Puzzles with numbers and things

[abextra]

Round 274 - Massive Mix-up

I like to give each ape a single Final Score for each match played. The Final Score is equal to the points that the ape scored in the final frame played, which I read off the scoreboard, plus 147 points for every previous frame won (because I never know how many points had been scored in previous frames), and I also give each ape a bonus 4 points just for playing the match.

......

Unfortunately, I mixed up the number of points that Gordon had got in the final frame with the number of frames he had previously won - i.e. I swapped them around, by mistake.

As a result, the Final Score I got for Gordon was the same as the Final Score I got for Oliver. If I had not made this mistake, Oliver's Final Score would have been twice Gordon's.

How many frames did each win, and what was the score in the final frame?

Well, I think that Gordon won the last frame 95 - 47,

in total Gordon won 48 frames and Oliver won 95 frames...

[davis_greatest]

and anyone who solves that too quickly can do

Round 272 - Pinks and Browns

Charlie goes to Barry The Baboon's Ball Shop to buy some balls for potting practice. Every ball he buys, he is equally likely to choose pink or brown (regardless of what he has chosen before).

The following day, Gordon comes in to Barry's shop and does the same - except that he buys one ball more than Charlie had.

How likely is it that Gordon has more pinks to pot than Charlie? Answers straight on the thread please.

Round 276 - More Pinks and Browns

This one is harder. Charlie again goes and buys some balls (each pink or brown, equally likely, as before) from Barry; and Gordon then does too - but this time Gordon buys the same number of balls as Charlie.

In an act of kindness, Charlie then gives one of his balls, chosen at random, to Gordon.

Again, how likely is it that Gordon has more pinks to pot than Charlie, if...

(a) Charlie bought 3 balls from Barry, or

(b) (if you can) Charlie bought 147 balls (you can give this as an expression rather than calculating it).

Answers / guesses to either part on the thread please...


PS Round 274 - Monique and snookersfun have sent answers by PM. Any other answers, please put on the thread also...

[Semih_Sayginer]

No, not wrong! Entirely correct!

Congratulations to Monique and semih!

thanks d_g....feel bad though cos it looked like id just replied the same as mon. thankfully i got the reasoning correct so may no appear so now

[davis_greatest]

wrong?

No, not wrong! Entirely correct!

Congratulations to Monique and semih!

[Semih_Sayginer]

Yes, 50%! And please explain Semih, first, as just pipped on the thread?

i got 50% by reasoning (though not 100% sure its correct) that:-

at 3-3, it can go 4-3 to either player.

at 4-3 to either player it can go 2 ways. 4-4, or 5-3, and so theres only a 50% chance of it going to a decider.

wrong?

(sorry Mon.....was still at the last page)

[davis_greatest]

Yes, 50%! And please explain Semih, first, as just pipped on the thread?

[Semih_Sayginer]

a 50% chance???

[Monique]

That's 50%

[davis_greatest]

Meanwhile, Monique is first in with a correct answer to round 274!

... and snookersfun has sent it too! Next answer on the thread please.

and a quick one...

Round 275 - Decider decision

I am watching a best-of-9 snooker match, and the score is now 3-3. If each player has a 50% chance of winning each frame, and all frames are independent, what is the chance that the match will go to the deciding frame?

Answers on the thread...

[davis_greatest]

x² = y (y+1) / 2
...


in total you potted 287 balls. (you're really good! )

Congratulations abextra, Monique and snookersfun! That is the correct answer.

From x² = y (y+1) / 2, there are a couple of ways to proceed:

either observe that y and y+1 are coprime (no common factor other than 1), meaning that either (i) y/2 and y+1 are both squares or (i) y and (y+1)/2 are both squares ... which doesn't leave many possibilities to try;

or write m=2x, n=2y+1 and the equation becomes n² - 2m² = 1, again a form of Pell's equation (details on request!)


Meanwhile, Monique is first in with a correct answer to round 274!

[abextra]

Round 270 - Pooker

I have just thrashed Charlie during a frame of pooker - it's a sort of cross between snooker and pool. There are many balls (between 50 and 500), numbered from 1 upwards. So the 1-ball is worth 1 point, the 2-ball is worth 2 points etc. Balls must be potted in ascending order.

I got in first and made a break. Then I missed and Charlie potted one ball, before missing. I then cleared the table.

The two breaks that I made during the frame were equal.

How many balls did I pot?

Let's say, the ball Charlie potted was x and the total number of balls was y.

Then your first break was (x-1) x / 2 and the second break was y (y+1) / 2 - x - (x-1) x / 2

(x-1) x / 2 = y (y+1) / 2 - x - (x-1) x / 2 ---> x² = y (y+1) / 2

The value of y was between 50 and 500, so I was looking for a number between 1275 and 125 250, which was both perfect square and triangular, the only one I found was 41 616 , it means that

x = 204 and y = 288

both of your breaks were 20 706 points

in the first break you potted 203 balls (1 - 203) and in the second break you potted 84 balls (205 - 288),

in total you potted 287 balls. (you're really good! )

[davis_greatest]

One of those 192 possible arrangements (can't believe there are so many of them! )

.......... 12

10... 7 .... 1 ... 8

.... 2 .......... 9

5 ... 11 ... 6 .... 4

............ 3

Congratulations again, abextra (and also snookersfun and Monique, mentioned previously)! Another correct answer... each line adding up to 26.

192 arrangements is not that many really... for each arrangement, you can rotate it around the 6 points of the star, or reflect each one in the mirror... so that already gives 12 arrangements for each essentially "different" one.

[abextra]

Round 269 - Even money

Each coin weighs a different amount, a whole number of grams from 1 to 12 grams.
There are 6 lines, each of 4 coins. Not only does each line add up to 4 pence, but each line also has the same weight.

Give a possible arrangement of the weights of the coins.

One of those 192 possible arrangements (can't believe there are so many of them! )

.......... 12

10... 7 .... 1 ... 8

.... 2 .......... 9

5 ... 11 ... 6 .... 4

............ 3

[davis_greatest]

Well, then I think

Round 268 - the first 6 balls were 1, 3, 7, 11, 13 and 49 and the 7th ball was 101... (and the 8th was 9901, in case you don't remember yourself ).

Congratulations, abextra (and previously Monique and snookersfun too). That is the correct answer!

The way to solve this was to find the factors of 147147147147

Obviously, it is divisible by 147 = 3 x 49 = 3 x 7 x 7.

Stripping out the 147, gives 147147147147 = 147 x 1001001001

and 1001001001 = 1000001 x 1001

You can see that 1001 is divisible by 11 because the sum of even digits is the same as the sum of the odd digits, i.e. both are 1 (a number is always divisible by 11 if the difference between the sum of even digits and the sum of the odd digits is 0 or another number divisible by 11).

In fact 1001 = 11 x 91. We can also factorise 91 as 7 x 13.

Similarly, you can write 1000001 = 101 x 9901.

So 147147147147 = 3 x 7 x 7 x 7 x 11 x 13 x 101 x 9901

and using the other information in the question, you find that you have to combine two of the 7s to get 49, and must also use the 1-ball.

[abextra]

Well, then I think

Round 268 - the first 6 balls were 1, 3, 7, 11, 13 and 49 and the 7th ball was 101... (and the 8th was 9901, in case you don't remember yourself ).