Puzzles with numbers and things

[davis_greatest]

Well, I don't think people will be able to use that site for this next one (even though it would have been pretty pointless doing so before, and some might say also pasting it here).

Round 282 - Twice Nightly Snooker Whiteley

Here are some snooker balls. You can add them, divide, multiply or subtract, and need to make the Target or as close as you can get to it.

You don't have to use all the balls, and can't use any more than once.

You cannot combine digits - so, for example, with yellow and green, you could make 2+3 = 5, or 3-2=1, or 2x3 = 6, or even 3/2=1½, but you can't make 23 or 32.

Answer on the thread - winner / joint winners for each round will be decided according to how quickly the answer is given and how close it is to the Target. Your answer must be a whole number, and you must state how you use the balls to get it.

...

Target this time is ... 1227

[Semih_Sayginer]

bid 1228

6 x 6 = 36

36-1 = 35

35 x 5 = 175

175 x 7 = 1225

1225 + 3 = 1228

[PaulTheSoave]

7*6= 42

42-1= 41

41 * 6 * 5 = 1230

1230 - 3 = 1227

Wrote it like this so everyone can understand

[davis_greatest]

Congratulations!

And Monique has joined snookersfun in solving round 281 and identifying how many plums Charlie was carrying...

[Monique]

Round 284 - more art work ...

In their constant effort to improve Barry the Baboon shop's decoration, Charlie and Oliver had another go yesterday (TSF was down ... so they had too much time on their hands!)

Charlie first sets up a plain hexagon of yellow snooker balls, each side of the hexagon having as many balls as he had pears in his bag ...
Oliver found it a bit dull, removed the balls that were inside the hexagon, leaving the perimeter in place, and started to build a six-legged star (David star shape) using that "perimeter" as a start and making the legs plain while the "centre" remainded empty.. However he did not have enough balls and had to use the balls of one yellow rectangle that was on display on the counter ... "Funny he said, that rectangle had exactly the number of balls I needed. In fact it was the smallest number of balls that could possibly have been arranged in a rectangle for improving this design ... any smaller hexagon would not have permitted that ..."

How many pears did Charlie have in his bag?

Edited in blue

[Monique]

Well done Snookersfun ... first in with a correct answer

[Monique]

And also abextra ... congratulations.

[davis_greatest]

Congratulations!

And Monique has joined snookersfun in solving round 281 and identifying how many plums Charlie was carrying...

... and there were 99 plums... explanations please on the thread.

[Semih_Sayginer]

.. congratulations.

yip, well done ABS and SF...

[snookersfun]

... and there were 99 plums... explanations please on the thread.

regressing
one of the 2 explanations:

p=number of plums
4p(p+1) = number of balls from rectangle
4p^4 as before, total number of balls in big square
m(m-398) or (n+199)(n-199) bucket/ball product

using d_g's sentence in front:
'Since Charlie was able to work out the difference between n and b, expression 1 (yep, which one now?) must be the only way of factorising the total number of balls delivered.'

so (n+199)(n-199)=n²-199² on one hand,
4p^4-(4p²+4p+1)= 4p^4-(2p+1)² on the other hand, these being of same form,

therefore (2p+1)=199
p=99

[Monique]

Mine is essentially the same

[Monique]

D_G has now solved R284 also ... well done. Answer on the thread anyone?

[davis_greatest]

D_G has now solved R284 also ... well done. Answer on the thread anyone?

It can be solved with algebra but a picture is simpler. As I can't draw pictures very well, however, I'll have to try to describe one!

If Charlie has p pears in his bag, then the hexagon of balls has sides with p balls. If Oliver takes out the centre ball, then the remaining shape can be cut into 6 equilateral triangles of p-1 rows. However, as he needs to leave the perimeter intact, the centre consists of the centre ball plus 6 equilateral triangles of only p-2 rows.

To make the Star of David, Oliver needs to have a triangle of p rows on each of the 6 edges. The bottom row with p balls is already there, from the perimeter of the hexagon. If he can find p-1 balls from somewhere, he can make these form the next row up on one edge, and then add one of those equilateral triangles of p-2 rows (which he took out of the centre of the hexagon) to make the full triangle of p rows. He can do this for all six edges.

So this means that Oliver needs to find 6(p-1) balls from somewhere, minus one for the ball that was in the centre of the hexagon.

So the number of balls in the rectangle is 6(p-1) - 1.

To be a rectangle, this number must not be prime. If we calculate 6(p-1) - 1 for p=2,3,4,5,6,7…. we get 5, 11, 17, 23, 29, 35,…. and 35 (=5x7) is the first one that isn't prime. This corresponds to Charlie transporting p=7 pears.

[snookersfun]

a nice and easy one, just to get going again:

R. 285: Trickshots (well, I tried to make it snooker related)

So, Steve Davis sets up one of his amazing trickshots, lays a cue down straight the length of the table, sets up the black ball at 2/5th the length of the cue and shoots the white ball against black, so that on impact white travels right back along the cue and black continues on along the cue, both at same speed. At the same time that the white contacts the black ball, Davis shoots the pink ball from the bottom cushion, so that it travels parallel to the other 2 balls in motion. He manages to judge the speed of the balls in such a way, that white ball reaches end of cue at exactly the time that pink ball whizzes by and moreover black ball reaches the other end of the cue at the same time that pink passes that end.
trick2.bmp
questions:
a) how much faster does the pink ball travel than white or black
b) (and haven't worked that out) give some range of possible cuesizes/cue positions (or state that it is impossible), given table and normal cuesize.

and just hoping this all makes sense (should have stuck with the original question it seems now)
answers everywhere except here for starters...

[snookersfun]

...and that might have been way too easy, as Monique is in with part a) in no time

therfore R 286: cross-number
cross number1.bmp
Across

1. Fifth power (11)
6. 6 across + 18 across = 21 across (4)
8. Decimal period of 1/7 (6)
9. A solution of the equations (4)
a + b - c = 0
3a - 5b + c = 3600
5a + 7b - 5c = 7026

11. 11 across, 15 across and 18 down are Pythagorean numbers whose right-angled triangle has area 1 down (3)
12. Year of Bonnie Prince Charlie's rebellion (4)
13. Sum of the proper factors of 14595 (5)
15. See 11 across (4)
16. Smallest 3 digit prime whose anagrams are all also prime (3)
18. See 6 across (4)
20. Product of 4 consecutive Fibonnacci numbers (6)
21. See 6 across (4)
22. Palindromic cube (11)


Down

1. 14 * 8 across / 3 (6)
2. abcd, where abcd = a^4 + b^4 + c^4 + d^4 (4) edit:sorry, powers were lost
3. A solution of the equations in 9 across (4)
4. Anagram of 8 across divisible by 367 (6)
5. 93rd triangular number (4)
7. Back 4.5 from WSW (3)
10. Power of 2 (4)
12. Year of the Battle of Waterloo (4)
13. Multiple of 16 down (6)
14. 9 seconds before 2 pm (6)
16. Treble top + treble prime + bullseye (3)
17. Multiple of 7 down (4)
18. See 11 across (4)
19. A solution of the equations in 9 across (4)