Puzzles with numbers and things

[davis_greatest]

For those who have asked, there are 100 pennies in one pound.

[snookersfun]

3 p shot in the dark (confused myself thoroughly)

[abextra]

Each ball cost 37 pence.

[abextra]

This means that Charlie bought 271 balls and could make 9 rings of the colour balls around the white ball.

[snookersfun]

Each ball cost 37 pence.

that's more like it!

[davis_greatest]

Each ball cost 37 pence.

Excellent! That's enough for the point. But can you give an explanation of how you got the answer?

[elvaago]

Could he not have bought 37 balls at 2.71 each, making 3 rings? How did I get to it? Since I'm not smart, I used Excel and Word. Excel to figure out which number, when you divide 100,27 by that number, gives a whole amount of pennies. The first number I got to was 37. So then I made circles with the drawing program in word. First the white ball, first circle is 6 balls, second circle is 12 balls, third circle is 18 balls. 37 balls total.

[davis_greatest]

That was solved faster than I thought it would be!

Could he not have bought 37 balls at 2.71 each, making 3 rings? How did I get to it? Since I'm not smart, I used Excel and Word. Excel to figure out which number, when you divide 100,27 by that number, gives a whole amount of pennies. The first number I got to was 37. So then I made circles with the drawing program in word. First the white ball, first circle is 6 balls, second circle is 12 balls, third circle is 18 balls. 37 balls total.

Yes, he could! That's one point for abextra and one for elvaago.

In fact, I realised slightly too late that I had very carelessly chosen one of the few sets of numbers that gave 2 possible solutions! Here's a full solution:

In the centre is 1 ball
In the first ring are 6 balls
In the 2nd ring are 6x2 = 12 balls
In the 3rd ring are 6x3 = 18 balls
...
In ring N are 6N balls.

The total number of balls in N rings is therefore
1 + 6 x (1+2+3+ ... + N)

In the brackets, the first term is 1 and the last term is N, so the average of the terms is (N+1)/2.

There are N terms, so the total in the bracket is then N(N+1)/2.

Therefore, the number of balls is 1 + 6 x N(N+1)/2 = 1 + 3N(N+1).

Charlie spent 10027 pence
If each ball costs K pence, we have K [1 + 3N(N+1)] = 10027

Note that 10027 = 37 x 271 (these are both prime numbers), so
the factors of 10027 are 1, 37, 271 or 10027.

Therefore 1 + 3N(N+1) = 1, 37, 271 or 10027

The only possibilities which give a whole number N (N>0) are
1 + 3N(N+1) = 37 or 271 (corresponding to N=3 or N=9)

which give K = 271 or 37 respectively.

So each ball could cost either £2.71 (3 rings) or 37p (9 rings) !


PS You don't need to draw circles in Word! To realise that there must be 6 balls in the first ring, just look at the white ball and any two touching balls in the first ring. If you join their centres, you must get an equilateral triangle. An equilateral triangle has angles of 60 degrees. Since there are 360 degrees in a circle, there must be 6 balls around the white (since 6 x 60 degrees = 360 degrees).


SO HERE IS THE SCOREBOARD AFTER ROUND 32

snookersfun……………………….…..15
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……3
elvaago...............................1

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[rambon]

Never try and understand a maths question when jet-lagged is the answer I think

[abextra]

Excellent! That's enough for the point. But can you give an explanation of how you got the answer?

Thank you for the point, davis_greatest! I'm afraid I have nothing to add to your explanation. At first I found out that Charlie had 10027 pennies ( ) . I knew, that there are 6 balls in the first ring around the white ball and there will be 6 more balls in every next ring. I used the same formula

N = 1 + 6 x ( 1 + 2 + ... + n ), where n is the number of rings.

Then I just tried to find the number of balls, which divides 10027 without a reminder.

[abextra]

Could he not have bought 37 balls at 2.71 each, making 3 rings? How did I get to it? Since I'm not smart, I used Excel and Word. Excel to figure out which number, when you divide 100,27 by that number, gives a whole amount of pennies. The first number I got to was 37. So then I made circles with the drawing program in word. First the white ball, first circle is 6 balls, second circle is 12 balls, third circle is 18 balls. 37 balls total.

I thought about 37 balls too, but as davis_greatest asked the price in pennies and mentioned a whole load of balls, I decided to go for 37 pence and 271 balls. Maybe I was too saving, I'm sure the balls that you let Charlie buy were much prettier.

[davis_greatest]

Warming up again with a relatively easy one

I was looking on the BBC thread a few minutes ago, as I understand it's going to disappear soon, and I came across an old problem that I had set. It was much easier than many of the ones I've set here on this thread recently, so was a bit shocked just now when I found that I couldn't see how to solve it myself.

I'm not sure whether this is because I have fever at the moment or whether dementia is kicking in, but I decided to post it again here to see if anyone else could solve it. (Actually, after a few minutes, I did manage to figure it out, but here it is anyway.)

ROUND 33 - Not in my league!

I play snooker in a league, and during the season each person plays every other person once. Always the same number of frames are played each match (with "dead frames" being played out), so that a player might win, lose or draw. In fact, we have arranged it so that the number of frames in a match is the same as the number of players in our league!

At the end of the season, while looking down the results, I was amazed to find that every possible match score appeared the same number of times. (A score of x-y is deemed to be the same as a score of y-x.)

What are all the possibilities for the number of players that might be in my league?

(And, I didn't ask for it originally, but you must prove that you have listed all possibilities, or you won't get the point.)

[snookersfun]

as nobody had a go yet...

I think there is only one possibility of number of players, which is 4 players. Those play 1+2+3=6 games with 3 possible different scorelines (4-0,3-1,2-2), which can therfore appear twice.

now the proof:
for n players the number of games are 1+2+3+...+n-1 or n(n-1)/2
possible scorelines are int(n/2+1) or (n+2)/2

deviding those thus: n(n-1)/(n+2) which will yield an integer only if (n-1)/(n+2) will give one, thus only for small numbers.

[snookersfun]

as nobody had a go yet...

I think there is only one possibility of number of players, which is 4 players. Those play 1+2+3=6 games with 3 possible different scorelines (4-0,3-1,2-2), which can therfore appear twice.

now the proof:
for n players the number of games are 1+2+3+...+n-1 or n(n-1)/2
possible scorelines are int(n/2+1) or (n+2)/2

deviding those thus: n(n-1)/(n+2) which will yield an integer only if (n-1)/(n+2) will give one, thus only for small numbers.

the last statement sounds a bit funny as it reads, what I meant is rather (n-1)/(n+2) should give 1 over an integer or (n+2) must be devisible by (n-1).

[davis_greatest]

Sunday morning kindness

Go on, have the point! The only possibility is indeed 4 players in my league.

I'm not really convinced that you got to the end in proving that this is the only possibility, but I'm awarding the point... perhaps I am getting too generous in old age.

Matches can be drawn, so we know that the number of frames per match (and number of players in the league), n, is even.

Here's the rest of the proof:

We got as far as the number of times that each score appears being:

n(n-1)/(n+2).

With a little bit of algebra, this can be re-arranged as

n - 3 + 6/(n+2)

For this to be an integer, 6 must be divisible by n+2, which means that n+2 must be one of:
-6, -3, -2, -1, 1, 2, 3 or 6.

The only one of these that gives n>=2 players is n+2 = 6, i.e. n=4.


SO HERE IS THE SCOREBOARD AFTER ROUND 33

snookersfun……………………….…..16
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……3
elvaago...............................1

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)