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What-is-the-minimum-value-obtained-when-an-arbitrary-number-of-three-different-non-zero-digits-is-divided-by-the-sum-of-its-digits-

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Question Number 111537 by Aina Samuel Temidayo last updated on 04/Sep/20
What is the minimum value obtained when an arbitrary number of three different non−zero digits is divided by the sum of its digits?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{obtained} \\ $$$$\mathrm{when}\:\mathrm{an}\:\mathrm{arbitrary}\:\mathrm{number}\:\mathrm{of}\:\mathrm{three} \\ $$$$\mathrm{different}\:\mathrm{non}−\mathrm{zero}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}? \\ $$$$ \\ $$
Answered by 1549442205PVT last updated on 04/Sep/20
Suppose abc^(−) =100a+20b+c.We need find the smallest value of P=((100a+10b+c)/(a+b+c)) where a,b,c∈{1,2,3,...,9};a≠b;a≠c;b;c P=1+((99a+9b)/(a+b+c)).We will prove that ((99a+9b)/(a+b+c))≥((171)/(18))⇔((11a+b)/(a+b+c))≥((19)/(18))(∗) ⇔198a+18b≥19a+19b+19c ⇔179a≥b+19c.Since b≠c≤9 This last inequality is always true since b,c≤9⇒b+20c≤8+19.9=179≤179a due to a≥1.Hence the inequality is proved.Consequently,P≥1+((171)/(18))=((21)/2) Which means ((abc^(−) )/(a+b+c)) has least value equal to ((21)/2)when abc^(−) =189
$$\mathrm{Suppose}\:\overline {\mathrm{abc}}=\mathrm{100a}+\mathrm{20b}+\mathrm{c}.\mathrm{We}\:\mathrm{need} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{P}=\frac{\mathrm{100a}+\mathrm{10b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:\mathrm{where} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{9}\right\};\mathrm{a}\neq\mathrm{b};\mathrm{a}\neq\mathrm{c};\mathrm{b};\mathrm{c} \\ $$$$\mathrm{P}=\mathrm{1}+\frac{\mathrm{99a}+\mathrm{9b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}.\mathrm{We}\:\mathrm{will}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{99a}+\mathrm{9b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\frac{\mathrm{171}}{\mathrm{18}}\Leftrightarrow\frac{\mathrm{11a}+\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\frac{\mathrm{19}}{\mathrm{18}}\left(\ast\right) \\ $$$$\Leftrightarrow\mathrm{198a}+\mathrm{18b}\geqslant\mathrm{19a}+\mathrm{19b}+\mathrm{19c} \\ $$$$\Leftrightarrow\mathrm{179a}\geqslant\mathrm{b}+\mathrm{19c}.\mathrm{Since}\:\mathrm{b}\neq\mathrm{c}\leqslant\mathrm{9} \\ $$$$\mathrm{This}\:\mathrm{last}\:\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{since} \\ $$$$\mathrm{b},\mathrm{c}\leqslant\mathrm{9}\Rightarrow\mathrm{b}+\mathrm{20c}\leqslant\mathrm{8}+\mathrm{19}.\mathrm{9}=\mathrm{179}\leqslant\mathrm{179a} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{a}\geqslant\mathrm{1}.\mathrm{Hence}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{is} \\ $$$$\mathrm{proved}.\mathrm{Consequently},\mathrm{P}\geqslant\mathrm{1}+\frac{\mathrm{171}}{\mathrm{18}}=\frac{\mathrm{21}}{\mathrm{2}} \\ $$$$\mathrm{Which}\:\mathrm{means}\:\frac{\overline {\mathrm{abc}}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:\mathrm{has}\:\mathrm{least}\:\mathrm{value}\: \\ $$$$\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{21}}{\mathrm{2}}\mathrm{when}\:\overline {\mathrm{abc}}=\mathrm{189} \\ $$
Commented by Her_Majesty last updated on 04/Sep/20
“three different digits”
$$“{three}\:{different}\:{digits}'' \\ $$
Commented by 1549442205PVT last updated on 04/Sep/20
Thank you,Sir.I missed that hypothesis and corrected
$$\mathrm{Thank}\:\mathrm{you},\mathrm{Sir}.\mathrm{I}\:\mathrm{missed}\:\mathrm{that}\:\mathrm{hypothesis} \\ $$$$\mathrm{and}\:\mathrm{corrected} \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
How did you know you are to prove that ((99a+9b)/(a+b+c))≥((171)/(18)) ?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know}\:\mathrm{you}\:\mathrm{are}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\frac{\mathrm{99a}+\mathrm{9b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\frac{\mathrm{171}}{\mathrm{18}}\:? \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 05/Sep/20
Had in the above solution since it is equivalent ((11a+b)/(a+b+c))≥((19)/(18)) ⇔198a+18b≥19a+19b+19c ⇔179a≥19c+b.The inequality is always true ∀a,b,c∈{1,2,...,9},a≠b, b≠c,c≠a.Indeed,19c+b≤19.9+8 =179.1≤179a due to a≥11
$$\mathrm{Had}\:\mathrm{in}\:\mathrm{the}\:\mathrm{above}\:\mathrm{solution}\:\mathrm{since}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{equivalent}\:\frac{\mathrm{11a}+\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\geqslant\frac{\mathrm{19}}{\mathrm{18}} \\ $$$$\Leftrightarrow\mathrm{198a}+\mathrm{18b}\geqslant\mathrm{19a}+\mathrm{19b}+\mathrm{19c} \\ $$$$\Leftrightarrow\mathrm{179a}\geqslant\mathrm{19c}+\mathrm{b}.\mathrm{The}\:\mathrm{inequality}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{true}\:\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\left\{\mathrm{1},\mathrm{2},…,\mathrm{9}\right\},\mathrm{a}\neq\mathrm{b}, \\ $$$$\mathrm{b}\neq\mathrm{c},\mathrm{c}\neq\mathrm{a}.\mathrm{Indeed},\mathrm{19c}+\mathrm{b}\leqslant\mathrm{19}.\mathrm{9}+\mathrm{8} \\ $$$$=\mathrm{179}.\mathrm{1}\leqslant\mathrm{179a}\:\mathrm{due}\:\mathrm{to}\:\mathrm{a}\geqslant\mathrm{11} \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 05/Sep/20
Thanks but you didn′t really answer my question.
$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{you}\:\mathrm{didn}'\mathrm{t}\:\mathrm{really}\:\mathrm{answer} \\ $$$$\mathrm{my}\:\mathrm{question}. \\ $$
Answered by Her_Majesty last updated on 04/Sep/20
((100a+10b+c)/(a+b+c)) (1) lowest number with a<b<c (2) a as low as possible ⇒ a=1 (3) c as high as possible ⇒ c=9 ((10b+109)/(b+10))=10 with b→∞ ⇒ b as high as possible ⇒ b=8 (a≠b≠c) ⇒ ((189)/(18))=10.5
$$\frac{\mathrm{100}{a}+\mathrm{10}{b}+{c}}{{a}+{b}+{c}} \\ $$$$\left(\mathrm{1}\right)\:{lowest}\:{number}\:{with}\:{a}<{b}<{c} \\ $$$$\left(\mathrm{2}\right)\:{a}\:{as}\:{low}\:{as}\:{possible}\:\Rightarrow\:{a}=\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{c}\:{as}\:{high}\:{as}\:{possible}\:\Rightarrow\:{c}=\mathrm{9} \\ $$$$\frac{\mathrm{10}{b}+\mathrm{109}}{{b}+\mathrm{10}}=\mathrm{10}\:{with}\:{b}\rightarrow\infty\:\Rightarrow\:{b}\:{as}\:{high}\:{as} \\ $$$${possible}\:\Rightarrow\:{b}=\mathrm{8}\:\left({a}\neq{b}\neq{c}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{189}}{\mathrm{18}}=\mathrm{10}.\mathrm{5} \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
How did you know ((10b+109)/(b+10)) is equal to 10?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know}\:\frac{\mathrm{10b}+\mathrm{109}}{\mathrm{b}+\mathrm{10}}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{10}? \\ $$

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