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Manager
Joined: 09 Jun 2009
Posts: 226
Followers: 2

Kudos [?]: 248 [0], given: 6

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29 Oct 2009, 13:11
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If we count in the following way on the fingers of our left hand such that we call the thumb 1, index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. Count upto 1994 then we'll end on

a) thumb
b) index finger
c) middle finger
d) ring finger

i calculated it and IMO it should end at the index finger.
SVP
Joined: 30 Apr 2008
Posts: 1887
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 570 [0], given: 32

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29 Oct 2009, 14:13
I agree with you, it should be index finger. Thumb = 1 and then when you come back again, thumb = 9. That's a difference of 8. So if you divide 1994 by 8, it leaves a remainder of .25 (or .25 of 8 = 2). so essentially, to determine if a number is on the thumb, subtract 1 and divide by 8. If it is an integer, then the answer is yes, that number is on the thumb. In order to determine if it is on the index finger, subtract 2 and divide by 8. here, $$\frac{1994-2}{8} = 249$$, an integer, so you know it's on the index finger.

papillon86 wrote:
If we count in the following way on the fingers of our left hand such that we call the thumb 1, index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finger for 11 and so on. Count upto 1994 then we'll end on

a) thumb
b) index finger
c) middle finger
d) ring finger

i calculated it and IMO it should end at the index finger.

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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