Easy math puzzles

Quote: Wizard

There are only 10 kinds of people -- Those who get binary arithmetic and those who don't.

Quote: charliepatrick

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It's all about the base!

Quote: DogHand

With thanks to Charlie for the hint, I get...

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220 = 24 in base 3. Each number is 24 in based decreasing from 12 to 2.

Dog Hand

Star Wars

Star Trek

Taylor Swift

Quote: Gialmere

It's toughie Tuesday. Let's go for a drive...



A 10kg box is placed on the back of a lorry. As the lorry accelerates uniformly from 0 to 36 kmh^-1 in 5 seconds, the box begins to slide. The lorry then continues to drive at a constant speed and the box comes to rest.

If the box slides 2 meters in total, and assuming g=9.8 ms^-2, find µ, the coefficient of friction between the box and the surface.

[For simplicity, assume throughout that the coefficient of friction remains constant and that the dynamic friction (once the box begins sliding) is the same as the static (limiting) friction.]

Quote: Gialmere

As the lorry accelerates uniformly from 0 to 36 kmh^-1 in 5 seconds,

Quote: Wizard

What does kmh^-1 mean?

Do you mean the rate of acceleration is constant, the lorry starts from a standing position and reaches 35 kilometers per hour in five seconds?

Also, what is a lorry?

Quote: Gialmere

Yes, and British for "truck".

Quote: Ace2

In Crapless Craps, what is the chance of winning the Fire Bet with all ten points?

Quote: miplet

spreadsheet

Quote: Ace2

I'm looking for a calculated (not simulated) answer expressed as a rational number

Quote: Wizard

Let the record show I beat Don by 13 seconds.

Quote: ThatDonGuy

But you didn't express it as a rational number, as requested

Quote: Ace2

I would be interested in seeing everyone's method...I'm guessing the Wizard integrated a Poisson formula like I did.

Quote: Ace2

I would be interested in seeing everyone's method...I'm guessing the Wizard integrated a Poisson formula like I did.

Quote: aceside

I haven’t got a correct confirmation, so I recalculated it. My new result is mu=0.18896447.

Quote: teliot

Okay ... new puzzle. A sum of consecutive squares equaling a sum of the following successive consecutive squares.

3^2 + 4^2 = 5^2.

10^2 + 11^2 + 12^2 = 13^2 + 14^2

21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2

18^2 + 19^2 + 20^2 + ... + 34^2 = 35^2 + 36^2 + 37^2 + ... + 42^2

What's the longest one you can find?

Quote: Ace2

Incidentally, my method uses the same concept as the Wizard's but it's not exactly the same. I took 1 minus the integral over all time of:

(1-((1-1/e^(t/210))*(1-1/e^(t/60))*(1-1/e^(t/30))*(1-1/e^(4t/75))*(1-1/e^(5t/66)))^2)*1/e^(7303t/11550)*7303/11550 dt

Quote: ThatDonGuy

I understand the first five terms, but where does 7303 / 11550 come from?

Quote: ThatDonGuy

Answer

Let there be A numbers on the left, and B on the right
(N - (A - 1))^2 + (N - (A - 2))^2 + ... + (N - 1)^2 + N^2 = (N + 1)^2 + (N + 2)^2 + ... + (N + B)^2
The left is the sum of 1^2 through N^2 - the sum of 1^2 through (N - A)^2
The right is the sum of 1^2 through (N + B)^2 - the sum of 1^2 through N^2
1^2 + 2^2 + ... + N^2 = N (N + 1) (2N + 1) / 6 for all positive integers N
1/6 N (N + 1) (2N + 1) - 1/6 (N - A) (N - A + 1) (2 (N - A) + 1) = 1/6 (N + B) (N + B + 1) (2 (N + B) + 1) - 1/6 N (N + 1) (2N + 1)
N (N + 1) (2N + 1) - (N - A) (N - A + 1) (2 (N - A) + 1) = (N + B) (N + B + 1) (2 (N + B) + 1) - N (N + 1) (2N + 1)

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

Quote: teliot

Quote: ThatDonGuy

Answer

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

Clever! And beats my largest by (ehem) a few terms.

Quote: Ace2

I owe all three of you a beer:

Quote: ThatDonGuy

Answer

Let there be A numbers on the left, and B on the right
(N - (A - 1))^2 + (N - (A - 2))^2 + ... + (N - 1)^2 + N^2 = (N + 1)^2 + (N + 2)^2 + ... + (N + B)^2
The left is the sum of 1^2 through N^2 - the sum of 1^2 through (N - A)^2
The right is the sum of 1^2 through (N + B)^2 - the sum of 1^2 through N^2
1^2 + 2^2 + ... + N^2 = N (N + 1) (2N + 1) / 6 for all positive integers N
1/6 N (N + 1) (2N + 1) - 1/6 (N - A) (N - A + 1) (2 (N - A) + 1) = 1/6 (N + B) (N + B + 1) (2 (N + B) + 1) - 1/6 N (N + 1) (2N + 1)
N (N + 1) (2N + 1) - (N - A) (N - A + 1) (2 (N - A) + 1) = (N + B) (N + B + 1) (2 (N + B) + 1) - N (N + 1) (2N + 1)

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

Quote: aceside

Can you also find some successive consecutive cubes for fun? Just replace all of the second power to the third power.

Quote: ThatDonGuy

I'll see what I can do.

Cubes should be easier, as 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2
which means if you start with
(n-(a-1))^3 + (n-(a-2))^3 + ... + n^3 = (n+1)^3 + (n+2)^3 + ... + (n+b)^3
you get:
(n (n + 1) / 2)^2 - ((n - a) (n - a + 1) / 2)^2 = ((n + b)(n + b + 1) / 2)^2 = (n (n + 1) / 2)^2
2 (n (n + 1))^2 = ((n + b) (n + b + 1))^2 + ((n-a) (n - a + 1))^2

So far, the only one I have found is 3^3 + 4^3 + 5^3 = 6^3 = 216

Quote: Ace2

Incidentally, my method uses the same concept as the Wizard's but it's not exactly the same. I took 1 minus the integral over all time of:

(1-((1-1/e^(t/210))*(1-1/e^(t/60))*(1-1/e^(t/30))*(1-1/e^(4t/75))*(1-1/e^(5t/66)))^2)*1/e^(7303t/11550)*7303/11550 dt

Quote: teliot

Within the range of the INT data size that's the only one I found as well.

Quote: Gialmere

Fill a cylindrical glass with water. A quick way to empty it to exactly half its volume is to pour out the water until the water line just touches the circular bottom of the glass. Suppose you tip the glass further, until the water line crosses the center of the bottom of the glass (forming a diameter of the bottom circle).

What fraction of the glass' volume is now filled with water?

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Let the height and radius of the glass equal 1. Then:

Quote: ThatDonGuy

Let me see if I have this straight:
1 - 1/e^(t/210) is the probability that at least one point of 2 will be rolled and then made in time t (probability = 1/30 that a 2 will be rolled (ignoring 7s on the comeout) x 1/7 that a 2 will be rolled again before a 7 = 1/210)
1 - 1/e^(t/60) is the probability that at least one point of 3 will be rolled and then made in time t
1 - 1/e^(t/30) is the probability that at least one point of 4 will be rolled and then made in time t
...
1 - 1/e^(t/60) is the probability that at least one point of 11 will be rolled and then made in time t
1 - 1/e^(t/210) is the probability that at least one point of 12 will be rolled and then made in time t
The product is the probabiity that all 10 points will be rolled and then made in time t, so 1 - the product is the probability that at least one point will not be rolled and then made
1/e^(t*7303/11550) is the probability that a point will be rolled and then missed (i.e. a seven-out), so multiply this by the previous product to get the probability of sevening out before making all ten points
But why do you then multiply by 7303/11550 (the probability of sevening out in a particular pass)?

Quote: Ace2

You have it mostly correct. Ignoring 7s on come out rolls, there are 11 possibilities: 1/210 winning a 2, 1/210 of winning a 12, 1/60 if winning a 3...etc and a 7303/11550 chance of sevening-out. As mentioned, these probabilities add up to 1.

The second to last term of 1/e^(7303t/11550) is the probability that zero seven-outs have happened. And, as you know, the previous terms represent the probability that at least 1 of the 10 points has not been won. Therefore, starting from this state, if the next decision is a seven-out then the fire bet loses at that point in time t. There is a 7303/11550 chance of that happening so we multiply by that factor. Combining all that and integrating gives the sum of the probabilities, at all times from zero to infinity, that the fire bet will lose at that point in time.

Quote: ChesterDog

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Let the height and radius of the glass equal 1. Then:

Quote: Wizard

I agree with Charlie on four cards. My answer is by trial and error.

Thoughts on nine cards

I think this is a trick question. I suspect it is impossible if you stick with the constraints of using integers and adding the two cards.

I've tried using negative numbers and adding, but can't find a solution.

I've tried multiplying, instead of adding, but still can't find a solution.

I suspect there is some other function of the two cards I must take.

That's about where I am now.

Quote: ThatDonGuy

...

Here's a suggestion

Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

Quote: ThatDonGuy

Here's a suggestion

Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

Quote: Wizard

...
Practical Question

In the land of Calizonia, casinos may not use dice in craps, but they may use cards. They must draw two cards, without replacement, from a seven card deck and add the results. The probabilities must match those with two dice. There are no regulations on what can be on the cards.

What should the Calizonia casino do?

Quote: charliepatrick

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Since if you only looked at the two cards there would be 21 permutations you must use the order. Then you have 42 combinations, so some of these must be "to try again".

The following method works.

The cards are labeled 1 2 3 4 5 6 "same". The first card drawn determine the roll of the first die. However if the first card drawn is "same" then you start over again. The second card drawn determines the roll of the second die, except if it's the "same" then you've thrown a repeat of the first roll. Using the "same" to replace the first roll means the second roll still has an equal chance for each outcome.

Quote: charliepatrick

for two sets of four cards

Each set is 0 1 2 4.

The way to show it is to consider the binary for the numbers 1 thru 6 as 001 010 011 100 101 and 110, some which use two digits and others which use one. So picking two from 1 2 4 covers the two digit ones while 0+ another cover the others. Thus each set create the numbers 1 thru 6, so are equivalent to a regular die.

Quote: charliepatrick

Quote: ThatDonGuy

...

Here's a suggestion

Who said anything about the numbers having to be integers? Try having all of the numbers be {integer} + 1/2.

For example, the lowest two can be 1/2 and 3/2, so you can "roll" a 2.

thanks

That helps as it had to be symmetric and having the middle one being 3.5 seemed to work. Working from the top the first two have to be .5 and 1.5 (to get one 2) then two 2.5 to (get two 3's). Similarly 6.5 5.5 4.5 4.5. The rest falls out.

Quote: Gialmere

But wait! You recount and find there are actually nine cards available. Hmm. It occurs to you that you could number the nine cards in such a way that randomly drawing two cards from the pile of nine cards also gives an integer from 2 to 12 with the same frequency of occurrence as rolling that sum on two standard dice.

Quote: Gialmere

There's only eight cards. Hmm. A ha! You can still divide the cards into two sets (piles) and number them in such a way that by drawing two cards from each pile, and adding the four cards together, you get an integer from 2 to 12 with the same frequency of occurrence as rolling that sum on two standard dice.