Question-41044

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 \equiv 5 + 2 \equiv 4 + 3 \equiv 0 (\mod 7)

so selecting a number divisible is not alowed in pair cause

0 + 0 \equiv 0 (\mod 7)

if we select a number, the complement is forbidden

so we have

(\begin{matrix} m_{1} \\ m_{6} \end{matrix}) + (\begin{matrix} m_{2} \\ m_{5} \end{matrix}) + (\begin{matrix} m_{3} \\ m_{4} \end{matrix}) + 1 m_{0}

= (\begin{matrix} 8 \\ 7 \end{matrix}) + (\begin{matrix} 7 \\ 7 \end{matrix}) + (\begin{matrix} 7 \\ 7 \end{matrix}) + 1

m a x p o s s i b l e w e c o u l d c h o o s e i s m_{1}, m_{2 / 5}, m_{3 / 4} and 1 m_{0}

8 + 7 + 7 + 1 = 23

Commented by Tawa1 last updated on 01/Aug/18

God bless you sir

Facebook Tweet Pin