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Question Number 199932 by hardmath last updated on 11/Nov/23

1. If 3 ∙ ab^(−) + bc^(−) = 115 Find: max(a+b+c)=? 2. a,b,c∈N If (a/2) + (b/3) = (c/4) Find: min(a+b+c)=?

$$\mathrm{1}. \\ $$$$\mathrm{If}\:\:\:\mathrm{3}\:\centerdot\:\overline {\mathrm{ab}}\:+\:\overline {\mathrm{bc}}\:=\:\mathrm{115} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{max}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=? \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{If}\:\:\:\frac{\mathrm{a}}{\mathrm{2}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{4}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=? \\ $$

Answered by Frix last updated on 11/Nov/23

1. Only 2 possbilities a=1 b=6 c=7 a+b+c=14 a=2 b=4 c=3 a+b+c=9 2. If a, b, c ≠0 c=2a+((4b)/3) a=m∧b=3n Minimum at a=1 b=3 c=6 a+b+c=10

$$\mathrm{1}.\:\mathrm{Only}\:\mathrm{2}\:\mathrm{possbilities} \\ $$$${a}=\mathrm{1}\:{b}=\mathrm{6}\:{c}=\mathrm{7}\:{a}+{b}+{c}=\mathrm{14} \\ $$$${a}=\mathrm{2}\:{b}=\mathrm{4}\:{c}=\mathrm{3}\:{a}+{b}+{c}=\mathrm{9} \\ $$$$\mathrm{2}.\:\mathrm{If}\:{a},\:{b},\:{c}\:\neq\mathrm{0} \\ $$$${c}=\mathrm{2}{a}+\frac{\mathrm{4}{b}}{\mathrm{3}} \\ $$$${a}={m}\wedge{b}=\mathrm{3}{n} \\ $$$$\mathrm{Minimum}\:\mathrm{at} \\ $$$${a}=\mathrm{1}\:{b}=\mathrm{3}\:{c}=\mathrm{6}\:{a}+{b}+{c}=\mathrm{10} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Nov/23

3 ∙ ab^(−) + bc^(−) = 115; a,b,c∈{0,1,...,9};a,c≠0 3(10a+b)+(10b+c) 30a+13b+c=115 115−13b−c=30a 115−13b−c≡0(mod 30) 13b+c≡25≡55≡85≡115(mod 30) (b,c)=(4,3),(6,7) Both pairs are successful to produce valid value of a (b,c)=(4,3): a=((115−13b−c)/(30)) a=((115−13(4)−3)/(30))=((60)/(30))=2 (b,c)=(6,7): a=((115−13(6)−7)/(30))=((30)/(30))=1 (a,b,c)=(2,4,3),(1,6,7)

$$\:\mathrm{3}\:\centerdot\:\overline {\mathrm{ab}}\:+\:\overline {\mathrm{bc}}\:=\:\mathrm{115}; \\ $$$$\:{a},{b},{c}\in\left\{\mathrm{0},\mathrm{1},...,\mathrm{9}\right\};{a},{c}\neq\mathrm{0} \\ $$$$\mathrm{3}\left(\mathrm{10a}+\mathrm{b}\right)+\left(\mathrm{10b}+\mathrm{c}\right) \\ $$$$\mathrm{30}{a}+\mathrm{13}{b}+{c}=\mathrm{115} \\ $$$$\mathrm{115}−\mathrm{13}{b}−{c}=\mathrm{30}{a} \\ $$$$\mathrm{115}−\mathrm{13}{b}−{c}\equiv\mathrm{0}\left({mod}\:\mathrm{30}\right) \\ $$$$\mathrm{13}{b}+{c}\equiv\mathrm{25}\equiv\mathrm{55}\equiv\mathrm{85}\equiv\mathrm{115}\left({mod}\:\mathrm{30}\right) \\ $$$$\left({b},{c}\right)=\left(\mathrm{4},\mathrm{3}\right),\left(\mathrm{6},\mathrm{7}\right) \\ $$$${Both}\:{pairs}\:{are}\:{successful}\:{to} \\ $$$${produce}\:{valid}\:{value}\:{of}\:{a} \\ $$$$\left({b},{c}\right)=\left(\mathrm{4},\mathrm{3}\right): \\ $$$${a}=\frac{\mathrm{115}−\mathrm{13}{b}−{c}}{\mathrm{30}} \\ $$$${a}=\frac{\mathrm{115}−\mathrm{13}\left(\mathrm{4}\right)−\mathrm{3}}{\mathrm{30}}=\frac{\mathrm{60}}{\mathrm{30}}=\mathrm{2} \\ $$$$\left({b},{c}\right)=\left(\mathrm{6},\mathrm{7}\right): \\ $$$${a}=\frac{\mathrm{115}−\mathrm{13}\left(\mathrm{6}\right)−\mathrm{7}}{\mathrm{30}}=\frac{\mathrm{30}}{\mathrm{30}}=\mathrm{1} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{2},\mathrm{4},\mathrm{3}\right),\left(\mathrm{1},\mathrm{6},\mathrm{7}\right) \\ $$$$ \\ $$

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