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If-97-19-a-1-b-1-c-where-a-b-and-c-are-positive-integers-then-what-is-the-sum-of-a-b-and-c-




Question Number 112804 by Aina Samuel Temidayo last updated on 09/Sep/20
If ((97)/(19))= a+(1/(b+(1/c))), where a,b and c are  positive integers, then what is the sum  of a,b and c?
$$\mathrm{If}\:\frac{\mathrm{97}}{\mathrm{19}}=\:\mathrm{a}+\frac{\mathrm{1}}{\mathrm{b}+\frac{\mathrm{1}}{\mathrm{c}}},\:\mathrm{where}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{are} \\ $$$$\mathrm{positive}\:\mathrm{integers},\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}? \\ $$
Answered by nimnim last updated on 09/Sep/20
((97)/(19))=5+(2/(19))=5+(1/((19)/2))=5+(1/(9+(1/2)))  ⇒a=5,b=9,c=2  ∴ (a+b+c)=16★
$$\frac{\mathrm{97}}{\mathrm{19}}=\mathrm{5}+\frac{\mathrm{2}}{\mathrm{19}}=\mathrm{5}+\frac{\mathrm{1}}{\frac{\mathrm{19}}{\mathrm{2}}}=\mathrm{5}+\frac{\mathrm{1}}{\mathrm{9}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{5},\mathrm{b}=\mathrm{9},\mathrm{c}=\mathrm{2} \\ $$$$\therefore\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=\mathrm{16}\bigstar \\ $$
Commented by Aina Samuel Temidayo last updated on 09/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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