21
Obviously a is one of {1,3,7,9} and a,b are coprime. Also, b can't be a multiple of 3 regardless (else our number is a multiple of 3). That leaves 23 possibilities (4 for a, 6 for b, but we can't have a=b=7), or 22 if you don't count 11111 as "limerick".
The only other trick I see is that 1001 = 7x11x13; so mod 1001, aabba = 11b-a. Clearly that isn't going ...
20
Here's the solution:
There's a very neat method for finding this, inspired by the no-computers way of solving another related puzzle. Namely,
More specifically, given the constraints of this problem:
How can we achieve this?
The way I used (unique up to swapping of rows and columns) is
That gives the following grid:
which is what I put at the top in ...
15
I think that it could be
Reasoning
9
I'd start with the last digit first:
Now we're looking at this subset of numbers:
With a little trick/some summations we can eliminate:
That leaves us with these 22 possibilities:
Then we can eliminate a couple more with another division rule (thanks @IronEagle)
To get the final 8 from these 20 we just...
Note: I assumed a!=b for it to be a limerick. I'...
8
Short solution with generalization potential:
6
I think the next word is
Because
Other options in English are
With this logic, the 6 letter word could be
No 7 letter words follow this rule though.
6
My guess:
Reasoning:
5
Not sure if this is optimal, but here's my solution
5
A solution using all vertical transpositions. All changes annotated and words defined.
5
One solution:
Strategy:
4
Here is the solution. I guess it would have been easier if I converted it to numbers, but it was fun to solve as is.
4
As others (ex: Gareth McCaughan) have mentioned, by looking at divisibility by primes up to 13 we can narrow the list down to 19 possibilities:
11111, 11441, 11551, 11771, 11881, 33113, 33223, 33443, 33773, 33883, 77117, 77227, 77447, 77557, 77887, 99119, 99559, 99779, 99889.
Now, we'd like to test for divisibility by higher primes. To do so, we can make ...
4
The number of candidates is very limited.
If "a" is even or 5, the resulting number is obviously a composite.
If "b" is a multiple of 3, the sum of digits is a multiple of 3, therefore the resulting number is a multiple of 3.
If "b" is "a", the resulting number is obviously a multiple of "a".
So you are left with a list of 23 candidates:
11bb1 with b = ...
3
"So, you got what I said about limerick numbers, right? Well, the thing is, some of them are probably primes. You know what, I'm going to pay for your next beer if you manage to guess how many. You have one minute!"
The gears in Brian's head started grinding.
One minute? I guess I could pull out my phone and look for "limerick primes" or something. But has ...
3
3
I think this is the solution
There isn't any particularly hard deduction here but if anybody has any queries, I can address them in the question.
2
Here's a solution. I don't have much to say about it other than the fact it looks elegant and is different from the others posted.
2
I believe the answer could be
Reason:
2
This will do it:
2
Another solution (ninjaed):
Strategy (not so illuminating nor useful I think):
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