# Intuitive Understanding of Distributive Multiplication Property for Modulo

I can follow an algebraic proof that

$$(ab)\textrm{ mod } n \equiv \big((a\textrm{ mod } n)\cdot(b\textrm{ mod } n)\big)\textrm{ mod } n$$

However, I'd like to find a way to "see" it that makes it really obvious, perhaps with reference to number lines or clocks.

One attempt of mine was

$$(22*15)\textrm{ mod } 8 \equiv \big((22\textrm{ mod } 8)\cdot(15\textrm{ mod } 8)\big)\textrm{ mod } 8$$

is like saying that advancing the hands on a clock whose numbers range form 0-7 22 positions 15 times is the same as advancing the hands on the same clock 22 spaces 15 times, but it seems a bit too tautological to add any insight.

Any ideas on how to fully "grock" this without using algebra?

Make an $$a \cdot b$$ rectangle of points. We are working modulo $$n$$, so partition the rows and columns as follows:

• starting from the left edge, into blocks of $$n$$ columns, with perhaps some small number of columns left in a single block on the right edge and
• starting from the bottom edge, into blocks of $$n$$ rows, with perhaps some small number of rows left in a single block on the top edge.

This divides the rectangle into four similar pieces

• the "bulk" of the rectangle, starting in the lower-left corner is made of $$n \cdot n$$ blocks,
• along the right edge of the rectangle, starting from the bottom, is some $$k \cdot n$$ blocks with $$0 \leq k < n$$,
• along the top edge of the rectangle, starting from the left, is some $$n \cdot \ell$$ blocks with $$0 \leq \ell < n$$, and
• finally, in the upper right is a single $$k \cdot \ell$$ block where $$0 \leq k < n$$ and $$0 \leq \ell < n$$. Note that we allow the dimensions of the non-$$n \cdot n$$ blocks to be zero, so from a particular choice of $$a$$ and $$b$$, any of the last three regions might contain zero points.

Now observe: $$k = (a \bmod n)$$, $$\ell = (b \bmod n)$$, and all the blocks in the first three regions contribute $$0 \bmod n$$ because they contain multiples of $$n$$ points. Only the $$k \cdot \ell$$ block in the fourth region can have a nonzero contribution. Since that block can contain more than $$n$$ points, we should reduce one last time modulo $$n$$.