BuilderBoi

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X

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X

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X    X

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Each X represents where one block would be. The other possible combinations would involve transforming the L \(1-3\) units to the right.

There are \(4\) ways to arrange the L so the bottom \(2\) peices are in the bottom row.

There are \(3\) rows that you can do this on, so there are \(12\) ways to arrange the L this way. 

There are \(3\) additional ways to rotate them \(90^\circ\) with the same concept. This means that there are a total of \(48\) ways with rotations.

There are \(3\) ways to reflect the L , with each one having \(12\) ways to arrange them. 

This makes for a total of \(\color{brown}\boxed {84}\) ways.

The 4 squares are unit squares that form an L shape that is 3 high and 2 wide right?

There are \(36\) possible outcomes (\(6\) for the first roll and \(6\) for the second roll).

The only square numbers in the list are \(1\) and \(4\).

If we don't want either of these, there is \(4\) sucsessful options for the first role and \(4\) sucsessful options for the second roll. This means that there is \(16\) succsessful outcomes, out of the \(36\) total outcomes, meaning that the answer is \(16\over 36\), which can be simplified to \(\color{brown}\boxed{4\over9}\)

For any positive integer to satisfy the conditions, it has to be a perfect power of \(2\), and it should be able to be equivalent to \(2^{42}\), with integer values.

The only possible options are:

\(2\), can be written as \(2^{42}\)

\(4\), can be written as \(({2^2)}^{21}\)

\(8\), can be written as \((2^3)^{14}\)

Thus... the answer is \(\color{brown}\boxed3\)

That's what we're here for!

The possible pairs of numbers are \((0,3), (1,4), (2,5), (3,6) ,(4, 7),(5,8),(6,9)\)

Each of the \(7\) pairs can be ordered in \(2\) ways, making for \(14 \) numbers.

However, we need to account for the fact that the number \(03\) is not a \(2\)-digit number, so we have to subtract \(1\), making our total \(\color{brown}\boxed{13}\)