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Ecoris
Promising
Supreme Hero
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posted August 26, 2006 10:06 PM |
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Edited by Ecoris at 22:11, 26 Aug 2006.
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Quote:
What is the minimum number of strikes that you must roll in order to achieve a score of 250?
I'll do it mathemtically instead of just giving the answer. I will introduce the following variables:
X = number of frames that result in a strike
~X number of frames that do not result in a strike.
Thus by definition we have
X + ~X = 10 <=> ~X = 10 - X
We know that a strike will score at most 30 points (when it's followed by two more strikes), and any non-strike will at most score 19 points; a spare followed by a "9". This implies that:
Total score (S) <= 30X + 19~X
If S >= 250 we have
30X + 19~X >= 250
substituting ~X with 10 - X and rearranging the inequality we get
30X + 19(10 - X) >= 250
11X >= 60
X >= 5,45..
And since X must be an integer we have to roll at least 6 strikes to score a total of 250 or more.
It can be done with only 6 strikes:
1: 9,1
2: 9,1
3: 9,1
4: 9,1
5: Strike
6: Strike
7: Strike
8: Strike
9: Strike
10: Strike
(11: 9,9)
The total score will be 3*(10+9) + 10+10 + 5*(10+10+10) + 10+9+9+ = 255
Quote:
What is the maximum number of strikes that you may roll without going higher than 250?
This one is easy:
1: Strike
2: Strike
3: 0,0
4: Strike
5: Strike
6: Strike
7: Strike
8: Strike
9: Strike
10: Strike
(11: 10,10)
Then S = 20 + 10 + 0 + 7*30 = 240 and we would have rolled 9 strikes (or 11 if the 10s from "frame 11" are counted as strikes).
If we rolled a strike in every frame then S = 8*30 + 10+10+? + 10+?+? which would be greater than 250 even though the two rolls of frame 11 were zeros.
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted August 26, 2006 10:53 PM |
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Good job 
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friendofgunnar
Honorable
Legendary Hero
able to speed up time
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posted August 29, 2006 07:28 AM |
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From Make Magazine:
Calendar Cubes:
"Each morning I come into work and arrange two normal cubes on my desk to display the current day of the month. For example, on the 1st of the month, the cubes have a 0 and a 1 on them. On the 16th day of the month, one cube has a 1 and the other a 6. On the 31st day of the month, a 3 and a 1 appear on the front of the cubes. Each cube has six sides and one number painted on each side. What are the numbers painted on each cube?"
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this_other_guy
Famous Hero
{0_o} heh...
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posted August 29, 2006 08:07 AM |
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cube 1: 1 2 3 4 5 0
cube 2: 1 2 6 7 8 0
I guess you can flip the 6 on cube 2 to make it a 9, so 09, 19 and 29 can be made.
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1f u c4n r34d th1s u r34lly n33d t0 g37 l41d
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Einhorn95
Tavern Dweller
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posted November 06, 2019 09:59 AM |
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Riddles
Hi Everyone,
Would it be alright if I add one or two riddles?
Both are rather related to mathematics.
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CountBezuhoff
Supreme Hero
Nihil sub sole novum
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posted November 06, 2019 02:00 PM |
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Please, go ahead.
The Count
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Einhorn95
Tavern Dweller
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posted November 08, 2019 06:16 PM |
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Riddles
Alright, thank you!
So the first one is: Connect all 9 dots using 3 reflector dishes and the absorber.
So you have a 5 by 5 grid. In it is a smaller 3 by 3 one.
Within the 3 by 3 grid is an Emitter, three reflectors and an Absorber. Originally there were 4 reflectors connecting all nine dots in an S.
However reflector C broke down. So now using the three remaining reflectors a new layout which connects the dots must be made. The Emitter cannot be moved but the remaining parts can be moved one cell to the left, right, up or down. All parts can rotate in 1/16 increments. Each cell of the grid has a set angle to which the dish can be set. Example: If you place a dish in cell 1 it's angle must be 22.5 degrees. If you place it in cell 2 it's angle will be 45 degrees. This continues to the last number in the grid.
So where must each movable part of the system have to be to connect all nine dots starting from the Emitter going through reflector A, then B then D and ending at the Absorber? Please note the answer has to be in coordinates.
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This thread is pages long: 
