)=11$$$${4\mathrm{n}}^2+3\mathrm{n}+2-{4\mathrm{n}}^2-\mathrm{n}-3=11$$$$3\mathrm{n}+2-\mathrm{n}-3=11$$$$2\mathrm{n}-1=11$$$$2\mathrm{n}=11+1$$$$2\mathrm{n}=12$$$$\mathrm{n}=\frac{12}{2}$$$$\mathrm{n}=6$$

Answered by mr W last updated on 08/Sep/19

$$4n^2+3n+2-(4n^2+n+3)=11$$$$2n-1=11$$$$\Rightarrow n=6$$

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