Puzzles with numbers and things

[Monique]

Tallguy already in with a perfect answer to R342 Very well done and welcome to the puzzles !!!

[Monique]

R342 High lowest break ...

And to celebrate the revival of the thread here comes Round 342 ...

Having just played a best of 11 friendly match, Rollie and Charlie are sitting at their favourite Banana bar, reminiscing about their first ever snooker game, while Oliver eavesdrops on their conversation, savouring his banana split.

"You know what occurred to me the other day, Charley?" asks Rollie
"What's that?" says Charlie.
"Well, when we played then, the square of your lowest break contained the same three digits as the square of my highest, just in a different order" Rollie continues.
"This will really blow your socks off then, Roll! If you take the square of the sum of your highest break and my lowest in that first game and split it into two 2-digit numbers, you'd have my lowest break then and your lowest break today!" Charlie exclaims.

"Ooh but that's an impressive lowest break Rollie!" interjects Oliver.

Indeed What's Rollie's lowest break today?

[Monique]

Abextra now sent me a perfect solution for Number Cruncher R336. Well done !!! Abextra will you put your solution on the thread? thanks!

[snookersfun]

and the solutions don't stop coming in now:

Abextra solved R. 326, 332, 337 and 341 this time

which allows me to close several rounds now:

R. 326 one for the long Friday

In the following grid all the five letters A,B,C,D,E are to appear in every column and row once and once only. The letters on the sides indicate first letter seen from that side.

[ATTACH]1264[/ATTACH]

and the solution is:
D B - E - A C
A E B - D C -
- D - C E B A
E - C D A - B
- A D B C E -
B C A - - D E
C - E A B - D

well done Mon and Abextra!

R. 332: counting balls

Oliver and Gordon, rather bored by now, sit together one day and start to play around with their snooker balls. Initially both start out with a certain amount of balls each and then each add one ball at a time (and kind of simultaneously) to their previous amount to try to built nice geometrical figures (as usual, mainly triangles or pyramids).

Charlie is observing them from the side and at one point comments to Gwendoline: Do you know what? Gordon right now has as many balls as Oliver will have, when Gordon will have twice as many balls as Oliver had when Gordon had half as many balls as the sum of their balls now. Also Oliver has as many balls now as Gordon had when Oliver had half as many balls as he will have when he takes 10 more balls.

How many balls do Gordon and Oliver have at the time that Charlie is making his observation?

solution: Oliver had 30 balls and Gordon had 40 balls

well done d_g, Robert, Mon and Abextra

R. 333: reds around blacks

The following drawing shows part of a snooker table partioned by a grid. Several black balls are placed on it.

Your task is to place additional red balls, exactly two reds surrounding each black (touching grid of black ball at edges or corners). Reds though are not allowed to touch (even diagonally).
The numbers on the side of rows and top of columns give total amount of reds in each respective row/column.

solution: one of those famous smiley pics: oh, darn, smiley limit... just a sec

smileys.bmp

congrats Mon and Abextra

R 337 another 'low breaks' day

Gwenny, Charlie, Oliver and Gordon are just back from this morning's snooker practice. Suffice it to say it didn't go very well. The recapitulate that during one particularly abysmal session in which each had three breaks, they scored breaks from 1-13 only (each number maximum once) but noticed that all ape's combined total score was the same.
Trying to recall the individual scores, they initially only remember that Gwenny had a break of 1, Oliver a break of 3 and Charlie a break of 11.
But actually that is enough information to figure all the breaks out:

So, who had the highest break in that session and who scored what?

Gwendoline - 1, 8, 12
Oliver - 3, 5, 13 - highest break
Charlie - 4, 6, 11
Gordon - 2, 9, 10

congrats to Mon, Ja, d_g and Abextra

and finally R. 341:

8 - 2 / 3 + 6 / 4 + 9 / 1 x 7 - 5 = 72

well done Mon, Moglet and Abextra

and I hope I didn't forget anybody...

Happy puzzling

[Monique]

And she also solved R335 perfectly! Hurray! Well done!!!
Time for closing this one I think ...
so here comes the solution for the "Pot Square" puzzle

9 4 6 15
7 14 12 1
16 5 3 10
2 11 13 8

Once again congratulations to all contenders

[snookersfun]

breaking news:

meanwhile I have the pleasure to announce that Abextra sent perfect solutions to R.338, 340 and even forgotten R.333 (with one of her famous smiley pictures). Well done!

Abextra, could you put the latter up on the thread?

[Kathrin]

Any other input Kathrin?

Sorry, that´s actually the whole thing. But if you make it right there have to be three couples walking over the bridge and 2 single walkers (those bringing back the light) if this is any help for you.

[snookersfun]

Well... I think The Captain and The Jester have to cross the bridge more than once... hopefully they like to walk together in the dark.

I've heard moglet agrees with that. Any other input Kathrin?

[abextra]

R. 342 (if I´m right): Bridge over troubled water

It´s an easy one, perfect for lunch

The Captain, The Nugget, The Jester and The Wizard have to cross a bridge. It´s dark and they only have one light. The bridge can be crossed by two of them at once and they only have 60 minutes.

The Captain needs 5 minutes to get to the other side.
The Jester needs 10 minutes.
The Wizard needs 20 minutes.
The Nugget needs 25 minutes.

In which order do they have to cross the bridge to make it in 60 minutes? Remember, they always have to bring back the light...

Well... I think The Captain and The Jester have to cross the bridge more than once... hopefully they like to walk together in the dark.

[snookersfun]

update: Moglet solved R 341 now as well

[Kathrin]

R. 342 (if I´m right): Bridge over troubled water

It´s an easy one, perfect for lunch

The Captain, The Nugget, The Jester and The Wizard have to cross a bridge. It´s dark and they only have one light. The bridge can be crossed by two of them at once and they only have 60 minutes.

The Captain needs 5 minutes to get to the other side.
The Jester needs 10 minutes.
The Wizard needs 20 minutes.
The Nugget needs 25 minutes.

In which order do they have to cross the bridge to make it in 60 minutes? Remember, they always have to bring back the light...

[snookersfun]

moglet just sent me two very nice pictures ... solving R339. Well done!

... and Mon solved R.340 and R.341 perfectly. Well done again!

and congrats moglet!

[Monique]

moglet just sent me two very nice pictures ... solving R339. Well done!

[snookersfun]

R.340 'similar to R.338 Balancing balls'

...using much more balls now (in fact 3 full sets (colours and reds) plus 15 reds), but again use all the reds and the colours, distribute them (all different number of balls) into the 12 bags and balance the mobile.

mobile-3.bmp

How many balls are in each specific bag?


and a short one:

R.341 'reach 72'

use the numbers from 1-9 (each digit once) in the blanks to equalise the equation (perform all operations strictly from left to right):

_-_/_+_/_+_/_x_-_=72

[abextra]

For round 335 - as an example, in the yellow square, numbers 1 and 2 are in the same row. Therefore, in the green square...

Thank you for the help, DG. It's very nice to see you around again, welcome back!
How are Charlie, Gordon, Gwenny and little Oliver?

... and as this one seems to have been open for a while, may as well give an answer directly on the thread - but in hidden text...

(9,7,2,14,11,5,4,12,13,3,6,10,15,1,8)

Thank you so much for the answer!!!