Challenge: An Ant on a Cube

Congratulations to Arbitrary Renaissance, fmakofmako, Minh Cong Nguyen, Kevin Tong, Kazi Abu Rousan, Aadi Karthik, Benjamin Wang, Laura Kuttnig, silverlizard, and Daniel Darroch for successfully solving the last week's math challenge question! Arbitrary Renaissance was the first person to solve the question.

Your support is truly a huge encouragement.
Please take a second to subscribe in order to send us your valuable support and receive notifications for new videos!

Every subscriber and every like are wholeheartedly appreciated.

Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

[ExplosiveBrohoof]

Here's a nifty way to solve the problem:

Note that if you plot out the factor tree of 30, you get a cube. In this sense, moving along an edge is like multiplying or dividing by a prime value (you move from vertex '3' to vertex '15' by multiplying by 5, for example). How do we get to 30 from 1 in 7 moves? This will be the same number as the ant problem.

Imagine we have three counters: one for 2, one for 3, and one for 5. We need to end up with exactly 1 on each of these. Another way to think of this is that we have switches for each of these numbers, and we need to flip each one an odd number of times.

For 7 switches, there are only two partition shapes that this will take: (1, 1, 5) and (1, 3, 3). For each of these, there are 3 different permutations that are unique. It remains to count how many ways we can order these switches.

For the (1, 1, 5) partition, we can think of it like choosing a spot for a and a spot for b between a bunch of 5 c's. We have 6 locations for the a, and 7 locations for the b, giving us 42 different combinations.

For the (1, 3, 3) partition, it's a bit trickier. We can scramble the threes in 6C3 = 20 different ways, leaving us with 7 locations for the one remaining. That leaves us with 140 different combinations.

Putting it all together, we get 3*140 + 3*42 = 546 different ways.

I really hope this is right. This is a roundabout way to come up with the answer but I think I have it correct.

[Quichey]

I believe the answer is 546.
I labelled the vertices even or odd depending on the number of moves it takes to get to that point. It can be shown that the parity of the starting point and the ending point are the same, in this case odd.
Thus, it will never be the case that an even number of moves will allow you to land on the opposite diagonal.

I then labelled each edge with a direction corresponding to an addition of 1 to x, y, and z. The only way to land on the vertex directly opposite of the start point is if the total number of additions to each x, y, z coordinate are congruent to 1 mod 2.

With 3 moves, there must be an addition of 1 x direction move, 1 y direction move, and 1 z direction move. There are 3! ways to achieve this, thus there are 6 ways to move to the end.
With 5 moves, there must be an addition of 3 x direction moves, 1 y direction move, and 1 z direction move, or 1 x direction move, 3 y direction moves, or 1 z direction move, or 1 x direction move, 1 y direction move, and 3 z direction moves. For each case, there are 5!/3! ways to arrange those moves, thus there are 20 moves for each case for a total number of 60 ways to get to the end.
With 7 moves, the combinations are 5x1y1y, 1x5y1z, 1x1y5z, 3x3y1z, 3x1y3z, and 1x3y3z. Thus, there are 3 x 7!/5! + 3 x 7!/(3!*3!) ways to arrange the moves for a total of 546.

[TheBetterVersion]

546 If we connect the dots on a graph, we can write two functions
a(x) = 3*b(x-1)
b(x)=2*b(x-1)+a(x-1)
Where a(x) is the moves to start/end, and b(x) is the moves to surrounding corners (when even/odd are considered to see which corners are this and which are 0)
1, 0 (0)
0, 1 (1)
3, 2 (2)
6, 7 (3)
21, 20 (4)
60, 61 (5)
183, 182 (6)
546, 547
So 546 is the correct answer

[joshuaryman]

Notice that the cube can be split into 2 subsets of 4 points within which all points are mutually not adjacent to eachother. Let's call the groups A and B. This also implies that every move must be from an A point to a B point or vice versa. Every A point can have two relationships with a B point: either it is adjacent (connected by a single edge, every A has 3 of these) or diagonally opposite (requires crossing three edges to reach). By exhaustively counting possibilities, it can be shown that, starting on an A point, there are 7 ways to finish on an adjacent B point after three moves; also by counting possibilities there are 6 ways for the ant to finish on the diagonally opposite corner after three moves. For the number of ways of landing on a particular A point after four moves, all we now need to do is add up all of the 'weightings' for any B points that are one move away (connected by an edge). This means that after four moves there are (7+7+6) =20ways to finish on any of the three A points which are adjacent to the final B 'target point'; there are (7+6+6)=21 ways of landing on the starting A point. There are now three moves left so, by symmetry, we can apply the same '6 or 7 ways' method to get from any of the four A points to the target B point. Thus there are (6*21)=126 ways of finishing in the target, starting from having finished on the initial A point after four moves; there are (7*20)=140 ways of achieving this for the remaining three (adjacent to the target) A points. Therefore the total possible ways of reaching the final B point after 7 moves is 126+3(140)=546 ways.

[JalebJay]

I found the answer to be 546:
Let (x, y, z)=(0, 1). Start by noticing that each direction of movement needs an odd number moves to start from 0 and end at 1. This means 7 needs to be split into a 3-long partition each containing an odd number. You can quickly find the only unique partitions that qualify for this are [5 1 1] and [3 3 1].

For the first case of [5 1 1] you can create (7 choose 5)*(2 choose 1)*(1 choose 1) = 21*2*1 = 42 different ways to arrange the 3 sets of elements in a unique order. You can also find that there are 3 different permutations of the values [5 1 1] thus you get the number of ways of using 5 of one direction, and 1 of each other gives: 42*3 = 126.

Similarly, for the second case of [3 3 1] you can create (7 choose 3)*(4 choose 3)*(1 choose 1) = 35*4 = 140 ways. This set also has 3 different permutations, thus there are 140*3=420 different arrangements.

There are no other cases which can create an odd partition so our solution is the sum of our two cases or 126+420=546

[srpenguinbr]

The answer is 5040=7!



I think it can be useful to think of the position of the in a 3d graph, where the start point is at (1, 1, 1), the end at (0, 0, 0) and 1 is the length of each vertex.
The ant needs to move so that the sum of all movements (if thought as vectors) gives you minus one unit on each axis. I'll define:
+x to add 1 to the x-direction
-c to decrease 1 in the x-axis
+y adds 1 in the y-axis
and so on.
The simplest path is -y-x-z or any of the other 5 permutations of that. They will always appear somehow in the steps. But there are 4 more movements to be done yet.
They can be something like +x-x, +y-y, -y+y, etc, that is, two movements that cancel each other out, one positive and one negative movement on the same axis, regardless of the order.

So, it can initially choose one of the 3 axis and then another one of the 3, they can surely be repeated (it is stated in the problem). It gives 3*3=9 possibilities.

Then you can sort the order of the movements. Í'll first consider the case in which the canceled axis does not repeat, such as in -x-y-z+x-x+y-y
You can choose the axis that will not appear or think of that as the permutation where the order does not matter:
3!/2!=3 ways
Then you sort the 7 movements: 7! *3
Now, divide by the number of ways 2 repeated vectors can be organized. Otherwise, you would be counting two times the same paths:
3*7!/2!
Do that again, because there is the downwards movement of the other axis to consider yet. Just the downwards, such as -x-x+x.
3*7!/(2^2)=3780 ways.


Now, the case where the canceled axis is repeated. I'll add these combinations later

You choose one of the 3 axis to be repeated, then sort all movements:
3*7!

Now, we need to divide by the different permutations of the repeated vectors. In this case, there is one group of 3 elements (such as -y-y-y) and other of 2 (e.g. +y+y). The # of combinations of the former is 3! and of latter is 2!:
3*7!/(3!*2!)
7!/4=1260

adding everything together:

1260+3780=5040.

[johnwroblewski]

answer: 546.

note that the sum of forward and backward moves ( and likewise for up, down and right, left) must always be odd to end up in the correct corner. then we get that (2k + 1) + (2m + 1) + (2n + 1) = 7 must hold, where are k, m, and n are arbitrary positive integers. Thus k + m + n = 2 must hold. There are 3 ways to have one int be 2 and the rest 0. Then in this instance we can count the number of paths with the expression: 3 * (7!/5!). And there are 3 ways for one int to be 0 and the rest 1. We can count the paths in this instance with the expression: 3 * (7!/3! * 3!). Summing these two expressions gives us 546. Also I'm sure I'm past the submission deadline for winning, but I just wanted to play!!

[dhruvanaidu]

I got 546
Lets assume the ant is crossing a square, not a cube. It has to get to the vertex diagonally across from where it is now. If it moves in the closer to the desired vertex, we will call that move a forward move. If it moves away from the desired vertex we will call that a backward move.
We notice that the ant has to move down as well as get across the top of the cube. The bottom square is identical to the top square. We also notice that we need 5 forward moves and 2 backward moves to complete the cube in 7 moves.

Case 1 : The ant moves down and never comes back up
If this were to happen We can refer to our square model.
We have 4 F(orward) moves and 2 B(ackward) moves, since 1 F move is taken by moving down.
The only ways to complete our square model in six moves are
FFBBFF
FBFBFF
FBFFBF
FFBFBF
(we have to start with a F move and end with and F move, plus other restrictions were present).
Each permutation has 8 ways for being completed
8x4=32
We need to travel down as well, there are 7 spots in each sequence to insert a down move
32x7=224

Case 2: The ant moves down, up, and down again.
We have 3 F's and 1 B (going down is an F, up is a B).
The only two combination that work are:
FFBF
FBFF
Both have 4 ways of being completed.
4x2=8
We have to insert a down, up, and a down into each a sequence. There are 35 ways to do this.
8x35=280

Case 3: The ant moves down, up, down, up, then down.
The only way to finish our square model with 2 F's is:
FF
There is 2 ways to achieve FF. There is 21 ways to put the the downs and ups.
2x21=42

224+280+42= 546

[kevincaotong]

The answer is 546:
Notice, there are 3 choose 1 choose 1 = 6 paths to go from the starting point to the end point (same idea as 2d, but with multinomial coefficients). Of these 6 paths, we can group them into 2 sets of equal size: One in which the ant retreats its move in one axis, and another in which it retreats in two axes. This equates to 7*6=42 ways for the one axis case, and 7*20=140 for the two axes case, making the total
3*7*6+3*7*20 = 126+420=546 ways.

[alperenaydin]

The answer is 546.

Let's first name the vertices. Let's call 1 the starting vertex and 8 the ending vertex, (2, 3, 4) the vertices directly adjacent to 1, and (5, 6, 7) the vertices adjacent to 8.
Then let us group the vertices together in the following way: A = {1}, B = { 2, 3, 4}, C = { 5, 6, 7} and D = {8}


Now notice that to there are 3 ways to pass from A -> B, 1 for B -> A, 2 for B -> C, 2 for C->B, 1 for C -> D, and 3 for D -> C.


We can us this graph to calculate the number of ways a path can be taken. For example, if we do A -> B -> C -> D, the value of the edges are 3, 2 and 1. We multiply these to get 6 which the number of ways we can do A -> B -> C -> D.

Then we notice that the number of ways we can reach a certain set in n-steps (starting from A) is the weighted sum of number of ways we can reach the adjacent set in the graph where the weights correspond the values of the edges. If we have x(n) be the number of ways we can reach X in n-steps we have the following formulas:

a(n+1) = b(n)
b(n+1) = 3*a(n) + 2*c(n)
c(n+1) = 2*b(n) + 3*d(n)
d(n+1) = c(n)

[Bani]

The answer is 546 I think. Let the coordinates of the ant at the starting position are (0, 0, 0) and at the end are (1, 1, 1).
The question is how many ways can we add or subtract 1 to the coordinates so that in 7 operations we get (1, 1, 1).
x1(1, 0, 0) + y1(0, 1, 0) + z1(0, 0, 1) + x2(-1, 0, 0) + y2(0, -1, 0) + z2(0, 0, -1) = (1, 1, 1) or (x1-x2, y1-y2, z1-z2)=(1, 1, 1) and x1+y1+z1+x2+y2+z2=7.
So after substituting 3 of them we get x2+1+y2+1+z2+1+x2+y2+z2=7 or x2+y2+z2=2.
This equation for (x2, y2, z2) has these solutions: (2, 0, 0) (0, 2, 0) (0, 0, 2) (1, 0, 1) (1, 1, 0) (0, 1, 1).
So the full solutions with all 6 variables are (3, 1, 1, 2, 0, 0) (1, 3, 1, 0, 2, 0) (1, 1, 3, 0, 0, 2) (2, 1, 2, 1, 0, 1) (2, 2, 1, 1, 1, 0) (1, 2, 2, 0, 1, 1).
For all the solutions the ant can't leave the boundaries of the cube so that means it can't make 2 consecutive choices in the same direction and axis and it must choose a move in the positive direction first.
From the 1st, 2nd and 3rd solution we get 7*6 for each one(for choosing when to do the "back-and-forth" in one axis), and from the 4th, 5th and 6th we get 7*20 for each one(also for choosing when to do the "back-and-forth" in two axes now).
So 3*7*6+3*7*20=546.
I hope that my logic is correct xD
Love your channel btw :D

[vivekkumarsingh]

A small variation of this problem with same answer came in indian national maths Olympiad 2014 and Singapore national maths Olympiad in 2006

[mrtourist]

Using the concept of graphs from data structures one can solve this

[fmakofmako]

Here's my attempt: I essentially used a dynamic programming style solution, but it's been a while and maybe I'll miss a case or three.

For simplicity call the start point S, the points one move away from the start point A (3), the end point E, and the vertices one move away from the end point B (3).

Then define a state [x, y] = [# of moves to reach this point, # of ways to reach this point in x moves]. Hence the start point S has state [0, 0]. Then we fill in the graph using the formula for a vertex with three adjacent vertices that can be reached in a, b, and c ways in x moves. Then the vertex has state [x+1, a+b+c].

S starts with state [0, 0]. Then A is [1, 1]. B is [2, 2]. and E is [3, 6]. As a check we note there are 6 ways to reach the end point in 3 moves, because there are 3 adjacent edges and for each square two paths along the edges.

Then S is adjacent to A with state [1, 1], so S is [2, 3]. A is [3, 7] (2 ways to 2 B points and 3 ways to S), B is [4, 20]. and E is [5, 60].

Then S has state [4, 21]. A is [5, 61]. B is [6, 181]. and E is [7, 543]. Hence there are 543 ways to reach the end point in 7 moves.