$(-4339 \cdot 8559) \text{ mod } 43$ without calculator

Question

How can one calculate $(-4339 \cdot 8559) \text{ mod } 43$ without a calculator?

I know that the solution is 8, but just because i used a calculator. What is the correct way when trying to calculate modulo with big numbers? I know Fermats little theorem, but we can't apply it here.

Answer 1

Look for large, but easy to find, multiples of 43 that are close to the numbers you have. Then you can reduce each factor mod 43 and multiply what's left.

Answer 2

You can rewrite it as follows:

$$-4339 \times 8559 = (-101 \times43+4) (199 \times 43 + 2)\equiv 4 \times 2 \pmod{43}$$

because $4339=4343-4=101 \times 43-4$ and $8559=8600-41=200 \times 43 - 41 = 199 \times 43 + 2$

Answer 3

You have that $a\cdot b\pmod{n} \equiv ((a\mod n)\cdot (b\mod n)) \pmod n$

Answer 4

Employing $\ a,b\, :=\, a\cdot 100+b =\,$ radix $100$ notation allowing negative digits $\,a,b\,$ makes it easy

$\!\!\bmod 43\!:\ {-}4339\, = -43,\color{#c00}{-39}\, \equiv\, 0,\:\!\color{#c00}4\ $

$\qquad\qquad\ \ 85,59\, =\ \ \ 86,\color{#0a0}{-41}\, \equiv\, 0,\:\!\color{#0a0}2,\ $ by carrying $1,\,$ i.e. $\ a,b\, =\, a\!+\!1,\,b\!-\!100$

Hence we infer: $\:\! \ {-}4339\cdot 8559\, \equiv\, \color{#c00}4\cdot \color{#0a0}2\,\equiv\, 8\ \,$ by Congruence Sum & Product Rules.

Remark $ $ This is a special case of the universal divisibility test.