Puzzles with numbers and things

[snookersfun]

well, nothing here yet, unless I can ditch a silly black ball from the picture

[Monique]

And here is mine

2,6,7,2,3,7,4,5,7,4,6,7,2,6,7,5 and the six "colours":snooker:

[snookersfun]

ooops, edited instead of quoted, so, that went all wrong, this top part had to move down, will do that in a minute, but what had been here before was my break:

Here's my break
2,2,4,7,6,5,7,6,2,7,5,6,7,3,4,7 and the final 6 colours :snooker:

[davis_greatest]

Round 369 - Countdown Breakdown

Use the 13 snooker balls below, and any combination of addition, subtraction, multiplication or division to make the following three target snooker numbers: 100, 147 and 155.

You must use every ball exactly once. (That is, once to make all three targets together - not once for each target. You must not have any balls unused.)

You may not combine balls without using +,-,x or /. For instance, you cannot simply put red and black together to make 17.

You may not multiply or divide by a red ball.

And those are all the rules! Good luck.

[davis_greatest]

Congratulations to snookersfun, Monique and abextra, who have all found the highest possible break in round 368! And, hopefully, no one peeked. Again, we'll leave the round open a little longer. :snooker:

Meanwhile, could those who have already bid (and any future bidders) now please put up the breaks too (just listing the colours is sufficient), again in hidden text, so they can be verified.

[abextra]

Thank you, Snookersfun!
Hopefully my 123 works too.

[snookersfun]

I'm not sure about this part -

at no point must the total points you have scored from any one colour (green upwards) exceed the combined total points you have scored from all lower valued colours.

Could the scores be equal?

exceed should mean bigger than..., so yes, equal is possible

[abextra]

I'm not sure about this part -

at no point must the total points you have scored from any one colour (green upwards) exceed the combined total points you have scored from all lower valued colours.

Could the scores be equal?

Abextra, the white shows

Sorry!

[Monique]

my bid 123:snooker:

[snookersfun]

oops, there we go, mistake on adding the reds
123 now

[snookersfun]

this one is tricky and I just ran out of time trying (and checking myself carefully), so first bit of
122 now

Abextra, the white shows

[abextra]

:snooker:First bid 115 :snooker:

[davis_greatest]

Round 368 - Breakdown Break

You are playing a normal frame of snooker. Normal, except that, during any break you make:

(a) from the third colour onwards, you never pot the same colour as your previous one, nor as the colour before that (you're not too fond of repetition); and

(b) at no point must the total points you have scored from any one colour (green upwards) exceed the combined total points you have scored from all lower valued colours. For example - there must never be a time when the total points you have scored from potting blues exceeds the total points you have scored from potting yellows + greens + browns.

What is the highest break you might get?

Please feel free to bid here!


We will perhaps trial a new system of public, hidden bids! I don’t know if this will work, but we can try! So please post bids in hidden text, if you can, like this (deleting the space after the two open square brackets):

([ COLOR=#f1f1f1]
Enter bid here, e.g. 147! [ /COLOR])

[davis_greatest]

We shall also close round 366 then. The answer - and congratulations to those who found it - is one, four, seven.

The crux to solving it is to note that:
(i) If Charlie's "rectangle" (which we shall see in a moment must turn out to be a square) has x rows and y columns, then the bottom layer of Oliver's triangle has (x+y) rows.
(ii) Charlie's two triangles then have (x+y)^2 balls. So xy must divide (x+y)^2.
(iii) If xy divides (x+y)^2, then x=y, so Charlie has a square, and Oliver has 4 times as many balls as Charlie.
(iv) Hence there are 4 crates of red balls - and this we know without even considering the total number of balls there are, nor have we yet considered Gordon. Also the number of balls in each crate is x^2 = y^2, hence a square.
(v) Then look at the highest square that divides 147 million to find how many balls in a crate. Our first try - 49 million - is too big, so look at the next highest square, by dividing by 4. Et voila.

Round 368 to follow....

Round 366 Barry's Crazed Crate Banana Bonanza

Charlie, Oliver and Gordon go to buy some balls from Barry the Baboon’s Ball Shop. Charlie buys a crate of pink balls, Oliver some crates of red balls and Gordon some crates of black balls – each crate contains the same number, and together they bring back 147 million balls to my house! And some bananas. Quite a sight!

Charlie lines his pink balls out in a grid formation, using them all – lining up rows and columns evenly in straight lines.

Oliver makes a great big triangular pack of red balls, and then places a second (slightly smaller) triangular pack of red balls on top - so the top level balls fill all the hollows made by the balls in the lower level. Oliver finds that he exactly uses all his balls to do this.

Then naughty Gordon steals the top row of Charlie’s balls, and then steals the end column from what is then left in Charlie’s rectangle. Gordon finds that the number of pink balls he has stolen is exactly the same as the number of rows in the top layer of Oliver’s triangle!

Oliver had bought the maximum possible number of crates consistent with the above information.

Gordon had bought the minimum possible number of crates consistent with the above information.

How many crates did (i) Charlie, (ii) Oliver, (iii) Gordon bring back to my house?

(And for a bonus point - what is amazing about this answer? )

Answers initially by Private Message…

[abextra]

Thank you all for explanations, very well done.