Question Number 189357 by BaliramKumar last updated on 15/Mar/23
$$ \\ $$How many pairs of positive integers x, y exist such that HCF (x, y) + LCM(x, y) = 91?
Commented by mr W last updated on 15/Mar/23
$$\mathrm{5}\:{pairs}: \\ $$$$\left(\mathrm{1},\mathrm{90}\right) \\ $$$$\left(\mathrm{2},\mathrm{45}\right) \\ $$$$\left(\mathrm{5},\mathrm{18}\right) \\ $$$$\left(\mathrm{7},\mathrm{84}\right) \\ $$$$\left(\mathrm{9},\mathrm{10}\right) \\ $$
Commented by BaliramKumar last updated on 15/Mar/23
$${sir},\:\:\:\:{answer}\:{is}\:\mathrm{8} \\ $$
Commented by som(math1967) last updated on 15/Mar/23
$$\left(\mathrm{13},\mathrm{78}\right)\:,\left(\mathrm{21},\mathrm{28}\right)\:,\left(\mathrm{26},\mathrm{39}\right) \\ $$$$\mathrm{91}=\mathrm{1}×\mathrm{7}×\mathrm{13}\:\therefore{H}.{C}.{F}\:{of}\:{x},{y} \\ $$$${either}\:\mathrm{1}{or}\:\mathrm{7}{or}\:\mathrm{13} \\ $$
Commented by mr W last updated on 15/Mar/23
$${thanks}\:{sirs}! \\ $$
Commented by BaliramKumar last updated on 15/Mar/23
$${nice}\:{sir} \\ $$
Answered by BaliramKumar last updated on 15/Mar/23
![Let HCF(x, y) = k x = k∙a, y = k∙b, LCM(x, y) = k∙a∙b HCF(x, y) + LCM(x, y) = 91 k + k∙a∙b = 91 k(1 + a∙b) = 91 ..........(i) k = 1, 7, 13, 91 put k = 1 in equ. (i) 1(1 + ab) = 91 ab = 90 = 2^1 ×3^2 ×5^1 [3 distinct prime factor of 90] No. of co−prime pair (a, b) = 2^(3−1) = 4 determinant ((((a, b)),((1, 90)),((2, 45)),((5, 18)),((9, 10))),(((x, y) = (ka, kb)),((1, 90)),((2, 45)),((5, 18)),((9, 10)))) put k = 7 in equ. (i) 7(1 + ab) = 91 ab = 12 = 2^2 ×3^1 [2 distinct prime factor of 12] No. of co−prime pair (a, b) = 2^(2−1) = 2 determinant ((((a, b)),((1, 12)),((3, 4))),(((x, y) = (ka, kb)),((7, 84)),((21, 28)))) put k = 13 in equ. (i) 13(1 + ab) = 91 ab = 6 = 2^1 ×3^1 [2 distinct prime factor of 6] No. of co−prime pair (a, b) = 2^(2−1) = 2 determinant ((((a, b)),((1, 6)),((2, 3))),(((x, y) = (ka, kb)),((13, 78)),((26, 39)))) Total pair = 4 + 2 + 2 = 8 Answer](https://www.tinkutara.com/question/Q189396.png)
$$ \\ $$$${Let}\:\:{HCF}\left({x},\:{y}\right)\:=\:{k} \\ $$$${x}\:=\:{k}\centerdot{a},\:\:\:\:\:{y}\:=\:{k}\centerdot{b},\:\:\:\:\:{LCM}\left({x},\:{y}\right)\:=\:{k}\centerdot{a}\centerdot{b} \\ $$$${HCF}\left({x},\:{y}\right)\:+\:{LCM}\left({x},\:{y}\right)\:=\:\mathrm{91} \\ $$$${k}\:+\:{k}\centerdot{a}\centerdot{b}\:=\:\mathrm{91} \\ $$$${k}\left(\mathrm{1}\:+\:{a}\centerdot{b}\right)\:=\:\mathrm{91}\:\:……….\left({i}\right) \\ $$$${k}\:=\:\mathrm{1},\:\mathrm{7},\:\mathrm{13},\:\cancel{\mathrm{91}}\: \\ $$$${put}\:\:\:\:{k}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:{in}\:{equ}.\:\left({i}\right) \\ $$$$\mathrm{1}\left(\mathrm{1}\:+\:{ab}\right)\:=\:\mathrm{91} \\ $$$${ab}\:=\:\mathrm{90}\:=\:\mathrm{2}^{\mathrm{1}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{5}^{\mathrm{1}} \:\:\:\:\:\left[\mathrm{3}\:{distinct}\:{prime}\:{factor}\:{of}\:\mathrm{90}\right] \\ $$$${No}.\:{of}\:{co}−{prime}\:{pair}\:\left({a},\:{b}\right)\:=\:\mathrm{2}^{\mathrm{3}−\mathrm{1}} \:=\:\mathrm{4} \\ $$$$\begin{array}{|c|c|}{\left({a},\:{b}\right)}&\hline{\left(\mathrm{1},\:\mathrm{90}\right)}&\hline{\left(\mathrm{2},\:\mathrm{45}\right)}&\hline{\left(\mathrm{5},\:\mathrm{18}\right)}&\hline{\left(\mathrm{9},\:\mathrm{10}\right)}\\{\left({x},\:{y}\right)\:=\:\left({ka},\:{kb}\right)}&\hline{\left(\mathrm{1},\:\mathrm{90}\right)}&\hline{\left(\mathrm{2},\:\mathrm{45}\right)}&\hline{\left(\mathrm{5},\:\mathrm{18}\right)}&\hline{\left(\mathrm{9},\:\mathrm{10}\right)}\\\hline\end{array} \\ $$$${put}\:\:\:\:{k}\:=\:\mathrm{7}\:\:\:\:\:\:\:\:{in}\:{equ}.\:\left({i}\right) \\ $$$$\mathrm{7}\left(\mathrm{1}\:+\:{ab}\right)\:=\:\mathrm{91} \\ $$$${ab}\:=\:\mathrm{12}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{1}} \:\:\:\:\:\left[\mathrm{2}\:{distinct}\:{prime}\:{factor}\:{of}\:\mathrm{12}\right] \\ $$$${No}.\:{of}\:{co}−{prime}\:{pair}\:\left({a},\:{b}\right)\:=\:\mathrm{2}^{\mathrm{2}−\mathrm{1}} \:=\:\mathrm{2} \\ $$$$\begin{array}{|c|c|}{\left({a},\:{b}\right)}&\hline{\left(\mathrm{1},\:\mathrm{12}\right)}&\hline{\left(\mathrm{3},\:\mathrm{4}\right)}\\{\left({x},\:{y}\right)\:=\:\left({ka},\:{kb}\right)}&\hline{\left(\mathrm{7},\:\mathrm{84}\right)}&\hline{\left(\mathrm{21},\:\mathrm{28}\right)}\\\hline\end{array} \\ $$$${put}\:\:\:\:{k}\:=\:\mathrm{13}\:\:\:\:\:\:\:\:{in}\:{equ}.\:\left({i}\right) \\ $$$$\mathrm{13}\left(\mathrm{1}\:+\:{ab}\right)\:=\:\mathrm{91} \\ $$$${ab}\:=\:\mathrm{6}\:=\:\mathrm{2}^{\mathrm{1}} ×\mathrm{3}^{\mathrm{1}} \:\:\:\:\:\left[\mathrm{2}\:{distinct}\:{prime}\:{factor}\:{of}\:\mathrm{6}\right] \\ $$$${No}.\:{of}\:{co}−{prime}\:{pair}\:\left({a},\:{b}\right)\:=\:\mathrm{2}^{\mathrm{2}−\mathrm{1}} \:=\:\mathrm{2} \\ $$$$\begin{array}{|c|c|}{\left({a},\:{b}\right)}&\hline{\left(\mathrm{1},\:\mathrm{6}\right)}&\hline{\left(\mathrm{2},\:\mathrm{3}\right)}\\{\left({x},\:{y}\right)\:=\:\left({ka},\:{kb}\right)}&\hline{\left(\mathrm{13},\:\mathrm{78}\right)}&\hline{\left(\mathrm{26},\:\mathrm{39}\right)}\\\hline\end{array} \\ $$$$ \\ $$$${Total}\:{pair}\:=\:\mathrm{4}\:+\:\mathrm{2}\:+\:\mathrm{2}\:=\:\mathrm{8}\:{Answer} \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 15/Mar/23
![hcf(x,y)+lcm(x,y)=91 lcm(x,y)=91−hcf(x,y) ((lcm(x,y))/(hcf(x,y)))=((91−hcf(x,y))/(hcf(x,y)))∈Z^+ [∵ hcf(x,y) ∣ lcm(x,y) ] ⇒hcf(x,y)=1,7,13 ⇒lcm(x,y)=90,84,78 Case1: hcf(x,y)=1 ∧ lcm(x,y)=90 hcf(x,y)×lcm(x,y)=x×y (1)×lcm(x,y)=x×y lcm(x,y)=x×y=90 (x,y)=(1,90),(2,45),(5,18),(9,10) (4 pairs) Case2: hcf(x,y)=7 ∧ lcm(x,y)=84 hcf(x,y)×lcm(x,y)=x×y (7)×lcm(x,y)=x×y lcm(x,y)=((x×y)/7)=84 x×y=588=2^2 ×3×7^2 (3×7,2^2 ×7)=(21,28) (7,4×3×7)=(7,84) (2 pairs) Case3: hcf(x,y)=13 ∧ lcm(x,y)=78 hcf(x,y)×lcm(x,y)=x×y (13)×lcm(x,y)=x×y lcm(x,y)=((x×y)/(13))=78 x×y=13×78=1014=2×3×13^2 (x,y) =(13,2×3×13)=(13,78) =(2×13,3×13)=(26,39) (2 pairs) 4+2+2=8 pairs But as if (x,y) is solution so as (y,x) is also solution. So Total pairs=16](https://www.tinkutara.com/question/Q189397.png)
$$\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)+\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{91} \\ $$$$\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{91}−\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\frac{\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)}{\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)}=\frac{\mathrm{91}−\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)}{\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)}\in\mathbb{Z}^{+} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)\:\mid\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)\:\right] \\ $$$$\Rightarrow\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{1},\mathrm{7},\mathrm{13} \\ $$$$\Rightarrow\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{90},\mathrm{84},\mathrm{78} \\ $$$$\mathrm{Case1}:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{1}\:\wedge\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{90} \\ $$$$\:\:\:\:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\:\left(\mathrm{1}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y}=\mathrm{90} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{1},\mathrm{90}\right),\left(\mathrm{2},\mathrm{45}\right),\left(\mathrm{5},\mathrm{18}\right),\left(\mathrm{9},\mathrm{10}\right) \\ $$$$\left(\mathrm{4}\:{pairs}\right) \\ $$$$\mathrm{Case2}:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{7}\:\wedge\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{84} \\ $$$$\:\:\:\:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\:\left(\mathrm{7}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{x}×\mathrm{y}}{\mathrm{7}}=\mathrm{84} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}×\mathrm{y}=\mathrm{588}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{7}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}×\mathrm{7},\mathrm{2}^{\mathrm{2}} ×\mathrm{7}\right)=\left(\mathrm{21},\mathrm{28}\right) \\ $$$$\left(\mathrm{7},\mathrm{4}×\mathrm{3}×\mathrm{7}\right)=\left(\mathrm{7},\mathrm{84}\right) \\ $$$$\left(\mathrm{2}\:{pairs}\right) \\ $$$$\mathrm{Case3}:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{13}\:\wedge\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{78} \\ $$$$\:\:\:\:\:\mathrm{hcf}\left(\mathrm{x},\mathrm{y}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\left(\mathrm{13}\right)×\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}×\mathrm{y} \\ $$$$\:\:\:\:\mathrm{lcm}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{x}×\mathrm{y}}{\mathrm{13}}=\mathrm{78} \\ $$$$\:\:\:\:\:\:\mathrm{x}×\mathrm{y}=\mathrm{13}×\mathrm{78}=\mathrm{1014}=\mathrm{2}×\mathrm{3}×\mathrm{13}^{\mathrm{2}} \\ $$$$\left(\mathrm{x},\mathrm{y}\right) \\ $$$$=\left(\mathrm{13},\mathrm{2}×\mathrm{3}×\mathrm{13}\right)=\left(\mathrm{13},\mathrm{78}\right) \\ $$$$=\left(\mathrm{2}×\mathrm{13},\mathrm{3}×\mathrm{13}\right)=\left(\mathrm{26},\mathrm{39}\right) \\ $$$$\left(\mathrm{2}\:{pairs}\right) \\ $$$$\: \\ $$$$\mathrm{4}+\mathrm{2}+\mathrm{2}=\mathrm{8}\:{pairs} \\ $$$${But}\:{as}\:{if}\:\left({x},{y}\right)\:{is}\:{solution}\:{so}\:{as} \\ $$$$\left({y},{x}\right)\:{is}\:{also}\:{solution}. \\ $$$${So}\:{Total}\:{pairs}=\mathrm{16} \\ $$
Commented by BaliramKumar last updated on 15/Mar/23
$${Nice} \\ $$