What is the simplest solution for this math pattern problem?
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The discussion revolves around identifying the next number in the sequence 3, 8, 17, 24, 49, 58, 117. Participants explore various methods, including polynomial fitting and differences between terms, to find a solution. One proposed pattern involves alternating operations, where odd-numbered terms follow a rule of doubling the previous term and adding an odd increment. The conversation highlights the complexity of such problems, emphasizing that while there are infinite solutions, a well-formed problem can yield a unique answer through simpler methods. Ultimately, the missing number is identified as 128, showcasing the importance of recognizing underlying patterns.
#1
lazyaditya
176
7
3,8,17,24,49,58,117,?
what is the missing number ? Also give the pattern you followed.
There are of course, an infinite number of solutions. One way, pretty much "brute strength" is to find a polynomial, P(n), such that P(0)= 3, P(1)= 8, P(2)= 17, P(3)= 24, P(4)= 49,P(5)= 58, P(6)=117, and then calculate P(7). Since we are given 8 values, there is a unique 7th degree polynomial satisfying that.
We can find that polynomial using the "difference". If we subtract each number in the sequence from the next we get the 7 differences 8- 3= 5, 17- 8= 9, 24- 17= 7, 49- 24= 25, 58- 49= 9, and 117- 58= 59. The second differences are 9- 5= 4, 7- 9= -2, 25- 7= 18, 9-25= -16, and 59- 9= 50. The third differences are -2- 4= -6, 18- (-2)= 20, -16- 18= -34, and 50- (-16)= 66. The fourth differences are 20- (-6)= 26, -34- 20= -54, 66- (-34)= 100. The fifth differences are -54- 26= -80 and 100- (-54)= 154. Finally, the sixth difference is 154- (-80)= 234.
Now, by "Newton's difference formula, the polynomial that, taking n to be 0 to 6, gives those values is
8+ 9n+ (4/2!)n(n- 1)- (6/3!)n(n-1)(n-2)+ (26/4!)n(n- 1)(n- 2)(n- 3)- (30/5!)n(n- 1)(n- 2)(n- 3)(n- 4)+ (184/6!)n(n- 1)(n- 2)(n- 3)(n- 4)(n- 5). Set n= 7 in that polynomial to determine the next number in that sequence.
Wow Halls impressive! I never knew about Newton's difference formula. I tried the differences to the 3rd order but still didn't see a pattern and began to think the sequence was wrong or from the Veritasium video something completely unexpected:
There are of course, an infinite number of solutions.
It is implicit in this kind of problem that although there are an infinite number of potential solutions, if the problem is well-formed the unique correct solution can be found by applying Occam's razor.
It is implicit in this kind of problem that although there are an infinite number of potential solutions, if the problem is well-formed the unique correct solution can be found by applying Occam's razor.
As you well know HallsofIvy
Interesting solution, with alternating rules.
Using Occam's razor is still an arbitrary choice for any problem with incomplete knowledge.
There may yet be an underlying pattern to the one you discovered that even more understandable using one rule instead of two.
The veratasium video highlights that premise where the rule was totally unexpected.
Also could we drop the sarcasm from your post? Halls is a respected contributor and mentor to this forum and part of his responsibility is to direct students along the path of solution but not actually solve it.
Using Occam's razor is still an arbitrary choice for any problem with incomplete knowledge.
I don't understand what you mean by this.
jedishrfu said:
There may yet be an underlying pattern to the one you discovered that even more understandable using one rule instead of two.
Yes there may: Occam's razor is an heuristic that may guide one towards a better solution, not a deterministic test of truth.
jedishrfu said:
Also could we drop the sarcasm from your post?
I think that fitting an nth order polynomial to an n-term series is sarcasm (albeit of a subtle kind not appreciable by all); I was merely returning in kind.
jedishrfu said:
Halls is a respected contributor and mentor to this forum and part of his responsibility is to direct students along the path of solution but not actually solve it.
I hold HallsOfIvy in the highest respect, and would not have responded in that way to someone for whom that was not the case.
In case you are still not getting the joke, directing students along the path of n-th degree polynomial fitting will not lead them to successful solutions to "find the next term" problems posed in exams or for recreational purposes.
What insight did you have or what method did you follow?
The same solution had occurred to me, but did not seem sufficiently simple, so I refrained from posting. There are a number of pairs where the first member is n and the second member was 2n+1. The final such pair involves large enough numbers to make the coincidence suspicious.
The pairs occur in a pattern (every odd numbered term is the first member of such a pair).
The first members of those pairs occur in a pattern (simple arithmetic sequence of differences).
My apologies, I didn't understand that your sarcasm was in response to Hall's solution which was perceived as sarcasm. Personally, I felt it was complex but it brought to light another way to solve these kinds of problems.
As I looked at it I did see the odd numbers 5, 7, and 9 so perhaps that's all that needed to solve it. However, I didn't see the succ = pred*2 + 1 expression though.
Can you tell us how you came to your solution?
It would help the OP understand the methods of solution better. I didn't see the complete solution either so I too would benefit.
Personally, I felt it was complex but it brought to light another way to solve these kinds of problems.
I can't think of any non-trivial "find the next number" problem where the solution can be found by polynomial fitting. Even in the "real" world, polynomial fitting may be useful for interpolation but is more or less guaranteed to fail when used for extrapolation.
jedishrfu said:
As I looked at it I did see the odd numbers 5, 7, and 9 so perhaps that's all that needed to solve it. However, I didn't see the succ = pred*2 + 1 expression though.
Can you tell us how you came to your solution?
It would help the OP understand the methods of solution better. I didn't see the complete solution either so I too would benefit.
Similar to jbriggs444, I saw 48 = 24 x 2 + 1 and 117 = 58 x 2 + 1 on first inspection. I then took diferrences between successive terms and the 5,7 and 9 pattern became clear. I juggled these two patterns mentally for a while and couldn't come up with anything to unify them..
In terms of useful general methods of solution, looking at differences between terms is always the place to start. When you are looking at this:
Code:
3 8 17 24 49 58 117
5 9 7 25 9 59
it is easy to notice 8,9 24,25 and 58,59. However like many puzzle-solvers my favourite "find the next numbers" are those that defy heuristic analysis and require a flash of inspiration: 3, 3, 5, 4, 4, 3, 5 is perhaps the most well known but has a flaw which my favourite, 1, 11, 21, 1211, 111221 does not (although I know some prefer the latter problem with the next term, 312211, included to reduce ambiguity).
I would stress that, like jbriggs444, I don't think this is the complete solution: if it is, it is rather unsatisfactory.
So you did the same analysis that we did and had a flash of insight.
WHats interesting is that while we saw the odd number progression we ignored it while we searched for a more inclusive algorithm.
Its a classic 'cant see the forest for the trees' problem we only needed the odd number progression to answer the problem.
It reminds me of a famous code breaking movie where the protagonist was trying to crack a Russian coded transmission. He used every sort of codebreakers trick but came up empty handed. The signal sounded like beads in a bottle sloshing around. His insight came when he gave up for the night and sat down with his infant son and shook the baby rattle. The code was the shakng of the rattle not the more intricate sounds of the beads shaking about.
The same solution had occurred to me, but did not seem sufficiently simple, so I refrained from posting. There are a number of pairs where the first member is n and the second member was 2n+1. The final such pair involves large enough numbers to make the coincidence suspicious.
The pairs occur in a pattern (every odd numbered term is the first member of such a pair).
The first members of those pairs occur in a pattern (simple arithmetic sequence of differences).
Model theory might help here. It allows models for functions to be compared in a way that penalises according to the number of arbitrary parameters introduced. Perhaps one could use a variant which measures the algorithmic complexity. The answer is then the one with the lowest price.
