Round 93 solution
I shall now close round 93.
The answer is 2 presents. Charlie can easily seat people to prevent me from matching 3 presents with their intended recipients (for example, by making everyone sit in the same order as the presents, but going round the other way, so clockwise rather than anticlockwise). However, no matter where everyone sits, I shall always be able to match at least 2 presents by rotating the table.
Here's why:
At first, not one of the 16 guests is matched to a present. I can therefore rotate the table through 15 other positions, and as it passes through all those positions, everyone must be matched up once with a present on the way. That's 15 positions, and 16 guests matched to a present - so at least one position must match at least 2 presents to their recipients!
I shall award a whole point to snookersfun, whose (final!) answer was good enough (although not as elegant as the above ); and half a point to each of abextra and Snooker Rocks!, who both said the answer was 2 but did not satisfactorily prove it.
SO HERE IS THE SCOREBOARD AFTER ROUND 93
snookersfun.........................47
abextra...............................31
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2½
Ginger_Freak.........................1½
April Madness........................1
ROUND 94 ...
... to follow.
I shall now close round 93.
The answer is 2 presents. Charlie can easily seat people to prevent me from matching 3 presents with their intended recipients (for example, by making everyone sit in the same order as the presents, but going round the other way, so clockwise rather than anticlockwise). However, no matter where everyone sits, I shall always be able to match at least 2 presents by rotating the table.
Here's why:
At first, not one of the 16 guests is matched to a present. I can therefore rotate the table through 15 other positions, and as it passes through all those positions, everyone must be matched up once with a present on the way. That's 15 positions, and 16 guests matched to a present - so at least one position must match at least 2 presents to their recipients!
I shall award a whole point to snookersfun, whose (final!) answer was good enough (although not as elegant as the above ); and half a point to each of abextra and Snooker Rocks!, who both said the answer was 2 but did not satisfactorily prove it.
SO HERE IS THE SCOREBOARD AFTER ROUND 93
snookersfun.........................47
abextra...............................31
davis_greatest.....................23½
Vidas..................................12½
chasmmi..............................12½
elvaago...............................10½
Sarmu..................................8
robert602.............................7
The Statman.........................5
Semih_Sayginer.....................2½
austrian_girl and her dad.........2½
Snooker Rocks! .....................2½
Ginger_Freak.........................1½
April Madness........................1
ROUND 94 ...
... to follow.
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