@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!

@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!)

@victoria_huang said in Solve this in 1 second😆 @dashingrhinoceros

"honeys"

lol

Another cool application of Euler's formula is taking powers of complex numbers;

For example if you had a complex number like (sqrt(3)/2 + 1/2 i)^20 you could easily figure this out with Euler's formula

(also a great way to farm views on youtube jkjk)

@the-blade-dancer

trigonometry is a very i n t e r e s t i n g subject, but my math teacher said that it's the main thing that "murders" high school students' grades and self-esteem. It's a good idea to start getting the concepts when you're in middle school.

@the-blade-dancer I suppose you can prove this with coordinate bashing, that it is impossible for a line to touch a circle at more than 2 points.

@mathnerd_101 Glad that you figured it out! 🙂

@bulba_bulbasaur 🙂

@energizedpanda In the March 26 Ask Math Anything livestream, Prof. Loh actually explained why the volume of a sphere is \(\frac{1}{3}^{\text{rd}}\) its surface area!

It has to do with the fact that you can chop the volume into infinitesimally-small pyramids where the bases lie on the surface of the sphere and the opposite corners all lie at the center of the sphere. Then, since the volume of a pyramid is just \(\frac{1}{3} \text{ ( area of base ) } \times \text{ height },\) you just take the surface area, multiply by the height of each pyramid, which is the same for each pyramid, \(r,\) and multiply by \(\frac{1}{3}\) to get the volume of the sphere!

20200326-sphere.png

AMC 8 2013 Problem 25 🙂 I remember doing this.