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Click to See Complete Forum and Search --> : The most fun - and most simple - Maths game in the world


Mack
Mar 7th, 2003, 11:53 AM
Okay, perhaps slightly false advertising, but here goes.

Person A picks a number between 1 and 1,000,000
Person B must ascertain what the number is by asking a maximum of twenty yes/no questions.

Very simple, but surprisingly addictive. But the thread does not stop there.

What I want to know is something much more complicated; is it possible to win every time and if so, how would would one go about it? What's the minimum number of guesses without any "luck" involved?

Thanks for your time.
Mack

opus
Mar 7th, 2003, 01:47 PM
If you can ask twenty times , you better win all the time!
Since 2^20>1 000 000

sql_lall
Mar 8th, 2003, 03:10 AM
To win, just do this:

Q1) Is the right-most digit of it's binary form 1??
Q2) Is the 2nd right digit of it's binary form 1??

etc..

then after 20 Qs, you should have it's binary form, convert it to decimal and Tada!

da_silvy
Mar 9th, 2003, 02:11 AM
And if the person you're playing with isn't a nerd :p?

jokes

DiGiTaIErRoR
Mar 9th, 2003, 04:28 AM
1000000/2^20

Is the number > 500,000? no

is the number > 250,000? no

is the number > 125,000? no

etc.

sql_lall
Mar 9th, 2003, 10:13 PM
unfortunately, this can get tricky when the number isn't divisible by two.

In that case, do what i said, but in reverse:
Q1) Is it >= 524288 (=2^19)
Q2) Is it >= 262144 (=2_18)
OR: Is it >= 786432 (=2^19 + 2^18)

etc

That way you can just keep on dividing by two, making it nice and simple :D

opus
Mar 10th, 2003, 12:39 AM
Looks like the starter of this threat left us!

da_silvy
Mar 10th, 2003, 03:45 AM
Originally posted by sql_lall
unfortunately, this can get tricky when the number isn't divisible by two.

In that case, do what i said, but in reverse:
Q1) Is it >= 524288 (=2^19)
Q2) Is it >= 262144 (=2_18)
OR: Is it >= 786432 (=2^19 + 2^18)

etc

That way you can just keep on dividing by two, making it nice and simple :D

You'd just split the range

e.g.

Q1. is it > 500,000

if yes, is it > 250,00 then range = 250,000 --> 500,000

then you'd ask if it was > 375,000 etc...

Another method a friend suggested (which I haven't tested) is to ask if each digit is prime and determine how many digits there are in the number.