Puzzles with numbers and things

[snookersfun]

Dan quite impressive! I didn't understand a thing

[abextra]

Dan quite impressive! I didn't understand a thing

Yeah, Dan, you're thinking on the level I will never reach...

Here's something easy for the weekend

Round 197

Replace each letter by a number:

...... A B . x . C D B . = . E F G B
........ + ............ + .............. -
...... F A . x ..... F A . = . E A F B
- ... = ......... = ........... =
... C D H . + . D H H . = ..... I D H

Answers by PM, please.

[davis_greatest]

If anyone wants to find out the answer to R196 use this code within Excel...

Public Function R196(intPlayers As Integer) As Long

Dim n As Long
Dim f As Long
Dim BE As Long

BE = 0

For n = intPlayers To 2 Step -1

f = 0.5 * n * (n - 1)
BE = BE + 1
R196 = R196 + f * BE

Next n

End Function


Sub CallR196()

Dim cntI As Integer

For cntI = 1 To 300

Cells(9 + cntI, 2).Value = R196(cntI)
If R196(cntI) = 19720001 Then MsgBox cntI & " Players in Tournament!"

Next cntI

End Sub

The code works... but it took me longer to open Excel, copy and paste that code and run it than to solve the puzzle in the first place!

[davis_greatest]

Round 195 - congratulations to snookersfun, abextra and Monique (actually, haven't checked Monique's answer above properly, but I'll trust that it is right!)

Below is the picture that abextra sent to me early this week.

[dantuck_7]

The code works... but it took me longer to open Excel, copy and paste that code and run it than to solve the puzzle in the first place!

Thats the way I'm going atm. Managed to work out the first part; that there are 0.5 * n * (n - 1) frames per round - where n is the number of players. Its probably possible to then sum that number of frames and multiply by 'BE'.

I just find it quicker to run it through Excel. Will hopefully got an answer to R197 in the not to distant future....

Dan

[davis_greatest]

Round 196 answer

I'll guess that no one else is trying R196, and as Monique is on holiday and dantuck_7 has given some nice code to solve it, I'll put Gordon my pet gorilla's answer up. So, congratulations snookersfun, abextra, dantuck_7 and Gordon my pet gorilla

A quick way of solving it is to notice that, if there are n apes in the tournament, then the number of bananeuros dispensed is the same as the number of ways of choosing 4 fruit from n+2 different fruit (where the order of selection does not matter).

The number of such ways is (n+2)(n+1)n(n-1) / (4x3x2x1),
which (for large n) is approximately (n+½)^4 / 24.

So you just need to calculate the 4th root of (24 x 19720001), which is approximately 147½, so there are n=147 apes.

You can check this by observing that 149 x 148 x 147 x 146 / 24 = 19720001

{ To prove the bit in red above, number the apes from 1 to n, where each ape is given the number of the round in which he is eliminated from the tournament, and the champion has number n.

The number of bananeuros given for each frame is the same as the number of that round (1 to n-1).

Then each bananeuro dispensed corresponds exactly to a set of four different numbers (p,q,r,s), where 1 <= p < q < r < s <= n+2; and

p is the number of the bananeuro given in the frame played in round number q-1 between apes r-2 and s-2. (p can take any value from 1 to q-1, because q-1 bananeuros are given for each frame in round q-1. Also, any ape numbered less than q-1 would have already been eliminated before round q-1.)

The number of sets (p,q,r,s) is exactly the number of ways of choosing 4 different numbers (or fruit) from n+2 numbers (or fruit), as claimed in red above. }

[Monique]

R196 closing...

Thanks and congratulatioms for everybody!
Here is yet another explanation ...

If we have n+1 apes, in round one we have (n+1)*n/2 frames played; in round 2, (n)(n-1)/2, in round 3, (n-1)(n-2)/2 etc... till round n where we have just one frame

So the total number of bananeuros is (n+1)*n/2 + 2*(n)(n-1)/2 + 3*(n-1)(n-2)/2 + ... + n

This in turn can be reworked in
S= sum(T(i) for i=1 to n) + sum(T(i) for i=1 to n-1) + sum(T(i) for i=1 to n-2) + ... + T(1) where T(i) = (i+1)*(i)/2
and sum(T(i) for i=1 to n) = (n+2)*(n+1)*(n)/3!
and S = (n+3)*(n+2)*(n+1)*(n)/4!

For those who want a proper demonstration of the formulas they are wellcome but will need some patience
For those who want to try to demonstrate them here is a clue: prove it's true for n=1, then prove that if it's true forn it' true for n+1

[davis_greatest]

Yes, although needs a few small adjustments (in red below), otherwise you'd end up with 148 apes instead of 147! With n apes, there are n-1 rounds, rather than n rounds...

Thanks and congratulatioms for everybody!
Here is yet another explanation ...

If we have n apes, in round one we have n*(n-1)/2 frames played; in round 2, (n-1)(n-2)/2, in round 3, (n-2)(n-3)/2 etc... till round n-1 where we have just one frame

So the total number of bananeuros is n*(n-1)/2 + 2*(n-1)(n-2)/2 + 3*(n-2)(n-3)/2 + ... + n-1

This in turn can be reworked in
S= sum(R(i) for i=1 to n) + sum(R(i) for i=1 to n-1) + sum(R(i) for i=1 to n-2) + ... + R(1)
where R(i) = i*(i-1)/2
or
S= sum(T(i) for i=1 to n-1) + sum(T(i) for i=1 to n-2) + sum(T(i) for i=1 to n-3) + ... + T(1) where T(i) = (i+1)*(i)/2
and sum(T(i) for i=1 to n) = (n+2)*(n+1)*(n)/3!
so sum(T(i) for i=1 to n-1) = (n+1)*n*(n-1)/3!

and S = (n+2)*(n+1)*(n)*(n-1)/4!

As before, you can recognise S as being the number of ways of choosing 4 objects from n+2 objects. So, if you were doing the lottery and had to choose 4 numbers between 1 and 149, there would be S=19720001 possible choices of ticket. This saves having to do any algebra

[Monique]

R196

Thanks D_G for the adjustement! I had come here this morning to do this in fact ... (just having realised that I had used n both for the number of apes in the beginning of the explanation and for the number of rounds after that...) Not easy to keep concentrated in an internet cafe ) See edited items in bold in my last post.

[abextra]

Closing round197 - congratulations for correct answers to Snookersfun, davis_greatest, dantuck_7 and Monique, well done!. .


...... 4 6 x 1 2 6 = 5 7 9 6 ....... or .... A = 4 ... B = 6
........ + ...... + ........ - .................. C = 1 ... D = 2
...... 7 4 x ..7 4 = 5 4 7 6 ................ E = 5 ... F = 7
-. .... = ...... = ........ = .................. G = 9 ... H = 0
... 1 2 0 + 2 0 0 = . 3 2 0 ................ I = 3

[davis_greatest]

Round 198 - Final black

(This is much easier than it might seem!)

Gordon and Charlie are playing a frame of snooker, the scores are level, and it has gone to the final black. On any shot, there are three possible outcomes:

- the black is potted; or
- there is a foul; or
- the black is not potted but there is no foul.

Gordon and Charlie play to exactly the same standard, which means that on any given shot they always have the same likelihood as each other of potting the black; and the same likelihood as each other of fouling; etc. I happen to know what these chances are, and I know that each time they come to the table, Gordon and Charlie each have a 10% chance of potting the black.

Gordon is about to play his shot, and I work out that Gordon and Charlie are equally likely to win the frame.

What is the chance that Gordon fouls on this shot?

Answers (or guesses) please on the thread only!

[chasmmi]

10% maybe?

[davis_greatest]

10% maybe?

Congratulations, chasmmi! You keep your place in the Hall of Frame!

Fancy a go at an explanation?

[chasmmi]

I just assumed that if both are equally likely to win the frame then the chances of fouling must be equal to the chances of potting otherwise somebody will always have an advantage.

[chasmmi]

Or the snookersfun answer:

Well if a=b2/Mc4 then 45.73% of 5/6 of the rate of pi to the eight percentile requires a defigulative 17% increase in the rate of velocity on the cloth varying in Newtons 4th law of esoterics and the wind skill factor of 17 to the fourth power.

Once this value has been reached you square it, multiply by 3.4% of zero and then add ten.