math problem

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mklim_irlpl
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Doesn't it cost anything from 0.01 to 0.1? :D
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gutuami
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You have Albert, Bernard and Cheryl. Cheryl says "my birthday is one of these ten dates"

May 15 May 16 May 19
June 17 June 18
July 14 July 16
August 14 August 15 August 17

She gives Albert the month of her birthday. And Bernard the number.

Then the following conversation occurs:

1) Albert: I don’t know when the birthday is, but I know Bernard doesn’t know too.
2) Bernard: At first I don’t know when the birthday is, but now I know.
3) Albert: Then I know the birthday too.

From that information, work out Cheryl's birthday.
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gutuami
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gutuami wrote:You have Albert, Bernard and Cheryl. Cheryl says "my birthday is one of these ten dates"

May 15 May 16 May 19
June 17 June 18
July 14 July 16
August 14 August 15 August 17

She gives Albert the month of her birthday. And Bernard the number.

Then the following conversation occurs:

1) Albert: I don’t know when the birthday is, but I know Bernard doesn’t know too.
2) Bernard: At first I don’t know when the birthday is, but now I know.
3) Albert: Then I know the birthday too.

From that information, work out Cheryl's birthday.
solution: https://www.youtube.com/watch?v=emiMj8cCL5E
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marksmeets302
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Awesome!
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marksmeets302
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problem.gif
You have to walk from point A to point B 10 times, blindfolded. Every time before you start walking, a fair coin is tossed, and if it comes out heads, a wall in the middle is present (situation 2 in the picture), while if it comes out tails it is not (situation 1 in the picture). As you are blindfolded, you do not know if the wall is present or not.
Even though you are blindfolded, you can move in perfectly straight lines towards any point you choose. Can you give a prescription how to walk from A to B so that the total distance you cover is as small as possible?
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gutuami
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if you take the round route that would be √2*2*10=~28.2m
if you take the 50% chance route then:
min=10*2=20m
max=(1+1+√2)*10=~34.1m
avg=27.05m

the standard deviation depends on sample size. For ten times coin toss it is √2.5=~1.58

so taking chances will give large deviation from 27.05m

on 10 walks only I would take the round routes.
on 1000 walks I would take the coin toss routes
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gutuami
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Imagine a scientist deals four cards out in front of you. Unlike normal playing cards, these have single numbers on one side and single colors on the other. You see from left to right a three, an eight, a red card, and a brown card. Now he says, “I have a deck full of these strange cards, and there is one rule at play. If a card has an even number on one side, then it must be red on the opposite side. Now, which card or cards must you flip to prove I’m telling the truth?”
Remember — three, eight, red, brown — which do you flip?
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marksmeets302
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Regarding the pop-up wall, there's another solution.
PHOTO_20150528_095503_resized.jpg
If you want to walk from A to B, not knowing if the wall pops up or not, walk straight to point X, taking path p0. If the wall is down, take path p3 to arrive at B. If the wall is up, walk around it, by taking path p1 followed by p2.

So it's a matter of minimizing half the length of p0 + p3 plus half the length of p0 + p1 + p2.

According to pythagoras the distance D(x) is

D(x) = 0.5 * 2 * sqrt(1+x^2) + 0.5*(sqrt(1+x^2) + 1 - x + sqrt(2) )

The minimum of D(x) can be found by differentiating and looking where D'(x) equals zero.

D'(x) = 3x / (2*sqrt(x^2 + 1)) - 1/2

D'(x) = 0 when x = 1 / (2*sqrt(2))
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marksmeets302
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The rule is of the form P implies Q. So if we check the 8 card (P), it must be red (Q). If it isn't you are not telling the truth. Given a brown card (not Q), we have P -> Q and (not Q), which leads to (not P). So if the brown card is even, you are not telling the truth.

Flipping the other cards gives no information.

If based upon the 8 and the brown card you are still telling the truth.. then I don't know how to definitively prove you are telling the truth. There may still be an even black card in the rest of the deck. I must be missing something here :-)
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gutuami
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The only answer is to turn over both the eight card and the brown card. If you replace the numbers and colors on the cards with a social situation, the test becomes much easier. Pretend the psychologist returns, and this time he says, “You are at a bar, and the law says you must be over twenty-one years old to drink alcohol. On each of these four cards a beverage is written on one side and the age of the person drinking it on the other. Which of these four cards must you turn over to see if the owner is obeying the law?” He then deals four cards which read:
23—beer—Coke—17
Now it seems much easier. Coke tells you nothing, and 23 tells you nothing. If the seventeen-year-old is drinking alcohol, the owner is breaking the law, but if the seventeen-year-old isn’t, you must check the age of the beer drinker. Now the two cards stick out—beer and 17.
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marksmeets302
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Finish this sequence:

10, 11, 12, 13, 14, 21, 100, 1001

This one is also fun. What comes next in this sequence:
1, 11, 21, 1211, 111221
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SeaHorseRacing
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As the racing is so crap.

Answer 1 = 10002

Answer 2= 312211
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jimibt
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Location: Narnia

SeaHorseRacing wrote:As the racing is so crap.

Answer 1 = 10002

Answer 2= 312211
yeah, agreed pretty crap racing, so google to the rescue:

https://oeis.org/A001731 (111111111 is next here mind you!!)
https://en.wikipedia.org/wiki/Look-and-say_sequence

:D :D
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marksmeets302
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Yep, the first sequence is all nines, but in different bases. 10 = 9 in base 9, 11 is 9 in base 8 etc. 111111111 is 9 in base 1.

For the second sequence: 11 is pronounced as "2 ones", therefore the next in line 21. This is read as "1 two and 1 one", so the next one is 1211, etc.
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jimibt
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Location: Narnia

Not a maths puzzle per se, but actually a maths family heirloom passed down from my father to me and subsequently to my 4 kids.

This little visual formula works on the mysterious qualities of the multiplier of 11 (eleven). Basically, I introduced it to my own kids by asking them to write down 5 random 3 or 4 figure numbers on a piece of paper and then told them to grab a calculator. My challenge was then for them to write down the value of the random numbers on the sheet when multiplied by 11 before I verbally announced the results of each. I of course won every time as the mystical visual maths when using 11 is actually pretty straightfwd.

Here is a typical example of the random numbers that I would be challenged with:

453, 1427, 4263, 363, 906 (etc..)

I would of course instantly (in those days!!) respond with:

4983, 15697, 46893, 3993, 9966

They thought I was some kind of savant until I explained the *secret* behind this wizardry. It's just a simple case of adding the digits of all the sequence together to form the result as follows:

1427 * 11 = 15697 (take the 1st digit of the challenge number [1], then, add together each pair of adjacent digits and append the last digit to finalize)


i.e. 1 followed by 1st + 2nd(5), 2nd + 3rd(6) and 3rd + 4th(9). finally, take the last digit (7) and put it on the end of the result.

Hopefully, you get the gist and will borrow this as your family maths heirloom :)

Caveat - in order to keep this little puzzle simple, always ensure that adjacent digits add up to no more than 9, otherwise you have to do some further mental add and carry arithmetic; ok if you're sub 35, not nowadays alas :D
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