## Numbers on a Dodecahedron

We have written the numbers from 1 to 12 on the faces of a regular dodecahedron. Then, we have written on each vertex the sum of the five numbers on the faces incident with it. Is it possible that 16 of these 20 sums are the same?

We color the vertices in five colors as shown in the image below, such that each face of the dodecahedron has 1 vertex of each color. Then, the sum of the four numbers from each color must be equal to the sum of all numbers written on the faces: 1 + 2 + … + 12 = 78.

If 16 of the vertices have the same number written on them, then by the pigeonhole principle there will be 4 vertices with identical colors and identical numbers. Since 78 is not divisible by 4, we conclude that this is impossible.

## Sum Up to 999

Can you find a triple of three-digit numbers that sum up to 999 and collectively contain all digits from 1 to 9 exactly once? How many such triples are there? What if the sum was 1000?

There are exactly 180 such triples that sum up to 999 and none that sum up to 1000.

In order to see that, notice that the sum of the first digits of the numbers can be no more than 9. Therefore, the sum of the middle and the sum of the last digits cannot be less than 10. We then see that the sum of the last digits should be exactly 19 and the sum of the middle digits should be exactly 18. The sum of the first digits should be 45-19-18=8.

There are 2 ways to get 8 using unique digits from 1 to 9: 1+2+5 and 1+3+4.

• If the first digits are {1, 2, 5}, the options for the middle digits are {3, 6, 9}, {3, 7, 8}, and {4, 6, 8}. The last digits end up {4, 7, 8}, {4, 6, 9}, and {3, 7, 9} respectively.
• If the first digits are {1, 3, 4}, the options for the middle digits are {2, 7, 8} and {5, 6, 7}. The last digits end up {5, 6, 9} and {2, 8, 9} respectively.

Since the set of the first digits, the set of the middle digits, and the set of the last digits of the numbers can be permuted in 6 ways each, we get a total of 5×6×6×6=1080 solutions, or 180 up to permutation of the 3 three-digit numbers.

In order to see that we cannot get a sum of 1000, we note that since the sum of the digits from 1 to 9 is divisible by 9, then the sum of the 3 three-digit numbers should be divisible by 9 as well. Since 1000 is not divisible by 9, the statement follows.

FEATURED

## A Broken Circle

There are N points on a circle. If we draw all the chords connecting these points and no three of them intersect at the same point, in how many parts will the interior of the circle get broken?

For example, when N is equal to 1, 2, 3, 4, and 5, we get 1, 2, 4, 8, and 16 parts respectively.

The answer, somewhat surprisingly, is not 2ᴺ⁻¹, but 1 + N(N-1)/2 + N(N-1)(N-2)(N-3)/24.

In order to see that, we start with a single sector, the interior of the circle, and keep successively drawing chords. Every time we draw a new chord, we increase the number of parts by 1 and then add 1 extra part for each intersection with previously drawn chords.

Therefore, the total number of parts at the end will be:

1 + the number of the chords + the number of the intersections of the chords

Each chord is determined by its 2 endpoints and therefore the number of chords is N(N-1)/2.

Each intersection is determined by the 4 endpoints of the two intersecting chords and therefore the number of intersections is N(N-1)(N-2)(N-3)/4!.

## The Die Game

You pick a number between 1 and 6 and keep throwing a die until you get it. Does it matter which number you pick for maximizing the total sum of the numbers in the resulting sequence?

In the example below, the picked number is 6 and the total sum of the numbers in the resulting sequence is 35.

No matter what number you pick, the expected value of each throw is the average of the numbers from 1 to 6 which is 3.5. The choice of the number also does not affect the odds for the number of throws until the game ends, which is 6. Therefore, the total sum is always 3.5 × 6 = 21 on average, regardless of the chosen number.

## Symmetry and Perplexion

Find an X > 1, such that:

If we pick X=√2, (using a somewhat ambiguous notation) we get:

To see that the equality above holds, simply note that 2=(√2)² and take √2 root of both sides.