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Old 18-05-2003, 16:12   #1
Melifluous
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Default Psychic or what?

Can someone tell me how this works?

http://mr-31238.mr.valuehost.co.uk/a...sh/psychic.swf

Thanks,
cos it's freaking me and my wife out...

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Old 18-05-2003, 16:24   #2
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Hehe, I just figured it!!!

it's a very clever trick, but it is a trick... I'll let you stew a little more before telling you
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Old 18-05-2003, 16:57   #3
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Okay, because I'm feeling generous (or is it because I've got no turns to play and am bored ) here's the first part of the solution. (This is the maths part, brace yourselves.)

Let's take a set of numbers and examine the answers we get:

52 => 52 - 07 = 45
51 => 51 - 06 = 45
50 => 50 - 05 = 45
49 => 49 - 13 = 36
48 => 48 - 12 = 36
47 => 47 - 11 = 36
46 => 46 - 10 = 36
45 => 45 - 09 = 36
44 => 44 - 08 = 36

Once you realise there are a quite limited set of finite answers (namely the 9x table) the next part of the trick should become obvious...

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Old 18-05-2003, 17:24   #4
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This one has been posted on CFC before. It was very simple but I cant remember anymore how it worked... (I did figured it out on my own though quite easy!)
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Old 18-05-2003, 19:44   #5
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hehehehehehe - look more closely Meli.
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Old 18-05-2003, 23:43   #6
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thank god i worked it out, i can go to bed now (teach me to look on here at bedtime)

ok then as i was nearly deprived of sleep; try this:

you have eight identical objects, but one weighs more. You have a pair of scales, but you are only allowed to use them twice. How do you find out the odd one?

This is not a trick, it can be done.
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Old 19-05-2003, 09:26   #7
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nicely done, if you still haven't figured this out, just 'think' of the same number a few times.
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Old 19-05-2003, 09:31   #8
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Quote:
quote:Originally posted by prettyvacant

thank god i worked it out, i can go to bed now (teach me to look on here at bedtime)

ok then as i was nearly deprived of sleep; try this:

you have eight identical objects, but one weighs more. You have a pair of scales, but you are only allowed to use them twice. How do you find out the odd one?

This is not a trick, it can be done.
Weigh three against three.
if they balance then the odd one is in other two. Balance those to see which is heavier.
if they dont then you can tell which side is heavier then balance one against one . If one side goes down, thats the heavy one, if they balance the third one is the heavy one.

Actually you can find it from 9 if you apply the same principle. 3 v 3. Then 1 v 1.
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Old 19-05-2003, 13:05   #9
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Quote:
quote:Originally posted by col

Actually you can find it from 9 if you apply the same principle. 3 v 3. Then 1 v 1.
Yes but i think its 8 in the question to make it less obvious - youre sposed to get stuck at 4 on each side!

If anyone likes this kinda thing i can recommend a prog on immediately after Univ Challenge, Monday evening, either on BBC 3 or 4 cant remember. Oops that 9 pm. Dont be put off by most of the panellists looking like they've just walked in from presenting OU progs.
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Old 19-05-2003, 17:52   #10
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So extending this principle

3 1 weighing 1 v 1
9 2 " 3 v 3 then 1 v 1
27 3 " 9 v 9 then 3 v 3 then 1 v 1
81 4 " etc

3^N N

Col - who can never see a dead horse without flogging it.

You figured the mind reading yet , Meli?
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