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Question Number 193180 by BaliramKumar last updated on 06/Jun/23

If a + b = 1001 & HCF(a, b) = 13 then how many set of a & b.

$$\mathrm{If}\:{a}\:+\:{b}\:=\:\mathrm{1001}\:\&\:\mathrm{HCF}\left({a},\:{b}\right)\:=\:\mathrm{13}\: \\ $$$$\mathrm{then}\:\mathrm{how}\:\mathrm{many}\:\mathrm{set}\:\mathrm{of}\:{a}\:\&\:{b}. \\ $$

Answered by Subhi last updated on 06/Jun/23

a = 13k b = 13m a+b = 1001 = 13(k+m) k+m = 77 there are 78 pairs (77,0),(76,1).............(1,76),(0,77) there are restricrions when m and k are divisible by 11 or 7 m = 7l k = 7n k+m = 77 = 7(l+n) l+n = 11 (l,n) = (10,1),(9,2),(8,3),(7,4),(6,5)........(1,10) 10 pairs are excluded (ii) m = 11g k = 11h k+m = 77 = 11(g+h) g+h = 7 (g,h) = (6,1),(5,2),(4,3).....(1,6) 6 pairs are excluded 78−6−10 = 62

$${a}\:=\:\mathrm{13}{k} \\ $$$${b}\:=\:\mathrm{13}{m} \\ $$$${a}+{b}\:=\:\mathrm{1001}\:=\:\mathrm{13}\left({k}+{m}\right) \\ $$$${k}+{m}\:=\:\mathrm{77} \\ $$$${there}\:{are}\:\mathrm{78}\:{pairs} \\ $$$$\left(\mathrm{77},\mathrm{0}\right),\left(\mathrm{76},\mathrm{1}\right).............\left(\mathrm{1},\mathrm{76}\right),\left(\mathrm{0},\mathrm{77}\right) \\ $$$${there}\:{are}\:{restricrions}\:{when}\:{m}\:{and}\:{k}\:{are}\:{divisible}\:{by}\:\mathrm{11}\:{or}\:\mathrm{7} \\ $$$${m}\:=\:\mathrm{7}{l} \\ $$$${k}\:=\:\mathrm{7}{n} \\ $$$${k}+{m}\:=\:\mathrm{77}\:=\:\mathrm{7}\left({l}+{n}\right) \\ $$$${l}+{n}\:=\:\mathrm{11} \\ $$$$\left({l},{n}\right)\:=\:\left(\mathrm{10},\mathrm{1}\right),\left(\mathrm{9},\mathrm{2}\right),\left(\mathrm{8},\mathrm{3}\right),\left(\mathrm{7},\mathrm{4}\right),\left(\mathrm{6},\mathrm{5}\right)........\left(\mathrm{1},\mathrm{10}\right) \\ $$$$\mathrm{10}\:{pairs}\:\:{are}\:{excluded} \\ $$$$\left({ii}\right)\:{m}\:=\:\mathrm{11}{g} \\ $$$${k}\:=\:\mathrm{11}{h} \\ $$$${k}+{m}\:=\:\mathrm{77}\:=\:\mathrm{11}\left({g}+{h}\right) \\ $$$${g}+{h}\:=\:\mathrm{7} \\ $$$$\left({g},{h}\right)\:=\:\left(\mathrm{6},\mathrm{1}\right),\left(\mathrm{5},\mathrm{2}\right),\left(\mathrm{4},\mathrm{3}\right).....\left(\mathrm{1},\mathrm{6}\right) \\ $$$$\mathrm{6}\:{pairs}\:{are}\:{excluded} \\ $$$$\mathrm{78}−\mathrm{6}−\mathrm{10}\:=\:\mathrm{62} \\ $$

Commented by MM42 last updated on 06/Jun/23

there is no diffrence between the two numbers a=1,b=76 and a=76 , b=1 and ...

$${there}\:{is}\:{no}\:{diffrence}\:{between}\:{the}\:{two}\:{numbers}\: \\ $$$${a}=\mathrm{1},{b}=\mathrm{76}\:{and}\:{a}=\mathrm{76}\:,\:{b}=\mathrm{1}\:{and}\:... \\ $$

Commented by Subhi last updated on 06/Jun/23

(1,76) is different to (76,1) values of a and b differ ,but you are right that there restrictions for some values for a,b when factors k,m are divisible by 11 or 7

$$\left(\mathrm{1},\mathrm{76}\right)\:{is}\:{different}\:{to}\:\left(\mathrm{76},\mathrm{1}\right) \\ $$$${values}\:{of}\:{a}\:{and}\:{b}\:\:{differ} \\ $$$$,{but}\:{you}\:{are}\:{right}\:{that}\:{there}\:{restrictions}\:{for}\:{some}\:{values}\:{for}\:{a},{b}\:{when}\:{factors}\:{k},{m}\:{are}\:{divisible}\:{by}\:\mathrm{11}\:{or}\:\mathrm{7} \\ $$

Commented by Subhi last updated on 06/Jun/23

Thank you for that point

$${Thank}\:{you}\:{for}\:{that}\:{point} \\ $$

Commented by BaliramKumar last updated on 07/Jun/23

Sir 0, 77⇒ 13×(0, 77) = (0, 1001) HCF(0, 1001) = ?

$$\mathrm{Sir} \\ $$$$\mathrm{0},\:\mathrm{77}\Rightarrow\:\mathrm{13}×\left(\mathrm{0},\:\mathrm{77}\right)\:=\:\left(\mathrm{0},\:\mathrm{1001}\right) \\ $$$$\mathrm{HCF}\left(\mathrm{0},\:\mathrm{1001}\right)\:=\:? \\ $$

Commented by MM42 last updated on 07/Jun/23

you are right

$${you}\:{are}\:{right} \\ $$

Answered by MM42 last updated on 07/Jun/23

a=13a′ & b=13b′ ; (a′,b′)=1 ⇒13a′+13b′=1001⇒a′+b′=77 W={{a′,b′} : a′+b′=77 , (a′,b′)=1} S={{m,n} : m+n=77 }={{1,76},{2,75},...,{38,39}}⇒∣S ∣= 38 A={{m,n} : m+n=77 , (m,n)=7 } ⇒A={{7,70},{14,63},{21,56},{28,49},{35,42}}⇒∣A∣= 5 B={{m,n} : m+n=77 , (m,n)=11 } ⇒A={{11,66},{22,55},{33,44}}⇒∣B∣= 3 ⇒∣W∣=38−8=30

$${a}=\mathrm{13}{a}'\:\&\:\:{b}=\mathrm{13}{b}'\:;\:\left({a}',{b}'\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{13}{a}'+\mathrm{13}{b}'=\mathrm{1001}\Rightarrow{a}'+{b}'=\mathrm{77} \\ $$$${W}=\left\{\left\{{a}',{b}'\right\}\::\:\:{a}'+{b}'=\mathrm{77}\:\:,\:\left({a}',{b}'\right)=\mathrm{1}\right\} \\ $$$${S}=\left\{\left\{{m},{n}\right\}\::\:{m}+{n}=\mathrm{77}\:\right\}=\left\{\left\{\mathrm{1},\mathrm{76}\right\},\left\{\mathrm{2},\mathrm{75}\right\},...,\left\{\mathrm{38},\mathrm{39}\right\}\right\}\Rightarrow\mid{S}\:\mid=\:\mathrm{38} \\ $$$${A}=\left\{\left\{{m},{n}\right\}\::\:{m}+{n}=\mathrm{77}\:,\:\left({m},{n}\right)=\mathrm{7}\:\right\} \\ $$$$\Rightarrow{A}=\left\{\left\{\mathrm{7},\mathrm{70}\right\},\left\{\mathrm{14},\mathrm{63}\right\},\left\{\mathrm{21},\mathrm{56}\right\},\left\{\mathrm{28},\mathrm{49}\right\},\left\{\mathrm{35},\mathrm{42}\right\}\right\}\Rightarrow\mid{A}\mid=\:\mathrm{5} \\ $$$${B}=\left\{\left\{{m},{n}\right\}\::\:{m}+{n}=\mathrm{77}\:,\:\left({m},{n}\right)=\mathrm{11}\:\right\} \\ $$$$\Rightarrow{A}=\left\{\left\{\mathrm{11},\mathrm{66}\right\},\left\{\mathrm{22},\mathrm{55}\right\},\left\{\mathrm{33},\mathrm{44}\right\}\right\}\Rightarrow\mid{B}\mid=\:\mathrm{3} \\ $$$$\Rightarrow\mid{W}\mid=\mathrm{38}−\mathrm{8}=\mathrm{30} \\ $$

Answered by BaliramKumar last updated on 18/Nov/23

((φ(((1001)/(13))))/2) = ((φ(((7×11×13)/(13))))/2) = ((φ(7×11))/2) ⇒ (((7−1)(11−1))/2) = ((6×10)/2) = 30 unordered pairs ⇒ 60 ordered pairs

$$\frac{\phi\left(\frac{\mathrm{1001}}{\mathrm{13}}\right)}{\mathrm{2}}\:=\:\frac{\phi\left(\frac{\mathrm{7}×\mathrm{11}×\cancel{\mathrm{13}}}{\cancel{\mathrm{13}}}\right)}{\mathrm{2}}\:=\:\frac{\phi\left(\mathrm{7}×\mathrm{11}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{7}−\mathrm{1}\right)\left(\mathrm{11}−\mathrm{1}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{6}×\mathrm{10}}{\mathrm{2}}\:=\:\mathrm{30}\:\mathrm{unordered}\:\mathrm{pairs} \\ $$$$\Rightarrow\:\mathrm{60}\:\mathrm{ordered}\:\mathrm{pairs} \\ $$

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