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Question Number 160487 by HongKing last updated on 30/Nov/21

Answered by Rasheed.Sindhi last updated on 30/Nov/21

$$(10a+b)+(10b+a)=10c+3$$$$11a+11b=10c+3$$$$11(a+b)=10c+3..............\left(i\right)$$$$11\mid 10c+3\Rightarrow 10c+3=33\Rightarrow c=3$$$$(2-digitmultipleof11withunit3$$$$isonly33)$$$$$$$$\left(i\right)\Rightarrow a+b=\frac{10c+3}{11}=\frac{33}{11}=3$$$$a+b+2c=3+2\left(3\right)=9$$

Answered by Rasheed.Sindhi last updated on 30/Nov/21

$$\mathbb{A}\mathrm{n}\mathbb{O}\mathrm{ther}\mathbb{R}\mathrm{oad}\mathrm{to}\mathbb{R}\mathrm{ome}$$$$\overline{\mathrm{ab}}+\overline{\mathrm{ba}}=\overline{\mathrm{c}3};\mathrm{a}+\mathrm{b}+2\mathrm{c}=?$$$$\overline{\mathrm{ab}}+\overline{\mathrm{ba}}=\overline{\mathrm{c}3}\Rightarrow \mathrm{sum}\mathrm{of}\mathrm{units}:\mathrm{a}+\mathrm{b}=3\mathrm{or}13$$$$\underset{―}{\mathrm{a}+\mathrm{b}=3}\left(\mathrm{no}\mathrm{carry}\right)$$$$\mathrm{sum}\mathrm{of}\mathrm{tens}:\mathrm{a}+\mathrm{b}=\mathrm{c}=3$$$$\mathrm{a}+\mathrm{b}+2\mathrm{c}=3+2\left(3\right)=9$$$$\underset{―}{\mathrm{a}+\mathrm{b}=13}\left(1\mathrm{carry}\right)$$$$\mathrm{a}+\mathrm{b}+1=\mathrm{c}\Rightarrow \mathrm{c}=13+1=14\mathrm{but}\mathrm{c}⩽9$$$$\therefore \mathrm{a}+\mathrm{b}\ne 13$$$$\mathcal{H}ence\mathrm{a}+\mathrm{b}=3=\mathrm{c}$$$$\mathrm{a}+\mathrm{b}+2\mathrm{c}=3+2\left(3\right)=9$$

Answered by Rasheed.Sindhi last updated on 30/Nov/21

$$\overline{\mathrm{ab}}+\overline{\mathrm{ba}}=\overline{\mathrm{c}3};\mathrm{a}+\mathrm{b}+2\mathrm{c}=?$$$$\mathrm{Carry}\mathrm{of}\mathrm{units}=\mathrm{r}\left(say\right)$$$$\mathrm{Sum}\mathrm{of}\mathrm{units}:\mathrm{a}+\mathrm{b}=3+10\mathrm{r}$$$$\mathrm{Sum}\mathrm{of}\mathrm{tens}:\mathrm{a}+\mathrm{b}+\mathrm{r}=\mathrm{c}=3+11\mathrm{r}\Rightarrow \mathrm{r}=0(\because \mathrm{c}⩽9)$$$$\mathrm{a}+\mathrm{b}=3\wedge \mathrm{c}=\mathrm{a}+\mathrm{b}=3$$$$\mathrm{a}+\mathrm{b}+2\mathrm{c}=3+2\left(3\right)=9$$

Commented by HongKing last updated on 30/Nov/21

$$\mathrm{very}\mathrm{nice}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

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