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Puzzle thread Nov. 17 |
Mon, 17 November 2003 21:58 |
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Ron | | Commander Forum Administrator Stars! AutoHost Administrator | Messages: 1231
Registered: October 2002 Location: Collegedale, TN | |
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Here is this weeks puzzles. A mix of regular math, IQ-style, and logic puzzles. As usual, 24 hour time-limit.
Send all 5 answers to me via Private Message.
1. There are six busybodies in town who like to share information. Whenever one of them calls the other, by the end of conversation they both know everything that the other knew beforehand. One day, each of the six picks up a juicy piece of gossip. What is the minimum number of phone calls required before all six of them know all six of these tidbits of gossip?
2. If a bunch of positive integers adds up to 20, what is the greatest possible product of these numbers?
3. Which one of the following numbers is the odd one out? (2 possible answers, I only need one, and specify why it is the odd one out)
563
572
671
594
916
945
832
829
298
635
752
196
294
176
283
4. What, in connection with this question, is the next number below?
4, 13, 19, 21, 29, ?
(this is an IQ-style puzzle, NOT a math puzzle)
5. The Rainbow Wedding Chapel is so conveniently located in Las Vegas, Nevada, that an impulsive couple can get their marriage license, hold the wedding ceremony, and enjoy a spontaneous wedding reception at an all-night diner, all without having to walk farther than across the street. The officiant at the chapel, Justice Elliot, performed five consecutive weddings last Saturday evening, at 7:30, 8:00, 8:30, 9:00, and 9:30. THe grooms were Bud, Felix, Nathan, Rodney, and Thaddeus. The modern brides at these weddings (Eva, Lucy, Nora, Rita, and Stella) all elected to keep their maiden names (one is Ms. Johnson). For the wedding at each time, match the bride (identified by first and last names) with her groom (identified by first and last names, one is Mr. White).
a. Thaddeus and his bride were married at some point after Mr. Bookman, but at some point before Felix.
b. Ms. Sternman was married at some point after Nora, who was married at some point after Nathan (who was not married at 7:30). Nora is not Ms.
...
[Updated on: Tue, 18 November 2003 16:56]
Ron Miller
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Re: Puzzle thread Nov. 17 |
Tue, 18 November 2003 22:06 |
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Ron | | Commander Forum Administrator Stars! AutoHost Administrator | Messages: 1231
Registered: October 2002 Location: Collegedale, TN | |
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Wow! I guess these *were* a bit harder...
Only one person submitted all 5 correct answers: overworked is this weeks Puzzle Master!
Here are the answers:
1. Minimum number of calls is 8.
Standard mathematics: f(n) = 2n-4 for n>=4. f(6) = 8
2. 3x3x3x3x3x3x2=1458
3+3+3+3+3+3+2=20
3. The answer I was looking for was 294
All the other numbers are in anagram pairs.
There is an alternate answer found by EDog: 196, its the only number that has an integral square root (14).
4. The answer is 35. The numbers represent the position of each letter T in the sentence.
Alternate answer brute forced by Mazda: 64 the series 4,13,19,21,29,64 are the first 6 values (x=1 to 6) of the quartic function (11pow(x,4) - 114pow(x,3) + 373pow(x,2) - 270x + 96 ) / 24.
5. 7:30 Bud White and Lucy Johnson
8:00 Rodney Jones and Stella Fahie
8:30 Nathan Bookman and Eva Hartley
9:00 Thaddeus Smith and Nora Cooper
9:30 Felix Flagg and Rita Sternman
5 people got #5 correct, but I still should have posted a link to instructions on how to solve logic puzzles.
3 or 4 people got #4 correct.
Thread unlocked.
[Updated on: Wed, 19 November 2003 07:56]
Ron Miller
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Re: Puzzle thread Nov. 17 |
Wed, 19 November 2003 00:49 |
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EDog | | Lt. Junior Grade | Messages: 417
Registered: November 2002 Location: Denver, Colorado, USA | |
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BlueTurbit wrote on Tue, 18 November 2003 22:35 | Yes congrats Overworked.
But I question the answer to puzzle one? I get nine not eight. Do you have a breakdown besides a math formula of how you get eight calls?
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I'd be interested in this as well. I'm not a math guy, but I do love spreadsheets. I named all six ladies and worked it out call by call. No matter how I worked it out, I came up with nine actual calls not eight.
To wit:
Amy, Brenda, Chloe, Dana, Evelyn, and Francine are the gossipers.
Call 1: Amy calls Brenda (A=AB, B=AB)
Call 2: Evelyn calls Francine (E=EF, F=EF)
Call 3: Chloe calls Dana (C=CD, D=CD)
Call 4: Amy calls Chloe (A=ABCD, C=ABCD)
Call 5: Brenda calls Evelyn (B=ABEF, E=ABEF)
Call 6: Dana calls Francine (D=CDEF, F=CDEF)
Call 7: Amy calls Evelyn (A=ABCDEF, E=ABCDEF)
Call 8: Brenda calls Dana (B=ABCDEF, D=ABCDEF)
Call 9: Chloe calls Francine (C=ABCDEF, F=ABCDEF)
How can this be done with one fewer call?
BTW, loved the marriages problem. Kept me amused for like an hour when I should have been studying.
EDog
http://ianthealy.com
Born, grew up, became an adventurer
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Re: Puzzle thread Nov. 17 |
Wed, 19 November 2003 01:17 |
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EDog wrote on Wed, 19 November 2003 06:49 |
BlueTurbit wrote on Tue, 18 November 2003 22:35 | Yes congrats Overworked.
But I question the answer to puzzle one? I get nine not eight. Do you have a breakdown besides a math formula of how you get eight calls?
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I'd be interested in this as well. I'm not a math guy, but I do love spreadsheets. I named all six ladies and worked it out call by call. No matter how I worked it out, I came up with nine actual calls not eight.
To wit:
Amy, Brenda, Chloe, Dana, Evelyn, and Francine are the gossipers.
Call 1: Amy calls Brenda (A=AB, B=AB)
Call 2: Evelyn calls Francine (E=EF, F=EF)
Call 3: Chloe calls Dana (C=CD, D=CD)
Call 4: Amy calls Chloe (A=ABCD, C=ABCD)
Call 5: Brenda calls Evelyn (B=ABEF, E=ABEF)
Call 6: Dana calls Francine (D=CDEF, F=CDEF)
Call 7: Amy calls Evelyn (A=ABCDEF, E=ABCDEF)
Call 8: Brenda calls Dana (B=ABCDEF, D=ABCDEF)
Call 9: Chloe calls Francine (C=ABCDEF, F=ABCDEF)
How can this be done with one fewer call?
BTW, loved the marriages problem. Kept me amused for like an hour when I should have been studying.
EDog
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Amy, Brenda, Chloe, Dana, Evelyn, and Francine
Amy calls Francine.
Amy calls Evelyn.
Amy calls Brenda.
Chloe calls Dana.
Amy calls Chloe.
Brenda calls Dana.
Amy calls Evelyn.
Amy calls Francine.
now everybody knows everything!!!!
...
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Re: Puzzle thread Nov. 17 |
Wed, 19 November 2003 07:59 |
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Ron | | Commander Forum Administrator Stars! AutoHost Administrator | Messages: 1231
Registered: October 2002 Location: Collegedale, TN | |
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mazda wrote on Wed, 19 November 2003 03:27 |
Ron wrote on Wed, 19 November 2003 03:06 |
1. Minimum number of calls is 8.
Standard mathematics: f(n) = 2n-4 for n>=4. f(6) = 8
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BTW, what "standard" mathematics are you referring to in question 1 ?
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That line was quoted straight from Ashlyn's PM to me. I'll let her explain it (or someone else who knows the math).
[Updated on: Wed, 19 November 2003 07:59]
Ron Miller
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