Mathematics marathon

[JonWhite]

The differences between terms are 7, 19, 37; the differences between these are 12 and 18; the difference between 37 and the next increased amount is thus 24, so the next increased amount is 61, which added to 63 gets 124. This is not a 5-10 second problem for 9-11 year olds, but it is certainly solvable within a minute or two by a determined individual in that age group.

I believe I know exactly where this problem came from. If I am correct, then this problem must be solved mentally without any scratchwork. It's possible, but indeed very hard to do it this way without scratchwork. I'm also pretty sure there's another simpler way to solve this problem.

I was watching Deal or No Deal a couple days ago, and an interesting probability situation came up. It's very similar to another problem I've asked on here before.

The lady playing had gotten down to only 2 cases, the one with $750,000 and the one with $1,000,000 dollars. Before she was to open her case, Howie gave her the option to switch cases and open the one still on stage rather than the one she picked at the beginning of the game. She had been having such good luck throughout the game that she decided not to switch cases.

Did she make the right decision? Or did it not matter one way or the other whether she swapped cases or not?

I haven't watched this show, so I have a question: Did the lady pick which cases to open, and thus was lucky (or unlucky?) to not have opened the million dollar case, or did the host open cases for her, intentionally picking cases that did not contain the million, and thus the lady was "fated" to be left with the million unopened?

 

[qqwref]

I'd consider it a stretch to say that the sequence "12, 18, ..." is obviously followed by 24. I mean, sure, that's arguably the most obvious choice, but at that point you only just have enough information to even figure out an arithmetic series, which means it's a pretty iffy solution. At least "perfect cube minus one" gives you enough terms to verify that the idea isn't coming out of nowhere.

(Incidentally, I gotta say that I'm not fond of "follow the sequence" questions at all, just because there are many possibilities - any sufficiently difficult one which doesn't give a huge number of terms will have more than one plausible solution.)

 

[LewisJ]

I'd consider it a stretch to say that the sequence "12, 18, ..." is obviously followed by 24. I mean, sure, that's arguably the most obvious choice, but at that point you only just have enough information to even figure out an arithmetic series, which means it's a pretty iffy solution. At least "perfect cube minus one" gives you enough terms to verify that the idea isn't coming out of nowhere.

(Incidentally, I gotta say that I'm not fond of "follow the sequence" questions at all, just because there are many possibilities - any sufficiently difficult one which doesn't give a huge number of terms will have more than one plausible solution.)

I feel really stupid now for not noticing the perfect cubes part. I did feel the 12, 18, therefore 24 thing was a stretch, but it fit the solution.

Although I sort of share your sentiment about sequence problems, I have a rather reasonable one:
Find the next term in
2, 3, 10, 21, 55, 104, 221, 399, ___

 

[scylla]

sorry I don't know how the spoiler function works

Michael, my hat's off. I initially thought of this as a Monty Hall style problem, but I can see the flaw in that now because of the same probability of ending up with the same two cases at the end even if they were swapped.

I'm imagining a different version of Deal or No Deal where there are 26 cases, one case contains $1,000,000 and the other 25 are empty. Imagine in this game if you pick a case at the beginning, then proceed to open 24 empty cases. At this point Howie asks if you would like to swap cases or not. Since you are down to two cases, one empty and one the $1,000,000 I wonder if this works the same way.

My impression is to try to calculate it.

The probability of winning the $1,000,000 case assuming you do not switch would be 1/26 since you open the case that you initially picked.

The probability of winning the $1,000,000 assuming that you do switch is:


Wow, so it does not matter whether you switch or not even in this variation, the probability of winning the million is still 1/26. Very neat result! I definitely got it wrong when watching the show. I thought she should have switched. Cool, you learn something new every day.

Yes it doesn't matter in this case. The problem looks very like the 3-doors problem (in which switching is better) but it's different at 1 point.

In the 3-doors problem the quizmaster opens a door without the jackpot (because he knows where the jackpot is), in this problem the person chooses the boxes to be opened without knowing where the jackpot is. That leaves just a simple propability calculation that every price has the same propability to be in every box. So the choice is indifferent.

This problem I invented by myself in a pub when I was playing with (round) coasters:

Suppose you have three coasters with a diameter of 1. What is the biggest circle you can totally cover with those 3 coasters?

 

[keemy]

Lewis:

Spoiler

782 (nth prime)*(nth Fibonacci)

 

[Rinfiyks]

Had this optional brain teaser at the end of a question sheet for theory of programming:

Spoiler

There are three gods, Angel, Demon and Chaos. Angel will always answer truthfully, Demon will always lie, and Chaos will answer your questions randomly. The gods are standing before you, but you do not know which is which. Your task is to determine the identity of the gods, by asking them three questions. To make matters worse, the gods will answer your questions only with 'da' or 'ja', which mean true and false in their own language, but you don't know which is which.

Clarifications:
Each question can be directed only at one god, but you may ask two (or three) of your questions to the same god, if you so wish.
The gods, being gods, are all-knowing and perfect logicians.
There's no "trick", like asking them to answer paradoxes.
The three gods all speak the same language.

Hint

Spoiler

You don't need to know what 'da' and 'ja' mean.

Bigger hint

Spoiler

You can word a question in such a way that if asked to Angel or Demon you will know if it is true or false without knowing da or ja.

 

[Ordos_Koala]

Spoiler

Once you've managed to open 24 empty cases, you are indeed less likely to have chosen the million dollars than in the beginning - in 25 of 26 possibilities the million dollar case will be in the last of the 25 non-chosen ones, but in only 1 possibility will you have chosen the million dollars. Of course, you're a lot more likely to have opened 24 empty cases if you started with the million, but once you've done it the probability of the past event isn't relevant anymore.

The difference between this version and the version you saw on the real show was that the $750,000 and $1,000,000 cases function identically, so neither is biased against by the act of removing every other case. In this example, the $1,000,000 case and empty case you are left with are NOT identical, because there were more cases just like the empty one at the start.

Spoiler

you're wrong... at first you have chance 1/26. But when you open first case, your chance incrase to 1/25 since there are now only 25 cases... at the end, it doesn't matter wheter you opened million, twenty-four or any of these empty cases, your chance to win at the end is 1/2
What you tink is common mistake...

 

[cmhardw]

This seemed the best place to post this question without starting a new thread.

What would your next move be in this sudoku, and more importantly why?

Spoiler

If we label the columns A-I going from left to right and the rows 1-9 going from bottom to top, then here is how I decided my next move:

I placed a "3" in cell I6, then continued to solve the puzzle correctly. I don't like how I reasoned that a 3 belonged here, it feels a bit like cheating.

From earlier studying of the puzzle I found that in the upper left 3x3 grid that the 6 and 8 must be in cells B8 and B9. This will be useful information in a moment. After some time, I came to study where the 3s could go in the puzzle as a whole. All possible options as seen on the picture are:

A2, A4, A6, B2, B4, I4, I6

I noticed that if there was a 3 in cell I4 that it would force the position of all the other 3s. At this point I decided to search to see if this would lead to a contradiction. I always do this mentally, and I never write down the possible values that would result from "guessing" a number like this. I consider guessing like this cheating in a sense, so I prefer to limit myself to what my short term memory can handle, rather than writing down what happens.

If I4 is a 3, then so too will be cells A6 and B2. Now, cells A2, B2, C2 must be the 3,5,8 in the lower left 3x3 grid. With this extra information I can reason that cell A2 must be an 8 and cell C2 must be a 5. However, let's look at the entirety of row 4. If cell I4 is a 3, then I could reason that since cells B8 and B9 are the 6 and 8, then cell A4 must be an 8. This places the 8 in two different places in column A. This is a contradiction. Since the 3 can only occupy either I4 or I6 in column I, then I knew I must play it to cell I6.

I don't like what I did here, it feels like cheating. How should I have decided what number to play next?

 

[qqwref]

Spoiler

Your solution does make sense - another way to see it is that I4=3 forces the rest of column 4 (B4=4, E4=6, A4=8), and A4=8 forces A2=3, but then there's no place to put 3 in column B.

Alternate way to get a clue, using your notation:
- Notice that the only places on row 8 that can hold the 2 are G8 and I8. This means cell I9 can't be a 2, and since it can't be 1,3,4,5,6,9 either it is either 7 or 8.
- In the bottom-right area, row 3 tells us that H3 and I3 are both either 5 or 6, and this means that G1 and I1 are both either 7 or 8.
- So now we have I1 and I9 are both either 7 or 8. This eliminates the possibility of being an 8 from cell I4. But now the only place on row 4 that an 8 can be is A4, so there you go. Then you can solve A2 and go from there.

BTW, the logic you (could have) used to get that B8 and B9 were 6 and 8, if you forgot: B8 can be 6 or 8 already, but B9 can be 5, 6, or 8. However, B7 and C7 are the only cells in row 7 that can be a 5, so B9 can't be a 5.

 

[cmhardw]

Spoiler

Alternate way to get a clue, using your notation:
- Notice that the only places on row 8 that can hold the 2 are G8 and I8. This means cell I9 can't be a 2, and since it can't be 1,3,4,5,6,9 either it is either 7 or 8.
- In the bottom-right area, row 3 tells us that H3 and I3 are both either 5 or 6, and this means that G1 and I1 are both either 7 or 8.
- So now we have I1 and I9 are both either 7 or 8. This eliminates the possibility of being an 8 from cell I4. But now the only place on row 4 that an 8 can be is A4, so there you go. Then you can solve A2 and go from there.

I see this now. I had already noticed that G1 and I1 were the 7 and 8 in the lower right grid, but I never ruled out the 2 as a possibility for cell I9. I like this line of logic better, and I see how the rest follows from there to get A4 as an 8 (assuming you remember that B8 and B9 are the 6 and 8).

Thanks Michael!

 

[qqwref]

Here's an interesting problem: How many playable chords are there on a piano, for a typical person, if they only use one hand? Okay, I know, that's really vague. Let's approximate, in rough order of difficulty/accuracy:

First approximation: A piano has 88 keys. How many possible chords are there if you can play any combination of keys? (Two combinations of the same keys in a different order count as the same.)

Second: A typical person can reach about an octave (two keys separated by exactly 11 keys in between), so let's use that. How many chords are there that are no more than an octave wide?

Third: A typical person has five fingers on each hand, and will only press one key with each finger. How many chords from part 2 have 5 or fewer keys pressed?

 

[mr. giggums]

Spoiler

There are 88 ways to get one key
87+86+85+84+83+82+81+80+79+78+77+76=978 ways for 2 keys
86+2*85+3*84+4*83+5*82+6*81+7*80+8*79+9*78+10*77=4400 for 3
85+3*84+6*83+10*82+15*81+21*80+28*79+36*78=9570 for 4
84+4*83+10*82+20*81+35*80+56*79=10782 for 5
88+978+4400+9570+10782=25818
My answer is 25818 combinations

 

[Rinfiyks]

Here's an interesting problem: How many playable chords are there on a piano, for a typical person, if they only use one hand? Okay, I know, that's really vague. Let's approximate, in rough order of difficulty/accuracy:

First approximation: A piano has 88 keys. How many possible chords are there if you can play any combination of keys? (Two combinations of the same keys in a different order count as the same.)

Spoiler

Pascal's triangle
Take 4 keys on a piano as an example. 4th row on Pascal's triangle (if the first row is the 0th row) is 1 4 6 4 1
1 way to play a chord with 0 keys, 4 ways to play a chord with 1 keys, 6 ways to play a chord with 2 keys, 4 ways to play a chord with 3 keys, 1 way to play a chord with 4 keys.
The sum of the nth row is 2^n.
But a chord with 0 keys doesn't really count
So it's 2^88 - 1.

Second: A typical person can reach about an octave (two keys separated by exactly 11 keys in between), so let's use that. How many chords are there that are no more than an octave wide?

Spoiler

For the first 11 notes, 2^11 - 1 chords.
Then if you move 1 note to the right, you need to add another 2^11 - 1, but discount all the chords that don't include the rightmost note, because they have been covered by the first 11 notes. The number of chords that don't include the rightmost note is 2^10 - 1. Then we move 1 to the right again, and more discounting by the same amount, and so on, right to the end. So you subtract (88 - 5)(2^10 - 1) from the final answer of (88 - 4)(2^11 - 1).
(88 - 4)(2^11 - 1) - (88 - 5)(2^10 - 1) = 87039.
I don't know if this is right so I won't attempt the final question until I know.

Third: A typical person has five fingers on each hand, and will only press one key with each finger. How many chords from part 2 have 5 or fewer keys pressed?

Spoiler

I don't know if I got part 2 right, but I'll assume I did
If we're limiting it to 5 keys, then we only add up the first 5 (not including the 1 at the start) values on the 11th row of Pascal's triangle.
11+55+165+330+462 = 1023 = 2^10 - 1.
To work out how many we discount, add up the first 5 values (ex. 1) on the 10th row.
10+45+120+210+252 = 637.
So we have
(88 - 4)(2^10 - 1) - (88 - 5)(637) = 33061.
I hope.

 

[Keroma12]

Problems I'm stuck on:

\( \int ln(sin(x))dx \)

\( \int ln(cos(x)+1)dx \)

My attempt:

Spoiler

For the first, I said \( sinx=u \) and \( cos(x)dx=du \). This eventually leads to \( \int \frac{ln(u)du}{sqrt[1-u^2]} \), but this doesn't really help as far as I can see.

Also, how do I make a square-root symbol?

 

[Sa967St]

Here are two questions from a math contest I did earlier today that I had trouble with.
1.

2.
A 75 year old has a 50% probability of living for another 10 years.
A 75 year old has a 20% probability of living for another 15 years.
An 80 year old has a 10% probability of living for another 10 years.
What is the probability that an 80 year old will live for another 5 years?


My fail attempt at the first one:

Spoiler

lol. I didn't get much further than this.

I had a few answers for the second one. This one was my best guess:

Spoiler

1.5(.50/.20)= 2(x/.10)
x=0.1875 or 18.75%

 

[Keroma12]

Here are two questions from a math contest I did earlier today that I had trouble with.
1.

I did the same contest.

Spoiler

From where you got to, replace \( 2^{logx} \) with \( n \) and you have a quadratic: \( n^2-16=6n \)
So \( (n-8)(n+2)=0, n=8 \) or \( n=-2 \)
Then \( 2^{logx}=8 \) (the other has no real solutions)
So \( logx=3, x=1000 \)

 

[cmhardw]

Problems I'm stuck on:

\( \int ln(sin(x))dx \)

\( \int ln(cos(x)+1)dx \)

My attempt:

Spoiler

For the first, I said \( sinx=u \) and \( cos(x)dx=du \). This eventually leads to \( \int \frac{ln(u)du}{sqrt[1-u^2]} \), but this doesn't really help as far as I can see.

Also, how do I make a square-root symbol?

Spoiler

I tried the same thing you did, as well as some integration by parts and I never got anywhere. I found this interesting though, perhaps this may be useful?

\( \int ln(sin(x))dx = \int \frac{1}{2}ln\left[sin^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[1-cos^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[(1+cos(x))(1-cos(x))\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2}\left[\int ln(1+cos(x))dx + \int ln(1-cos(x))dx\right] \)

I don't see where to go from here, unless we could figure out the right hand side integrals relatively easily. I will try to keep looking into this as well. This is interesting.

P.S.

\( \sqrt{5} \)

\( \sqrt[n]{(x+1)^{2n+1}} \)

 

[Sa967St]

I did the same contest.

Spoiler

From where you got to, replace \( 2^{logx} \) with \( n \) and you have a quadratic: \( n^2-16=6n \)
So \( (n-8)(n+2)=0, n=8 \) or \( n=-2 \)
Then \( 2^{logx}=8 \) (the other has no real solutions)
So \( logx=3, x=1000 \)

Wow, I definitely went the wrong direction with this. I tried stuff like taking the log of both sides, which got me no where. I didn't even think of changing 3(2^(1+logx)) to 3(2^1)(2^logx)=6(2^logx). :fp Thanks.
My math teacher couldn't even get it btw.

 

[qqwref]

1.

Spoiler

Let \( y = \log_{10}x \). Then the equation can be simplified to \( 2^{2y} = 6 * 2^y + 16 \). Now let \( z = 2^y \) and this becomes \( z^2 = 6z + 16 \), which is a quadratic, so we get that z must be -2 or 8, and -2 doesn't make sense, so z = 8. Then y = 3 and x = 1000.

.

2.
A 75 year old has a 50% probability of living for another 10 years.
A 75 year old has a 20% probability of living for another 15 years.
An 80 year old has a 10% probability of living for another 10 years.
What is the probability that an 80 year old will live for another 5 years?

Spoiler

Wait a minute. 20% of 75-year-olds live to age 90, but only 10% of 80-year-olds do? That means, for every living 90-year-old, there were 10 living 80-year-olds 10 years ago, but only 5 living 75-year-olds 15 years ago. How do you get to be 80 without getting to be 75 first?

 

[Keroma12]

Spoiler

I tried the same thing you did, as well as some integration by parts and I never got anywhere. I found this interesting though, perhaps this may be useful?

\( \int ln(sin(x))dx = \int \frac{1}{2}ln\left[sin^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[1-cos^2(x)\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2} \int ln\left[(1+cos(x))(1-cos(x))\right]dx \)
\( \int ln(sin(x))dx = \frac{1}{2}\left[\int ln(1+cos(x))dx + \int ln(1-cos(x))dx\right] \)

I don't see where to go from here, unless we could figure out the right hand side integrals relatively easily. I will try to keep looking into this as well. This is interesting.

P.S.

\( \sqrt{5} \)

\( \sqrt[n]{(x+1)^{2n+1}} \)

Thanks for the square root. Well we've only done u substitutions at school so I don't know many techniques. Actually if you could find the sum of the two, if each own its own is too difficult, that would be just as useful.