Math Is Fun Forum

That's the formula for Combinations, not Permutations. The difference is that with permutations, order is important. This means that, for example, when choosing all the members of a set, there are many ways of doing so instead of just one.

The formula for permutations is:

Maybe I'm missing something.

The spaceship approaches the planet, but we're not told how far away it was to start with.
If it's sufficiently far away, then no matter what its movements are, it will always be looking at approximately the same point on the planet.

In this case,  the inhabitant can easily move to the other side to avoid its gaze.

I think the key to this is that the polynomial has factors of (x-a), (x-b), (x-c), (x-d).

13 only has two factors - 1 and 13 - but here we've shown that any number produced using this polynomial will have at least 4.
(The exception being when you use a value of n that causes P(n) = 0, but P(n) isn't 13 in that case either)

This sequence can also be written as:

1x0, 1x1, 2x1, 2x2, 3x2, 3x3, 4x3, 4x4, ...

So you go from term to term by adding one to alternating sides of the product.

How about this as a variant?

Remove the time limit, and focus on accuracy more than speed.
You get shown a pair of containers like you are now, and asked to guess the volume in the same way.

If you get at least one tick, you score a point. If you don't, the game ends. (Or maybe you lose a life)

As you score more points, you have to get more and more accurate to avoid being penalised.
(For example, once you have 5 points you'd need to get at least 2 ticks to be rewarded)

Maybe there should still be a timer to stop people working it out on their calculators (it's an estimation game), but it should be lenient enough to give people a bit of thinking time, and fill up again for each new question.

Well that must be a bug. Fixed for you.

It looks good though. I can't think of anything I'd like changed.

Edit: When I look at my "You" page, it tells me "Challenges: 8".
Clicking on the 8 takes me to the scoreboard of the volume challenge. What does the 8 mean?

Depending on what exactly the levers do, it might not always be possible to get the slots to do what you want. (Trivial example: If all 5 levers move every slot 0 places, you can't get anything to happen)

But in general, the strategy would be to try to work out combinations of levers that only affect one slot. The ultimate goal should be to get a combination that moves the first slot one place, and 2 more combinations that do the same for the second and third slot.

I'll use my calculator to make some random lever functions and see what I can do with them.

A: (2,5,0)
B: (7,6,7)
C: (9,8,7)
D: (4,9,5)
E: (3,6,4)

A looks like it could be useful, since it doesn't affect the third slot.
We should be looking for sets that add up to 10 for one of the slots, since that would mean that combination doesn't change that slot.

For example, pulling B and E would leave the first slot alone, and pulling D twice wouldn't change the third.

Eventually we'll find something useful.

Here, pulling A once and D twice has a final result of (0,3,0).
It's good that the second slot is moved 3 places, because that means that repeating this combination can get us any position we want on the second slot.
For example, (ADD)x7 gives (0,1,0).

We can now effectively ignore the second slot in our calculations, since we have full power over it anyway.

Here's the important stuff:

A: (2,0)
B: (7,7)
C: (9,7)
D: (4,5)
E: (3,4)

A only affects one slot, which is good, but we can only get half of the positions from it, so more work is needed.

Bingo! B and E give a result of (0,1), which is exactly what we're looking for.
Now that we can ignore the third slot as well, the last part is easy. For example, B three times moves the first slot once.

We've now shown theoretically that any position can be made, and to find how to get to a specific one, you backtrack through the working we've just done.

Let's say the slots start at (a,a,a), and you want them to get to (f,c,j).
In number form, this is equivalent to (5,2,9).

Since we solved it last, we'll get the first slot to the right position first.
According to the calculations, this is done by pulling B 3*5 = 15 times.
However, pulling any lever 10 times will result in nothing, so we can actually pull B just 5 times if we want.

The slots now sit at (f,a,f). Next we sort out the third.
We want to move it four times, so we do (BE)x4.

This results in (f,i,j).

Finally we need to move the second slot forward four times.
We can do this by (ADD)x36 = (ADD)x6 = A6D2.

So an overall command for (a,a,a) -> (f,c,j) would be A6B9D2E4.

Note that there may be (and probably are) ways to do this involving less lever pulls.
To try to improve it, I'd try to find combinations of levers that affect nothing overall. Then you can add or take away those combinations to a solution to get less overall pulls.

The fence costs £2 per metre for three of the sides, and £3 per metre for the other one.

<------- x --------->

----------------------   ^
|                         |   |
|                         |   y
|                         |   |
============   V

This bad drawing shows that the bottom side has fancy double-thickness, and that's why it's more expensive.

If the area has a length of x metres, then to enclose it you'll need x metres of normal fence and x metres of expensive fence. This comes to £5x altogether.

Similarly, if the area has a width of y metres, then the enclosure will need y metres of normal fence for one side, and another y metres for the other side, making the total cost £4y.

You have £120 to spend, so the equation is 5x+4y = 120.

The area A is given by xy.
To get the answer in b), just arrange the equation from a) until you can remove y from the area's expression.

You can do differentiation problems, so once you've got this far, c) shouldn't be a problem.

Let's say for the moment that AD has a length of 1 and AC has a length of x.
Then CD has a length of (1-x).

Using the matching ratios, we can write the equation 1/x = x/(1-x).

Rearrange to form a quadratic, and this solves to give x = (√5-1)/2.

The diagram is symmetric, in that AC = BD.
Therefore, we can work out distance BC by finding C's distance from the halfway point and doubling it.

The distance from the halfway point will be (√5-2)/2, and naturally doubling that gets us √5-2.

So we now know that 1 is to √5-2 as AD is to BC.

From there, we do 8314/(√5-2) to get the final answer of ~35219cm.