Can You Crack These Challenging Number Puzzles?

[wonderful]

Dear all,

I would highly appreciate if you could help me solving the following sequences:

Question 1: 3,5,8,24,209,3591,?
33811
34308
35534
35200
35010

Question 2: 4,7,13,21,34,55,88,?
110
148
123
138

Question 3: 8,12,18,27,42,70,126,241,?
478
441
503
488
486

Question 4: 3,8,23,60,137,256,?
500
513
507
511
512

Question 5: 5,10,19,32,49,70,?
89
95
121
135
99

 

[Evo]

Have you tried to find the pattern for each of these problems? Try Question 5, you should recognize the pattern by the time you get to 19.

Finding the pattern is the fun in doing these types of problems.

May I ask what those numbers listed under the question are? The question is asking what comes after 70, but none of the numbers below are correct following the pattern.

 

[Rogerio]

Question 5: 5,10,19,32,49,70,?
89
95
121
135
99

Try Question 5...
... The question is asking what comes after 70, but none of the numbers below are correct following the pattern.

?
Of course, 95 is the correct answer!

 

[Evo]

?
Of course, 95 is the correct answer!

Duh, I missed the 95. Completely skipped over it.

And you gave away the answer!

 

[wonderful]

Thanks! I figured it out why 95 is the answer for question 5:

10 - 5 = 5
19-10 = 9
32-19= 13
49 - 32 = 17
70 - 49 = 21

Consider the sequence : 5, 9, 13, 17, 21 the increment is 4. Thus, the next number is something that is greater than 70 by 21+4=25.

How about questions 1 - 4? Can anyone find the solution?

Have A Great Day!

 

[coomast]

I have solved the third question. The others are not easy. If anybody wants a hint, just ask :-)

@ wonderful: Is there a change that you have gotten these questions from the following site? http://www.allthetests.com/quiz18/quizpu.php?testid=1141425609 If so then you have forgotten a possible solution at question 2, namely 140. Whether this is the right solution or not I can't tell because I haven't been able to solve this one so far.

There are also 5 other questions, 6 to 10 of which I have solved 9 and 10. If wanted you might post them, I'm not going to because I don't want to mess up your post.

Anybody else solved one of the others?

 

[wonderful]

I had been out of town in the last two days. Coomast is correct in saying that 140 is also one of the choices in question 2. Coomast can you give us a hint for question 3? These questions are very interesting.

Have A Great Day!

 

[coomast]

A hint for question 3: use the same principle as for solving question 5. Write line per line the differences of two adjacent numbers. Can you see a known sequence after 4 lines of differences? Hope this is clear, otherwise please ask.

I'm working on question 1 and I think I've got something. I'm now working on the general formula, but it isn't straightforward.

These are very strange as a IQ test. One can't possibly have the time to solve all of them within the time frame of such a test. On the other hand they are very nice brain teasers indeed :-)

 

[wonderful]

Thanks coomast for the hint. As you suggested, I looked at the sequence of difference between two adjacent numbers: 4, 6, 9, 15 ... but saw no clear pattern. Can you help? It seems to mention on the related website that there is no time limit on these kind of questions.

 

[coomast]

Thanks coomast for the hint. As you suggested, I looked at the sequence of difference between two adjacent numbers: 4, 6, 9, 15 ... but saw no clear pattern. Can you help? It seems to mention on the related website that there is no time limit on these kind of questions.

Take the differences again for this newly created line, and again, and again. Then you can see a pattern related to some powers.

I've solved the first one as well, this is an even more complicated sequence. A hint: take the differences as done before a few times. Then a pattern involving 13 appears together with the adding of another sequence. This is a complicated one indeed. The general formula looks awesome.

The general formula for question 5 can be written as:

$$x_0=5$$
$$x_1=10$$
$$x_n=4+2 \cdot x_{n-1}-x_{n-2}$$

It is a nice exercise to look for these as well.

 

[wonderful]

Cool! Thanks coomast. I found the answer for question 3 is 478. I am going to solve question 1. These questions are very interesting and great for brain.

 

[wonderful]

Thanks to coomast hint, I've found the solution for question 1: 35534. Thus, we have two not-yet-solved questions 2, 4. After solving these two, I will post new questions that are as intersting as the above 5.

 

[coomast]

Indeed wonderful, that's the one. I will try to look for the second and fourth sequence.

 

[wonderful]

Following the same line of logic, I found the solution for question 2 is 140

Have A Great Day!

 

[coomast]

Can you give an additional hint on sequence 2? I don't see a pattern.

[Edit] Is it possible that the next number after 140 equal is to 219?

 

[wonderful]

I am very happy to share ideas with you. Take the first difference we have: 3, 6, 8, 13, 21, 33. Take the second difference we have 3, 2, 5, 7, ... Here, we can see: 3+2=5, 2+5=7, and so on...

Please feel free to let me know if this helps.
Have A Great Day!

 

[coomast]

Hello wonderfull, I looked at this for some time and didn't see any order in it. Finally I figured it out, I think you made an error. The sequence of the differences of the differences reads: 3 2 5 8 12, you seem to have a 7 instead of the 8. Or am I missing something?

 

[wonderful]

Hi coomast
Thanks for pointing that out. Could you give us a hint on this question.
Have A Great Day!

 

[coomast]

I didn't solve this one yet, also number 4 seems almost a random number writing. I keep working on it.

 

[coomast]

Sorry wonderful, I can't find the others.

 

[ƒ(x)]

Hey guys, I think I found something for number 4; but there's a catch, my solution isn't one of the given answers...but it makes sense.

Start with 3,8,23,60,137,256

3+8=11
8+23=31
23+60=83
60+137=197
137+256=393

So you have 11,31,83,197,393

Take those differences

31-11=20
83-31=52
197-83=114
393-197=196

Now you have 20,52,114,196

Take those differences, which gives you 32,62,82, and take those differences. You should, if neither of us made a mistake, get 30,40, and the next logical one is 50.

Now all you have to is work backwards.

82+50=132
196+132=328
393+328=721
721-256=465

465, by this logic, is the next term in the sequence. Unfortunately this isn't one of the given answers.

Any ideas?

 

[ƒ(x)]

AHHH someone reply, I know this is an old post. But, I feel like I found something.

 

[Gokul43201]

When you have a set of (effectively) 5 numbers, it becomes meaningless to go through 3 repetitions of differences, leaving you with just two numbers as you have. To extrapolate 30, 40 ... to 50 is seeing a pattern where one doesn't really exist.

There's a simpler way to say all this: you can fit an n'th degree polynomial to any set of (n+1) points. You can - in puzzles like this - claim to have found a "pattern" only if you fit an n'th degree polynom to a set of at least (n+2) points. The lower the order of the pattern you fit, the better is the pattern.

Fitting a biquadratic to 5 points, as you have done, is what I'd call the trivial solution. Such a solution exists for any finite set of numbers. You need to find a more clever pattern.

PS: Notice that you can get your second set of numbers (20, 52, ...) from the original numbers in a single step, by subtracting from every number, the number 2 places to its left.

 

[ƒ(x)]

Hmm...anyone have any ideas about sequence #4?

 

[coomast]

Hey guys, I think I found something for number 4; but there's a catch, my solution isn't one of the given answers...but it makes sense.

Start with 3,8,23,60,137,256

3+8=11
8+23=31
23+60=83
60+137=197
137+256=393

So you have 11,31,83,197,393

Take those differences

31-11=20
83-31=52
197-83=114
393-197=196

Now you have 20,52,114,196

Take those differences, which gives you 32,62,82, and take those differences. You should, if neither of us made a mistake, get 30,40, and the next logical one is 50.

Now all you have to is work backwards.

82+50=132
196+132=328
393+328=721
721-256=465

465, by this logic, is the next term in the sequence. Unfortunately this isn't one of the given answers.

Any ideas?

The last step should be 30,20 and thus 10, however it still does not give you the right answer, meaning one that is in the list. Good attempt though.
I almost forgot this question. I did look for this and the others for a while but finally gave up, maybe when I have some more time in the future. These tend to take up a bit of time

 

[ƒ(x)]

The last step should be 30,20 and thus 10, however it still does not give you the right answer, meaning one that is in the list. Good attempt though.
I almost forgot this question. I did look for this and the others for a while but finally gave up, maybe when I have some more time in the future. These tend to take up a bit of time

Oh, haha wow...

That's particularly embarrassing.