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Toumeng
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« on: July 15, 2010, 05:48:50 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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humpty dumpty
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« Reply #1 on: July 25, 2010, 06:05:01 PM »
2519 guests.

Microsoft Excel saves a lot of time.
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Toumeng
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« Reply #2 on: July 26, 2010, 06:56:28 PM »
Correct!

2519 guests / 11 guests = 229 tables. all guests seated.
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dreamlogic
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« Reply #3 on: November 15, 2010, 11:34:46 PM »
Quote from: humpty dumpty on July 25, 2010, 06:05:01 PM
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 15, 2010, 11:36:56 PM by dreamlogic » Logged
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YAX
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« Reply #4 on: November 16, 2010, 04:22:05 PM »
darn.  excel, why didn't I think of that.  Grin
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passingby#2
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« Reply #5 on: April 28, 2011, 06:52:51 PM »
Could've been exactly 2200 guest as well...

2200 guest / 11 per guest = 220 tables.  ThumbUp
« Last Edit: April 28, 2011, 06:58:59 PM by passingby#2 » Logged
evil-in-deed
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« Reply #6 on: September 01, 2011, 04:38:30 PM »
Quote from: passingby#2 on April 28, 2011, 06:52:51 PM
Could've been exactly 2200 guest as well...

2200 guest / 11 per guest = 220 tables.  ThumbUp

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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