crusty math questions

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Re: crusty math questions

MeDeFe wrote:
LYR wrote:
HardAttack wrote:nice

q.2. a chess tournament,
every played game means 1 player is eleminated and winner promotes.
in a 100 player chess tournament (assume no stale happens) how many chess games should be completed to call the winner/champion ?

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They wouldn't, however. You can get a free pass in the first round of a knock-out tournament, but after that, everyone plays.

Indeed.

Could you explain the method you used? I was wondering why you started out with 36 games.

LYR

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Re: crusty math questions

LYR wrote:
MeDeFe wrote:
LYR wrote:
HardAttack wrote:nice

q.2. a chess tournament,
every played game means 1 player is eleminated and winner promotes.
in a 100 player chess tournament (assume no stale happens) how many chess games should be completed to call the winner/champion ?

show

They wouldn't, however. You can get a free pass in the first round of a knock-out tournament, but after that, everyone plays.

Indeed.

Could you explain the method you used? I was wondering why you started out with 36 games.

To cut the number of participants down to 64 and avoid free passes in the subsequent rounds. As you said in your solution, in rounds 3 through 5 someone gets a free pass, those are the quarter-final, semi-final and final rounds. You just don't do free pases at that stage in any tournament. (BTW, looking at it again I realise that my alternate solution doesn't work, so disregard that one.) In the given setup you have to start with 36 games. In the first round you reduce the number of participants to the nearest number that can be written as 2^n where n is a natural number. After that everyone plays in every round.
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MeDeFe

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Re: crusty math questions

HardAttack wrote:Q.4.

Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers?

I have Googled the answer but I agree, it's way too hard for this forum! Someone used scripts and functions to do the calculations - I'd suggest you stick to logical/ lateral thinking puzzles if you want any response.
clangfield

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Re: crusty math questions

HardAttack wrote:Q.3.

2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72,
7+8=??

98 I think.

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Re: crusty math questions

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Re: crusty math questions

Two trains leave different stations 65km apart at the same time. One travels at 9m/s and the other at 17m/s.

A motorbike rider who travels at 37m/s starts with the slow train, rides along a road beside the track until he reaches the fast train, then turns around and goes back to the slow train, before turning back towards the fast train, and so on.

How far has the motorbike travelled when the two trains crash head on into each other?

(assume instant acceleration/direction changes for all three moving objects)

(If you end up with an infinite series of ever decreasing fractions you're doing it wrong)

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Re: crusty math questions

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HA, I think nobody is able to answer your question number 4. Could you give us some help?

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Re: crusty math questions

HardAttack wrote:Q.4.

Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers?

The numbers are 4 and 13. I don't know how this could have been solved with just logic. My solution was not completely brute force, but it required a lot of excel calculations. I don't think I can adequately explain how I did it - even justifying the answer was pretty difficult.

First, I generated a list of all possible starting number combinations (there are 2304), and calculated the product and sum for each possible pair.

In another column, I added a count for the number of times each product appears in the list, and eliminated all S values for which there was a P value with only one possible set of factors.

From that list, I filtered out all combinations where a particular P value appeared more than once, and then filtered out all combinations where a particular S value appeared more than once.

The only remaining combination was S=17 and P=52, so the numbers are 4 and 13.

To confirm that this works:
4+13 = 17, which is given to S.
4*13 = 52, which is given to P.

1. There are two ways for P to have been given 52 (4*13, or 2*26), but he doesn’t know which, so he tells S that he cannot determine the numbers.

2. There are seven ways for S to have been given 17 (2+15, 3+14, …, 8+9). None of these pairs consist of two prime numbers, so S knows that P could not determine the numbers, and he tells this to P. (If P was given the product of two primes, then it would have been easy for him to factor it to determine the numbers.)

3. P knows that S was given either 17 or 28. If S was given 28, then some of the possible pairs (5+23, 11+17) consist of two primes, so S would not have known that P couldn’t determine the numbers. Therefore, S has 17, and P now knows that the numbers are 4 and 13, and tells S that.

4. S knows that P must have been given 30, 42, 52, 60, 66, 70 or 72. And since P was able to determine the numbers after eliminating one possibility, his number must have had only two possible sets of factors. Only 52 meets the criteria, so S now knows that the numbers are 4 and 13.
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