# Ponder This

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## August 2018 - Solution

August 2018 Challenge

We first present Hao Zheng's solution (thanks, Hao!):

Start with a set of prime numbers (for example, the first 60 primes) as the candidate pool for the water droplets and each time choose nine of these primes. The key is to filter the candidate pool first. Let S be the prime set of the first 60 primes. Here we use two steps to filter S:
1. Let p1 be the smallest number from S and p2 be the 2nd smallest, and so on. We can safely eliminate p1=2. We must have p1> (p1+p2+...+p9)/10 or 10*p1>p1+p2+...+p9>p1+(p1+2*1)+(p1+2*2)+...+(p1+2*8) and this gives us p1>72. So the minimum prime of the set S is 73.
2. After applying the above step to filter the prime set, we end up with a new set. Let's still call it S and still assume p1 is the smallest, p2 is the 2nd smallest, and so on. We should still have
p1> (p1+p2+...+p9)/10. If not, then we can remove p1 from S. Repeat this step until we cannot remove any more numbers from S.

We end up with the following prime set: S = {191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281}
So there are C(18,9) = 48620 different combinations of choosing 9 numbers from 18 numbers. It's fairly quick to verify these 48620 cases using a program.
Here is one of the solutions: (w1, w2, w3, w4, w5, w6, w7, w8, w9) = (223, 227, 229, 233, 239, 251, 257, 277, 281) and (g1, g2, g3, g4, g5, g6, g7, g8, g9) = (13, 53, 73, 113, 173, 293, 353, 553, 593). These are all primes except 553.

Oleg Vlasii (thanks!) sent us a nice solution, inspired by Trinity Hall Prime number:

```W_1 = 1111111111111111113989; G_1 = 1111111111111111139891;
W_2 = 1121222112111112113953; G_2 = 1212221121111121139531;
W_3 = 1121211212211122114003; G_3 = 1212112122111221140031;
W_4 = 1121211212211122113693; G_4 = 1212112122111221136931;
W_5 = 1121222112121212115303; G_5 = 1212221121212121153031;
W_6 = 1121211212121212118547; G_6 = 1212112121212121185471;
W_7 = 1121211212112112117037; G_7 = 1212112121121121170371;
W_8 = 1121222112112112113457; G_8 = 1212221121121121134571;
W_9 = 1040377703888884080017; G_9 = 0403777038888840800171.
```

When focus on digits '2' we can see "IBM" in both W and G.

Several of you found the wrong (but interesting) solution where the sum of G_i is one-fifth of the sum of W_i.

 WATER CONTR GOLD 269 271 277 281 283 293 307 313 331 6.5 8.5 14.5 18.5 20.5 30.5 44.5 50.5 68.5 13 17 29 37 41 61 89 101 137

Amos Guler found a quite large solution: W = 53129 53171 53267 53309 53327 53411 53441 53591 103407 and G = 1237 1657 2617 3037 3217 4057 4357 5857 504017 where the only non-prime is W_9=103407.

And Kipp Johnson found a nice way to convert a solution without the primality condition into a valid solution.

The real story is about 19 (3+7+9) rings:
"Three Rings for the Elven-kings under the sky,
Seven for the Dwarf-lords in their halls of stone,
Nine for Mortal Men doomed to die"
So Radu-Alexandru Todor found a solution for all 19 rings holders:
W = [6737, 6761, 6791, 6827, 6857, 6863, 6899, 6971, 6997, 7001, 7019, 7109, 7127, 7349, 7417, 7433, 7451, 7481, 7499] and
G = [151, 631, 1231, 1951, 2551, 2671, 3391, 4831, 5351, 5431, 5791, 7591, 7951, 12391, 13751, 14071, 14431, 15031, 15391]
All these numbers are prime.