# Challenge Math and Brain Teaser Problems!

Ooh I see you like these, (I’m really happy to hear that haha) I have quite a few more!

And nice wondering (if you look back to the question, this triangle isn’t specified as equilateral either… )

1 Like

Honestly the only reason I'll probably never be able to do this is because i'm probably the youngest person here XD

2 Likes

Really? Awesome! I'm gonna think of some more problems, myself

(I probably gave you like 20 notifications by now)

1 Like

Correct! That game sounds cool, I want to try it later. I have saved some math puzzles (which I can post here), one of them I have is similar where you have to make the number 6 with the numbers from 0 to 9

BTW @JonnyGamer, would you recommend any good online resources (preferably free) that teach advanced math? I would love to get into it, but I don't really know where to start. We should probably talk about this in the Numberphile topic.

3 Likes

7, because 1 did nothing and fish can't drown in water or swim away in an enclosed tank.

1 Like

Hey! I changed the topic so there can be brain teasers! Have fun!

3 Likes

6/6+6
6/6=1
1+6=7

1st Timothy 4:something
Do not let people look down on you because you are young.

(Yes, I know this has very little to do with the post I'm replying to unkess you're good at relating things to other things.)

2 Likes
Hmm I am guessing some combination of...

(888 + 88) = 976
976 + 8 + 8 = 992
Okay I have it, I think

888 + 88 + 8 + 8 + 8 = 1000

Great question that was really fun!!

1 Like
Hmm... so this is what I have been doing

so I decided to split `30!` up into:

`1*2*3*4*5*...*28*29*30`

then from there I tried to put it in its prime factorisation:

226 * 314 * 57 * 74 * 112 * 132 * 17 * 19 * 23 * 29

(I don't know if this is accurate even after I checked over roughly)

And then I was thinking about trying to see how many factors there are in total, and how many factors can be made without using 2 as a factor (using combinatorics or whatever it is that is used for combinations/permutations, I don't know what it's called), but my brain is a bit like this right now

^this part was quoted but [details] show in quotes so I took it out.

This is for @JonnyGamer's friend's problem! (What is the chance that a chosen factor of `30!` is odd?)

For the problem of not knowing how many combinations of factors can be made, I think I could look at it like this:

Out of the 26 twos, I can pick 0 twos, 1 two, 2 twos, .... or 26 twos in making a factor. So that is 26+1 choices for the twos.

Soo for all of these:

226 * 314 * 57 * 74 * 112 * 132 * 17 * 19 * 23 * 29

From that, there would be:

27*15*7*3*3*2*2*2*2 factors.

And the number of odd factors would be those that don't have 2 in them. Which would be:

15*7*3*3*2*2*2*2 factors

So that would mean there is a 1/27 chance of a factor being odd....

I think my thinking here is not quite thorough but I tried it for smaller numbers (e.g. 36 = 22 * 32, gives 3*3 factors = 9 factors) and it seemed to work.

I would be curious to hear from anyone hehe

2 Likes

### Yes! You got it!

Yeah, that was a very very very very very.........

confusing one.

Edit: My friend is really impressed. Number theory is awesome

I'm gotta go, so tommorow, I'll be creating a ton of new problems (hopefully)

1 Like

Ooh thanks for letting me know! I did not know if that was it or not. Did your friend get the same answer? and yes that question was awesome! And okay bye for now!

2 Likes

This challenge sounds simple enough.
See if you can solve this.
I’m going to try and make a project using Construct 2 that will perform this task automatically.
A user can enter any size number variable and click a button, then the project checks if the variable is odd or even and follows the rules continuously non stop until the variable reaches 1.
You could make this project using Hopscotch or another program you like using.

5 Likes

I will have more riddles coming tonight...

Right now I have to go to school.

1 Like

i guess I'm still gonna try this

Person1 has 45 bananas, Person2 has 87 bananas, Person3 has 36 bananas, and Person4 has 184 bananas. Person1 ate 4 bananas, and Person4 got 32 more bananas. How many bananas do all people have? After you find that, subtract it by 72

1 Like

Ooh I feel inspired to try replicate that in Hopscotch I think I will use recursion

Wait, before I do that, I was just thinking about it more and that you will end up getting 1 if you run into any power of 2 (because you will keep dividing by 2)

And from that, the question is, if you have a multiple of 3 and add 1 to it, will it always eventually reach a power of 2? hmm…

Edit: I'll be going off soon so I just decided to post pseudocode here for the time being
``````
CollatzConjecture(number):

output number

If number mod 2 == 0 then // number is even
set number to number divided by 2
CollatzConjecture(number)
else if not(number == 1) then // number is odd and is not 1
set number to (number*3) + 1
CollatzConjecture(number)
End
``````

Oooh okay I will try this

Hmm so it sounds like we have 45 + 87 + 36 + 184 bananas at first. Then - 4 + 32 bananas. I am going to grab my virtual working out paper…

2 Likes

Ooh, yes! The Colleta Conjecture is one of my favorites!
Here's a Numberphile video

3 Likes

uhhhh why did I watch the things people post it's getting so boring for me.. lol

@TheRealBlah made a project on this:

2 Likes

And… 100th post!

Ok, @t1_hopscotch, I made another geometry problem! I’m gonna do this one with you since I haven’t worked out the answer yet. Here you go!

Find the area of the green shaded areas
(Pentagon has a side length of 1)

###/!\ Warning /!
This problem is a tough cookie. And believe me.

3 Likes