Author Topic: Lunch Plans  (Read 2155 times)

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Offline Toumeng

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Lunch Plans
« on: July 15, 2010, 06:48:50 PM »
Between 2200 and 2600 guests were invited to the Mayor's luncheon. The organizer decided to arrange the seating so that each table would accommodate an equal number of guests, and the number of guests at each table would be an odd number.

The organizer determined that the following arrangements would not work:
3 guests per table (2 would be left without seating); 5 guests per table (4 left out); 7 guests per table (6 left out); and 9 guests per table (8 left out).

However, when he tried seating 11 guests per table, no guests were left without a place to sit.

How many guests in total were invited to the luncheon?



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Offline humpty dumpty

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Re: Lunch Plans
« Reply #1 on: July 25, 2010, 07:05:01 PM »
2519 guests.

Microsoft Excel saves a lot of time.



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Offline Toumeng

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Re: Lunch Plans
« Reply #2 on: July 26, 2010, 07:56:28 PM »
Correct!

2519 guests / 11 guests = 229 tables. all guests seated.



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Offline dreamlogic

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Re: Lunch Plans
« Reply #3 on: November 16, 2010, 12:34:46 AM »
2519 guests.

Microsoft Excel saves a lot of time.

LOL....man, hilarious.  You ought to sell this quote to MS.  By the way, you look like my sister.


« Last Edit: November 16, 2010, 12:36:56 AM by dreamlogic »

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Offline YAX

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Re: Lunch Plans
« Reply #4 on: November 16, 2010, 05:22:05 PM »
darn.  excel, why didn't I think of that.  ;D



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Offline passingby#2

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Re: Lunch Plans
« Reply #5 on: April 28, 2011, 07:52:51 PM »
Could've been exactly 2200 guest as well...

2200 guest / 11 per guest = 220 tables.  O0


« Last Edit: April 28, 2011, 07:58:59 PM by passingby#2 »

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Offline evil-in-deed

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Re: Lunch Plans
« Reply #6 on: September 01, 2011, 05:38:30 PM »
Could've been exactly 2200 guest as well...

2200 guest / 11 per guest = 220 tables.  O0

then they couldve seat 5 per table without any guest left out also...

you have to have 4 guest left out when seating 5 per table... and so on with 3, 7, and 9 per table...



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