Puzzles with numbers and things

[Monique]

Well so ... listening to the experts on this forum Gordon can have 145 bananas if he fancies Snookersfun's design, or 163 if he prefers D_G proposal
(both solutions are valid well done!)

Anybody with more suggestions ... for more bananas?

[davis_greatest]

Nice question Monique. I'll bid 163 bananas for dear Gordon.

[Monique]

Yes exactly. Sorry if that was unclear.

[davis_greatest]

Now Round 246 inspired Barry the Baboon ... What a bright idea for his shop window! A nice rectangle of snooker balls
So Barry went to see his good friend Gordon and told him: "Here I have this complete set of snooker balls, plus one pink , two blues , three browns O, four greens , five yellows , seven reds O and 200 cue balls.
Can you arrange them in a rectangular formation (no holes allowed) so that the total value of balls in each row is the same, and the total value of balls in each column is the same? (e.g. value of red is 1, yellow is 2 etc. A cueball has a value of 0) But please try to keep the number of cue balls minimal ... You'll get one banana for each unused cue ball"

1. Will Gordon be able to oblige?
2a. If not why?
2b. If yes how many bananas will he get? Propose a design for Barry's shop window.

Please, bids on the thread. Solutions if any initially by PM ...

Edited: seven reds! Sorry for that ...

Monique - you mean that all the reds and colours have to be used, but the whites don't?

[snookersfun]

here it goes (after having confused myself nicely all afternoon, especially after the change to 7 reds):

145 bananas?

PM on the way...

[Monique]

Round 247

Now Round 246 inspired Barry the Baboon ... What a bright idea for his shop window! A nice rectangle of snooker balls
So Barry went to see his good friend Gordon and told him: "Here I have this complete set of snooker balls, plus one pink , two blues , three browns O, four greens , five yellows , seven reds O and 200 cue balls.
Can you arrange them in a rectangular formation (no holes allowed) so that the total value of balls in each row is the same, and the total value of balls in each column is the same? (e.g. value of red is 1, yellow is 2 etc. A cueball has a value of 0) But please try to keep the number of cue balls minimal ... You'll get one banana for each unused cue ball"

1. Will Gordon be able to oblige?
2a. If not why?
2b. If yes how many bananas will he get? Propose a design for Barry's shop window.

Please, bids on the thread. Solutions if any initially by PM ...

Edited: seven reds! Sorry for that ...

[April madness]

You are definitely going bananas - a puzzle with no answer?

better than a puzzle with no bananas

saying that, our quizmaster was very generous to me for my "answer" to round 246

[Monique]

You are definitely going bananas - a puzzle with no answer?

Well that's precisely the answer !

cuesamples1_025.jpg

[dantuck_7]

You are definitely going bananas - a puzzle with no answer?

[davis_greatest]

Round 246 - congratulations to snookersfun, Monique and dantuck_7, who each got the maximum 3 bananas, and also to April Madness who correctly got 0 bananas

(a) There are no possible arrangements of the balls in a rectangle. As there are 21 balls, these must be in a 7x3 rectangle (say 7 rows - a 1x21 rectangle wouldn't work).

Since the total value of the balls (15 reds, & colours worth 27) is 42, each row would have to add up to 6 points (42/7). But the black alone is more than that, so it is impossible.

(b) Therefore, no bananas are given under part a, so if you answer 0 to part b you get 2 bananas.

(c) If you answer 3 bananas to part c, you are correct as you get 2 bananas for part b and 1 banana for part c - total 3 bananas.

Over and out!

[davis_greatest]

Perfect, Monique. Thank you!

Round 246 – Going bananas

Can you take a complete set of snooker balls (excluding the cue ball) and arrange them all in rectangular formation (no holes allowed) so that the total value of balls in each row is the same, and the total value of balls in each column is the same? (e.g. value of red is 1, yellow is 2 etc)

Answers by Private Message please, by midnight BST tonight, and you can get bananas for these three parts:

(a) 1 banana for each different formation you find;
(b) 2 additional bananas if you guess correctly the total number of bananas awarded to all TSF members by midnight tonight under part (a) of this question; and
(c) 1 additional banana if you guess correctly the total number of bananas that you will get for answering all three parts of this round.

[Monique]

As an intro the value of a whole number a modulo another whole number b is the value of the remainder of the division a/b. So 4 modulo 8 is 4; 11 modulo 8 is 3; 48 modulo 8 is 0.


Round 244
I first had a 113 break: start with 2 reds, then black-red-black ... until exhaustion of reds. All breaks yield 1 or 2 modulo 8. All sums of three breaks must be 3, 4, 5 or 6 modulo 8. So no "collision". The yellow yields 115 = 25+41+49 so you can't go further.
For the 114: replace last black with a pink, you get 112 (0 modulo 8) then add yellow for 114 (2 modulo 8)

Round 245
For 106: 3 reds then black-red-black ... until exhaustion of reds. All breaks yield 3 or 2 modulo 8. Sums of two breaks yield 4, 5 or 6 modulo 8. Sums of three breaks yield 6, 7, 0 or 1 modulo 8. Again no "collision"

[davis_greatest]

Rounds 244 and 245 closed!

Congratulations, Monique, on the 114 and 106 breaks!

Please explain on the thread how they are made with your proof that they are valid breaks.

[April madness]

hmm, the breaks seem getting smaller and smaller. wish I could improve that, but I keep missing the first red all the time

[Monique]

Ok. 106.