The worlds first riddle!

Mark, I hear that; that is why the quote was in bold as the wording leaves no room for creativity.



Well shiver me timbers and swash my buckle the Silver Surfer; one of the noblest of cosmic entities in the universe rides to the rescue… Or does he?





Thank you for your view of the problem ID, I think many would agree with your way of thinking.

BTW: Talking of 911 – I just put a deposit down on a brand new Porsche & mentioned it on Facebook, I said, "I can't wait for the new 911 to arrive!", next thing I know, 60,000 Muslims have added me as a friend!



I have just invented a die, bearing the numbers 0,1,2,3,4,5 on its faces. The more astute amongst you will have noticed the absence of a 6, and the inclusion of 0.

Now if I repeatedly throw the die until the total of the throws first exceeds 12.

What is the most likely total that will be thus obtained?


Edit: Thank you Missy, it is not 69 sweetie… But what the heck – that’s close enough!

die:
I get 14 by a whisker over 13. The probability for a given number is roughly 1/3 when you get past 7. For 14, I get 0.334882859581440. For 13, I get 0.334508409651200.
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die correction: I messed up. I was computing the probabilities of getting to 13-17 with no other constraints. For instance, I failed to exclude getting to 14 from 13. My new answer is 13:
13: 0.3345084097
14: 0.2679811777
15: 0.1996684993
16: 0.1315172852
17: 0.0663246283

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By a Felis catus whisker Mark redeems one of his seven remaining lives. His ‘old’ answer was close; but his new answer is in the pocket!

Consider the throw before the last. The total after that throw must be 12, 11, 10, 9, or 8. If the total is 12, the final result must be 13, 14, 15, 16 or 17, with an equal chance for each.

Similarly, if the total is 11, the final result must be 13, 14, 15, 16 with an equal chance for each and so on.

It is now clear that the most likely final result is 13.
If 12 is replaced by N>3, the most likely total is N+1.


I was lying in bed this morning cogitating on whether it was worth gnawing through the retaining straps, when by remarkable coincidence I found that the probability of me getting to first base with Missy is exactly the same as that if the digits 0, 1, 2, …, 9 were placed in RANDOM order in the blank spaces of:

5_383_8_2_936_5_8_203_9_3_76

The resulting number would be divisible by 396.

Without asking Missy, what is that probability?

Missy:
Take your base. The probability is 1.
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Geez Louise! Don’t ya just hate it when a guy can answer a question faster than the time it takes me to type it!

Slow down already; you are making it look like these are easy questions.

Mark wrote, “Take your base.”
We sit down and await a musical interlude…
Or possibly a jam session …

The number 76 formed by the last two digits is divisible by 4.
The difference between 73, the sum of the even-placed digits and 17+45, the sum of the odd-placed digits, is divisible by 11, regardless of the order in which the blanks are filled.

The sum of all the digits, 90 + 45, is divisible by 9.
It follows that the number is divisible by (4) (11) (9) or 396. The probability is therefore 1.

And what did the mysterious Mark have to say on the matter, “The probability is 1.”

We can’t both be wrong, so you can take that one to the bank.



I want to share with y’all a polynomial f (x) with integer coefficients and tell you that today is my friends daughter’s birthday.

When her age A is substituted for x, then f (A) = A.

You will note also that f (0) = P and that P is a prime number greater than A.

How old is his daughter?


WHAT! You don’t think that’s enough information to formulate an answer!
Ok then; he lives in Chicago and his daughter is a real sweetie…

daughter's age:
She's 1.
f(x) can be rewritten as x*g(x) + P
so, A*g(A) + P = A
P = A*(1-g(A))
since P is prime, A must be 1 or P
since it was given that P>A, A must be 1
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That sounds like way too much work for just first base. I think your aim should be further.

Try: The first base coach is waving you on. Will you slide into second, or try to stretch it into a triple or a home run?

Try is smart to play to play it safe. He never knows what to expect.

Since Marks answer is sublime (just like Missy) it must see the light of day so others may bask in its glory.

She's 1.
f(x) can be rewritten as x*g(x) + P
so, A*g(A) + P = A
P = A*(1-g(A))
since P is prime, A must be 1 or P
since it was given that P>A, A must be 1

________________________________________

Eye on the other hand have reached the same conclusion…

Since f (0) = P,
f(x) = x.q(x) +p and f(A) = A.q(A) +P =A.
Consequently, A divides P. Since P>A and P is prime, A = 1year.

x 3 – 3x2 + 3 is one of an infinite class of polynomials that could have been written.






Questions, questions; Mark, always you are making with the questions already! Are you trying to set me on the Road to Perdition!

Hell, it’s bad enough being in a state of eternal punishment and damnation into which a sinful and unpenitent person passes after death. I had no idea A2K was the waiting room!

As a child of God and born of his image – although why he should want to look like a corpulent Hillbilly I have no idea – I must beware the sins of the flesh and the temptation of lust.

I shall consult the Holy Scriptures for divine guidance…
The book of Exodus describes the Ten Commandments as being spoken by God, inscribed on two stone tablets by the finger of God – Well that’s good enough for me!

You shall not covet your neighbor's house. = Check!
His manservant or maidservant. = Check!
His ox or donkey. = Check!
— Exodus 20:17

So, no mention of not coveting your neighbor’s ass! – Houston, I’m good to go…

However, I had better double check; what say you Mark?

“The second is this: ‘You shall love your neighbor as yourself.’
There is no other commandment greater than these.” - Mark 12:31

Hallelujah! Thanks Mark, my poor eyesight is testament to how much I love myself!!!

If its Gods will; I must obey – I’ll slide into second base to get my breath back before I start to begat …


Butt first; what is going on with this weather! First it was global warming, then climate change…

Well, during a period of days I observed that when it rained in the afternoon, it had been clear in the morning, and when it rained in the morning, it was clear in the afternoon.

It rained on 9 days, and was clear on 6 afternoons and 7 mornings.

How long was this period?

rain:
11 days
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You are so hot when you start spouting scripture. Seriously.

It Never Rains in Southern California…
http://youtu.be/-pyC7WnvLT4

But when it does Mark is there to count them…

There must have been …
1 ( 6 + 7 – 9) or 2 days.
2
So there were 9 + 2 or 11 days in the period.

Oh Mis, a little chanting to get you in the mood…
http://youtu.be/ojI3QB-U0Nw

Then we can turn our attention to the Kama Sutra – Hallelujah!

Meanwhile, through a cacophonie of sound I ponder if there is an integral arithmetic progression with an arbitrarily large number of terms such that no term is a perfect r th power for r = 2, 3, …, n?

What with all the chanting and reading, Mis and I will be somewhat busy over the weekend; so iffin’ y’all ain’t so lucky as what I iz –

You may wish to peruse my most problematical enigma ever!





Have a great weekend guys.

arithmetic progression
There are no non-trivial powers of integers that are equal to 3 mod 9. Therefore, such a progression is:
3, 12, 21, 30, ...
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polygon:
Using Pick's Theorem, I get:
42 + 14/2 - 1 = 48 square units
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arithmetic progression:
An even denser one: There are no powers of integers that are equal to 2 mod 4.
Therefore, this is a solution:
2, 6, 10, 14, 18, ...
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Yet again Mark hits the ball right outta the park.
Add two home runs to his score!

The progression 2, 6, 10, … ,(4k + 2), … can contain no perfect powers whatsoever.

For a power of an odd integer is odd, and a power of an even integer must be divisible by 4.

(There is also an obvious trivial solution in which the first term is a nonpower and the common difference is zero.)



As for the departed Polygon:

Mark is on track because Pick’s theorem states that the area of any simple polygon whose vertices are lattice points is given by the formula:
b/2 + c – 1

Where b is the number of lattice points on the boundary while c is the number of lattice points inside.
Thus the area of the polygon in the figure is 14/2 + 42 – 1, or 48.

Congrats to Mark.



Can you help me find it, I know it’s around here someplace; it is a positive number such that
1/5 of it multiplied by 1/7 of it equals the number?

found:
35
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I get up in the middle of the night to post so Mark won’t see it, and what happens!

He takes a whole nine minutes to find it, read it, think, type the correct answer and post it!!!

If (N/5) (N/7) = N, then N (N-35) = 0, so N = 35.
Otherwise, if instead of multiplying 1/5 of the number by 1/7 of the number, the number were multiplied by itself, the product would be 35 times as great as by the other method, so the number is indeed 35.

Is this guy a machine or what!


Perhaps he is a farmer; that would explain why he is up so early.
In which case he must buy 100 head of animals with $100.
Calves cost $10 each, lambs cost $3 each and pigs cost 50 cents each.

If he buys at least one of each kind of animal, how many of each does he buy?