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Question Number 103209 by nimnim last updated on 13/Jul/20

Answered by bramlex last updated on 13/Jul/20

$$\mathrm{n}=7\mathrm{k}+1...\left(1\right)\times 6$$$$\mathrm{n}=6\mathrm{p}+3...\left(2\right)\times 7$$$$\mathrm{n}=5\mathrm{q}+2...\left(3\right)$$$$\Rightarrow \left(2\right)-\left(1\right)$$$$\Rightarrow \mathrm{n}=42(\mathrm{p}-\mathrm{k})+15...\left(4\right)$$$$\mathrm{set}\mathrm{p}-\mathrm{k}=l$$$$\mathrm{consider}$$$$5\mathrm{q}+2=42l+15$$$$\Rightarrow 2\left(mod5\right)=42l+15$$$$\Rightarrow 2l=2\left(mod5\right),l=1\left(mod5\right)$$$$\iff l=5t+1$$$$n=42l+15=42(5t+1)+15$$$$n=210t+57$$

Commented by nimnim last updated on 13/Jul/20

$$thanks$$

Answered by floor(10²Eta[1]) last updated on 13/Jul/20

$$\left(1\right)\mathrm{x}\equiv 1\left(\mathrm{mod}7\right)\Rightarrow \mathrm{x}=7\mathrm{a}+1$$$$\left(2\right)\mathrm{x}\equiv 3\left(\mathrm{mod}6\right)$$$$\left(3\right)\mathrm{x}\equiv 2\left(\mathrm{mod}5\right)$$$$\left(2\right):7\mathrm{a}+1\equiv 3\left(\mathrm{mod}6\right)\Rightarrow \mathrm{a}\equiv 2\left(\mathrm{mod}6\right)$$$$\Rightarrow \mathrm{a}=6\mathrm{b}+2$$$$\Rightarrow \mathrm{x}=7(6\mathrm{b}+2)+1=42\mathrm{b}+15$$$$\left(3\right):42\mathrm{b}+15\equiv 2\left(\mathrm{mod}5\right)\Rightarrow 2\mathrm{b}\equiv 2\left(\mathrm{mod}5\right)$$$$\Rightarrow \mathrm{b}\equiv 1\left(\mathrm{mod}5\right)\Rightarrow \mathrm{b}=5\mathrm{c}+1$$$$\Rightarrow \mathrm{x}=42(5\mathrm{c}+1)+15$$$$\Rightarrow \mathrm{x}=210\mathrm{c}+57,\mathrm{c}\in \mathbb{Z}$$

Commented by nimnim last updated on 13/Jul/20

$$Thanks$$

Answered by 1549442205 last updated on 13/Jul/20

$$\mathrm{From}\mathrm{the}\mathrm{condition}\mathrm{that}\mathrm{number}\mathrm{when}$$$$\mathrm{divided}\mathrm{by}6\mathrm{gives}\mathrm{remainder}3\mathrm{and}$$$$\mathrm{when}\mathrm{divided}\mathrm{by}5\mathrm{gives}\mathrm{remainder}2$$$$\mathrm{it}\mathrm{follows}\mathrm{that}\mathrm{if}\mathrm{adding}\mathrm{that}\mathrm{number}$$$$\mathrm{to}3\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{multiple}\mathrm{of}30.\mathrm{Hence}$$$$\mathrm{Denoting}\mathrm{by}\mathrm{n}\mathrm{the}\mathrm{number}\mathrm{we}\mathrm{need}\mathrm{find}$$$$\mathrm{then}\mathrm{n}=30\mathrm{k}-3.\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},\mathrm{n}=7\mathrm{m}+1.\mathrm{We}$$$$\mathrm{infer}30\mathrm{k}+3=7\mathrm{m}+1\Rightarrow \mathrm{m}=\frac{30\mathrm{k}-4}{7}=4\mathrm{k}+\frac{2\mathrm{k}-4}{7}$$$$\mathrm{Put}\frac{2\mathrm{k}-4}{7}=\mathrm{p}\Rightarrow \mathrm{k}=\frac{7\mathrm{p}+4}{2}=3\mathrm{p}+2+\frac{\mathrm{p}}{2}$$$$\mathrm{Put}\frac{\mathrm{p}}{2}=\mathrm{q}\Rightarrow \mathrm{p}=2\mathrm{q}\Rightarrow \mathrm{k}=7\mathrm{q}+2$$$$\mathrm{m}=30\mathrm{q}+8\Rightarrow \mathrm{n}=210\mathrm{q}+57.\mathrm{Thus},\mathrm{the}$$$$\mathrm{number}\mathrm{we}\mathrm{need}\mathrm{find}\mathrm{is}$$$$\mathbf{n}=210\mathbf{q}+57(\mathbf{q}\in \mathbb{N})$$$$$$

Commented by nimnim last updated on 13/Jul/20

$$Thanks$$

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