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Final Scores at the PrimeBowl (Posted on 2007-11-28) Difficulty: 3 of 5
1. What is the highest score you can obtain in a standard game of bowling if your cumulative score in each of the ten frames is required to be a prime number?

2. What is the lowest score you can obtain in a standard game of bowling if your cumulative score in each of the ten frames is required to be a prime number and for each throw you knock down a prime number of pins?

The next two questions refer to 3-6-9 bowling. The difference in 3-6-9 bowling and the standard game is the 3rd, 6th and 9th frame already have strikes recorded.

3. What is the highest score you can obtain in a game of 3-6-9 bowling if your cumulative score in each of the ten frames is required to be a prime number?

4. What is the lowest score you can obtain if your cumulative score in each of the ten frames is required to be a prime number and for each throw (excluding the pre-recorded throws) you knock down a prime number of pins?

The first score sheet represents Games #1 and #2.
The second is for use with Games #3 and #4.

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

X

X

X

One Scoring reference is here should you need one.

See The Solution Submitted by Dej Mar    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): and just part 3 (and part 4). -- same scores, different ways, other alternates Comment 10 of 10 |
(In reply to re(2): and just part 3 (and part 4). -- same scores, different ways, other alternates by Dej Mar)

I have found the bug.

For the record:

        Frame(Fr - 2) = Frame(Fr - 2) + n
        Frame(Fr - 1) = Frame(Fr - 1) + n + n2
        Frame(Fr) = Frame(Fr - 1) + n + n2

adds the two throws, n and n2 into each of the two strike frames that preceded it, but neglects to add into the immediately preceding frame (Fr-1) the extra amount that was added to the strike before that (Fr-2), which would, if correct, have added to the base on which Fr - 1 was built.  It should have been:

        Frame(Fr - 2) = Frame(Fr - 2) + n
        Frame(Fr - 1) = Frame(Fr - 1) + 2*n + n2
        Frame(Fr) = Frame(Fr - 1) + n + n2

to account in frame Fr - 1 for the increase to frame Fr - 2.


  Posted by Charlie on 2007-11-29 14:19:52
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