The Maths Topic

Nope. The last digit in the mentioned number is 1.

Any solution is better accompanied by a demonstration. I’ll put it myself for this one:
The series starts with 3, 9, 81, 6561, 43046721… But the next number’s units digit is only influenced by the units digit of the last number, as any other calculus would work torwards the bigger digits. IE. 8181= 11 + 801 + 180 + 8080= 1 + 80 + 80 + 6400 = 6461. And since 11 = 1, the lastr digit will always be 1 from that moment on.

Good luck with the other problems ^^

III: The first letter we can choose in two ways. The second we can choose in three ways, the third in two ways, the fourth in one way and the fifth in one way. The answer is 23211 = 12.

Great! *gives bombax cookie :cookiemon: Now for another 2 problems to keep 3 of them up:
2a) On a chessboard (8x8 squares) Paul puts tokens on each square following this rule: both rows and columns are numbered, form 1 to 8, and for each square, he lookt up the corresponding row and column number, then he sums them up, and puts that many tokens on that square. How many total tokens has Paul put on the chessboard?
3a) That one’s a little more difficult… see if you can do it ^^
Samantha has 2006 equilateral triangles of the same size, and wants to put them all on a table, in a way they don’t overlap, and each one is sharing 2 sides with other 2 triangles. Can she make it? Can still she do it if she has 2005 of them? Is it possible for any number of them >12 ?

2a) the chessboard has 8x8=64 squares. He puts 2 tokens in 1 corner, and 16 in the opposite. The alternate corners are 9 and 9. Similarly each pair of squares totals 18 tokens, and the board has 32 pairs, resulting in 18x32= 576 tokens total. This feels sloppy though, is there a better way to arrive at the answer?

3a) in order to achieve 2 and only 2 shared sides, the triangles need to loop in some manner. The closed loop would be 6 triangles in a hexagon, 12 triangles to encompass a 1 triangle space, 14 for 2, 16 for 3, etc. etc… I would venture a guess 2005 would NOT work, but any even number 12 and above, as well as 6, would meet the req’s… assuming of course the table is large enough and/or the triangles small enough :smile:

2a) *gives Phoenyx a cookie :cookiemon: The counting can be separated into rows and coloumns: tokens total = (tokens form row counting) + (tokens form columns) = 8*( 1+2+…+8 ) + 8*( 1+2+…+8 )= 8362=576 indeed. ^^
3a) Nice, but I still need to know that it’s impossible for any odd number of triangles to be used successfully (it’s easy: try visualizing some examples)

Now for some counting! (seems difficult, but it’s not :wink: )
2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?

GAHHH! :lol:

Working on this for quite awhile now, I keep coming up with ways to explain why it needs 2 at a time, but can’t seem to figure how to explain why it CANT be an odd number.

I have a feeling it is something to do with it being impossible for two odd numbers to sum to an odd number, but… :confused: :help: :eek: I give up

Guess I’l do it :wink:
First things first, a graph where all dots (representing the triangles) are connected to two and two only of the other elements is necessarily made of separate loops (convince yourself of this)
Now, in any single loop of triangles, look at them on the table: there are traingles with one side facing up and others with one side facing down. The only way a triangle can connect to a side-up to share a side is to be side-down, and vice versa. So it’s necessary for every loop to have an even number of triangles, since if it was odd, at one point we’d necessarily have 2 connected triangles that are oriented in the same way (and that’s impossible ;D)

Now for another one!
3b) A, B and C are playing with 2008 skittles. ‘A’ brings down three times the skittles brought down by ‘B’, and ‘B’ him/herself brings down double the number of C’s skittles. (They are all playing the same match, so the skittles they have brought down as a total can’t exeed 2008)
What is the maximum number of skittles A could bring down?

And come on guys, 1) is not that hard, there is a nice formula to calculate the number of divisors, look “divisor” up on Wikipeida :wink:

  1. We have 1-5, all possible combinations of 2, 3, 4, and 5 of those numbers. There’s only one possible combination for all 5. For 4, we have 5 combinations. For 3, 10; for 2, I got 10 as well. But there’s some overlap; 2-5 can also be expressed as combinations of 1*(2-5), so we can ignore those. For similar reasons, we can drop 5 from the group of 3’s, 2 from the 4’s, and 1 from the group of 5. So 5+1+5+10+10-1-4-5-2 = 19.

3b. Is it really that easy? a+b+c=2008. a=3b. b=2c a=3(2c)=6c. 6c+2c+c=2008. 9c=2008. c=223.111 a = 1338.666. Of course, you can’t have a fractional number of Skittles, so 1338.

I really didn’t understand the explanation, however the answer is wrong. :tongue: I’ll give an hint: start writing 5!=120 as 2^3 * 3 * 5;

Yes, it was. ^^ Cookie earned :cookiemon: and new problem!!
3c) (I’ll leave the 3rd problems easy) In a stadium there must be a precise order to enter through the 5 gates: first 1 person goes in the 1st gate, then 2 in the 2nd, then 3 in the 3rd, 4 in the 4th, and 5 in the 5th, and then the cycle starts over. Which gate will the 2007-th person pass through?

  1. Let’s see, we have 2, 3, 5, 23, 25, 235, 2^2, 2^23, 2^25, 2^235, 2^3, 2^33, 2^35, 2^335, and 3*5, and 1. That makes 16.

:cookiemon: Counting is good for little numbers, but now go for 10! = 10 * 9 * … * 2 * 1. I want a formula! :tongue:

Anyone tried doing math/homework in a dream? :razz:

It was in a past challenge: do all 4 operations with numbers bigger than 5 digits or so (writerscube was in it, but I can’t really find it right now).
Some other also tried to go by the powers of 2, and didn’t manage to get to his usual big numbers.
I’m guessing math is a good way to get the left hemisphere (rational thinking) back in gear before starting the real LD, so you can have more control and stability ;D

So if I give you a formula, do I get a cookie recipe?

Let’s see. 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 * 3^2 * 2 * 5=
2^8 * 3^4 * 5^2 * 7
I want to say it’s just 953*2 .

Here’s a math-related puzzle. You’ll have to think very creatively to solve it. I lol’d when I found the secret.

Draw that without lifting your pencil, going over a line twice, or manipulating the paper (i.e. folding it).

Well I’ll just give you 2 cookies, beacuse you didn’t explain the formula. :cookiemon: :cookiemon: That makes three cookies for a single problem, you should be more than happy :lol: I’ll demonstrate it in your place.

We have a number of prime divisors, (namely 2, 3, 5 and 7), each wth his own exponent, and to form a divisor, we can combine any available power of those divisors freely. We can take 2^3, 3 and 5 to form 835=120, take only 2^7 and form 128, and so on.

To form all possible divisors, we just have to multiply all possibilities for each prime divisor, and if that divisor has n as its maximum exponent, that means we have n+1 choices for it. (IE for 5^2, we can ignore it and take 5^0, take a single 5, or take both and get 5^2=25, three total choices). Multiplying all choices (that are completely independent) we get (8+1)(4+1)(2+1)(1+1) = 9532 = 270 total divisors. (Notice how it is completely irrelevant from the kind of prime divisors we have)

It is indeed mathematically impossible to do it the standard way, because there are 4 points from which an odd number of lines is leaving. (And the maximum number to make it possible is 2). So, we could draw extra lines here and there to help us, perhaps over another paper so we don’t ruin the drawing. Is it so? :tongue:
But if you didn’t say not to fold the paper, I would have immediately thought of it first :tongue:

Actually, I was just going for “it’s impossible.”

I found this one…

uppsidedown L = gamma
gamma(n+1)=n!

read more on en.wikipedia.org/wiki/Gamma_function

I definitely think too hard into stuff. I seriously LOL’d when I saw your answer :rofl: [com]and what, no cookie? :smile: [/com]

Quite right, but we were talking more about a formula to calculate the munber of divisors of any given number :tongue:

Ok so, 2 problems are still up, feel free to add a 3rd one to the mix, or I will post one tonight.

2b) How many numbers of five digits (from 10000 to 99999) that are without zeros AND divisible by 12?

Not going to work on this until I’m actually awake, but I’d say it’s somewhere around 5,250. Close?

And fine, here’s your cookie. :cookiemon: