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Question Number 157701 by MathSh last updated on 26/Oct/21
a;b;c∈N (1/(a + (1/(b + (1/c))))) = ((16)/(37)) ⇒ a+b+c=?
$$\mathrm{a};\mathrm{b};\mathrm{c}\in\mathbb{N} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{b}\:+\:\frac{\mathrm{1}}{\mathrm{c}}}}\:=\:\frac{\mathrm{16}}{\mathrm{37}}\:\:\:\Rightarrow\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$
Answered by mindispower last updated on 26/Oct/21
(1/(2+(5/(16))))=(1/(2+(1/(3+(1/5)))))
$$\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{5}}{\mathrm{16}}}=\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{5}}}} \\ $$
Commented by MathSh last updated on 26/Oct/21
thank you so much dear Ser very nice
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{very}\:\mathrm{nice} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/21
(1/(a + (1/(b + (1/c))))) = ((16)/(37));a,b,c∈N ⇒a+b+c=? (1/(a + (1/( ((bc+1)/c))))) = ((16)/(37)) (1/(a + (c/( bc+1)))) = ((16)/(37)) (1/( ((abc+a+c)/( bc+1)))) = ((16)/(37)) ((bc+1)/(a(bc+1)+c))=((16)/(37)) 37(bc+1)=16( a(bc+1)+c ) 16∣(bc+1)⇒bc+1=16,32,48,... bc+1=16⇒bc=15=3×5 37(16)=16(16a+5) 16a+5=37⇒a=2 (a,b,c)=(2,3,5) (One of the solutions) I think no other solution.
$$\frac{\mathrm{1}}{\mathrm{a}\:+\:\frac{\mathrm{1}}{\mathrm{b}\:+\:\frac{\mathrm{1}}{\mathrm{c}}}}\:=\:\frac{\mathrm{16}}{\mathrm{37}};\mathrm{a},\mathrm{b},{c}\in\mathbb{N}\:\Rightarrow\mathrm{a}+\mathrm{b}+\mathrm{c}=? \\ $$$$\frac{\mathrm{1}}{\mathrm{a}\:+\:\frac{\mathrm{1}}{\:\frac{\mathrm{bc}+\mathrm{1}}{\mathrm{c}}}}\:=\:\frac{\mathrm{16}}{\mathrm{37}} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}\:+\:\frac{\mathrm{c}}{\:\mathrm{bc}+\mathrm{1}}}\:=\:\frac{\mathrm{16}}{\mathrm{37}} \\ $$$$\frac{\mathrm{1}}{\:\frac{\mathrm{abc}+\mathrm{a}+\mathrm{c}}{\:\mathrm{bc}+\mathrm{1}}}\:=\:\frac{\mathrm{16}}{\mathrm{37}} \\ $$$$\frac{\mathrm{bc}+\mathrm{1}}{\mathrm{a}\left(\mathrm{bc}+\mathrm{1}\right)+\mathrm{c}}=\frac{\mathrm{16}}{\mathrm{37}} \\ $$$$\mathrm{37}\left(\mathrm{bc}+\mathrm{1}\right)=\mathrm{16}\left(\:\mathrm{a}\left(\mathrm{bc}+\mathrm{1}\right)+\mathrm{c}\:\right) \\ $$$$\mathrm{16}\mid\left(\mathrm{bc}+\mathrm{1}\right)\Rightarrow\mathrm{bc}+\mathrm{1}=\mathrm{16},\mathrm{32},\mathrm{48},… \\ $$$$\mathrm{bc}+\mathrm{1}=\mathrm{16}\Rightarrow\mathrm{bc}=\mathrm{15}=\mathrm{3}×\mathrm{5} \\ $$$$\mathrm{37}\left(\mathrm{16}\right)=\mathrm{16}\left(\mathrm{16a}+\mathrm{5}\right) \\ $$$$\mathrm{16a}+\mathrm{5}=\mathrm{37}\Rightarrow\mathrm{a}=\mathrm{2} \\ $$$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=\left(\mathrm{2},\mathrm{3},\mathrm{5}\right)\:\left({One}\:{of}\:{the}\:{solutions}\right) \\ $$$${I}\:{think}\:{no}\:{other}\:{solution}. \\ $$
Commented by MathSh last updated on 26/Oct/21
thank you so much dear Ser very nice
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{very}\:\mathrm{nice}\: \\ $$

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