Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Answered by candre last updated on 01/Aug/18

we have that remainder of sum is sum of remakder  then, making a list  7,14,21,28,35,42,49 (7 m_0 )  1,8,15,22,29,36,43,50 (8 m_1 )  2,9,16,23,30,37,44 (7 m_2 )  3,10,17,24,31,38,45 (7 m_3 )  4,11,18,25,32,39,46 (7 m_4 )  5,12,19,26,33,40,47 (7 m_5 )  6,13,20,27,34,41,48 (7 m_6 )  congruence tells  6+1≡5+2≡4+3≡0(mod7)  so selecting a number divisible is not alowed in pair cause  0+0≡0(mod 7)  if we select a number, the complement is forbidden  so we have   ((m_1 ),(m_6 ) )+ ((m_2 ),(m_5 ) )+ ((m_3 ),(m_4 ) )+1m_0   = ((8),(7) )+ ((7),(7) )+ ((7),(7) )+1  max possible we could choose is m_1 ,m_(2/5) ,m_(3/4)  and 1 m_0   8+7+7+1=23

$$\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{remakder} \\ $$$$\mathrm{then},\:\mathrm{making}\:\mathrm{a}\:\mathrm{list} \\ $$$$\mathrm{7},\mathrm{14},\mathrm{21},\mathrm{28},\mathrm{35},\mathrm{42},\mathrm{49}\:\left(\mathrm{7}\:{m}_{\mathrm{0}} \right) \\ $$$$\mathrm{1},\mathrm{8},\mathrm{15},\mathrm{22},\mathrm{29},\mathrm{36},\mathrm{43},\mathrm{50}\:\left(\mathrm{8}\:{m}_{\mathrm{1}} \right) \\ $$$$\mathrm{2},\mathrm{9},\mathrm{16},\mathrm{23},\mathrm{30},\mathrm{37},\mathrm{44}\:\left(\mathrm{7}\:{m}_{\mathrm{2}} \right) \\ $$$$\mathrm{3},\mathrm{10},\mathrm{17},\mathrm{24},\mathrm{31},\mathrm{38},\mathrm{45}\:\left(\mathrm{7}\:{m}_{\mathrm{3}} \right) \\ $$$$\mathrm{4},\mathrm{11},\mathrm{18},\mathrm{25},\mathrm{32},\mathrm{39},\mathrm{46}\:\left(\mathrm{7}\:{m}_{\mathrm{4}} \right) \\ $$$$\mathrm{5},\mathrm{12},\mathrm{19},\mathrm{26},\mathrm{33},\mathrm{40},\mathrm{47}\:\left(\mathrm{7}\:{m}_{\mathrm{5}} \right) \\ $$$$\mathrm{6},\mathrm{13},\mathrm{20},\mathrm{27},\mathrm{34},\mathrm{41},\mathrm{48}\:\left(\mathrm{7}\:{m}_{\mathrm{6}} \right) \\ $$$$\mathrm{congruence}\:\mathrm{tells} \\ $$$$\mathrm{6}+\mathrm{1}\equiv\mathrm{5}+\mathrm{2}\equiv\mathrm{4}+\mathrm{3}\equiv\mathrm{0}\left(\mathrm{mod7}\right) \\ $$$$\mathrm{so}\:\mathrm{selecting}\:\mathrm{a}\:\mathrm{number}\:\mathrm{divisible}\:\mathrm{is}\:\mathrm{not}\:\mathrm{alowed}\:\mathrm{in}\:\mathrm{pair}\:\mathrm{cause} \\ $$$$\mathrm{0}+\mathrm{0}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{select}\:\mathrm{a}\:\mathrm{number},\:\mathrm{the}\:\mathrm{complement}\:\mathrm{is}\:\mathrm{forbidden} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\begin{pmatrix}{\mathrm{m}_{\mathrm{1}} }\\{\mathrm{m}_{\mathrm{6}} }\end{pmatrix}+\begin{pmatrix}{\mathrm{m}_{\mathrm{2}} }\\{\mathrm{m}_{\mathrm{5}} }\end{pmatrix}+\begin{pmatrix}{\mathrm{m}_{\mathrm{3}} }\\{\mathrm{m}_{\mathrm{4}} }\end{pmatrix}+\mathrm{1}{m}_{\mathrm{0}} \\ $$$$=\begin{pmatrix}{\mathrm{8}}\\{\mathrm{7}}\end{pmatrix}+\begin{pmatrix}{\mathrm{7}}\\{\mathrm{7}}\end{pmatrix}+\begin{pmatrix}{\mathrm{7}}\\{\mathrm{7}}\end{pmatrix}+\mathrm{1} \\ $$$${max}\:{possible}\:{we}\:{could}\:{choose}\:{is}\:{m}_{\mathrm{1}} ,{m}_{\mathrm{2}/\mathrm{5}} ,{m}_{\mathrm{3}/\mathrm{4}} \:\mathrm{and}\:\mathrm{1}\:{m}_{\mathrm{0}} \\ $$$$\mathrm{8}+\mathrm{7}+\mathrm{7}+\mathrm{1}=\mathrm{23} \\ $$

Commented by Tawa1 last updated on 01/Aug/18

God bless you sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com