Enigma 1305 - Buzz

Enigma 1305 - Buzz New Scientist magazine, 4 September 2004. by Richard England.

In the game of buzz the players form a circle and count in turn, the first saying '1', the next '2', the next '3' and so on. But every time the next number is a multiple of 7 or contains a 7 the player whose turn it is must say 'buzz', and then the direction in which play is going round the circle is reversed; so after the sixth player says '6', the seventh says 'buzz', then the one who said 6 says '8' and the one who said 5 says '9' and so on until a player says 'buzz' rather than 14, whereupon the one who said 13 must say '15'.

At 27 and 28, and again at 56 and 57, the direction of play is reversed twice in succession, and through the 70s two players must each say 'buzz' five times alternately.

In a game which only ended when a player said '97' instead of 'buzz' my only contribution was to say 'buzz' twice.

(a) Which two numbers did I say 'buzz' for?

(b) How many players took part?

Ciao,

Spoiler - I hope.

Position 14 - ie thirteen people speak first in group of 19

Regards

Well you got the same answer as me, as the two numbers they say 'buzz' for are 42 and 91.

Unfortunately I just brute-forced it with Excel, but I'd be interested in seeing an elegant way to solve it.

S p o i l e r

S p a c e

S p o i l e r

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First, list all the numbers when the Buzzes occur – these are 7, 14, 17, 21, 27, 28, 35, 37, 42, 47, 49, 56, 57, 63, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 84, 87 and 91.

Next we take 0 as the origin, and work out how far the counting moves from the origin. At each reversal, the distance from the origin is…

Buzz Number Distance 7 7 14 0 17 3 21 -1 27 5 28 4 35 11 37 9 42 14 47 9 49 11 56 4 57 5 63 -1 67 3 70 0 71 1 72 0 73 1 74 0 75 1 76 0 77 1 78 0 79 1 84 -4 87 -1 91 -5

Since the furthest we move from origin is –5 to 14, the maximum number of people is 20 (further people are redundant).

Since the puzzle says the only contribution our player made was to say 'Buzz' twice, it must mean the user was at 1 or both of the extremes from the origin. In the above table, 14 occurs only once and –5 appears once, and there are no other possible places the player could have sat without counting (e.g. if he sat 11 away from origin, he would have 'Buzzed' at 35 and 49, but also counted 39 and 51) thus –5 and 14 must be the same player.

Thus the game had 19 players, the player 'Buzzed' at 42 and 91.

puzzle as spoiler

This one surely *screams out* for computer help with the mechanics and machinations.

I set up a spreadsheet listing in a column the numbers n from 1 to BIG. The next column holds the truthvalue of [the decimal presentation of n contains a '7']. The next column holds the truthvalue of [n mod 7 = 0]. The fourth column is the ORing of the previous two; it then holds 'this n should be BUZZed'. The next column holds a number p from 0 to P-1, where P is the number of players and p is the player who says this n. This is the interesting column.

I set literal 0 and 1 for the first two rows of this column. For all subsequent rows, the idea is: { if the previous n was BUZZed, then reverse direction (ie this row's p = last row's p MINUS [last row's p less the prior row's p]); else continue in the same direction (ie this row's p = last row's p PLUS [last row's p less the prior row's p])}, the whole being subject to a mod p operation since we are moving in a circle. This mod operation conveniently smooths over end-cases

Now we have the entire history of the game laid out for us in terms of the variable P. We can visually check the correct behaviour at 27-28 and through the 70s.

Now to solve the puzzle as given, consider only rows up to n=97. Set up additional columns as necessary to count individual player's 'n's and buzzes. Set up a column which is 2^(number of 'n's).3^(number of buzzes). Set up a cell that searches this last column for 2^0.3^2 = 9 and return 0 if not found, non-zero if found.

Then we just need to find a value of P that sets our target cell non-zero. Press the magic button and...

Number of players = 19. I was the 14th player and I said BUZZ for 42 = 6.7 and 91 = 13.7

Note that it took me longer to *write-up* this post than it did to create the spreadsheet it describes, so *elite* are my Excel skillz

ObPuzzle (hard): Increasing P in my model showed that if the game ends at 97, then if P>20, players 15 through (P-6) are never involved. So we might say that we have a function value, MaxInvolvedPlayers(97) = 20. What interesting facts can be derived about the MIP function as n

I've often wondered if there is a finite number of players that the game can sustain? I.e. is it possible that there will come a point when your shot will never come up, owing to the swapping of directions?