Riddler Classic - 2/11/2022
From Dean Ballard comes a cubical conundrum:
Consider a cube, which has eight vertices, or corners. Suppose I assign a prime number to each vertex. A “face sum” is the value I get when I add up all four prime numbers on one of the six faces.
Can you find eight primes and arrange them on a cube so that the six face sums are all equal?
Extra credit: Can you find another set of eight primes that can similarly be arranged on the vertices of a cube? How many more can you find?
I'm speculating that this requires finding four sets of twin primes.
To try an example, take the first 4 distinct twin prime pairs (meaning each prime is in exactly one pair, as is specified in the riddle). These primes are: (3, 5, 11, 13, 17, 19, 29, 31).
For clarity, consider these primes in the pairings below:
1. (3, 5)
2. (11, 13)
3. (17, 19)
4. (29, 31)
Next, orient each prime on a vertex of a cube. Ensure that each member of a pair lies on the vertex opposite its partner.
E.g.:
31 ------------ 17
/| /|
/ | / |
/ | / |
/ | / |
3 ------------- 13 |
| 11 --------|--- 5
| / | /
| / | /
| / | /
|/ |/
19 ------------ 29
For clarity, represent each cube face as a 4-tuple:
1. (3, 13, 19, 29)
2. (5, 11, 17, 31)
3. (3, 13, 17, 31)
4. (5, 11, 19, 29)
5. (3, 11, 19, 31)
6. (5, 13, 17, 29)
Now, sum each 4-tuple:
1. 3 + 13 + 19 + 29 = 64
2. 5 + 11 + 17 + 31 = 64
3. 3 + 13 + 17 + 31 = 64
4. 5 + 11 + 19 + 29 = 64
5. 3 + 11 + 19 + 31 = 64
6. 5 + 13 + 17 + 29 = 64
As all 6 sums are equal, (3, 5, 11, 13, 17, 19, 29, 31) is an answer!
Try another selection:
(5, 7, 59, 61, 107, 109, 137, 139)
Pairs:
1. (5, 7)
2. (59, 61)
3. (107, 109)
4. (137, 139)
4-tuples:
1. (5, 61, 109, 137)
2. (7, 59, 107, 139)
3. (5, 61, 107, 139)
4. (7, 59, 109, 137)
5. (5, 59, 109, 139)
6. (7, 61, 107, 137)
Sums:
1. 5 + 61 + 109 + 137 = 312
2. 7 + 59 + 107 + 139 = 312
3. 5 + 61 + 107 + 139 = 312
4. 7 + 59 + 109 + 137 = 312
5. 5 + 59 + 109 + 139 = 312
6. 7 + 61 + 107 + 137)= 312
Again, as all 6 sums are equal, (5, 7, 59, 61, 107, 109, 137, 139) is another answer!
Consider the general case:
Take 4 pairs of twin primes:
(a, a + 2)or(a, z)such thatz = a + 2(b, b + 2)or(b, y)such thaty = b + 2(c, c + 2)or(c, x)such thatx = c + 2(d, d + 2)or(d ,w)such tahtw = d + 2
Consider the following general cube representation:
w ------------- c
/| /|
/ | / |
/ | / |
/ | / |
a ------------- y |
| b ---------|--- z
| / | /
| / | /
| / | /
|/ |/
x ------------- d
The 4-tuples of each face:
1. (a, y, x, d)
2. (z, b, c, w)
3. (a, y, c, w)
4. (z, b, x, d)
5. (a, b, x, w)
6. (z, y, c, d)
As established above:
- z = a + 2
- y = b + 2
- x = c + 2
- w = d + 2
Using these equalities, rewrite the 4-tuples as below:
1. (a, b + 2, c + 2, d)
2. (a + 2, b, c, d + 2)
3. (a, b + 2, c, d + 2)
4. (a + 2, b, c + 2, d)
5. (a, b, c + 2, d + 2)
6. (a + 2, b + 2, c, d)
With these tuples, compute the general sums:
1. a + (b + 2) + (c + 2) + d = a + b + c + d + 4
2. (a + 2) + b + c + (d + 2) = a + b + c + d + 4
3. a + (b + 2) + c + (d + 2) = a + b + c + d + 4
4. (a + 2) + b + (c + 2) + d = a + b + c + d + 4
5. a + b + (c + 2) + (d + 2) = a + b + c + d + 4
6. (a + 2) + (b + 2) + c + d = a + b + c + d + 4
The sum of each face is a + b + c + d + 4. As all sums are equal, we can deduce that there are as many sets of 8 primes meeting the riddle's criteria as there are distint groupings of 4 sets of twin primes.
Whether or not there are infinitely many twin primes is an open conjecture, but mathematician Yitang Zhang has made progress towards proving it in recent years.
Twin primes are not the only primes that work. Any common difference shared by 4 pairs of primes (E.g. 4, 34234234234, or 22) would work.
I solved this cubical riddle in a cubicle.