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Question Number 74703 by Mr. K last updated on 29/Nov/19

let b and r be two positive prime   numbers such that b≠r and b×r is  a divisor of 138. Consider an   arithmetic progression in which  the first term is b, the ratio is r  and the fourth term is 71. What is the  value of b+r?

$${let}\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{r}}\:{be}\:{two}\:{positive}\:{prime}\: \\ $$$${numbers}\:{such}\:{that}\:{b}\neq{r}\:{and}\:{b}×{r}\:{is} \\ $$$${a}\:{divisor}\:{of}\:\mathrm{138}.\:{Consider}\:{an}\: \\ $$$${arithmetic}\:{progression}\:{in}\:{which} \\ $$$${the}\:{first}\:{term}\:{is}\:\boldsymbol{{b}},\:{the}\:{ratio}\:{is}\:\boldsymbol{{r}} \\ $$$${and}\:{the}\:{fourth}\:{term}\:{is}\:\mathrm{71}.\:{What}\:{is}\:{the} \\ $$$${value}\:{of}\:\boldsymbol{{b}}+\boldsymbol{{r}}? \\ $$

Answered by mind is power last updated on 29/Nov/19

br∣138  138=2.69=2.3.23    b+3r=71⇒b≡−1(3)  ⇒b∈{2,23}  b+r=71−2r⇒b+r=1(2)⇒b,r are in different parite  one of them is 2  b=71−3r≤23⇒r≥((71−23)/2)=((48)/3)=16⇒r=23,b=2  b+r=25

$$\mathrm{br}\mid\mathrm{138} \\ $$$$\mathrm{138}=\mathrm{2}.\mathrm{69}=\mathrm{2}.\mathrm{3}.\mathrm{23} \\ $$$$ \\ $$$$\mathrm{b}+\mathrm{3r}=\mathrm{71}\Rightarrow\mathrm{b}\equiv−\mathrm{1}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{b}\in\left\{\mathrm{2},\mathrm{23}\right\} \\ $$$$\mathrm{b}+\mathrm{r}=\mathrm{71}−\mathrm{2r}\Rightarrow\mathrm{b}+\mathrm{r}=\mathrm{1}\left(\mathrm{2}\right)\Rightarrow\mathrm{b},\mathrm{r}\:\mathrm{are}\:\mathrm{in}\:\mathrm{different}\:\mathrm{parite} \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{is}\:\mathrm{2} \\ $$$$\mathrm{b}=\mathrm{71}−\mathrm{3r}\leqslant\mathrm{23}\Rightarrow\mathrm{r}\geqslant\frac{\mathrm{71}−\mathrm{23}}{\mathrm{2}}=\frac{\mathrm{48}}{\mathrm{3}}=\mathrm{16}\Rightarrow\mathrm{r}=\mathrm{23},\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{b}+\mathrm{r}=\mathrm{25} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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