@the-blade-dancer said in Is it possible to...:

Find the Mean Absolute Deviation without having to calculate the mean and the difference of each piece of data from the mean? In other words, can MAD problems be solved simply, or is it just something you have to grind through?

calm the problem down first or else your bones may break

@victorioussheep Here's a diagram that I drew:

(only drew half of it because that's all we need)
So, first off, the length of \(\overline{AD}\) comes from the Pythagorean Theorem. Or, by chance, if you happen to remember that \(119-120-169\) is a Pythagorean triple, \(1428-1440-2028\) is exactly the same triangle, except all the dimensions are multiplied by \(12\). What the problem states is that \(\triangle BOP\) happens to be similar to a \(5-12-13\) triangle. The solution took a huge shortcut, but the idea is that you'll eventually find out that the dimensions of \(\triangle BOP\) are \(420-1008-1092\), which is exactly \(84\) times the dimensions of a \(5-12-13\) triangle. I know that the solution says to use angle bisector, but I can't think of a way to do it that isn't tedious, maybe the other mods could weigh in on that.

Here's how I did it though-- as long as you get that \(\overline{AD}\) is \(1440\), recall that two tangents of a circle that meet have the same length.
Let's redraw the diagram:

So, \(\overline{BP_1} \cong \overline{BP_2}, \overline{AP_2} \cong \overline{AP_3}, \overline{DP_1} \cong \overline{DP_3},\) and I labeled them \(a, b, c\).
If you want to prove them, here's how:
To prove that \(\overline{BP_1} \cong \overline{BP_2}\), draw line \(\overline{BO}\) and prove that triangles \(\triangle{BOP_1}\) and \(\triangle{BOP_2}\) are congruent (hint: they're both right triangles!).
You can do the same with all of the other lines.

Anyways, knowing this, we can set up a system of equations, like the one I did to the right of the triangle. Then, notice that the radius of the circle is equivalent to \(c\) because quadrilateral \(OP_1DP_3\) is a square. Solving for \(c\) should give you \(\boxed{420}\).

Hope this makes sense! I know that it isn't the method that the solution used, but hopefully at least the part about multiplying the dimensions is clearer.
Also, if the other mods want to weigh in on the angle bisector method, that'd be great

@西瓜 @audrey Thank you both for the awesome explanations

@pioneeringemu I ended up finishing a more legit solution, here it is just in case other people stop by this post and want to see it too:

A number \(N\) is green if \(6N\) has none of the digits \(0,1,2,3,4\).

Looking at one-digit numbers, right away we know that \(1\) is a green number. That doesn't help though, since the problem already tells us that every green number has either a \(1\) or some other digit in it. Now, let's consider two-digit numbers.

Let's start off by considering the units digits of \(6N\).

If \(N\) has units digit 1, then \(6N\) has units digit 6.
If \(N\) has units digit 2, then \(6N\) has units digit 2.
Filling it out for all the digits, we have
1 -> 6
2 -> 2
3 -> 8
4 -> 4
5 -> 0
6 -> 6
7 -> 2
8 -> 8
9 -> 4

Using this list, we know that all green numbers must end in 1, 3, 6, or 8.

A quick check though shows that out of 1, 3, 6, and 8, only 1 is a green number. So this confirms that green numbers do indeed have to contain a 1 or some other number, but the problem already gave us that.
So, let's start looking through the two-digit numbers.

The possible two-digit green numbers are going to look like \(A1, A3, A6, A8\).
*note: whenever I use \(A\) here, I mean it as a digit. \(A1\) is a number with \(A\) in the tens place and \(1\) in the ones place, not \(A\cdot1\).

Now, let's see what \(6\) times these numbers gives us.

This is what \(A1 \cdot 6\) looks like. There is no digit \(A\) for which this works, because no value of \(6A\) has none of the digits \(0,1,2,3,4\). Let's move on to the case of \(A3 \cdot 6\).

So, we need a digit \(A\) such that \(6A+1\) has none of the digits \(0\to4\). Wait, \(A=9\) works! \(93\cdot6=558\), which has no digits \(0\to4\), so \(93\) is a green number (note that \(A=1\) also works in this case too, but that doesn't help us because the problem already tells us that \(1\) is one of the two numbers).

We already have two green numbers now: \(1\) and \(93\). Since the problem says that every green number has either a \(1\) or some other digit in it, we now know that the other digit has to be \(9\) or \(3\). Since the answer choices has no option for \(3\), the answer has to be \(\boxed{(E)\text{ } 9}\).

But, what if they didn't give us the answer choices? Well, I guess we just have to keep going through the cases.

Here's what \(A6 \cdot 6\) looks like. Again, the only possible \(A\) are \(A=1,9\). And we can confirm, \(16\) and \(96\) are green numbers, since \(16\cdot6=96\) and \(96\cdot6=576\). That means that the two numbers are indeed \(1\) and \(9\), and we can be safe with E as our answer.

@victorioussheep Yeah exactly! Always keep your eyes peeled for patterns when doing counting problems

@victorioussheep Glad that it helped!

@西瓜 No problem!

@victorioussheep Glad that you understood everything

@victorioussheep The best way is to always just practice. The more you're exposed to, the more types of problems you'll be able to tackle! And also try to understand every problem that you do. That way, when you see something completely new, you'll be able to use problem-solving skills to figure it out using what you already know. Those are just a few helpful tips

Hey @gentlegorilla! Unforunately, there's no good formula to tackle this problem. I started off by trying to just make a pattern by myself and here's what I noticed:

In every 2x2 square, there can only be one of each color (this makes sense since colors can't touch each other, even diagonally). It might seem obvious, but this is really important! So, I started off by filling in the colors in the middle 2x2 square and going off from there.

Let's say I start off with
B G
Y R
in the middle 2x2. This means that the colors above B G have to be R & Y, in some order. With the same logic, the colors below Y R have to be B & G in some order. From there, you just have to pick one of each for those two and keep going.
For example, one way to continue the pattern:
R Y
B G
Y R
G B
(I just added to the top and bottom)
Now what you'll notice is that to the right of the Y & G in the top right, there has to be an R and a B. However, R can't go directly next to G because then they will touch diagonally!
R Y
B G R
Y R
G B
(i'm talking about that)
So, R has to go next to Y. And actually, once we have the 4x2 arrangement, there's only one way to complete the pattern.
Y R Y R
G B G B
R Y R Y
B G B G
That's an arrangement that works!
Just to recap the process: I picked a random 2x2 square to start off with, then I added two colors above and below it, and from there there was only one way to complete the pattern. Using this, are you able to come up with the answer now?

Hope this helps!

Also, counting is always confusing in general, so feel free to ask any more questions if you need to.

@gentlegorilla The way I would approach this problem is to set a variable. Let there be n drivers that work on all 3 shifts-- we're trying to maximize n. Now, we can separate the drivers into 2 categories: there are n drivers that work on all 3 shifts, and 21-n drivers that do not.

Notice that there are 15 + 12 + 9 = 36 shifts total. How many of those shifts are taken up by the drivers that work all 3? How many shifts are remaining for the other 21-n drivers? Can you write an equation or inequality now using the restriction that every driver needs to work on at least one shift?

Hope this helped, and let me know if you have questions!

@gentlegorilla This one calls for a bit of 3D visualization! Based on the diagram, you can imagine three sets of 9 "bars" running through the cube. (One set running front to back, one left to right, and one from top to bottom) Can you find the volume of all those "bars"?

Unfortunately, this isn't exactly the answer... if you imagine the cube, some of the bars will "overlap," meaning that we subtracted too much. So the next step would be to figure out what bars overlap, and what the volume of the overlap is.

Hope that helped, and feel free to respond here with any other questions/answers!

(By the way, this problem is interesting because, even though it's geometry, it's very similar to a counting and probability technique called PIE. We use PIE to count the number of elements in some sets by first adding up the elements in all the sets, then subtracting and adding to correct for over/undercounting until we get to the right answer.)

No problem! Thanks for asking them!

Hey @victorioussheep! They happen to be exact opposites of each other.

This is three squared: \(3^2\)
This is the square root of three: \(\sqrt{3}\)

\(3^2 = 3 \cdot 3 = 9\)
\(\sqrt{9}=3\)

\(5^2=25\)
\(\sqrt{25}=5\)

Hope this helps! Feel free to ask if you still have any questions

@quacker88 this helps me a lot! Thank you so much!

@coherentmango thank you so much!

The answer should be 13, you find the answer by counting the number of ways to get to each tile, kind of like how you derive the next digits in Pascal's triangle, but you might have not gotten to that yet.

Basically, here's what you do:

There is 1 way for you to get to box 1 (since you start there).

There is 1 way for you to get to box 2 (since you start at 1).

There are 2 ways for you to get to box 3 (one way from 1 and another way from 2).

There are 3 ways for you to get to box 4 (one way from 2 and two ways from 3).

There are 5 ways for you to get to box 5 (two ways from 3 and three ways from 4)

There are 8 ways for you to get to box 6 (three ways from 4 and five ways from 5)

And lastly, there are 13 ways (the answer!!) for you to get to box 7 (five ways from 5 and eight ways from 6)

Reminder that you can't go to a square with a smaller number, which is why box 2 only had 1 possibility, etc.

Hope this helped.

Happy learning,
(Unofficial) The Daily Challenge Team

@modestwallaby Here's a hint to start- if two triangles have the same altitude, and you know the lengths of both of their bases, what can you say about their areas? Using this, try to find [EHC]....

I did some searching around and here's what I concluded:

For some functions, if

$$f(x)=f(y), \text{this does NOT mean that } x=y $$

.
For example, look at

$$f(x)=x^2 $$ $$f(-1)=f(1) \text{ , however} -1 \neq 1 $$

Same follows for sin(x)

$$\sin(0)=\sin(2\pi)=\sin(4\pi)=\sin(1290\pi)=0 $$

and obviously the values inside are not equal to each other.

This is the case with

$$f(x)=e^{ix} $$

You cannot deduce that just because

$$e^{ 2\pi \cdot i} = e^{0\cdot i} \implies 2\pi=0 $$

This is almost like saying

$$1^{100} = 1^1 \implies 100=1 $$

Exponentiation simply just doesn't work that way. I hope this explanation helps you! Let me know if you have any more questions (this is pretty confusing, so don't worry!)