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\[

{\text{As we know that according to Euclid's Division Lemma any number can be written as,}} \\

\Rightarrow a = bq + r{\text{ (1)}} \\

{\text{Where }}a{\text{ is dividend, }}b{\text{ is divisor, }}q{\text{ is quotient and }}r{\text{ is remainder }} \\

{\text{And as we see here }}117 > 65 \\

{\text{So, writing }}117{\text{ in the form of equation }}1{\text{ we get,}} \\

\Rightarrow 117 = 65*1 + 52{\text{ (2)}} \\

{\text{Now 65 will be dividend and 52 will be divisor}} \\

\Rightarrow 65 = 52*1 + 13{\text{ (3)}} \\

{\text{Now 52 will be dividend and 13 will be divisor}} \\

\Rightarrow 52 = 13*4 + 0 \\

{\text{As now the remainder is }}0{\text{ So, HCF will be the last divisor}}{\text{.}} \\

{\text{So, HCF of }}117{\text{ and }}65{\text{ will be }}13. \\

{\text{According to the question,}} \\

\Rightarrow 13 = 65m + 117n{\text{ (4)}} \\

{\text{So, we need to find value of }}m{\text{ and }}n \\

{\text{From equation 3 we get,}} \\

\Rightarrow 13 = 65 - 52*1{\text{ (5)}} \\

{\text{From equation 2 we get,}} \\

\Rightarrow 52 = 117 - 65*1{\text{ (6)}} \\

{\text{Putting value of }}52{\text{ from equation 6 to equation 5 we will get,}} \\

\Rightarrow 13 = 65 - \left( {117 - 65*1} \right){\text{ (7)}} \\

{\text{Solving equation 7 we get,}} \\

\Rightarrow 13 = 62*(2) + 117*( - 1){\text{ (8)}} \\

{\text{On comparing equation }}8{\text{ and }}4{\text{ we get value of }}m = 2{\text{ and }}n = - 1 \\

{\text{Now HCF of 65 and 117 in form of }}65m + 117n{\text{ is,}} \\

\Rightarrow 13 = 62*(2) + 117*( - 1){\text{ }} \\

{\text{NOTE: - Whenever you came up with this type of problem then best way is to calculate HCF}}{\text{ and}} \\

{\text{then solve all equations for calculating unknown variables}}{\text{. According to Euclid's Division Lemma}} \\

{\text{ any number can be written in the form of }}a = bq + r{\text{ where }}a > b. \\

\]

{\text{As we know that according to Euclid's Division Lemma any number can be written as,}} \\

\Rightarrow a = bq + r{\text{ (1)}} \\

{\text{Where }}a{\text{ is dividend, }}b{\text{ is divisor, }}q{\text{ is quotient and }}r{\text{ is remainder }} \\

{\text{And as we see here }}117 > 65 \\

{\text{So, writing }}117{\text{ in the form of equation }}1{\text{ we get,}} \\

\Rightarrow 117 = 65*1 + 52{\text{ (2)}} \\

{\text{Now 65 will be dividend and 52 will be divisor}} \\

\Rightarrow 65 = 52*1 + 13{\text{ (3)}} \\

{\text{Now 52 will be dividend and 13 will be divisor}} \\

\Rightarrow 52 = 13*4 + 0 \\

{\text{As now the remainder is }}0{\text{ So, HCF will be the last divisor}}{\text{.}} \\

{\text{So, HCF of }}117{\text{ and }}65{\text{ will be }}13. \\

{\text{According to the question,}} \\

\Rightarrow 13 = 65m + 117n{\text{ (4)}} \\

{\text{So, we need to find value of }}m{\text{ and }}n \\

{\text{From equation 3 we get,}} \\

\Rightarrow 13 = 65 - 52*1{\text{ (5)}} \\

{\text{From equation 2 we get,}} \\

\Rightarrow 52 = 117 - 65*1{\text{ (6)}} \\

{\text{Putting value of }}52{\text{ from equation 6 to equation 5 we will get,}} \\

\Rightarrow 13 = 65 - \left( {117 - 65*1} \right){\text{ (7)}} \\

{\text{Solving equation 7 we get,}} \\

\Rightarrow 13 = 62*(2) + 117*( - 1){\text{ (8)}} \\

{\text{On comparing equation }}8{\text{ and }}4{\text{ we get value of }}m = 2{\text{ and }}n = - 1 \\

{\text{Now HCF of 65 and 117 in form of }}65m + 117n{\text{ is,}} \\

\Rightarrow 13 = 62*(2) + 117*( - 1){\text{ }} \\

{\text{NOTE: - Whenever you came up with this type of problem then best way is to calculate HCF}}{\text{ and}} \\

{\text{then solve all equations for calculating unknown variables}}{\text{. According to Euclid's Division Lemma}} \\

{\text{ any number can be written in the form of }}a = bq + r{\text{ where }}a > b. \\

\]

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