July 2018 - Page 2

petrw1 · 2018-07-06, 15:08

Quote:

Originally Posted by henryzz

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

That's probably it.

14,628 candidates with a<=b<=c<=100

henryzz · 2018-07-06, 18:37

It should be < not <= for the same reason.

Dieter · 2018-07-07, 06:56

Quote:

Originally Posted by petrw1

That's probably it.

14,628 candidates with a<=b<=c<=100

There are ((100+3-1) choose 3) = 171700 triplets (a,b,c) with a<=b<=c<=100 ("combinations with repetitions").
18953 of these 171700 are obscure, but it is reasonable that you have found only 14628:
For (a,b,c) being obscure there must be a "partner of obscurity" (a1,b1,c1) with
a+b+c=a1+b1+c1 and a*b*c=a1*b1*c1.
With a1<=b1<=c1<=100 you get indeed 14628 obscure triplets.
But there is no reason for subjecting the "partners" to these limits.
It seems that there are 18953-14628 obscure triplets (a,b,c) with a<=b<=c<=100 and a1>100 or b1>100 or c1>100.

Dieter · 2018-07-07, 12:43

An example:

(1,23,100) is obscure, because (4,5,115) has the same sum and product. If you confine yourself to numbers <=100, you don't see the obycurity of (1,23,100).

science_man_88 · 2018-07-07, 14:00

you can also note that for:
{a,b,c}=abc
{a+1,b+1,c+1}=abc+ab+bc+b+ac+a+c+1
{a+2,b+2,c+2}=abc+2ab+2bc+4b+2ac+4a+4c+8
{a+3,b+3,c+3}=abc+3ab+3bc+9b+3ac+9a+9c+27
{a+4,b+4,c+4}=abc+4ab+4bc+16b+4ac+16a+16c+64
{a+5,b+5,c+5}=abc+5ab+5bc+25b+5ac+25a+25c+125

the set sums are all the same mod 3. and the products are above. three above is same mod 3.

Dr Sardonicus · 2018-07-07, 18:54

I doubt I'll be able to devise a good way to solve the stated problem, but I was able to find a reasonable way to generate "obscure triples." I was curious about finding small ones. The method I devised was, to start with the product n. Then I would go through triples d, d2, d3 with d <= d2 <= d3 and d*d2*d3 = n, looking for triples with the same sum.

The smallest product n producing "obscure triples" is 36, with [1, 6, 6] and [2, 2, 9], both having product 36 and sum 13.

If you insist that the numbers in the triple all be different, the smallest product is 90, with the triples [1, 9, 10] and [2, 3, 15], both having product 90 and sum 20.

The smallest product yielding a "triplet" of obscure triples is 1200, with

[4, 15, 20], [5, 10, 24], and [6, 8, 25], all having sum 39 and product 1200.

The smallest product yielding a "quadruplet" is 25200, with

[6, 56, 75], [7, 40, 90], [9, 28, 100], and [12, 20, 105], all having sum 137 and product 25200.

I'm sure there are k-tuples of obscure triples with the same sum and product for any positive integer k, but I am too lazy even to find a "quintuplet."

EDIT: Quintuplet found: We have the five triples

[11, 84, 90], [12, 63, 110], [15, 44, 126], [18, 35, 132], and [22, 28, 135]

with product 83160 and sum 185.

henryzz · 2018-07-07, 21:22

It looks like you are most of the way there.

Dieter · 2018-07-29, 06:24

There aretwo solutions for 6 consecutive years in the domain1<=a<=b<=c<=1500 with b<=900 for both solutions. So you shouldhave found these both combinations. A possible explanation could be that in onecase b1>1000 (if I denote the “partner of obscurity” of abc with a1b1c1).
Currently Icheck 1<=a<=b<=c<=2000. There are 2000*2001*2002:6 = 1.335.334.000 triples.For each triple I store only 0 for “not obscure” and 1 for “obscure”. So myfield containing the full information of obscurity has only 1.335.334.000: 8 = 166.916.750 bytes:
111 bit 0 of byte 0
112 bit 1 of byte 0
….
118 bit 7 of byte 0
119 bit 0 of byte 1
…
2000 20002000 bit 7 of byte166916749
For findingthe bit corresponding to a given triplet abc I use two tables built so that Ineed only multiplications by powers of 2.

Xyzzy · 2018-08-01, 22:10

http://www.research.ibm.com/haifa/po.../July2018.html

henryzz · 2018-08-02, 07:58

I was hoping that someone would have pushed it to find 7 consecutive.

EdH · 2018-08-02, 14:13

I guess I didn't have the horsepower and my c++ programs were way too inefficient. I found lots of fives, but was nowhere near the region where the sixes resided. I even dusted off my archaic assembly skills. At one point I had a file of saved info that was over 18GB, but I couldn't efficiently use the data. I ran several machines to the end, though.

2018-07-07, 18:54	#17
Dr Sardonicus Feb 2017 Nowhere 4,643 Posts	I doubt I'll be able to devise a good way to solve the stated problem, but I was able to find a reasonable way to generate "obscure triples." I was curious about finding small ones. The method I devised was, to start with the product n. Then I would go through triples d, d2, d3 with d <= d2 <= d3 and dd2d3 = n, looking for triples with the same sum. The smallest product n producing "obscure triples" is 36, with [1, 6, 6] and [2, 2, 9], both having product 36 and sum 13. If you insist that the numbers in the triple all be different, the smallest product is 90, with the triples [1, 9, 10] and [2, 3, 15], both having product 90 and sum 20. The smallest product yielding a "triplet" of obscure triples is 1200, with [4, 15, 20], [5, 10, 24], and [6, 8, 25], all having sum 39 and product 1200. The smallest product yielding a "quadruplet" is 25200, with [6, 56, 75], [7, 40, 90], [9, 28, 100], and [12, 20, 105], all having sum 137 and product 25200. I'm sure there are k-tuples of obscure triples with the same sum and product for any positive integer k, but I am too lazy even to find a "quintuplet." EDIT: Quintuplet found: We have the five triples [11, 84, 90], [12, 63, 110], [15, 44, 126], [18, 35, 132], and [22, 28, 135] with product 83160 and sum 185. Last fiddled with by Dr Sardonicus on 2018-07-07 at 19:22 Reason: New info added

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2018-07-06, 18:37	#13
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 2³×3×5×7² Posts	It should be < not <= for the same reason.

2018-07-07, 12:43	#15
Dieter Oct 2017 163₈ Posts	An example: (1,23,100) is obscure, because (4,5,115) has the same sum and product. If you confine yourself to numbers <=100, you don't see the obycurity of (1,23,100).

2018-07-07, 14:00	#16
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	you can also note that for: {a,b,c}=abc {a+1,b+1,c+1}=abc+ab+bc+b+ac+a+c+1 {a+2,b+2,c+2}=abc+2ab+2bc+4b+2ac+4a+4c+8 {a+3,b+3,c+3}=abc+3ab+3bc+9b+3ac+9a+9c+27 {a+4,b+4,c+4}=abc+4ab+4bc+16b+4ac+16a+16c+64 {a+5,b+5,c+5}=abc+5ab+5bc+25b+5ac+25a+25c+125 the set sums are all the same mod 3. and the products are above. three above is same mod 3.

2018-07-07, 21:22	#18
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 2³·3·5·7² Posts	It looks like you are most of the way there.

2018-07-29, 06:24	#19
Dieter Oct 2017 5·23 Posts	There aretwo solutions for 6 consecutive years in the domain1<=a<=b<=c<=1500 with b<=900 for both solutions. So you shouldhave found these both combinations. A possible explanation could be that in onecase b1>1000 (if I denote the “partner of obscurity” of abc with a1b1c1). Currently Icheck 1<=a<=b<=c<=2000. There are 200020012002:6 = 1.335.334.000 triples.For each triple I store only 0 for “not obscure” and 1 for “obscure”. So myfield containing the full information of obscurity has only 1.335.334.000: 8 = 166.916.750 bytes: 111 bit 0 of byte 0 112 bit 1 of byte 0 …. 118 bit 7 of byte 0 119 bit 0 of byte 1 … 2000 20002000 bit 7 of byte166916749 For findingthe bit corresponding to a given triplet abc I use two tables built so that Ineed only multiplications by powers of 2.

2018-08-01, 22:10	#20
Xyzzy "Mike" Aug 2002 10000000100000₂ Posts	http://www.research.ibm.com/haifa/po.../July2018.html

2018-08-02, 07:58	#21
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 2³·3·5·7² Posts	I was hoping that someone would have pushed it to find 7 consecutive.

2018-08-02, 14:13	#22
EdH "Ed Hall" Dec 2009 Adirondack Mtns EE9₁₆ Posts	I guess I didn't have the horsepower and my c++ programs were way too inefficient. I found lots of fives, but was nowhere near the region where the sixes resided. I even dusted off my archaic assembly skills. At one point I had a file of saved info that was over 18GB, but I couldn't efficiently use the data. I ran several machines to the end, though.