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The worlds first riddle!

 
 
Adrian
 
  1  
Reply Thu 28 Sep, 2006 09:09 pm
8.46666667 x 10^10 zeptometers per shake. Which sounds fast but is actually quite slow...unless you're a snail.
0 Replies
 
markr
 
  1  
Reply Thu 28 Sep, 2006 09:57 pm
[size=8]SPIDER & FLY
10 feet in 6 minutes is 1/3 inch per second.
[/size]
0 Replies
 
thoh13
 
  1  
Reply Thu 28 Sep, 2006 10:17 pm
markr is correct....
was the shortest distance really that obvious? lol
0 Replies
 
markr
 
  1  
Reply Fri 29 Sep, 2006 12:50 am
Yes. Although I noticed that you tried to disguise it by stating the distance from the far wall.
0 Replies
 
Tryagain
 
  1  
Reply Fri 29 Sep, 2006 09:41 am
Mark:

TESTS
I get:

M = 1.5 * (N+1) * (G - Q/3) - (N*T)

[I am not saying I am right, but see below how close we are and decide.]


M = 71 for the given situation. Cool



For an overall grade of 83, and a quiz average of 89, the student must have an 80 test average. Since the student had an 83 test average before, she must have gotten a 71 on the next test:
(3 * 83 + M)/4 = 80

The general formula would be:
M = ( (N+1)(3G - Q) - (2NT) ) / 2

So in the problem above, where N = 3, G = 83, Q = 89, and T = 83,
M = ( (3+1)(3*83 - 89) - (2*3*83) ) /2 = 71




Quote, "markr is correct.... was the shortest distance really that obvious?"

Reply quote, "Yes. Although I noticed that you tried to disguise it by stating the distance from the far wall."


Good sneaky move Thoh, although you will have to kick it up another notch to put one over these Guys. Laughing





I am feeling lazy today, so can you simplify the following product (Answer must be left in simplest form )

(x - a)*(x - b)*(x- c)*...*(x - z) = Question





Two motorists travel from Winchester to Berryville, a distance of 15 miles. One travels at the legal speed limit of 55 mph while the other motorist travels at 75 mph.

If they both left Winchester at the same time, how much sooner would the law-breaking motorist arrive in Berryville Question
0 Replies
 
thoh13
 
  1  
Reply Fri 29 Sep, 2006 12:40 pm
simplify:

0 because u get to (x-x) which is 0


motorists:

the one who speeds will arrive after the slower one since he gets a speeding violation and/or jail time
0 Replies
 
markr
 
  1  
Reply Sat 30 Sep, 2006 01:46 am
Tryagain wrote:
I am not saying I am right, but see below how close we are and decide.


You are right. So am I. The formulas are equivalent.
0 Replies
 
Tryagain
 
  1  
Reply Sat 30 Sep, 2006 10:47 am
0 Replies
 
markr
 
  1  
Reply Sat 30 Sep, 2006 02:51 pm
[size=8]CROSSNUMBER

X94X
1357
2622
5X49
[/size]
0 Replies
 
Tryagain
 
  1  
Reply Sun 1 Oct, 2006 08:09 am
Mark:

CROSSNUMBER

X94X
1357
2622
5X49
XCool CoolX


You need to use some logic to solve the puzzle.
Begin with 3 ACROSS:
The numbers must be 2468, 1357, 3579, 9753, 7531, or 8642.
Look at 5 ACROSS: The digits must include one of the following:
one 6 and three 2's, or two 4's and two 2's.
Clue #3 DOWN must be either 125, 216, 343, 512, or 729.

Or wait for Mark to post.



The cop's were convinced that one of the following A2K members -- Dick Clarke, Jack Hand, Sally Shera or Jimmy Hood -- had stolen the keys to the executive washroom.

Each of the suspects made a statement, but only one of the statements was true.

Dick said, "I didn't do it."
Jack said, "Dick is lying."
Sally said, "Jack is lying."
Jimmy said, "Jack did it."

Who committed the crime Question





A B C = C^4, B C A = D^4 Question
0 Replies
 
markr
 
  1  
Reply Sun 1 Oct, 2006 06:21 pm
[size=8]MISSING KEYS
Jack is telling the truth. Dick did it.


Well, given that there are only two 3-digit fourth powers, this was rather trivial.

625 = 5^4, 256 = 4^4
[/size]
0 Replies
 
thoh13
 
  1  
Reply Sun 1 Oct, 2006 06:23 pm
625=5^4; 256=4^4
0 Replies
 
Tryagain
 
  1  
Reply Mon 2 Oct, 2006 09:07 am
Mark:

MISSING KEYS
Jack is telling the truth. Dick did it. Cool Cool


You have one of two ways to approach this problem.

One way is to assume that one of the statements is true and that the others are false, and see what you can conclude from that. For example, suppose that Dick's statement is true. Then, Jack's is false, but Sally's is true, which contradicts the information that only one statement is true. So some other statement must be the true one, and Dick's is false. So, Dick committed the crime!

Another way to attack the problem is to assume that one of the suspects committed the crime and determine which statements are true and false. Suppose Dick was the criminal. Then Dick was lying. So his statement was false, Jack's was true, Sally's was false, and Jimmy's was false. Since there are three false statements and one true, this fits and Dick was the criminal! (Of course, you should check the other possibilities in case there were two criminals!)



Well, given that there are only two 3-digit fourth powers, this was rather trivial.

625 = 5^4, 256 = 4^4 Cool

Any difficult questions are easy if you know the answer, just as the easiest question is difficult if you don't know the answer.


Thoh;

625=5^4; 256=4^4 Cool






Can you reconstruct the multiplication, in which all the A=3 are given Question

Code:


. . . . . A
A . . . . . .
---------------
A . A . A . J
. . . . . A
. . . A . . N
. . . A . V
. A . . . . I
. . . . . A E
J A N V I E R
---------------------------
. . . . . . . . . . . . .








Something far easier:

In the traditional holiday carol called " The Twelve Days of Christmas," what is the total number of gifts given Question
How many of each gift were given in this song Question

Here is the carol:

On the first day of Christmas my true love gave to me a partidge in a pear tree.

On the second day of Christmas my true love gave to me two turtle doves and a partidge in a pear tree.

On the third day of Christmas my true love gave to me three French hens, two turtle doves and a partidge in a pear tree.

On the fourth day of Christmas my true love gave to me four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the fifth day of Christmas my true love gave to me five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the sixth day of Christmas my true love gave to me six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the seventh day of Christmas my true love gave to me seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the eighth day of Christmas my true love gave to me eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the ninth day of Christmas my true love gave to me nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the tenth day of Christmas my true love gave to me ten lords-a-leaping, nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the eleventh day of Christmas my true love gave to me eleven pipers piping, ten lords-a-leaping, nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.

On the twelfth day of Christmas my true love gave to me twelve drummers drumming, eleven pipers piping, ten lords-a-leaping, nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five gold rings, four calling birds, three French hens, two turtle doves and a partidge in a pear tree.
0 Replies
 
thoh13
 
  1  
Reply Mon 2 Oct, 2006 12:01 pm
total: 364 gifts:
12 partridges
22 doves
30 hens
36 calling birds
40 rings
42 geese
42 swans
40 swans
36 maids
30 lords
22 pipers
12 drumers
0 Replies
 
markr
 
  1  
Reply Mon 2 Oct, 2006 10:45 pm
[size=7]JANVIER
1,826,871,909,081

GIFTS
Total = the 12th tetrahedral number.
[/size]
0 Replies
 
Tryagain
 
  1  
Reply Tue 3 Oct, 2006 02:47 pm
Code:


G H O S T S
G H O U L S
+ G O B L I N S
---------------
O O O O O O S

0 Replies
 
Adrian
 
  1  
Reply Tue 3 Oct, 2006 10:21 pm
That 12 days of christmas thing is a load of rubbish.

Quote:


Beta-2-microglobulin [Homo sapiens] = Alpha-2-macroglobulin pseudogene [Homo sapiens] x alanine-glyoxylate aminotransferase (oxalosis I; hyperoxaluria I; glycolicaciduria; serine-pyruvate aminotransferase) [Homo sapiens]

Consecutive digits makes it pretty simple.

The other one will take a bit longer.
0 Replies
 
thoh13
 
  1  
Reply Wed 4 Oct, 2006 12:34 am
567 = 3 x 189
0 Replies
 
Tryagain
 
  1  
Reply Wed 4 Oct, 2006 10:14 am
567 updated 02-Oct-2006

Summary
Official Symbol B2M
Official Full Name beta-2-microglobulin
Primary source HGNC:914
See related HPRD:00189; MIM:109700
Gene type protein coding
RefSeq status Validated
Organism Homo sapiens
Lineage Eukaryota; Metazoa; Chordata; Craniata; Vertebrata; Euteleostomi; Mammalia; Eutheria; Euarchontoglires; Primates; Haplorrhini; Catarrhini; Hominidae; Homo


The rest is also correct, believe me. Laughing




Thoh:

Comes to the rescue with the English version:

567 = 3 x 189 Cool





( G H O S T S)

"….will take a bit longer."


Woohoo!!! At last, about time too.




Using only the symbols for addition, subtraction, multiplication, division, and parentheses:
+, - , x, /, ( ), arrange four fours to form all the numbers from 1 to 10 Question

For example, 0 = 4 + 4 - 4 - 4




Make yourself a neat 3x3 square and place the nine numbers 335, 418, 501, 584, 667, 750, 833, 916, and 999 in the nine squares so that each row, each column, and each diagonal adds up to 2001 Question
0 Replies
 
thoh13
 
  1  
Reply Wed 4 Oct, 2006 05:46 pm

1 = 4/4+4-4
2 = 4*4/(4+4)
3 = (4+4+4)/4
4 = (4-4)/4+4
5 = (4+4*4)/4
6 = 4+(4+4)/4
7 = 4+4-(4/4)
8 = 4+4+4-4
9 = 4+4+4/4
10 =(44-4)/4
0 Replies
 
 

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