MiniQuestion: why don't ii.) and iv.) work?

Module 4 Week 1 Day 2 Your Turn Part 1 MiniQuestion
Why is i.) the correct answer? Since the question asks for another multiplication, should the answer be anything other than i.)? (Since he did that in the video.) Also, why don't ii.) and iv.) work?

Haha, yes, you're right! Thanks for noticing that the \(66 \times 64\) is an example from the video! That's my fault... I forgot that Prof. Loh had covered this already. I'll try to fix this at some point.
As for your question about why choices ii.) and iv.) aren't correct, I'm happy to answer that!
Well, what's the pattern that we saw from the \(\textcolor{purple}{11}\textcolor{red}{7} \times \textcolor{purple}{11}\textcolor{red}{3}\) and the \( 6\textcolor{red}{6} \times 6\textcolor{red}{4}\) in the video?
\( \textcolor{red}{7} + \textcolor{red}{3} = 10\) \( \textcolor{red}{6} + \textcolor{red}{4} = 10\)
In this trick, the units digits of the numbers in the product must add to \(10,\) but why?Let's think about what the \( ( \textcolor{red}{7} + \textcolor{red}{3}) \) gets multiplied to when we express this as a product of two sums, like so:
\( (\textcolor{purple}{11}0 + \textcolor{red}{7})(\textcolor{purple}{11}0 + \textcolor{red}{3}) \)
One of our terms will be \(\textcolor{purple}{11}0 \times \textcolor{red}{3}.\)
\( (\textcolor{purple}{11}0 + \textcolor{red}{7})(\textcolor{purple}{11}0 + \textcolor{red}{3}) \)
And another crossterm is \(\textcolor{purple}{11}0 \times \textcolor{red}{7}.\)
We can add these two crossterm together to get \(\textcolor{purple}{11}0 \times 10.\) Let's express that as \(\textcolor{purple}{11} \times 100.\)
Why is this nice? It's because there is another term from this product which looks a lot like \(11 \times 100.\)
Which term is it?
\( (\textcolor{purple}{11}0 + \textcolor{red}{7})(\textcolor{purple}{11}0 + \textcolor{red}{3}) \)
It's the \(\textcolor{purple}{11}0 \times \textcolor{purple}{11}0\) term!
This can be written as \(\textcolor{purple}{11}^2 \times 100.\)
This is now really nice, because we can combine the \(\textcolor{purple}{11}^2 \times 100 \) and the \(\textcolor{purple}{11} \times 100\) terms together.
\( (\textcolor{purple}{11}^2 + \textcolor{purple}{11})(100) \)
\(= (\textcolor{purple}{11})(\textcolor{purple}{11} + 1)(100) \)
\( = ( \textcolor{purple}{11})(12)(100) \)
\( \text{The first three digits come from here} \)
This was the trick for how to get the first three digits of the answer. And the beauty is that since those terms above were multiples of \(100,\) we didn't mess with the last digits at all. So the last two digits come directly from the \(\textcolor{red}{7} \times \textcolor{red}{3}.\)
\( (\textcolor{purple}{11}0 + \textcolor{red}{7})(\textcolor{purple}{11}0 + \textcolor{red}{3}) \)
\( \text{ The last two digits come from here.} \)
Now, if we had something like \(22\textcolor{blue}{4} \times 22\textcolor{blue}{8}\) or \(11\textcolor{blue}{1} \times 11\textcolor{blue}{5},\) where the units digits don't add up to \(10,\) then we won't be able to nicely say that the first three digits are equal to \( (22)(23)(100).\) We would instead get \( (22)(10)(12),\) which doesn't have two zeroes at the end, and so it would "mess up" our tens digit!That's why choices ii.) and iv.) aren't good candidates for this trick.