Acrux
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A boy counts on his fingers, backward and forwards across his right -hand as follows: thumb, 1st finger, 2nd finger, 3rd finger, little finger,3rd finger, 2nd finger, 1st finger, thumb, 1st finger, … and so on.If he starts counting at one, on his thumb, which finger will he be onwhen he reaches two thousand and thirteen

Can you solve this using Sn=n/2(2a+(n-1)d)
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TeeEm
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(Original post by Acrux)
A boy counts on his fingers, backward and forwards across his right -hand as follows: thumb, 1st finger, 2nd finger, 3rd finger, little finger,3rd finger, 2nd finger, 1st finger, thumb, 1st finger, … and so on.If he starts counting at one, on his thumb, which finger will he be onwhen he reaches two thousand and thirteen

Can you solve this using Sn=n/2(2a+(n-1)d)


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Acrux
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(Original post by TeeEm)
I meant that
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TeeEm
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(Original post by Acrux)
??
From your previous threads one would assume that your are a 100% candidate...

From your comment
using the summation formula for an arithmetic progression to solve this problem is as relevant as using Pythagoras' theorem to calculate the fuel consumption in a car.
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Jeremy Ong
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Is the answer 2nd finger?
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Acrux
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(Original post by TeeEm)
From your previous threads one would assume that your are a 100% candidate...

From your comment
using the summation formula for an arithmetic progression to solve this problem is as relevant as using Pythagoras' theorem to calculate the fuel consumption in a car.
What? being a candidate is irrelevant.

I am not sure how to work it out that is the only relevance mate. I guessed if that formula would work not sure how to do it.
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TeeEm
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(Original post by Acrux)
What? being a candidate is irrelevant.

I am not sure how to work it out that is the only relevance mate. I guessed if that formula would work not sure how to do it
Maybe you have a point ...

My apologies for my limitations, inadequacies and disabilities ...

I hope you solve this problem.
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Acrux
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(Original post by TeeEm)
Maybe you have a point ...

My apologies for my limitations, inadequacies and disabilities ...

I hope you solve this problem.
So you dont know how to solve this?
Any suggestions on how to try and work it out?
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TeeEm
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(Original post by Acrux)
So you dont know how to solve this?
Any suggestions on how to try and work it out?
look for a simple pattern
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Aspinboy
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If you extrapolate the pattern a bit while focusing on a certain area, I think you can establish enough to get you to the solution.

For example, if you look at the numbers in the count where you are hitting the 'first finger', you get the following pattern:

2,8,10,16,18,24....

You can see that every second number here represents the eight-times-table.

So:

2,8,10,16,18,24....
(8x_) 1 2 3.....

Each odd multiple of eight corresponds to the sweep FROM little-finger TO thumb, and each even multiple of eight corresponds FROM thumb, TO little finger.

2013/8 = 251.625

251 * 8 = 2008

So the number 2008 hits the first finger, and as it is an ODD multiple of eight it hits the first finger whilst moving in the direction from little finger to thumb. Therefore,

2008: 1st finger
2009: Thumb
2010: 1st finger
2011: 2nd finger
2012: 3rd finger
2013: little finger.

I'm sure there are simpler or more elegant formulaic ways to establish this, but unless I'm mistaken this does the job.
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TenOfThem
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(Original post by Aspinboy)
If you extrapolate the pattern a bit while focusing on a certain area, I think you can establish enough to get you to the solution.

For example, if you look at the numbers in the count where you are hitting the 'first finger', you get the following pattern:

2,8,10,16,18,24....


I'm sure there are simpler or more elegant formulaic ways to establish this, but unless I'm mistaken this does the job.
I think it is easier to look at the numbers that land on the little finger (mainly because this only happens once per run)

They form the pattern 8n-3 and 2013 is a term in this pattern
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Acrux
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(Original post by TenOfThem)
I think it is easier to look at the numbers that land on the little finger (mainly because this only happens once per run)

They form the pattern 8n-3 and 2013 is a term in this pattern
Is there a formula that can be used for this question?
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Aspinboy
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(Original post by TenOfThem)
I think it is easier to look at the numbers that land on the little finger (mainly because this only happens once per run)

They form the pattern 8n-3 and 2013 is a term in this pattern
Sure, the key to it is to focus on any one finger and scrutinise the sequence formed around it.

The thing that might put people off a problem like this trying to establish a sort of grand unifying theory of the problem as a whole.

If you sort of 'hone in' on a smaller segment then you can usually find the solution to the problem (or at least a route to the solution) pretty quickly.
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TenOfThem
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(Original post by Aspinboy)
Sure, the key to it is to focus on any one finger and scrutinise the sequence formed around it.

The thing that might put people off a problem like this trying to establish a sort of grand unifying theory of the problem as a whole.

If you sort of 'hone in' on a smaller segment then you can usually find the solution to the problem (or at least a route to the solution) pretty quickly.
I am just saying that a focus on the thumb or little finger makes more sense as they only appear once each run, the others have two patterns associated with them


I am not saying you are wrong, I am agreeing with you that there is a more elegant method
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