Pizza Hut Meth Contest

[a1call]

Originally Posted by WMHalsdorf

3816547290 is not divisible by 7

1234759680 meets all constraints. Note there may be larger numbers.

3816547290 meets all constraints:

n=1
3/1=3

n=2
38/2=19

n=3
381/3=127

n=4
3816/4=954

n=5
38165/5=7633

n=6
381654/6=63609

n=7
3816547/7=545221

n=8
38165472/8=4770684

n=9
381654729/9=42406081

n=10
3816547290/10=381654729

[R. Gerbicz]

Originally Posted by axn

92165438 is not divisible by 8

There is only one solution, and is of the format OEOE5EOE0, which means you have to brute force 4!*4! = 576 permutations.

Nice observation, I've generalized this for general base, currently it is in draft at the corresponding sequence https://oeis.org/A111456 and searched for larger bases: now we know that there is no solution for base=15,16,..57. (for the largest known solution n=7838911147538198 the base is 14).

[LaurV]

You should report, if you went to 57. The oeis only mentions 40.
(edit: we found the sequence in oeis before, by searching for a1call's number, but didn't do any work in this direction).

[R. Gerbicz]

Originally Posted by LaurV

You should report, if you went to 57. The oeis only mentions 40.

My sent draft is:
"There is no more term up to base=56; and no solution for base=60. Furthermore all base is even: if the number formed by the first (base-1) digits is x, then x is divisible by (base-1) and x==base*(base-1)/2 mod (base-1), because the base-th digit is zero. From this the base is even. We can also see that if the i-th leftmost digit is d, then gcd(base,i)=gcd(base,d). To see this let g=gcd(base,i) and the number formed by the first i digit is x, then i divides x=k*base+d for some k, from this g divides d. And obviously g divides base, so g divides gcd(base,d), but it can't be larger than g, otherwise say gcd(base,d)=h>g, then in every h-th position we see a digit divisible by h, and the i-th digit is also divisible by h. This is a contradiction, there would be more than base/h digits divisible by h."

[a1call]

Originally Posted by richs

Win pizza for 3.14 years:


3. My key-rings are metal circles of diameter about two inches. They are all linked together in a strange jumble, so that try as I might, I can’t tell any pair from any other pair.

However, I can tell some triple from other triples, even though I’ve never been able to distinguish left from right. What are the possible numbers of key-rings in this jumble?

Well if each ring is connected to 2 neighboring rings in a circular configuration and every other ring is connected to 1 of their neighbor's neighbor, then there could be nx4 rings for any integers n (i.e. 4, 8, 12, ...)
If the rings don't form a circular configuration then the number would be nx4+1 for any integers n. (I.e. 5, 9, 13, ....)
That might satisfy the constraints.

[LaurV]

For the third problem the answer is zero. The condition of being impossible to distinguish between any pairs implies that all the rings are either not connected at all, either all pairwise connected (i.e. any ring goes through all the others). Because otherwise, you could carefully pick a pair of rings which are connected, and a pair of rings which are not connected (not necessarily disjunct, for example AB and CD, or AB and AC), and here is your differentiation: one pair is connected, one is not. So, you either have a lot of separate rings, not connected at all, in which case there is no way to make any distinction between triplets, or you have all rings in a big chinese-rings-fashion ball (think ball of yarn) in which case there is no way to distinguish triplets either.

A more relaxed interpretation of the "left-right" condition allows a solution with either 4 or 7 rings, and one with infinity (or well, depending on the diameter of the wire from which the rings are made, you still connect all such each go through all the others, but some go circularly, some go straight - think about in how many ways you can connect three rings together, then you have your triplets distinction).

The problem is intentionally vague formulated, they won't willingly give you one year worth of pizza, they want only the advertising part of it, but not many winners.

edit: the second problem is still interesting, tho.

[PawnProver44]

Thanks! This forum was very helpful! I'm getting free pizza for 3.14 years thanks to all of you!

[axn]

Originally Posted by LaurV

edit: the second problem is still interesting, tho.

partitions

[a1call]

Originally Posted by 0PolarBearsHere

Well...
OEOE5EOEO0
But yes, 4!*4!

I think the actual number of trials is substantially less, since a foreach function will exit the loop as soon as the 1st test fails. So the loop will be exited as soon as the test 127/3 is not integer and 1274, 1276,... won't be tested.

[PawnProver44]

Just fooling around here. lol. You guys don't really care anyway.

[science_man_88]

Originally Posted by PawnProver44

Just fooling around here. lol. You guys don't really care anyway.

for fooling around not really. people here also don't take nicely to people who claim to know things they don't have the potential ability to prove.