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Question Number 115058 by mnjuly1970 last updated on 23/Sep/20

$$....nicecalculus...$$$$a,b,c,d\in \mathbb{N}and$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$$$$find$$$$$$$$max(a+b+c+d)=???$$$$...m.n.july.1970...$$

Answered by 1549442205PVT last updated on 25/Sep/20

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}\left(1\right)$$$$\mathrm{Since}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\mathrm{are}\mathrm{equal}\mathrm{in}\mathrm{role}\mathrm{WLOG}$$$$\mathrm{we}\mathrm{can}\mathrm{suppose}\mathrm{that}\mathrm{a}⩾\mathrm{b}⩾\mathrm{c}⩾\mathrm{d}$$$$\Rightarrow \frac{4}{\mathrm{a}}⩽\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}⩽\frac{4}{\mathrm{d}}$$$$\Rightarrow \mathrm{a}⩾8⩾\mathrm{d}⩾3$$$$\mathrm{i})\mathrm{Case}\mathrm{d}=3\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$$$$\Rightarrow \frac{3}{\mathrm{a}}⩽\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6}⩽\frac{3}{\mathrm{c}}$$$$\Rightarrow \mathrm{a}⩾18⩾\mathrm{c}⩾7$$$$\mathrm{a})\mathrm{For}\mathrm{c}=7\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}$$$$\Rightarrow \frac{2}{\mathrm{a}}⩽\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=\frac{1}{42}⩽\frac{2}{\mathrm{b}}$$$$\Rightarrow \mathrm{a}⩾84⩾\mathrm{b}⩾43$$$$\bullet \mathrm{if}\mathrm{b}=43\mathrm{then}\frac{1}{\mathrm{a}}=\frac{1}{42}-\frac{1}{43}=\frac{1}{1806}$$$$\Rightarrow \mathrm{a}=1806$$$$\mathrm{Thus},\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=1806+43+7+3=1859$$$$\bullet \mathrm{If}\mathrm{b}=44\Rightarrow \frac{1}{\mathrm{a}}=\frac{1}{42}-\frac{1}{44}=\frac{1}{924}\Rightarrow \mathrm{a}=924$$$$42(\mathrm{a}+\mathrm{b})=\mathrm{ab}\Rightarrow (\mathrm{a}-42)(\mathrm{b}-42)=1764$$$$=4.{21}^2=2^2.3^2.7^2$$$$\mathrm{b})\mathrm{For}\mathrm{c}=8\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}$$$$\Rightarrow \mathrm{ab}=24(\mathrm{a}+\mathrm{b})\iff (\mathrm{a}-24)(\mathrm{b}-24)=576$$$$={24}^2=2^6.3^2$$$$\mathrm{b}=25\Rightarrow \mathrm{a}=600\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=636<1859$$$$\mathrm{c})\mathrm{For}\mathrm{c}=9,10,...,18\mathrm{by}\mathrm{similar}\mathrm{argument}$$$$\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}<1859$$$$\mathrm{ii})\mathrm{d}=4\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$$$\Rightarrow \frac{3}{\mathrm{a}}⩽\frac{1}{4}⩽\frac{3}{\mathrm{c}}\Rightarrow \mathrm{a}⩾12⩾\mathrm{c}⩾5$$$$\mathrm{a})\mathrm{For}\mathrm{c}=5\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}$$$$\Rightarrow \mathrm{ab}=20(\mathrm{a}+\mathrm{b})\iff (\mathrm{a}-20)(\mathrm{b}-20)=400=2^4.5^2$$$$\bullet \mathrm{if}\mathrm{b}=21\Rightarrow \mathrm{a}=420\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=450<1859$$$$\mathrm{b})\mathrm{For}\mathrm{c}=\overline{6...12},\mathrm{by}\mathrm{similar}\mathrm{argument}$$$$\mathrm{we}\mathrm{get}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}<1859$$$$\mathrm{ii})\mathrm{Cases}\mathrm{d}=\overline{5...8},\mathrm{repeating}\mathrm{above}\mathrm{argument}$$$$\mathrm{we}\mathrm{obtain}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}<1859$$$$\mathrm{Thus},\mathrm{if}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$$$$\mathrm{with}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{N}\mathrm{then}$$$$\max (\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})=1859$$

Commented by mnjuly1970 last updated on 23/Sep/20

$$verhexcellent.thankyou..$$

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