1, 3, 13, 39, 89, ??

Options (183, 267, 359, 479) .

I have the answer.

What I want to know is:-

Is there a way to solve such series if you don't know what the series is beforehand?

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- Apr 14th 2010, 04:40 AMontherocksComplete the series
1, 3, 13, 39, 89, ??

Options (183, 267, 359, 479) .

I have the answer.

What I want to know is:-

Is there a way to solve such series if you don't know what the series is beforehand? - Apr 14th 2010, 05:08 AMTKHunny
"solve"? This word makes no sense.

"have the answer"? No, you don't. There are infinitely many answers. It is easily shown that there is an expression to produce all four of those answers, and again, infintely many others.

You must learn the concept of a "difference". It is a common expected method for such games.

Subtracting successive values from each other gives first differences.

3 - 1 = 2

13 - 3 = 10

39 - 13 = 26

89 - 39 = 50

Do it again to obtain the second differences

10 - 2 = 8

26 - 10 = 16

50 - 26 = 24

Do it again to obtain the third differences

16 - 8 = 8

24 - 16 = 8

From here, it is likely that you are supposed to assume ALL third difference are identical. This is actually quite a conceptual leap, given that we've only two of them.

Anyway, working backwards:

The next third difference = 8

The next second difference = 24 + 8 = 32

The next first difference = 50 + 32 = 82

The next value = 89 + 82 = 171

I'm not sure what this says about "the answer" that you have, but as long as your answer can be supported, it should be acceptable. 183, for example, might be the result of constant 5th differences of 12. It may be a bit silly, but it is supportable.

The next two values in the sequence I have defined are 293 and 463, oddly interspersed with your "answer" list. - Apr 14th 2010, 05:11 PMontherocks
Firstly, I didn't cook up this question. It was in a formal written aptitude test.

Secondly, how could you say "There are infinitely many answers." when you have been given four options to choose. You are supposed to find "the pattern" in the question to match any of the options, not any of your "may be a pattern".

Is this the way you solve multiple choice questions, by arguing over the options. Your answer is obviously wrong (just on the fact that it is not even any of the options). So what would you do now? write an essay somewhere the paper showing your infinite answers that you can obtain by some pattern that you may "think" exists. This is sheer arrogance.

And for the record I do know, "the concept of a difference". Please don't assume others are village idiots. - Apr 14th 2010, 10:11 PMTKHunny
What an odd response. Take a deep breath and let's see if we can learn some mathematics.

Please don't confuse confidence with arrogance. In my case, it could be arrogance, but it is not fair to project this to anyone else with confidence.

I have changed many answers on many tests. A very favorite thing to do is to work the problem incorrectly to show the answer that is supposed to be correct and then work the problem correctly. Most test authors appreciate this sort of feedback. On some occasions, the aptitude sought in the exam is the confidence to determine that the appropriate solution is not offered in the list. It is an old joke, "What is the capital of Wyoming, LARE-A-MEE or LAHR-A-MEE?" A very, very large percentage of people get this wrong. Most pick the first, as that is the proper pronunciation of the fine city in Wyoming. The right answer is, of course, "Cheyenne."

I did not accuse you of cooking up anything. Examinations can be misprinted. This should not shock you. Humans build them.

Where did you tell me that you already were acquainted with the "difference"? Had you provided that information, there would have been no misunderstanding. There was no accusation of "village idiot". There was only a brief presentation of a necessary concept.

There**are**infinitely many answers. There is not "**the**pattern". It is a commonly held fallacy that there is such a pattern. Some have attempted to define a "simplest pattern", but I have found no such exploration particularly satisfactory. I can match any number that is provided on the exam. Anyone who can support ANY of the answers should be given credit for correctly answering the question.

My answer is**not**obviously wrong. I took the time to provide a complete and thorough response, rather than simply checking a provided box. Obviously, this is not what the exam intended, but the intent of this sort of problem is almost always misguided. Personally, I would take my arrogance into the exam presenters or writers and tell them what I think of their misleading and silly question.

I know a very capable mathematician who answers "-2" EVERY TIME he is presented with one of these silly what-is-the-next-number-in-the-sequence problems. I wish I had thought of it. - Apr 14th 2010, 11:37 PMontherocks
I guess you live in a perfect world where the test author who actually framed the questions is the one who sits and corrects thousands of answer papers, and then takes time to read students scribblings and gives them credit and where there is so much empty space in the test paper to write your calculations and what you think should be the answer and on top of that you have the time to do so.

You are absolutely right. I completely agree with you. But in this case its not so.

I failed to do so in this particular question in the stipulated amount of time. If you can match any of the numbers given,**please let know how. That was exactly and the only intention of my post in the first place.**All these time wasting ramblings are purely consequential.

(I completely fail to understand your argument about 183)

Thats again, how it "should" be (in a perfect world though). You may be lucky enough to have had such an experience.

You are really lucky to get to meet the actual authors of the test papers after your exam and speak to them about their whatever.

And this person scored full marks in all his tests...right?

Lets get back on track pls. I really want an how to find the answer to this particular series question. - Apr 15th 2010, 10:33 AMTKHunny
Wow, you are stubborn, aren't you? You should direct your valuable energy more toward learning and less toward hopeless arguing. Anyway...

Since you know differences, let's wade through it. Insert 183 as the next value and calculate the first differences.

3 - 1 = 2

13 - 3 = 10

39 - 13 = 26

89 - 39 = 50

183 - 89 = 94

Do it again to obtain the second differences

10 - 2 = 8

26 - 10 = 16

50 - 26 = 24

94 - 50 = 44

Do it again to obtain the third differences

16 - 8 = 8

24 - 16 = 8

44 - 24 = 20

Do it again to obtain the fourth differences

8 - 8 = 0

20 - 8 = 12

Do it one last time to obtain the fifth differences

12 - 0 = 12

There you have it. Make the fifth differences a constant 12 and you get 183.

Note: Make the fifth differences a constant 96 and you get 267.

Note: Make the fifth differences a constant 188 and you get 359.

Note: Make the fifth differences a constant 308 and you get 479.

Note: Really, infinitely many. How many constant 5th differences can you imagine?

Note: If I use successive ratios, rather than differences, a constant 4th ratio of 2.292483 produces a value of 389.1627 - again in the neighborhood of the provided values.

Final Note: Since there is a solution with constant THIRD differences, most feel that this is more satisfactory than a solution with constant 5th differences or 4th ratios. Therefore, except for the possibility that the test creators have pre-selected THE correct answer, which is inappropriate, 171 really is a better answer than any provided. There might be an awesome rational function or something else really interesting, but I didn't see anything right off. - Apr 15th 2010, 08:37 PMontherocks
- Apr 15th 2010, 10:09 PMTKHunny
Besides this one?

Note: Make the fifth differences a constant 96 and you get 267. - Apr 15th 2010, 11:04 PMontherocks
Yep besides using "nth order difference theory", which seems could be used to get any answer you desire (just by getting nth difference if there are n numbers and then adding to it the difference of the answer to support your logic)

But here in this case the reason was:

All the numbers are divisors of 3471. - Apr 16th 2010, 02:21 AMDeadstar
$\displaystyle f(x) = \frac{4x^3 - 12x^2 + 14x - 3}{3}$

when x=6, f(x)=... - Apr 16th 2010, 02:33 AMWilmer
- Apr 16th 2010, 03:05 AMDeadstar
- Apr 16th 2010, 03:47 AMTKHunny
We've been over this. The nth order "difference" is one way to proceed. Any value can be selected. There is no "[my] logic" to support. The examiner has pre-selected a "correct" answer, and this is patently unfair and mathematically inappropriate.

You have brought up another point against such problems. Very often, they are not even math problems. They are just silly puzzles that happen to use numbers. - Apr 16th 2010, 03:52 AMontherocks
- Apr 16th 2010, 04:05 AMontherocks