July 2018

a1call · 2018-07-02, 14:24

SmartMersenne · 2018-07-04, 17:14

I can find a solution with 3 more years and 4 more, but 5 more seems impossible for ages <= 122 (this is the maximum I heard of any age: Jeanne Calment).

Did anyone managed to find a triplet that is obsucre and stays obscure for 5 more years?

a1call · 2018-07-04, 18:08

I thought there bonus was for 5 consecutive years.
5 more years can be interpreted as 6 or even 9 consecutive years.
A solution for a total of 5 consecutive years is easy to find.
I have not attempted to find anything further than that.

KangJ · 2018-07-05, 02:21

Quote:

Originally Posted by SmartMersenne

I can find a solution with 3 more years and 4 more, but 5 more seems impossible for ages <= 122 (this is the maximum I heard of any age: Jeanne Calment).

Did anyone managed to find a triplet that is obsucre and stays obscure for 5 more years?

I tried all combination cases for ages <= 800, and there is no solution obscure with 5 more years.

Is there any rules for the obscure triplets?

henryzz · 2018-07-05, 11:22

Am doing a search for a and b less than 1000 and c less than 2000, a < b < c
I have noticed that c is often 1.5-2x as big as b in the larger runs.
My code is reaching its limits. I think I am going to have to rethink to go much further. I currently store all the tuples with common sums and products.

For a b c and a' b' c' to have the same product and sum, I don't think max(c, c') can be prime if a < b < c and a' < b' < c'. Not sure this helps much.

science_man_88 · 2018-07-05, 16:37

Quote:

Originally Posted by henryzz

Am doing a search for a and b less than 1000 and c less than 2000, a < b < c
I have noticed that c is often 1.5-2x as big as b in the larger runs.
My code is reaching its limits. I think I am going to have to rethink to go much further. I currently store all the tuples with common sums and products.

For a b c and a' b' c' to have the same product and sum, I don't think max(c, c') can be prime if a < b < c and a' < b' < c'. Not sure this helps much.

same product = same geometric mean. same sum= same arithmetic mean. the first can be acheived by rearranging the factors. second takes same average difference. for the sum parity if there was an odd number of odds before there still needs to be after.

petrw1 · 2018-07-06, 05:25

Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure,
Every other triplet has at least one other triplet with the same sum and product.

a1call · 2018-07-06, 06:01

Quote:

Originally Posted by petrw1

Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure,
Every other triplet has at least one other triplet with the same sum and product.

That does not add up.
That would mean that just about any triplet would be s solution to the bonus problem.
I think that would be in conflict with what has been posted here so far.
I did not run for loops for 1 to 100 but to a value smaller than 100 and found much less obscure-s than that. Then again, perhaps if I extend the loops to 100 steps, I would get the same results as you. But intuitively, I doubt that. FWIW.

henryzz · 2018-07-06, 08:46

Quote:

Originally Posted by petrw1

Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure,
Every other triplet has at least one other triplet with the same sum and product.

You need to add the restriction a <= b <= c otherwise a b c will match b a c etc

axn · 2018-07-06, 08:51

Quote:

Originally Posted by petrw1

Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure,
Every other triplet has at least one other triplet with the same sum and product.

Something is definitely wrong. Check your logic again.

With 1<=a<=b<=c<=100, I get 1 triplet the leads to 5 consecutive "obscures" (which is unfortunately not enough to earn a star), and 6 triplets with 4 consecutive "obscures" -- although one of them is merely the second term of the 5-er mentioned.

With 1<=a<=b<=c<=500, I get 6 additional triplets with 5 terms (in addition to the one from above), but no 6 term ones yet.

Incidentally, here are a few ones with 3 terms where c<=40.

Code:

2:24:25 
4:20:26
8:14:25
8:14:34
10:24:26

henryzz · 2018-07-06, 09:09

c a decent amount larger was the trick that found it for me. Although my 1000 1000 2000 search found only one solution for 6 consecutive years it was overkill.

I think I have been using the wrong method storing all the obscure tuples. There are too many of them. The correct method is probably to go up through the products distributing factors.

2018-07-02, 14:24	#1
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2⁶×31 Posts	July 2018 July 5.11.4_19

2018-07-05, 11:22	#5
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 2·2,909 Posts	Am doing a search for a and b less than 1000 and c less than 2000, a < b < c I have noticed that c is often 1.5-2x as big as b in the larger runs. My code is reaching its limits. I think I am going to have to rethink to go much further. I currently store all the tuples with common sums and products. For a b c and a' b' c' to have the same product and sum, I don't think max(c, c') can be prime if a < b < c and a' < b' < c'. Not sure this helps much. Last fiddled with by henryzz on 2018-07-05 at 11:26

2018-07-06, 05:25	#7
petrw1 1976 Toyota Corona years forever! "Wayne" Nov 2006 Saskatchewan, Canada 1000111001000₂ Posts	I must be doing something wrong... Taking every a,b,c from 1 to 100 I only get 98 of the 1,000,000 that are NOT obscure, Every other triplet has at least one other triplet with the same sum and product.

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2018-07-04, 17:14	#2
SmartMersenne Sep 2017 2·7² Posts	I can find a solution with 3 more years and 4 more, but 5 more seems impossible for ages <= 122 (this is the maximum I heard of any age: Jeanne Calment). Did anyone managed to find a triplet that is obsucre and stays obscure for 5 more years?

2018-07-04, 18:08	#3
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2⁶·31 Posts	I thought there bonus was for 5 consecutive years. 5 more years can be interpreted as 6 or even 9 consecutive years. A solution for a total of 5 consecutive years is easy to find. I have not attempted to find anything further than that.

2018-07-06, 09:09	#11
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 1011010111010₂ Posts	c a decent amount larger was the trick that found it for me. Although my 1000 1000 2000 search found only one solution for 6 consecutive years it was overkill. I think I have been using the wrong method storing all the obscure tuples. There are too many of them. The correct method is probably to go up through the products distributing factors.