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Question Number 157701 by MathSh last updated on 26/Oct/21

$$\mathrm{a};\mathrm{b};\mathrm{c}\in \mathbb{N}$$$$\frac{1}{\mathrm{a}+\frac{1}{\mathrm{b}+\frac{1}{\mathrm{c}}}}=\frac{16}{37}\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=?$$

Answered by mindispower last updated on 26/Oct/21

$$\frac{1}{2+\frac{5}{16}}=\frac{1}{2+\frac{1}{3+\frac{1}{5}}}$$

Commented by MathSh last updated on 26/Oct/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{very}\mathrm{nice}$$

Answered by Rasheed.Sindhi last updated on 26/Oct/21

$$\frac{1}{\mathrm{a}+\frac{1}{\mathrm{b}+\frac{1}{\mathrm{c}}}}=\frac{16}{37};\mathrm{a},\mathrm{b},c\in \mathbb{N}\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=?$$$$\frac{1}{\mathrm{a}+\frac{1}{\frac{\mathrm{bc}+1}{\mathrm{c}}}}=\frac{16}{37}$$$$\frac{1}{\mathrm{a}+\frac{\mathrm{c}}{\mathrm{bc}+1}}=\frac{16}{37}$$$$\frac{1}{\frac{\mathrm{abc}+\mathrm{a}+\mathrm{c}}{\mathrm{bc}+1}}=\frac{16}{37}$$$$\frac{\mathrm{bc}+1}{\mathrm{a}(\mathrm{bc}+1)+\mathrm{c}}=\frac{16}{37}$$$$37(\mathrm{bc}+1)=16(\mathrm{a}(\mathrm{bc}+1)+\mathrm{c})$$$$16\mid (\mathrm{bc}+1)\Rightarrow \mathrm{bc}+1=16,32,48,...$$$$\mathrm{bc}+1=16\Rightarrow \mathrm{bc}=15=3\times 5$$$$37\left(16\right)=16(16\mathrm{a}+5)$$$$16\mathrm{a}+5=37\Rightarrow \mathrm{a}=2$$$$(\mathrm{a},\mathrm{b},\mathrm{c})=(2,3,5)\left(Oneofthesolutions\right)$$$$Ithinknoothersolution.$$

Commented by MathSh last updated on 26/Oct/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{very}\mathrm{nice}$$

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