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THXTHXTHXTHX!!!!!

someone has posted it before, but there was no clear response. I was hoping for a very clear and thorough explanation.

This has been posted before.  Try searching for it.

Actually it is supposed to read 5/2 not 5/7

Crossproduct. Im sorry that i didn't specify

No problem!

Thank you so much. :)

I am really, sorry it was an acident, I will try to delet this post. Sorry.

Ok, Thankyou so much for your help :)

His account is still existing, but I think he somehow got removed from the leaderboard/ top users. Maybe his account is in the process of getting deleted?

The ending point of the path Franklin made depends on all the moves Franklin made, not the order. Consider a random 10-step path Franklin could take. NESWNNESEW (N for North, S for south, W for west, E for East.  Reduce our path by cancelling each E-W and each N-S pair, which will not change our endpoint. So our example reduces to NE. Notice that our reduced path must have an even number of moves, and it is thus possible to end at any point (x,y) where x+y is even and |x| + |y| is no greater than 10.'

If |x| + |y| = 0, there is one possibility for (x,y).

If |x| + |y|  <= (less than equal to) 2, there are 9 possible paths ((0,0) (-+1,-+1) (0, +-2) (+-2,0)); these solutions will form a 3x3 diagonal grid

The solutions with |x| + |y| <= 10 will form an 11x11 grid, so there are 11^2= 121 ways

Thankyou so much :)

It doesn't show me how many cubes its cut into, that's all that I see, sorry

https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_6

Go to this site, please do not post aops questions, please ask for hints and tell us your steps. This can not be a place where you just get free answers, please read through the link and understand! Ask the message board for help

PLease dont double post, I already answer your question once!

Let M and N be the base in which 1st and 2nd speaker talked, respectively. Then according to the 1st speaker

2M+6 residents use base M, and

2M+2 residents use base M+4,

which means that N = M+4. According to the second speaker

2(M+4) + 5 people reside in Abaze,

(M+4) + 3 people use both bases, and

1 resident uses neither base.

Now

(# of residents) = (# of M-users) + (# of (M+4)-users) – (# of users of both) + (# of users of neither).

Therefore

2(M+4) + 5 = (2M+6) + (2M+2) – ((M+4) + 3) + 1,

so 11 = M, whereby 2(M+4) + 5 =

@jibz yahoo,

This is the answer please figure out the rest, 

Last one, part of solution, aops,

Rather than going through casework of the different kinds of cubes that could be selected and then rolled, we instead count the total possible number of faces that could come up and count the number of them that are black. There are 125 smaller cubes, each with 6 sides, so there are $125(6) = 750$ faces that could come up. Each face of the original cube has a $5\times 5$ square of black faces that might come up. Therefore, there are 25 black faces per side of the original cube, for a total of $25(6)=150$ black faces. Hence, our desired probability is figure out the rest.... 

Unfortunately, most calculators on the Internet are faulty when handling very large numbers.

2017^(2017^2017). It is very easy to get the first 10 digits of big numbers such as this one:

1 - You must start from the RHS, what is inside the brackets: (2017^2017) =3.906 579 755 433 725 744 549 802 E+6665

2 - Take the log(base 10) of the first 2017 from the LHS and multiply by the number in (1) above:

Log(2017) x 3.906 579 755 433 725 744 549 802 E+6665 =1.291 009 715 962 041 616 156 219 E+6666. That is how this number begins.

3 - The last 10 digits are also very easy to get provided you have an accurate enough calculator that can handle very large numbers such as this one. The one "Big Calculator" is ,of course, WolframAlpha!. In this case, all you have to do is enter this into the "box" and you get the last 10 digits: 2017^(2017^2017) mod 10^10 and presto!: Last 10 digits =2,055,543,777

Next one figure

The cubes that are not painted black on any sides are on the inside of the original cube. These cubes make up a $3\times 3\times 3$ cube,

Figure out the rest

The cubes with one black face are those on the outside of the cube which are not on an edge. Omitting the outside squares on each face leaves a $3\times 3$ square of cubes which are each painted on exactly one face. Hence, there are $9$ cubes per face which are painted black on exactly one face, for a total of ....Figure out the rest, part of solution from aops for your first question.

This problem has also already been answered. please do not just give us 3 questions to answer. This is an aops question. Please tell us what you have tried and we can guide you, not give you the answer! You have to copy the aops question on to google then copy it again in the web2.0calc question page!