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Question Number 41044 by Tawa1 last updated on 31/Jul/18

Answered by candre last updated on 01/Aug/18

$$\mathrm{we}\mathrm{have}\mathrm{that}\mathrm{remainder}\mathrm{of}\mathrm{sum}\mathrm{is}\mathrm{sum}\mathrm{of}\mathrm{remakder}$$$$\mathrm{then},\mathrm{making}\mathrm{a}\mathrm{list}$$$$7,14,21,28,35,42,49\left(7m_0\right)$$$$1,8,15,22,29,36,43,50\left(8m_1\right)$$$$2,9,16,23,30,37,44\left(7m_2\right)$$$$3,10,17,24,31,38,45\left(7m_3\right)$$$$4,11,18,25,32,39,46\left(7m_4\right)$$$$5,12,19,26,33,40,47\left(7m_5\right)$$$$6,13,20,27,34,41,48\left(7m_6\right)$$$$\mathrm{congruence}\mathrm{tells}$$$$6+1\equiv 5+2\equiv 4+3\equiv 0\left(\mathrm{mod}7\right)$$$$\mathrm{so}\mathrm{selecting}\mathrm{a}\mathrm{number}\mathrm{divisible}\mathrm{is}\mathrm{not}\mathrm{alowed}\mathrm{in}\mathrm{pair}\mathrm{cause}$$$$0+0\equiv 0\left(\mathrm{mod}7\right)$$$$\mathrm{if}\mathrm{we}\mathrm{select}\mathrm{a}\mathrm{number},\mathrm{the}\mathrm{complement}\mathrm{is}\mathrm{forbidden}$$$$\mathrm{so}\mathrm{we}\mathrm{have}$$$$\left(\begin{array}{c}{\mathrm{m}}_1\\ {\mathrm{m}}_6\end{array}\right)+\left(\begin{array}{c}{\mathrm{m}}_2\\ {\mathrm{m}}_5\end{array}\right)+\left(\begin{array}{c}{\mathrm{m}}_3\\ {\mathrm{m}}_4\end{array}\right)+1m_0$$$$=\left(\begin{array}{c}8\\ 7\end{array}\right)+\left(\begin{array}{c}7\\ 7\end{array}\right)+\left(\begin{array}{c}7\\ 7\end{array}\right)+1$$$$maxpossiblewecouldchooseis{m}_1,m_{2/5},m_{3/4}\mathrm{and}1m_0$$$$8+7+7+1=23$$

Commented by Tawa1 last updated on 01/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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