0
$\begingroup$

I dont understand any step of the solution in this question below. I know that they are allowed to bring 7 to the other side but anything after that I dont understand. It would help if greater detail on each step can be provided so it can make sense as to what is happening and why it is.

enter image description here

I am not understanding what is going on in the table they have made...Any idea what they are doing in that table?

2 Answers 2

0

Recall $\ a\equiv b\pmod{5}\iff 5\ |\ a-b,\ $ so $\ -7\equiv 3\pmod{5}\ $ by $\ 5\ |\ {-}7-3$.

Simpler than that above, to solve $\rm\:x(x+3)\equiv 3\pmod 5\:$ note that

$\rm\ \ \ mod\ 5\!:\ \{(x,x+3)\} \equiv \{(0,3),(1,4),(2,0),(3,1),(4,2)\}$

and only the final two pairs have product $\equiv 3\pmod 5$

  • 0
    @Raynos Recall that $\rm\:A\equiv a,\ B\equiv b\ \Rightarrow\ A+B\equiv a+b,\ AB \equiv ab$. Thus in expressions composed of sums and products, one can replace any argument of a sum or product by any congruent integer. By division with remainder every integer is congruent mod $5$ to its remainder mod $5$, i.e. to one of $0,1,2,3,4$. The integers congruent to $\rm j$ are $\rm\:j + 5\: \mathbb Z = \{\ldots,\: j-10,\: j-5,\: j,\: j+5,\: j+10,\:\ldots\}.\qquad $2012-02-25
1

Nothing fancy. $-7 + 10 = 3$, so $-7 \equiv 3 \pmod 5$.