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Question Number 208342 by hardmath last updated on 13/Jun/24

a,b,c∈N x = 4(2a+5) = 6(b+9) = 9(c−1) find: min(x+a+b+c) = ?

$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{x}\:=\:\mathrm{4}\left(\mathrm{2a}+\mathrm{5}\right)\:=\:\mathrm{6}\left(\mathrm{b}+\mathrm{9}\right)\:=\:\mathrm{9}\left(\mathrm{c}−\mathrm{1}\right) \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\left(\mathrm{x}+\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\:=\:? \\ $$

Answered by A5T last updated on 13/Jun/24

c=((6(b+9))/9)+1=((2b)/3)+7⇒b=3k a=(1/2)[((6(b+9))/4)−5]=((3b+17)/4)⇒a=((9k+17)/4) ⇒k+1≡0(mod 4)⇒k≡3(mod 4) ⇒b=3(8q+3)⇒b=24q+9⇒min(b)=9 ⇒min(a)=11⇒min(c)=13 ⇒min(x+a+b+c)=108+11+9+13=141

$${c}=\frac{\mathrm{6}\left({b}+\mathrm{9}\right)}{\mathrm{9}}+\mathrm{1}=\frac{\mathrm{2}{b}}{\mathrm{3}}+\mathrm{7}\Rightarrow{b}=\mathrm{3}{k} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{6}\left({b}+\mathrm{9}\right)}{\mathrm{4}}−\mathrm{5}\right]=\frac{\mathrm{3}{b}+\mathrm{17}}{\mathrm{4}}\Rightarrow{a}=\frac{\mathrm{9}{k}+\mathrm{17}}{\mathrm{4}} \\ $$$$\Rightarrow{k}+\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{4}\right)\Rightarrow{k}\equiv\mathrm{3}\left({mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow{b}=\mathrm{3}\left(\mathrm{8}{q}+\mathrm{3}\right)\Rightarrow{b}=\mathrm{24}{q}+\mathrm{9}\Rightarrow{min}\left({b}\right)=\mathrm{9} \\ $$$$\Rightarrow{min}\left({a}\right)=\mathrm{11}\Rightarrow{min}\left({c}\right)=\mathrm{13} \\ $$$$\Rightarrow{min}\left({x}+{a}+{b}+{c}\right)=\mathrm{108}+\mathrm{11}+\mathrm{9}+\mathrm{13}=\mathrm{141} \\ $$

Commented by hardmath last updated on 13/Jun/24

thankyou dear professor

$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Jun/24

a,b,c∈N x = 4(2a+5) = 6(b+9) = 9(c−1) find: min(x+a+b+c) = ? lcm(4,6,9) ∣ x⇒36 ∣ x⇒x=36k a=((x−20)/8)=((36k−20)/8)=((9k−5)/2)⇒k∈O b=(x/6)−9=((36k)/6)−9=6k−9⇒k≥2 c=(x/9)+1=((36k)/9)+1=4k+1 k=3: x=36k=36(3)=108 a=((9k−5)/2)=((9(3)−5)/2)=11 b=6k−9=6(3)−9=9 c=4k+1=4(3)+1=13 min(x+a+b+c) =108+11+9+13=141

$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{x}\:=\:\mathrm{4}\left(\mathrm{2a}+\mathrm{5}\right)\:=\:\mathrm{6}\left(\mathrm{b}+\mathrm{9}\right)\:=\:\mathrm{9}\left(\mathrm{c}−\mathrm{1}\right) \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\left(\mathrm{x}+\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\:=\:? \\ $$$$\mathrm{lcm}\left(\mathrm{4},\mathrm{6},\mathrm{9}\right)\:\mid\:\mathrm{x}\Rightarrow\mathrm{36}\:\mid\:\mathrm{x}\Rightarrow\mathrm{x}=\mathrm{36k} \\ $$$$\mathrm{a}=\frac{\mathrm{x}−\mathrm{20}}{\mathrm{8}}=\frac{\mathrm{36k}−\mathrm{20}}{\mathrm{8}}=\frac{\mathrm{9k}−\mathrm{5}}{\mathrm{2}}\Rightarrow\mathrm{k}\in\mathbb{O} \\ $$$$\mathrm{b}=\frac{\mathrm{x}}{\mathrm{6}}−\mathrm{9}=\frac{\mathrm{36k}}{\mathrm{6}}−\mathrm{9}=\mathrm{6k}−\mathrm{9}\Rightarrow\mathrm{k}\geqslant\mathrm{2} \\ $$$$\mathrm{c}=\frac{\mathrm{x}}{\mathrm{9}}+\mathrm{1}=\frac{\mathrm{36k}}{\mathrm{9}}+\mathrm{1}=\mathrm{4k}+\mathrm{1} \\ $$$$\mathrm{k}=\mathrm{3}: \\ $$$$\mathrm{x}=\mathrm{36k}=\mathrm{36}\left(\mathrm{3}\right)=\mathrm{108} \\ $$$$\mathrm{a}=\frac{\mathrm{9k}−\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{9}\left(\mathrm{3}\right)−\mathrm{5}}{\mathrm{2}}=\mathrm{11} \\ $$$$\mathrm{b}=\mathrm{6k}−\mathrm{9}=\mathrm{6}\left(\mathrm{3}\right)−\mathrm{9}=\mathrm{9} \\ $$$$\mathrm{c}=\mathrm{4k}+\mathrm{1}=\mathrm{4}\left(\mathrm{3}\right)+\mathrm{1}=\mathrm{13} \\ $$$$\boldsymbol{\mathrm{min}}\left(\mathrm{x}+\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{108}+\mathrm{11}+\mathrm{9}+\mathrm{13}=\mathrm{141} \\ $$$$ \\ $$

Commented by hardmath last updated on 13/Jun/24

thankyou dear professor

$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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