Monday Night Bowling
CJ and his team(The Head Pins)played two bowling games against another team(The Misfits).For the first game,all the players sat at three tables with the same number of people at each table.At the end of the first game,it was found that The Misfits were cheating and an argument broke out.During the second game(with the two teams mad at each other)The Head Pins sat at one table while The Misfits sat at another.During both games,everyone bought a round of drinks for the table they were at.But during the second game,38 more drinks were bought(in the same manner)than the first game.
What is the number of people on each team?
How many drinks were bought during each game?
(there are more than 1 person on each team)
What is the number of people on each team?
How many drinks were bought during each game?
(there are more than 1 person on each team)
posted by Knightmare at
8:08 PM
25 Comments:
When the argument erupted, did anyone STRIKE another person? ... or did they SPARE them the embarrassment?
mmmmm ... I'd like apologize for the first comment (groan) :-)
LOL...cheeseball king
hey ... I apologized :)
Knightmare ... I don't have a clue
9 people
10 people
1st game - 19 drinks
2nd game - 38 drinks
no Zaux...the difference in drinks in game 1 and game 2 is 38
this maybe a hint but the answer is kinda long and has some (x-y)^2+xy and such
That's a relief. I've rejected 15.09... as an answer.
7 and 8 members on each team.
very good Chris
7 and 8 is right
now,how many drinks total per game?
and to win full prize,you need need to tell us how you got the answers...you know...all that math stuff
btw...Chris...how would you rate the difficulty of this question-from a scale of 1 to 114.
i would like Cam(and anyone else) to state this as well
just so i have an idea of what it takes to stump you
Let m,n be the team sizes. If p peple buy each other a round,
tht'll be p² drinks altogether. We are given that p = (m+n)/3
for the first game.
Need to find m² + n² - 3(p/3)² = 38 => m² + n² - mn = 57
This may be written as:
(m-n)² + mn = 57 = 3*19 or (m+n)² - 3mn = 57 = 3*19
Sadly, at this point, I can't think of a sweet argument.
So, I'll post the followinng sad apology of a solution.
Try m = n to enable getting an estimate.
Then (2n)² -3n² = n² = 57 => n, m ≈ 7.55.
Because m + n is divisible by 3, m and n must both be divisible
by 3 or the have a remainders 1 and 2, after divison by 3.
So try 7 and 8, and voila! (7+8)² - 3*7*8 = 57.
But that's cheating. I hope I come back with something better,
possibly involving mod 19 arithmetic.
So I'd better rate the problem as hard, even though the numbers
are easy enough to stumble upon.
This post has been removed by the author.
I'm a twerp. m and n cannot both be divisible by 3 as 57 only has one factor of 3. So one of m and n must have remainders 1 and 2 or 2 and 1 after dividing by 3. But that still doesn't lead to anything really slick. But I can at least find all solutions almost methodically.
As m+n is divisible by 3, let m+n = 3p, where p = 1,2,3,...
Substitute into (m+n)² -3mn = 57 = 3*19 => (3p)² - 3mn = 3*19
=> 3p² - mn = 19 => mn = 3p² - 19
For each p and mn product, need m and n to be factors of mn,
and m+n = 3p.
p = 1 or 2, => m or n < 0 so no good.
p = 3 => m+n = 9, mn = 8. Only m,n = 1,8 does it.
p = 4 => m+n = 12. mn = 29. Only choose m,n = 1,29 so no go.
p = 5 => m+n = 15, mn = 56. Only m,n = 7,8 fits.
p = 6 => m+n = 18, mn = 89. This is insoluble in the integers.
p = 7 and up are also insoluble in the integers.
So there are (only) two solutions. m,n = 1,8 or m,n = 7,8
Not very nice, but not really hard. So despite struggling (which can be easy to do with Diophantine problems), I can only give it 20% on a difficulty level.
I've just noticed that I hadn't answered the question fully.
You have disallowed 1 and 8, but despite that, that would correspond to 27 (=3*3²) drinks in the first game and 65 (=1²+8²) in the second game. Check, 65-27 = 38.
For the 7,8 solution have 75 (=3*5²) drinks in the first game and 113 (7²+8²) in the second game. Check, 113-75 = 38.
I only usually buy 3 cokes in all when I take my kids.
Consider m+n = 18, mn = 89. m = 89/n => 89/n + n = 18
=> n² - 18n + 89 = 0 => n = 9 +/-i 2√2.
So there isn't even a solution in the reals.
I did a quick search based on a approximation that the teams would be closed to balanced.
Then for X divisible by 3 i.e. 3,6,9,12,15,18,21.....
I plugged them into 2*(X/2)^2-3*(X/3)^2
X=12 gives 24
X=15 gives 37.5
X=18 yields 54
37.5 is really close, so try different sums of 15=x+y where x and y are integers
1st try 7,8 gives answer of 38.
7,8 people on each team
75 drinks first game
103 drinks second game
Cam
Hardness level scale 1 to 10:
1 is someone asking my name
2 is adding or subtracting
3 is multiplication and division
4 is any of the below in combination
5 is simple algebra
6 is more complex algebra, or basic logic problems
7 is calculus, OR complex algebra + logic combined
8 is a genuine head scratcher. New methods must be invented, or old methods applied in an innovative way.
9 immediately dismiss the problem as impossible. Start to dig through text books. Make a small break through after a few days, and dismiss problem as unsolvable after a week. After a few iterations of the above and a dozen pots of coffee finally solve.
10 Like 9, but solved after at least 1 month. Publish thesis on solution afterwords.
11 After going to school for PhD in math, solve one of the 6 unsolved Millennium Prize Problems set by the Clay Mathematics Institute:
-P versus NP
-The Hodge conjecture
-The Riemann hypothesis
-Yang–Mills existence and mass gap
-Navier-Stokes existence and smoothness
-The Birch and Swinnerton-Dyer conjecture.
Collect $1 million and Fields Medal.
Note: not a linear scale.
e.g. This problem sits on ~6
Cam
LOL. That was an excellent post Cam.
You can't get a Fields Medal once you're hit 40. So that's why I haven't bothered to solve all of them ;)
that was good...thanx
Knightmare, I'll give your problem 5.7 on the Cam scale. The Cam scale looks to be roughly logarithmic.
I think I'll keeep a copy. It's a prety good scale. As a guide, I won't tackle problems with a rating higher than a 7.5 unless they are really nice problems. It would have to be a very beautiful problem to make me go to an 8.7 (I'd mull over a 9 or more, but would probably fail to solve).
thanks for the 5.7 Chris
on my scale,basic logic is way easier than algebra
A problem with the scale is that it's objective, it really is classifying the problem, rather than saying how hard I found it.
I'd call a Cam scale 5, a Chris 1 and a Cam scale 9 a Chris 10. So the curent problem would be about 2.
Chris!...stop changing your mind!
i guess it's all about what your good at :]
Tee hee. I'll stick with my 8:22 PM post.
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