Jigal van Hemert wrote:
Table:
CREATE TABLE `age1` (
`age1` tinyint(2) NOT NULL auto_increment,
PRIMARY KEY (`age1`)
) TYPE=MyISAM AUTO_INCREMENT=21 ;
Data:
INSERT INTO `age1` VALUES (1), (2), (3), (4), (5), (6), (7), (8), (9), (10),
(11), (12), (13), (14), (15), (16), (17), (18), (19), (20);
Query:
SELECT * , a.age1 + b.age1 + c.age1 as sumage
FROM `age1` a
JOIN `age1` b
JOIN `age1` c
WHERE a.age1 * b.age1 * c.age1 =36
ORDER BY sumage
We now have to search for combinations that
- have more than one unique set of sorted ages for a certain sum (since
Boris didn't know the ages while he knew the sum)
- have only one highest age (since there is an "eldest")
Looking at the output this would only be true for ages 9, 2 and 2
Yes but 6,6,1 doesn't necessarily mean that the two children
aged 6 are twins - one could have been born in January and one
in December in the same year and hence have the same age (in
years) on December 31st - but the one born in January would
clearly be the eldest.
Regards, Jigal.
----- Original Message -----
From: "Tom Roos" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, March 08, 2004 9:31 AM
Subject: SQ puzzle
SQL puzzle from the Informix mailing list:
Two mathematicians (Boris and Vladimir) met accidently for the first time in
20 years.
They greet each other and begin catching up on their respective lives.
Boris asks Vladi "Do you have any children?"
"Yes" replies Vladimir, "I have three."
"How old are they?", asks Boris.
"The product of their ages is 36 and the sum of their ages is equal to the
number of windows on that building across the street."
Boris looks at the building, counts the windows then says "Vladi, that still
doesn't tell me the ages."
"Ah, says Vladi, then I must tell you that the eldest has red hair."
"Oh", says Boris, "now I know their ages."
What are the ages of Boris' children?
Create a table, load it with data, and write a single SQL statement to
produce the data set required to deduce the answer.
Of course the problem cannot be solved with M$ SQLServer :-)
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