Submitted by Heirio on Thu, 10/12/2017 - 20:55

I'm not sure I get some parts of the modulo function. I'll run through what I did in the first part and then go through what I don't get about the second.

Okay, so for the first part I put that xmoda = (x+k)moda, where k is any multiple of a. Later, we get N1 + N2 = a(m1 + m2) + n1 + n2.

Therefore, if we put N1 + N2 into the modulo function under a, we get the same function as if we put n1 + n2. This worked for N1N2 as well.

For the second part, I just didn't know what to do. Since 10mod3 = 1, I put that 10amod3 = a. Hence, via the results we got prior, we should get bmod3 = b, which I didn't trust. So I went to the hints and found this:

10a + b = (10mod3) * (amod3) + bmod3

I'm not sure how we get this. Could someone go through how we come to this from what we're given?

With the last part, we have (amod3) + (bmod3) = (a + b), which I'm also not sure about.

## Working (mod 3)

You are working (mod 3) so you really need to write these at the end of the lines. So it would be wrong to write 10 (mod 3) = 10 as this is saying 1 = 10. However you can write:

\[

10 \, \text{mod } 3 \equiv 10 \quad (\text{mod } 3)

\]

Using this idea we have:

\begin{align*}

10 a + b &\equiv (10 \, \text{mod }3)\times (a \, \text{mod }3) + b \, \text{mod }3 \quad & (\text{mod }3)\\

&\equiv 1 \times a \, \text{mod }3 + b \, \text{mod }3 \quad &(\text{mod }3)\\

&\equiv (a+b) \quad &(\text{mod }3)

\end{align*}

There fore we have $(10 a + b) \equiv (a+b) \, \text{mod }3$ and so $10a+b$ is divisible by 3 if and only if $a+b$ is divisible by 3.

## So where do we get (10a + b =

So where do we get (10a + b = (10mod3) * (amod3) + bmod3) from? Sorry, but I still don't understand where those things came from.

## "Equivalent to" is different to "Equal to"

You need to be quite careful with which signs you are using. We are not saying at any point that $10=1$, but instead $10 \equiv 1 \, (\text{mod }3)$.

We have:

\begin{align*}

10 &\equiv 1 \quad (\text{mod }3)\\

5 &\equiv 2 \quad (\text{mod }3)\\

8 &\equiv 2 \quad (\text{mod }3)

\end{align*}

Note that none of these are "="!

Using "$N_1N_2 \equiv n_1n_2 \, (\text{mod }a)$" and "$N_1+N_2 \equiv n_1+n_2 \, (\text{mod }a)$" we have:

\begin{align*}

58 &= 5 \times 10 + 8 \quad \text{note this is not working in mod 3 yet, hence equals sign!}\\

58 &\equiv (5 \, \text{mod }3) \times (10 \, \text{mod }3)+ (8 \, \text{mod }3) \quad (\text{mod }3)\\

&\equiv 2 \times 1 + 2 \quad (\text{mod }3)\\

&\equiv 4 \quad (\text{mod }3)\\

&\equiv 1 \quad (\text{mod }3)

\end{align*}

Which is what we expect has $58$ has a remainder of 1 when divided by 3.