This is a great way! And the minus signs all cancel out very nicely.

You asked for another way, and here is a slightly different way which really isn't all that different from the original method, but it simplifies some of the intermediate calculation a little bit. You can multiply the first few together, and then find the remainder of that product at each step before multiplying the other numbers.

$$ 71 \text{ is } 1 \text{ mod } 7 $$

$$ 72 \text{ is } 2 \text{ mod } 7 $$

$$ 73 \text{ is } 3 \text{ mod } 7 $$

$$ \rightarrow 71 \times 72 \times 73 = 1 \times 2 \times 3 = \textcolor{red}{-1 \text{ mod } 7} $$

Now, let's multiply the other numbers!

$$ 74 \text{ is } 4 \text{ mod } 7 $$

$$ \rightarrow \textcolor{blue}{ 71 \times 72 \times 73 \times 74 } = -1 \times 4 = -4 = -4 + 7 = \textcolor{red}{3 \text{ mod } 7} $$

Now let's multiply the \(75.\)

$$ 75 \text{ is } 5 \text{ mod } 7 $$

$$ \rightarrow \textcolor{blue}{71 \times 72 \times 73 \times 74 \times 75} = 3 \times 5 = 14 + 1 = \textcolor{red}{1 \text{ mod } 7} $$

And finally, we've got the \(76.\)

$$ 76 \text{ is } 6 \text{ mod } 7 $$

$$ ( 1 \text{ mod } 7 ) (6 \text{ mod } 7 ) = \boxed{ \textcolor{red}{6 \text{ mod } 7 }}. $$

This way, your "running total of your remainder" is always less than \(7,\) so there are just smaller numbers to deal with. The are so many little shortcuts to calculating this, and it's kind of like a puzzle to find them!

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