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Lots of questions on different subjects where you have to try to answer the latest question and if you are right then its your turn to ask the next question.

Moderators: playgirl, JUS, rodbox

#105141 by wildrover
Mon Feb 08, 2010 7:52 am
Yep - correct.

#105293 by tonto
Wed Feb 17, 2010 11:27 am
If 4928 is to 74,

and 3556 is to 58,

what is 2142 to?

#105304 by wildrover
Wed Feb 17, 2010 11:58 pm
36 - divide each pair by 7

#105306 by tonto
Thu Feb 18, 2010 9:09 am
Very Good. :D I could have looked at it for a year and not got the answer. :?

#105318 by pulse876
Fri Feb 19, 2010 12:57 am
What is the next number in the following sequence:

2, 3, 7, 43, 13, 53

And if you are feeling very brave what is the number after that?

#105357 by pulse876
Thu Feb 25, 2010 10:39 am
Hint - to determine any number in the series start by multiplying all the previous terms together. And what is a common feature of all the numbes in the sequence.

#105358 by wildrover
Thu Feb 25, 2010 11:01 am
They're all prime which struck me immediately and I can follow the sequence to 43 but I can't then see how to get to 13....I end up with 1807.

#105365 by pulse876
Fri Feb 26, 2010 12:57 am
You're right about the multiplication and then add one gets to 1807. But is 1807 prime? And if not....

#105501 by pulse876
Fri Mar 12, 2010 2:40 pm
This seems to have you stumped you all so the answer is:

Each term is the lowest prime factor of one more than the product of the previous terms.

[So 2*3*7*43+1 = 1807 = 13*139 (both primes) and hence 13 is the fifth term in the sequence.]

So the seventh term in the sequence is obtained by considering 2*3*7*43*13*53+1 = 1244335 = 5*248867 (both primes). Hence the seventh term is 5.

And the next term in the sequence is obtained by considering 2*3*7*43*13*53*5+1 = 6221671 which is prime and is hence the eighth term.

It is conjectured that if you keep repeating this process then eventually all the prime numbers would feature in the sequence (if you can have 'all' of an infinite number of primes). But of course the numbers involved in the multiplications would be rather large.

#107389 by wildrover
Fri Jun 18, 2010 4:28 pm
Topical one - in the World Cup each group contains four teams who play each other once with 3 points for a win and one for a draw. What's the maximum number of points a team can get and not qualify; and what's the minimum a team can get and still qualify for the next round. (Top2 qualify)

#107398 by pulse876
Sat Jun 19, 2010 12:23 am
The highest possible total number of points in the group is 18 (none of the six matches end in draws). So three teams could each end up with six points and one of them would be eliminated. So the most points you can get and fail to qualify is six.

If one team wins all three matches while the matches between the other three teams all end in draws then the group winner has nine points and each of the other three teams has two points. And one of those teams would qualify. But you can't qualify with one point so two is the minimum.

#107402 by wildrover
Sat Jun 19, 2010 9:45 am
Correct - and an extra mark for showing all your working out Pulse. :D

#108834 by wildrover
Fri Aug 20, 2010 10:51 am
You're down to the last 5 boxes on Deal or No Deal and you still have the £100K and £250K boxes. If you choose to go on, what are your chances of having either one or both of those boxes left when you get down to the last two boxes?

#108838 by JUS
Fri Aug 20, 2010 1:56 pm
12/25


1/5 +1/5 + (2!x(1/5x1/5)) ?????????

#108842 by wildrover
Fri Aug 20, 2010 6:52 pm
Nope - a good starting point is to work out how many different ways there are of picking 2 boxes from 5...
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