Announcement

Collapse
No announcement yet.

Puzzles with numbers and things

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • davis_greatest
    replied
    Originally Posted by Monique
    Here is my reasoning ...
    Gordon had 1 apple when entering the shop. So he gained 2499 apples.
    2499 = 3*7*7*17

    If you have an equilateral triangle you have basically 3 directions.
    So if your triangle has n rows and you chose m balls
    you will gain 3*m*(n-m+1) bananas

    this is maximised when m=n-m+1 or 2m=n+1
    So the only possibility is
    m=7 and n=13.
    Gordon got 3*7*7=137 bananas.
    And Charlie gave him 17 apples for each banana.
    Congratulations Monique (apart from the typo there... it's 147 bananas, not 137 )

    The maximum number of bananas is 3m² = 147 and there are 13 rows.

    Leave a comment:


  • Monique
    replied
    R257

    Here is my reasoning ...
    Gordon had 1 apple when entering the shop. So he gained 2499 apples.
    2499 = 3*7*7*17

    If you have an equilateral triangle you have basically 3 directions.
    So if your triangle has n rows and you chose m balls
    you will gain 3*m*(n-m+1) bananas

    this is maximised when m=n-m+1 or 2m=n+1
    So the only possibility is
    m=7 and n=13.
    Gordon got 3*7*7=137 bananas.
    And Charlie gave him 17 apples for each banana.

    Leave a comment:


  • davis_greatest
    replied
    Round 257 update...

    A few attempts and re-attempts have come in!

    Congratulations to abextra and snookersfun for the first correct answers!

    Next person to answer, please do so on the thread.

    Leave a comment:


  • abextra
    replied
    Originally Posted by snookersfun

    R.254: either this or that...

    same as usual, just two clue options for each line, so you have to choose the correct clue. Have fun
    Here we go (sorry for delay ).
    Attached Files

    Leave a comment:


  • davis_greatest
    replied
    and also a quick one...

    Round 257... The Great Banana Giveaway

    Barry the Baboon is doing a special Banana Giveaway Promotion in his shop. He lays out in the entrance a big equilateral triangle of red snooker balls, touching as they would in a pack in snooker, with an odd number of rows.

    Gordon comes into the shop, holding an apple, ready to win some bananas. The deal is like this - for every line of touching red balls, going in any direction, Gordon can win some bananas.... and he can choose (and choose once only) how many bananas he wins for each line! The catch is this: every line must have at least as many balls as the number of bananas that Gordon chooses to win for each line.

    For example, if Gordon chooses to win 2 balls per line, then the only lines that count are those with 2 or more balls. If he chooses to win 100 balls per line, then the only lines that count are those with 100 or more balls. Each line only counts once (e.g. a line of 3 balls cannot be counted as 2 lines of 2 balls).

    Gordon, being a canny little gorilla, chooses wisely of course, and gets the maximum possible number of bananas. Just as Gordon is about to leave with his bananas, Charlie comes in and offers to swap him a number of apples for each banana. Gordon likes apples so does the swap... and leaves with 2,500 apples!

    How many bananas did Charlie receive in exchange for his apples?
    And how many rows of balls did Barry lay out?


    PS No fruit were eaten inside Barry's shop.

    Leave a comment:


  • snookersfun
    replied
    Originally Posted by dantuck_7
    Were you once a submarine commander?

    Looks like this could be tricky.
    1) nah, just discovered a great website and also subs are too deep in the water for personal comfort. Saw 'the Boat' though, great movie!

    2) yes, it is a bit tricky and time consuming, but do try

    Leave a comment:


  • dantuck_7
    replied
    Originally Posted by snookersfun
    Round 256:
    This time the regular fleet has to be arranged into 2 grids, one representing the initial situation and the second the position of ships after each moved 3 units either forward or backward
    Were you once a submarine commander?

    Looks like this could be tricky.

    Leave a comment:


  • snookersfun
    replied
    Round 256:

    battleships (what else?): before and after

    this time the regular fleet has to be arranged into 2 grids, one representing the initial situation and the second the position of ships after each moved 3 units either forward or backward. (Meaning bigger ships can only move in 2 possible directions, subs in 4). During moving (imagine the intermittent stepwise moves of 1 and 2 units) ships may touch but never overlap.



    answers to me please initially

    Leave a comment:


  • snookersfun
    replied
    Originally Posted by davis_greatest

    Round 253: (snookersfun no doubt has one but I think that I only saw a shorthand version which didn't state all the pockets; ...
    ok, to the 'long hand'ish version:
    all unbracketed balls into their own holes:
    11 (9 in 3or6) 10 (8 in 6or3) 11 (9 in 3) 10 (8 in 6) 11 (9 in 3) 10 8 7 9 11
    (2 in 5) (3 in 6) (4 in 2) (5 in 4) 6 7 8 (9 in 5) (10 in 6) 11

    and can somebody please put up the solution to the last battleship?

    Leave a comment:


  • Monique
    replied
    Originally Posted by davis_greatest
    [B]and Monique's, I believe, needed a tweak to avoid having two consecutive pots near the end along the same edge of the table?).
    Indeed I did ... only a 280 break for me. Congrats to abextra and snookersfun!
    And thanks to D_G ... for all the checking; also quite a lot of work but then he looked for it himself ...

    Leave a comment:


  • davis_greatest
    replied
    Rounds 248 to 253 closed... at last

    Finally, I have checked the answers received to rounds 248 to 253! The results are as follows. If I have missed anyone's answer (quite possible), please protest here!

    Round 248: Correctly answered by abextra, Monique and snookersfun

    Round 249: Correctly answered by abextra, Monique and snookersfun

    Round 250: Correctly answered by abextra, Monique and snookersfun

    Round 251: the highest possible break (323) was found by abextra, Monique snookersfun and dantuck_7

    Round 252: Correctly answered by abextra, Monique and snookersfun (dantuck_7, did you get this one too?)

    Round 253: the highest possible break (281) was claimed by abextra, Monique and snookersfun. abextra gave a complete, valid solution (snookersfun no doubt has one but I think that I only saw a shorthand version which didn't state all the pockets; and Monique's, I believe, needed a tweak to avoid having two consecutive pots near the end along the same edge of the table?).

    Congratulations to all! Some of them needed quite a bit of work

    PS Answer to round 252 was 9/20. The return to the hotel and back to where he is adds 18 minutes to the journey, so 9 minutes each way, or 9/20 of the 20-minute journey from hotel to venue.

    Round 252 - late in China

    Damon Grott is rushing to his first round match against Maun Shirty in China, knowing that it will take him 20 minutes from his hotel. On his way, he remembers that he has forgotten his chalk. He knows that his opponent won't lend him one - but he cannot possibly play the whole match with no chalk!

    He calculates that if he were to continue to the venue, he would be 8 minutes early. However, if he goes back to his hotel for his chalk, he will be 10 minutes late and be docked a frame.

    What fraction of the way to the venue is he at that moment?

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by snookersfun
    ...and making up for the extremely difficult missing addition:
    [ATTACH]1001[/ATTACH]
    attention not drawn to scale
    Very pretty!!

    (You could probably get the initial pack of reds closer to the pink though )

    Leave a comment:


  • snookersfun
    replied
    ...and making up for the extremely difficult missing addition:
    francoiscue1%20064.jpg
    attention (slightly less) not drawn to scale
    Attached Files

    Leave a comment:


  • Monique
    replied
    The design can be achieved moving 11 balls.

    creating a new row at the bottom of the existing triangle.
    yellow at the left of the new first row
    red moved from the left of the now second row to the rigth of yellow, then green in its place
    red moved from the left of the now third row to the rigth of red next to yellow, then brown in its place
    ... etc
    Black in edge red place

    So in the end
    yellow left to first row (5 reds)
    green left to second
    brown left to third
    blue left to fourth
    pink left to fifth
    black as a sixth

    All rows worth 7.

    Leave a comment:


  • davis_greatest
    replied
    Originally Posted by davis_greatest
    OK, a very easy one – with no probability!

    Round 255 (=1+2+4+8+16+32+64+128) – Snooker Triangle

    Set up the snooker balls as usual for the start of a frame (ignore the cue ball – leave it in your pocket / handbag). Now, move as few balls as possible in order to rearrange the balls into one equilateral triangle, with every horizontal row* worth the same number of points. The triangle must have its apex at the same place as it would usually be at the start of a frame, i.e. as close as possible to (without touching) a hypothetical ball on the pink spot.

    *a horizontal row means a row parallel to the top cushion

    How many balls do you move, and where?

    Answers by Private Message / pager please
    ... correctly answered by Monique, snookersfun and abextra. Can we have the answer on the thread please...

    Leave a comment:

Working...
X