I'm pretty sure I have the right solution - code runs fast, but apparently there is a bug somewhere...
If anyone wants to have a look at my code, I'd appreciate it. (Using C#)
-- NM figured it out.
Problem 118
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See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
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RishadanPort
- Posts: 79
- Joined: Mon Jun 10, 2013 7:31 am
Re: Problem 118
hmmpf
I'm pretty sure I have the right solution - code runs fast, but apparently there is a bug somewhere...
If anyone wants to have a look at my code, I'd appreciate it. (Using C#)
-- NM figured it out.
I'm pretty sure I have the right solution - code runs fast, but apparently there is a bug somewhere...
If anyone wants to have a look at my code, I'd appreciate it. (Using C#)
-- NM figured it out.
Rishada is the gateway to free trade—but the key will cost you.
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laogui
- Posts: 7
- Joined: Tue Jun 06, 2023 3:05 pm
Re: Problem 118
@francky @spacetweek a single number is clearly impossible since it will be divisible by 9;
two is the minimum and five is the maximum corresponding to partitions of 9 like (1,8), (4,5) and (1,1,1,1,5) or(1,2,2,2,2)
Looking at the (1,2,2,2,2) partition, the single digit must be 2 or 5 (since 2 4 5 6 8 require 5, not 4 2-digit primes
and we get (2,41/47,53/59,67/61,89/83) or (5,23/29,41/47,67/61,89/83) or 8 sets of (1,2,2,2,2). That's about the only partition of 9 I can do by hand!
two is the minimum and five is the maximum corresponding to partitions of 9 like (1,8), (4,5) and (1,1,1,1,5) or(1,2,2,2,2)
Looking at the (1,2,2,2,2) partition, the single digit must be 2 or 5 (since 2 4 5 6 8 require 5, not 4 2-digit primes
and we get (2,41/47,53/59,67/61,89/83) or (5,23/29,41/47,67/61,89/83) or 8 sets of (1,2,2,2,2). That's about the only partition of 9 I can do by hand!