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The worlds first riddle!

 
 
Stormwatch
 
  1  
Reply Thu 7 Sep, 2006 09:17 am
The clock at the station at the end of the first leg was five minutes fast.
0 Replies
 
Tryagain
 
  1  
Reply Thu 7 Sep, 2006 03:11 pm
Mark:


In general, the probability of selecting exactly K of R items chosen from a set of N items is:

C(R,K) * C(N-R,R-K) / C(N,R)

Note that when K=R, the numerator is 1, and we have the probability for the original problem.

In this case, the answer is:
C(6,3) * C(30,3) / C(36,6) = 0.0417 Cool




The number of ways that 3 numbers can be chosen from 6 is 20. The number of ways that 3 numbers can be chosen from the remaining 30 is 4,060. Multiplying these 2 numbers we obtain 81,200, which is the total number of ways 6 numbers can be chosen from 36 so that exactly 3 numbers will match the winning six. There are 1,947,792 possible combinations of choosing 6 numbers from 36 so the probability of having a ticket that matches 3 of the winning numbers is 81,200/1,947,792 which equals .04168822954. This means that the odds are about 1 in 24.



Stormy:
The clock at the station at the end of the first leg was five minutes fast. Razz

Good effort Stormy, this is how I see it, which due to the confusion amongst the boys has made me doubt the answer and you may be right.
The station clock is 3 minutes fast. The morning journey took 65 minutes, and the evening journey therefore took 52 minutes, and the train arrived 57 minutes after it should have left, that is, 3 minutes early.



Summing the powers of 2, using each power only once, any positive integer may be obtained. For example using the numbers 1, 2, 4, 8, 16, and 32 any number from 1 through 63 can be formed. For example, 9= 8+1, 10= 8+2, 11= 8+3, and so on.

(Remember that each power of 2 can be used only once in each sum and a sum such as 4 + 4 + 2 =10 would not be allowed).

So, given these rules, how would the number 1,000,000 be obtained Question
0 Replies
 
stuh505
 
  1  
Reply Thu 7 Sep, 2006 04:42 pm
1000000 base 10
= 11110100001000000 base 2
= 2^6+2^11+2^13+2^14+2^15+2^16
0 Replies
 
Tryagain
 
  1  
Reply Fri 8 Sep, 2006 08:13 am
Stu:

1000000 base 10
= 11110100001000000 base 2 Cool
= 2^6+2^11+2^13+2^14+2^15+2^16





The easiest way to solve this (as Stu points out) would be to convert 1,000,000 to binary which is 11110100001001000000. Now, wherever a '1' appears, convert that to its power of 2. The first '1' equals 219, the next 218, then 217, 216, 214, 29 and 26. These numbers equal:
524,288
262,144
131,072
65,536
16,384
512
64
The sum of the above 7 numbers equals 1,000,000






What 2 numbers (neither of which contain any zeroes) Shocked when multiplied together equal 1,000,000,000 Question
0 Replies
 
thoh13
 
  1  
Reply Fri 8 Sep, 2006 08:34 am
2^9 * 2^9 = 512 * 1953125 = 1000000000
0 Replies
 
Tryagain
 
  1  
Reply Fri 8 Sep, 2006 10:19 am
How did you do that so quick, I thought it was rather difficult, but then, I have been wrong before Laughing


There recently was a total lunar eclipse, so here is something appropriate. The Sun has a diameter of 864,000 miles, the Earth to Sun distance is 93,000,000 miles, and the Earth's diameter is 8,000 miles.

What is the length of the Earth's shadow Question
0 Replies
 
markr
 
  1  
Reply Fri 8 Sep, 2006 10:56 am
[size=8]SHADOW
Ignoring the bending of light due to gravity, 10,568,182 miles.
[/size]
0 Replies
 
markr
 
  1  
Reply Sat 9 Sep, 2006 01:16 am
For what prime values of p is (4^p) + (p^4) prime?
0 Replies
 
Tryagain
 
  1  
Reply Sat 9 Sep, 2006 03:02 pm
markr wrote:
For what prime values of p is (4^p) + (p^4) prime?
Shocked



The expression (4^p) + (p^4) is never prime Laughing
0 Replies
 
markr
 
  1  
Reply Sat 9 Sep, 2006 03:33 pm
I'm from Missouri. Show me (prove it).
0 Replies
 
Tryagain
 
  1  
Reply Sat 9 Sep, 2006 03:37 pm
Woops, Where did this go?


Thoh:


What 2 numbers
2^9 * 2^9 = 512 * 1953125 = 1000000000 Cool




Mark:

SHADOW
Ignoring the bending of light due to gravity, 10,568,182 miles.



This is how I have it:

Draw a line from the 'top' of the Sun to the 'top' of the Earth. Drop a perpendicular from the 'top' point on the Sun. From the 'top' point of the Earth, draw a perpendicular to the line just drawn on the Sun. This forms a right triangle whose sides are 428,000 and 93,000,000 miles. (Notice that one side is NOT 432,000 miles (the Sun's radius) because the radius of the Earth must be subtracted from this distance. The ratio of the sides of the triangle just formed is .004602150538 (428,000/93,000,000).

Now draw a perpendicular from the center of the Earth. Extend the 'top of the Sun Earth line' until it intersects the perpendicular just drawn. This forms a triangle that is similar to the triangle in the previous paragraph. (In geometry, similar triangles have the same shape and angles but different sizes). One side of this right triangle is 4,000 miles and the other side will be 'x'. Since these triangles are similar, then 428,000/93,000,000 = 4,000/x. Therefore 'x' = 869,158.878 miles.

This is the length of the Earth's shadow as measured from the Center of the Earth. Subtracting the radius of the Earth (4,000 miles), the Earth's shadow measured from the Earth's surface is 865,158.878 miles.

It's just that easy! Or was it?




An A2K Lottery game is played by choosing a 4 digit number. (Any number from 0000 through 9999). On one night, the winning number was exactly the same as the previous night's number. This prompted the A2K Lottery Director to say "Since the odds of picking the correct 4 digit number are 1 in 10,000, then the odds of the same number occurring 2 nights in a row are 1 in 100,000,000" (10,000 times 10,000).

Was the Director's statement correct Question
0 Replies
 
Tryagain
 
  1  
Reply Sat 9 Sep, 2006 03:48 pm
(For what prime values of p is (4^p) + (p^4) prime?)


Mark wrote, "I'm from Missouri. Show me (prove it)."


If you are from Missouri you would already know the answer to that. Therefore, you must be from further west. Laughing

Time is tight, but I think I can come up with a convincing prognosis.
0 Replies
 
markr
 
  1  
Reply Sat 9 Sep, 2006 07:02 pm
SHADOW
Embarrassed Boy did I mess that up! I used 43200 (instead of 432000) AND I used 8000 (instead of 4000) in one place and 4000 in another. Your answer is, of course, correct.

[size=8]LOTTERY
The director's statement is incorrect. Whatever number is chosen one night has a 1 in 10,000 chance of being chosen the next night.
[/size]
0 Replies
 
Tryagain
 
  1  
Reply Sun 10 Sep, 2006 02:54 am
(For what prime values of p is (4^p) + (p^4) prime?)

(prove it)."


Okay doky, how about:

Code:
For any prime, p 5, p 1, 3, 7, 9 mod 10.
If p 1, 9 mod 10, then p^2 1 mod 10, and p^4 1 mod 10.
If p 3, 7 mod 10, then p^2 -1 mod 10, and p^4 1 mod 10.
Therefore, for p 5, p^4 1 mod 5.

Note that 4^1 4 mod 5. Now if 4^m 4 mod 5, then multiplying both sides by 4^2 gives 4^m+2 4 mod 5. That is, if n is odd then 4^n 4 mod 5.
Hence, for p 5, 4^p + p^4 0 mod 5, and cannot be prime.

When p = 2, 4^p + p^4 = 32.
When p = 3, 4^p + p^4 = 145.
When p = 5, 4^p + p^4 = 1649 = 17 97.

Ipso facto (4^p) + (p^4) is never prime.


That's it, I'm off to the bunkhouse for some shuteye.
0 Replies
 
stuh505
 
  1  
Reply Sun 10 Sep, 2006 08:03 am
Your proof seems to be missing some symbols...
0 Replies
 
markr
 
  1  
Reply Sun 10 Sep, 2006 10:53 am
It does, but the line of reasoning is correct.
0 Replies
 
Tryagain
 
  1  
Reply Sun 10 Sep, 2006 11:12 am
0 Replies
 
Tryagain
 
  1  
Reply Sun 10 Sep, 2006 11:58 am
Mark:

LOTTERY
The director's statement is incorrect. Whatever number is chosen one night has a 1 in 10,000 chance of being chosen the next night. Cool


The statement was wrong for 2 reasons.

1) The first number drawn would not have any restrictions on it. In other words the first number could be ANYTHING from 0000 through 9999. The number drawn on the next night would then have to match the previous night's number and so the odds would only be 1 in 10,000.

2) Depending on how long the A2K Lottery Daily Number has been in existence, the odds of the same number occurring on consecutive nights decreases from 1 in 10,000.



SHADOW


To err is human, that is why I am far more human than you. Laughing




A small boat is floating in a swimming pool. You have a rock which you will throw into the pool. Will the water level rise higher if:

a) The rock is thrown into the boat Question
b) The rock is thrown into the pool Question
c) The level would rise equally if the rock were thrown into the boat or directly into the pool Question
0 Replies
 
stuh505
 
  1  
Reply Sun 10 Sep, 2006 02:46 pm
Tryagain wrote:
A small boat is floating in a swimming pool. You have a rock which you will throw into the pool. Will the water level rise higher if:

a) The rock is thrown into the boat Question
b) The rock is thrown into the pool Question
c) The level would rise equally if the rock were thrown into the boat or directly into the pool Question


The difference in displacement due to the boat having additional weight of the rock is dependent on many factors about the boat, but the rock will contribute a known amount of displacement. So it depends on the boat I believe.
0 Replies
 
thoh13
 
  1  
Reply Sun 10 Sep, 2006 04:57 pm
in the boat....because in the boat it will displace the mass of the rock while in the pool will just dispace its volume
0 Replies
 
 

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