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Re: Problem 061
Posted: Fri Sep 09, 2011 7:27 am
by mdean
Finally decided to bite the bullet and try this one. My jaw dropped when I found the answer in 16ms. I'm thinking that was just luck, but I know one way to find out.
Update: Nope. Still in the blink of an eye, though I'm not sure how that's even possible...
Problem 61 -reg
Posted: Thu Feb 14, 2013 12:59 pm
by gnanasenthil654321
Hello
what is An ordered set?
does it mean that , the first number of the set is triangular , the second is square,the third pentagonal , the forth hexgonal, the fifth heptagonal , the sixth octagonal ...
kindly explain
Re: Problem 61 -reg
Posted: Thu Feb 14, 2013 2:13 pm
by TheEvil
Hello.
There is an existing topic for that problem:
viewtopic.php?f=50&t=841&hilit=problem+061
Problems under 100 have a leading zero.
Ordered set means that there is exactly one triangular, square, etc. number in some order. That particular order is part of the question.
Re: Problem 061
Posted: Mon Jun 03, 2013 7:54 pm
by Flood
Nevermind, stupid me...
Re: Problem 061
Posted: Wed Jun 12, 2013 12:44 am
by RishadanPort
I did this problem a while ago.
-- The first number, needs to point to the second number.... The second number needs to point to the third.... The Last number needs to point to the first.
-- One of the numbers in the list must be a triangle number
-- One of the numbers in the list must be a square..
etc.
Does not need to be that the First is a square, the Second is a triangle, etc.
(It may also be the case that a number is both a triangle and a square number... -- But as long as ONE of each Triangle, Square, ... Octogonal.. is represented... that is fine.
Re: Problem 061
Posted: Wed Oct 16, 2013 10:01 am
by Himanshu-Mishra
I am getting total no of possibilities to be cyclic. Here each number are from different number(tri,sqr,...) no same list repeated. But I cannot understand "only ordered set " here. Pls explain me.
Code: Select all
2145 ---> 4510 ---> 1089 ---> 8911 ---> 1128 ---> 2821
2145 ---> 4510 ---> 1024 ---> 2465 ---> 6533 ---> 3321
2145 ---> 4510 ---> 1089 ---> 8965 ---> 6533 ---> 3321
6786 ---> 8646 ---> 4624 ---> 2465 ---> 6533 ---> 3367
6441 ---> 4186 ---> 8626 ---> 2640 ---> 4033 ---> 3364
4095 ---> 9517 ---> 1764 ---> 6441 ---> 4186 ---> 8640
6441 ---> 4141 ---> 4186 ---> 8640 ---> 4033 ---> 3364
1540 ---> 4033 ---> 3364 ---> 6426 ---> 2673 ---> 7315
1596 ---> 9633 ---> 3364 ---> 6426 ---> 2673 ---> 7315
2415 ---> 1541 ---> 4141 ---> 4186 ---> 8646 ---> 4624
8281 ---> 8128 ---> 2850 ---> 5017 ---> 1717 ---> 1782
8281 ---> 8128 ---> 2850 ---> 5017 ---> 1717 ---> 1782
5625 ---> 2512 ---> 1275 ---> 7526 ---> 2625 ---> 2556
9216 ---> 1651 ---> 5151 ---> 5151 ---> 5151 ---> 5192
2116 ---> 1651 ---> 5192 ---> 9296 ---> 9633 ---> 3321
2116 ---> 1651 ---> 5192 ---> 9296 ---> 9633 ---> 3321
9216 ---> 1651 ---> 5151 ---> 5151 ---> 5151 ---> 5192
5192 ---> 9216 ---> 1651 ---> 5151 ---> 5151 ---> 5151
5192 ---> 9216 ---> 1651 ---> 5151 ---> 5151 ---> 5151
2147 ---> 4753 ---> 5329 ---> 2926 ---> 2628 ---> 2821
4030 ---> 3010 ---> 1024 ---> 2485 ---> 8515 ---> 1540
5017 ---> 1782 ---> 8281 ---> 8128 ---> 2850 ---> 5050
6441 ---> 4186 ---> 8626 ---> 2640 ---> 4033 ---> 3364
3321 ---> 2116 ---> 1617 ---> 1770 ---> 7021 ---> 2133
2145 ---> 4510 ---> 1089 ---> 8911 ---> 1128 ---> 2821
2145 ---> 4510 ---> 1024 ---> 2465 ---> 6533 ---> 3321
2145 ---> 4510 ---> 1089 ---> 8965 ---> 6533 ---> 3321
6441 ---> 4141 ---> 4186 ---> 8640 ---> 4033 ---> 3364
4186 ---> 8614 ---> 1426 ---> 2628 ---> 2850 ---> 5041
1128 ---> 2821 ---> 2116 ---> 1617 ---> 1717 ---> 1711
1540 ---> 4033 ---> 3364 ---> 6426 ---> 2673 ---> 7315
2415 ---> 1541 ---> 4141 ---> 4186 ---> 8646 ---> 4624
6426 ---> 2628 ---> 2850 ---> 5017 ---> 1717 ---> 1764
1071 ---> 7140 ---> 4033 ---> 3321 ---> 2145 ---> 4510
1782 ---> 8281 ---> 8128 ---> 2850 ---> 5050 ---> 5017
4141 ---> 4186 ---> 8626 ---> 2628 ---> 2850 ---> 5041
1071 ---> 7140 ---> 4033 ---> 3321 ---> 2145 ---> 4510
2640 ---> 4095 ---> 9560 ---> 6084 ---> 8464 ---> 6426
4033 ---> 3367 ---> 6724 ---> 2485 ---> 8515 ---> 1540
Re: Problem 061
Posted: Wed Oct 16, 2013 1:18 pm
by nicolas.patrois
I think that an ordered set is a list.
The first one is triangular, the second square and so on.
Re: Problem 061
Posted: Wed Oct 16, 2013 1:31 pm
by Himanshu-Mishra
oppss .. I did a typing mistake in my own script. Thank you for response nicolas.patrois
Re: Problem 061
Posted: Tue Feb 25, 2014 4:47 pm
by satyres
I have a question please it may be it has already been discussed but it's not clear for me !
the number should be triangle, square, pentagonal, hexagonal, heptagonal, and octagonal.
this mean that first we check if the 4 digits number is triangle and than square, and so on.
Thanks so much
Re: Problem 061
Posted: Tue Feb 25, 2014 6:03 pm
by nicolas.patrois
The first number is triangular, the second is a square, the third is pentagonal and so on.
Re: Problem 061
Posted: Tue Feb 25, 2014 6:27 pm
by mdean
Just as in the shown example, except with 6 numbers instead of 3. Order doesn't matter, just one must be triangular, a different number square, etc. If a number is both square and triangular say, you may count it as triangular, but another number must be square, and vice-versa.
Re: Problem 061
Posted: Tue Feb 25, 2014 7:59 pm
by dawghaus4
mdean has the correct interpretation of the problem.
You must come up with 6 numbers. One must be triangular, one square, one pentagonal, one hexagonal, one heptagonal, and one octagonal. But not necessarily in that order. If one number happens to fall into two categories, then you may considered to be whichever one you need; but it can be counted in only one category.
The order referred to in the problem is that one number "points" to a second (its last two digits are the first two digits of the second) - the second number "points" to the third (it last two digits are the first two digits of the third) - and so forth until the last number points to the first. Since it is cyclic, it doesn't matter which one is first.
Re: Problem 061
Posted: Thu Feb 05, 2015 2:02 pm
by Oliver1978
[edit on]
*deleted my previous crap
[edit off]
I've reworked my approach. It's not much yet, but maybe somebody could say yes or no about the number of 255 potential candidates from which to build that cycle.
Re: Problem 061
Posted: Thu Feb 05, 2015 6:23 pm
by dawghaus4
If by "225" you mean the number of distinct figurates in the problem, you are probably correct. I found it more useful to deal with all 302 figurates, distinct or not. For example, all hexagonal numbers are also triangle numbers, but I needed to look at them as separate numbers, with the exception that if a triangle number is used in the cycle, the duplicate hexagonal can not be used.
Tom
Re: Problem 061
Posted: Fri Feb 27, 2015 1:09 am
by Oliver1978
Finally I did it. But I still don't see the ordered-ness...
Re: Problem 061
Posted: Fri Feb 27, 2015 8:57 am
by Georg
Required Property 1: The set has to be cyclic.
Only the elements of an ordered set have a "next" element to be cyclic.
These ordered sets are distinct:
(8128, 2882, 8281) ≠ (8281, 8128, 2882) ≠ (2882, 8281, 8128) ≠ (8128, 2882, 8281)
Each of them has the three interesting properties from the problem description.
(2882, 8128, 8281) has not, because it is not cyclic.
These sets are not ordered:
{8128, 2882, 8281} = {8281, 8128, 2882} = {2882, 8281, 8128} = {2882, 8128, 8281}
There is no "next number" for an element.
"Find the sum of the only ordered set of six cyclic 4-digit numbers ..."
should be understood as
"Find the sum of any set of the six ordered sets of six cyclic 4-digit numbers ..."
as long as it is not required to start or end a set with a distinguished element.
Re: Problem 061
Posted: Wed Feb 17, 2016 7:21 am
by LaughingSkull
RishadanPort wrote:
Does not need to be that the First is a square, the Second is a triangle, etc.
oooooooh, so that's what it was that prevented me from solving this problem for 12 hours! Way to go, Mr. Euler, your insufficient explanations stump me again!
Re: Problem 061
Posted: Sun Dec 09, 2018 10:27 am
by Vitaminol
The 'ordered set' phrase is ambiguous. I came up with at least these possibilities:
1. The set should be ordered by polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal - in that order.
2. Each number having index 'n' in its own type sequence, the resulting set should have descending order by 'n'. Say, first number is 56th heptagonal, second - 48th square, third - 35th octagonal, etc.
I could infer the correct meaning only after perusing in detail the forum posts. It's a waste of time! Judging by some posts, not only me confronted the problem.
I suggest to include the post of dawghaus4, published Tue Feb 25, 2014 10:59 pm, in the text of the problem:
Quote:
You must come up with 6 numbers. One must be triangular, one square, one pentagonal, one hexagonal, one heptagonal, and one octagonal. But not necessarily in that order. If one number happens to fall into two categories, then you may considered to be whichever one you need; but it can be counted in only one category.
The order referred to in the problem is that one number "points" to a second (its last two digits are the first two digits of the second) - the second number "points" to the third (it last two digits are the first two digits of the third) - and so forth until the last number points to the first. Since it is cyclic, it doesn't matter which one is first.
End quote.
It gives a very clear idea and prevents any misunderstanding