Number Sequence Puzzle - Try to solve this!

The answer is 6.

Check out the sequence.

1, ,2, ,3, ,4, ,5, ,6, ,7, ,8, ,9, ,10, ,11, ,12, ,13, ,14, ,15, ,16, ,17, ,18, ,19, ,20, ,21, ,22, ,23, ,24
while the spaces to the left of even numbers is a number half of it.
No. to left of 2 is 1, left of 4 is 2, left of 6 is 3 etc....
So we have: 1,1,2,A,3,2,4, A,5,3,6,A ,7,4,8,A,9,5,10,A ,11,6,12,A,13,7,14, ,15,8,16,A ,17,9,18,A,19,10,20,A,21,11,22
Now let us assume the blanks as A

Every A is half of second digit to the digit right for every even pair.

So we have 1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 7, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22

Continuing the same sequence we will have ................19, 10, 20, 3, 21, 11,22,6,23,12,24

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1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 7, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22, x
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The highlited digit is correct in series? I am not sure about that digit in series.

-Regards
RajdeepCE . . .

@Rajdeep: That digit is wrong. It should be 17 and NOT 7. Try your answer assuming 17. I hope you will get 6 too. 😀

The puzzle is simple.

ishutopreCheck out the sequence.

1, ,2, ,3, ,4, ,5, ,6, ,7, ,8, ,9, ,10, ,11, ,12, ,13, ,14, ,15, ,16, ,17, ,18, ,19, ,20, ,21, ,22, ,23, ,24
while the spaces to the left of even numbers is a number half of it.
No. to left of 2 is 1, left of 4 is 2, left of 6 is 3 etc....
So we have: 1,1,2,A,3,2,4, A,5,3,6,A ,7,4,8,A,9,5,10,A ,11,6,12,A,13,7,14, ,15,8,16,A ,17,9,18,A,19,10,20,A,21,11,22
Now let us assume the blanks as A

Every A is half of second digit to the digit right for every even pair.

So we have 1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 7, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22

Continuing the same sequence we will have ................19, 10, 20, 3, 21, 11,22,6,23,12,24

I have doubt about your method man. Here is why from what I understand of your methods.
The main problem is in the second part of your method. You are saying that every A is half of second digit to the right for every even pair.
So from that can we say that from first to last here what I get the double part of A,
1st A -> 2, 2nd A -> 3, 3rd A -> 4, 4th A -> 5, 6th A -> 7, 7th A -> 8, 8th A -> 9, so on. SO basically you are telling to floor the digit when the number is odd for the A. Is it so? Than can you please check the values for all the A's according to your method, cause I am getting false values after few starting values.

Correct me if I am wrong. 😀

I might have created confusion.

1,1,2,A,3,2,4, A,5,3,6,A ,7,4,8,A,9,5,10,A ,11,6,12,A,13,7,14,A ,15,8,16,A,17,9,18,A,19,10,20,A,21,11,22

Starting from the left most A,
A is half of 2
A is half of 6,
A is half of 8
A is half of 10,
and so on.
Continuing the sequence as we have 24 the last number, the number to its left is 12, while number to left (leaving one place in between) Will be half of 12 i.e.; 6.
Check this-
1, ,2, ,3, ,4, ,5, ,6, ,7, ,8, ,9, ,10, ,11, ,12, ,13, ,14, ,15, ,16, ,17, ,18, ,19, ,20, ,21, ,22,6 ,23,12 ,24

Here is my method for solving the series. Fortunately the answer is same as CE-an ishutopre.

First write the series in simple way,

1, ,2, ,3, ,4, ,5, ,6, ,7, ,8, ,9, ,10, ,11, ,12, ,13, ,14, ,15, ,16, ,17, ,18, ,19, ,20, ,21, ,22, ,23, ,24,... ( 😉 wew its pain in my fingers to write such long)

Than you can do as isutopre said, half of every even digit to the right,

1, 1,2, ,3, 2,4, ,5, 3,6, ,7, 4,8, ,9, 5,10, ,11, 6,12, ,13, 7,14, ,15, 8,16, ,17, 9,18, ,19, 10,20, ,21, 11,22, ,23, 12,24, ..

now the main trick part is to find the 3rd missing series combined. Isn't? Well, according to my logic the 3rd sequence is hidden with the sequenc.
What? 😲
Let me make it simple, here is the sequnce till now.
1, 1 ,2,___,3, 2 ,4,___,5, 3 ,6,___,7, 4 ,8,___,9, 5 ,10,___,11, 6 ,12,___,13, 7 ,14,___,15, 8 ,16,___,17, 9 ,18,___,19, 10 ,20,___,21, 11 ,22,___,23, 12 ,24,...

Now you have to just start filling all the blanks with the sequnce itself in correct order.

so let's start filling the blanks ( 😉 yeah...),
1,1,2,1 (cause it is the first number of sequnce),3,2,4,1,...

Same way we can manage to write the whole sequnce. And also don't forget to add the filled blanks in your sequnce, cause once it is filled than it is part of the sequnce itself.

1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 17, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22, 6, 12, 24, 2, 25,..

Tell me if still have any doubt cause it is hard to "dumb down" my explanation this much.

-Thanks

ishutopreI might have created confusion.

1,1,2,A,3,2,4, A,5,3,6,A ,7,4,8,A,9,5,10,A ,11,6,12,A,13,7,14,A ,15,8,16,A,17,9,18,A,19,10,20,A,21,11,22

Starting from the left most A,
A is half of 2
A is half of 6, ( If A is half of 2 than this A should be half of 3)
A is half of 8, (same way this A should be half of of 4)
A is half of 10,
and so on.
Continuing the sequence as we have 24 the last number, the number to its left is 12, while number to left (leaving one place in between) Will be half of 12 i.e.; 6.
Check this-
1, ,2, ,3, ,4, ,5, ,6, ,7, ,8, ,9, ,10, ,11, ,12, ,13, ,14, ,15, ,16, ,17, ,18, ,19, ,20, ,21, ,22,6 ,23,12 ,24

And even though I tried putting values of all A according to your method but no matter what I do, it does not matches with the original series at all. I think you see ur calcuation right through the "right end" but it is not working from "left end".