Poll

7 votes (53.84%)
5 votes (38.46%)
3 votes (23.07%)
2 votes (15.38%)
6 votes (46.15%)
1 vote (7.69%)
2 votes (15.38%)
2 votes (15.38%)
7 votes (53.84%)
5 votes (38.46%)

13 members have voted

Ace2
Ace2
Joined: Oct 2, 2017
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May 21st, 2020 at 8:22:21 PM permalink
I used Poisson. Take the integral over all time of:

(ax)^9*e^(-ax)/9!*e^(-bx)*((bx)^9/9!+(bx)^8/8!+(bx)^7/7!+(bx)^6/6!+(bx)^5/5!+(bx)^4/4!+(bx)^3/3!+(bx)^2/2+(bx)+1)*a dx

Where a = 976/1925 and b = 949/1925

Which comes to:

2419531625796339034080102292958091287906159377026607445180416 / 4611277450446459784127710060894226116943173110485076904296875 =~ 0.52
Itís all about making that GTA
Gialmere
Gialmere
Joined: Nov 26, 2018
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May 22nd, 2020 at 8:05:54 AM permalink
BOB x BOB = MARLEY

For no particular reason, this easy math puzzle celebrates Bob Marley. Ya mon.



In the multiplication problem below, each digit has been replaced by a letter. Can you determine the value of each letter?



B = ?
O = ?
M = ?
A = ?
R = ?
L = ?
E = ?
Y = ?
I = ?

Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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Thanks for this post from:
Gialmere
May 22nd, 2020 at 10:17:39 AM permalink

Note that neither B nor O are 0 or 1, as none of the intermediate lines is 0 or BOB.
Also, B cannot be 5 or 6, as B x B would end in B, but B x B ends in Y.
O x B ends in O
The (O,B) pairs where O and B are distinct digits from 2 to 9 and this is true are (2,6), (4,6), (5,3), (5,7), (5,9), and (8,6).
Since B is not 6, O must be 5.
BOB x B and BOB x O are both four-digit numbers where the first digit is the same.
757 x 5 = 3785, but 757 x 7 = 5299, so (5,7) is incorrect.
959 x 5 = 4795, but 959 x 9 = 8631, so (5,9) is incorrect.
Therefore, B = 3, and the product is 353 x 353; MEOY = 1059, MILO = 1765, and MARLEY = 124,609.

B,O,M,A,R,L,E,Y are 3,5,1,2,4,6,0,9 (and I is 7)

Gialmere
Gialmere
Joined: Nov 26, 2018
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May 22nd, 2020 at 4:22:53 PM permalink
Quote: ThatDonGuy


Note that neither B nor O are 0 or 1, as none of the intermediate lines is 0 or BOB.
Also, B cannot be 5 or 6, as B x B would end in B, but B x B ends in Y.
O x B ends in O
The (O,B) pairs where O and B are distinct digits from 2 to 9 and this is true are (2,6), (4,6), (5,3), (5,7), (5,9), and (8,6).
Since B is not 6, O must be 5.
BOB x B and BOB x O are both four-digit numbers where the first digit is the same.
757 x 5 = 3785, but 757 x 7 = 5299, so (5,7) is incorrect.
959 x 5 = 4795, but 959 x 9 = 8631, so (5,9) is incorrect.
Therefore, B = 3, and the product is 353 x 353; MEOY = 1059, MILO = 1765, and MARLEY = 124,609.

B,O,M,A,R,L,E,Y are 3,5,1,2,4,6,0,9 (and I is 7)




Correct!

Good job mon! You be jammin!
--------------------------

Some historians speculate that instead of cancer, Bob Marley died of an acute case of ganjarea.
Have you tried 22 tonight? I said 22.
gamerfreak
gamerfreak
Joined: Dec 28, 2014
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May 23rd, 2020 at 9:06:01 AM permalink
Probably too easy, but I like the question.

What is the probability of a broken clock being correct?
Ace2
Ace2
Joined: Oct 2, 2017
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May 23rd, 2020 at 9:10:12 AM permalink
Quote: gamerfreak

Probably too easy, but I like the question.

What is the probability of a broken clock being correct?

Zero. The level of precision is infinite. Like the probability of two people being exactly the same height
Itís all about making that GTA
unJon
unJon
Joined: Jul 1, 2018
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May 23rd, 2020 at 9:13:50 AM permalink
Quote: Ace2

Zero. The level of precision is infinite. Like the probability of two people being exactly the same height

And yet it happens twice a day.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
EdCollins
EdCollins
Joined: Oct 21, 2011
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May 23rd, 2020 at 9:40:44 AM permalink
Quote: gamerfreak

What is the probability of a broken clock being correct?



My answer would be 0.0000231481481481481

24 hours x 60 minutes x 60 seconds = 86,400 seconds each day.

86,400 / 2 = 0.0000231481481481481
CrystalMath
CrystalMath
Joined: May 10, 2011
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May 23rd, 2020 at 9:52:16 AM permalink
While a broken clock is correct twice a day, most working clocks are always wrong.
I heart Crystal Math.
Gialmere
Gialmere
Joined: Nov 26, 2018
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May 24th, 2020 at 4:17:43 AM permalink


Meanwhile, at the pool hall, you place 15 balls into a snug, exactly fitting triangle rack...


You remove one of the balls and notice that all the balls left in the rack are still immobile and locked in place. How many balls can you remove from the rack and still have all the balls remaining stay (at least theoretically) locked in place.




When you set up to re-rack the balls, you're stunned to see that the side lengths of the triangle rack have all magically decreased by 20%. (This is what happens when you shoot pool after visiting the NuWu Marketplace.) How many normal size balls will fit into this smaller rack?



Have you tried 22 tonight? I said 22.

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