# Dodecahedron

• Oct 10th 2010, 07:09 AM
atreyyu
Dodecahedron
Every edge of a regular dodecahedron is assigned a number from $1, 2, ..., 30.$ We use each number exactly once. Determine whether it is possible to do it in such way that the sum of edges that come out of every vertex is (a) even; (b) divisible by 4.

For (a), I figured that among the three edges that come out of each vertex, either all must be even, or one even and two odd. So instead of assigning number, I just try to color the edges on the net of a dodecahedron with two colors, say black and red, where red = odd number, so that every vertex has two red edges or no red edges, and that there are 15 red edges in total. But I have trouble arriving at the final solution.
Your help will be appreciated. (Wink)
• Oct 10th 2010, 09:06 AM
Traveller
a)Attachment 19257

b) Hint : Look at the sum over the dodecahedron.
• Oct 10th 2010, 11:17 AM
atreyyu
Thanks, but I have trouble interpreting the picture you posted. Shouldn't a dodecahedron have 12 faces? Unless I'm missing something.
• Oct 10th 2010, 12:27 PM
Traveller
That is how we represent polyhedrons on planar surfaces. Consider the whole region outside of the polygon to be another face of the polyhedron.
• Oct 10th 2010, 01:38 PM
atreyyu
Thanks a lot!
• Oct 10th 2010, 02:26 PM
Traveller
No prob ! :)
• Oct 11th 2010, 06:59 AM
atreyyu
Hm... I still have trouble on (b)... been trying to consider assigning 0, 1, 2, 3 (mod 4). Would you mind a follow-up? (Happy)
Here's what I have: we assign $0,1,2,3 \pmod{4}$ to the edges. We have seven 0s, eight 1s, eight 2s, seven 3s. For total sum $S$ holds $S\equiv 7*0+8*1+8*2+8*3\equiv 90\equiv 2 \pmod{4}$. But then, the sum of $S_n$ (where $S_n$ is the sum of edges joined by the nth vertex) for $n=1,2,...,30$ equals $2S$ as each edge is counted twice. Thus $2S\equiv 2\cdot 2\equiv 0 \pmod{4}$.
• Oct 11th 2010, 08:47 AM
Traveller
7*0 + 8*1 + 8*2 + 7*3 = 45
• Oct 11th 2010, 09:42 AM
atreyyu
Thanks again.
• Oct 16th 2010, 02:17 AM
Glyper
Wow, that's what I call an interesting problem! :)

However, I don't get one thing in the solution and that is: why did you represent the numbers as 0, 1, 2 and 3 to solve the b? Are the zeroes the numbers which are divisible by 4, twos the numbers divisible by 2 while ones and threes the odd numbers? If so, why do we use 1s and 3s at all instead of just 1s? I mean: odd numbers is an odd numbers, isn't it?
• Oct 16th 2010, 04:23 AM
Traveller
The numbers were represented by their remainders when divided by 4.
• Oct 16th 2010, 08:06 AM
Glyper
Ooh.. I now get it :) Thank you a lot :)

So there isn't such way of numbering to make it divisible by 4, right?
• Oct 16th 2010, 08:48 AM
Traveller
Right.
• Oct 21st 2010, 06:23 AM
Glyper
Hmm.. and is there any way of describing the a? I mean: any way to show that it's true apart from drawing the dodecahedron.
• Oct 29th 2010, 07:25 AM
Glyper
I thought of proving it just like b but it doesn't work. I mean - if we know that te overall sum isn't divisible by 4, we know that there is no such way of numeration to make every triple divisible but when it comes to a, proving the sum is divisible by 2 doesn't help us at all since it doesn't mean we can assign such triplets so that every of them is even.

How could I prove this? Do you have any ideas?