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 Post subject: City of London questionPosted: Thu Jan 22, 2009 2:12 pm

Joined: Sun Nov 30, 2008 6:38 pm
Posts: 19
This is from a 10+ paper. Can anyone help me explain to a 9 year old without the formula 2^n -1. No calculator is allowed. Is there a quick way to get to the 50th student?

The number 1 is written on a piece of paper and then passed round the
room to each of 50 students in turn. Each student must cross out the
number seen and replace it with its double plus 1.

b. The last student does this then announces the final number.
What should the final number be?

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 Post subject: Posted: Thu Jan 22, 2009 2:32 pm

Joined: Fri Nov 17, 2006 8:54 pm
Posts: 1769
Location: caversham
That looks tough and working it out by hand very time consuming.

Is it a multiple-choice question, so you can guess the best answer by getting the right order of magnitude?

steve

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 Post subject: Posted: Thu Jan 22, 2009 2:41 pm

Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1287
Location: Birmingham
seems a silly question - the only way you could express the answer is as (2^51)-1 or Hex 4 0000 0000 0000 - 1 ie 3 FFFF FFFF FFFF

Not sure what this is trying to prove given that the answer is:-

1,125,899,906,842,623

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 Post subject: Posted: Thu Jan 22, 2009 2:50 pm

Joined: Sun Nov 30, 2008 6:38 pm
Posts: 19
No multiple choice, youngest just got cross working it out (she gave up after the 10th person), the paper is only 45 minutes long and this was question 19 out of 19.

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 Post subject: Posted: Thu Jan 22, 2009 2:53 pm

Joined: Fri Nov 17, 2006 8:54 pm
Posts: 1769
Location: caversham
It might be a "time filler" question at the end of the exam.

They do not expect you to complete it but look at how far you got and what method you used.

Does seem a bit cruel.

steve

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 Post subject: Posted: Thu Jan 22, 2009 3:02 pm

Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1287
Location: Birmingham
Perhas this is a scholarship question - I assume this is an independent number.

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 Post subject: Posted: Thu Jan 22, 2009 11:04 pm

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1192
This threw me to start with... the only easy technique is to change it to binary!

Then each pass of the card results in a left shift (a double in base 2) and a 1 being added on the right

i.e.

initial 1
next 11
next 111
next 1111

which as said above leads to.... 2^n-1

Tricky question... assume a filler to see what people write.

Regards
SVE

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