Can Brain Teasers Enhance Problem-Solving Skills?

[siddharth]

http://www.iucaa.ernet.in/~paddy/playthemes/playthemes.htm"

That's a nice collection of difficult(IMO) brain teasers and math problems (especially geometry). Till now, I haven't got any

 

[davee123]

Show that the 13th of a month falls more frequently on Friday than on any other day.

Is it just me, or is this incorrect? I wrote a script that started Jan 13th, 1970, and went through until Dec 13th, 2037, and came up with the following counts for 13ths of the month:

Mon - 116
Tue - 118
Fri - 117
Wed - 115
Sun - 117
Thu - 117
Sat - 116

Am I missing something? Do I need a far more significant dataset before the trend becomes visible?

DaveE

 

[NateTG]

Is it just me, or is this incorrect?

You'll want to run through at least 400 years to account for leap year cycles.
Since 7 divides the number of days in 400 consecutive years (365*400+97), that's a complete cycle.

 

[davee123]

You'll want to run through at least 400 years to account for leap year cycles.
Since 7 divides the number of days in 400 consecutive years (365*400+97), that's a complete cycle.

Yep, THAT did it, thanks! Had to modify the program to include leap year logic rather than use existing functions. Ran it on 146097 days (the complete cycle) and came up with:

Mon - 685
Tue - 685
Fri - 688
Wed - 687
Sun - 687
Sat - 684
Thu - 684

DaveE

 

[Jimmy Snyder]

I got a few of the quickies.

4. [itex]-1 = e^{i\pi}[/itex]
5. [itex]\left(\frac{499}{500}\right)^{500}[/itex]
9. o{ne}, t{wo}, t{hree}, f{our}, f{ive}, s{ix}, s{even}, e{ight} and so the answer is n, t, e, t, t, f, etc.
16. Footnote 4 is the same as footnote 3, but in English.

P.S.
I don't know why the first two answers aren't hidden.

 

[Euclid]

I'm not sure if [tex]-i\pi[/tex] counts as irrational, so I decided to give an alternate demonstration:

We know that [tex] 2 = e^x[/tex] for some x. If x were rational (say x= p/q clearly neq 0), then we have [tex]2^q=e^p[/tex] but no positive integer power of e can be an integer, so this is a contradiction.
(If we did have [tex] n = e^p[/tex] for positive integers n,p, then e would be a root of [tex] x^p -n \in \mathbb{Q}[x][/tex])

 

[NateTG]

Resorting to 'e is transcendental' without proof is IMHO unsatisfying.

I'd prefer something like:
Consider the real solution to:
[tex]x^x=2[/tex]

Clearly, [itex]x[/itex] is not an integer. Now, if [itex]x[/itex] is rational, then we have:
[tex]x=\frac{a}{b}[/tex]
expressed as a fraction in lowest terms, and with [itex]b\neq1[/itex].
But then we can do the following:
[tex](x^x)^b=2^b[/tex]
[tex]\left(\frac{a}{b}\right)^a=2^b[/tex]

Since [itex]a[/itex] and [itex]b[/itex] are integers, we have the LHS is a non-integer, and the RHS is, which is a contradiction - so the assumption that [itex]x[/itex] is rational is clearly false.

 

[VietDao29]

For Quicky #10:
Here are sets of numbers enclosed by square brackets. Fill the next set of numbers. [1],[1,1],[2,1],[1,2,1,1],[1,1,1,2,2,1],[3,1,2,2,1,1], [....]
Start with [1], we count 1 number 1, so we write [1, 1], then we continue counting 2 numbers 1, so we write [2, 1], now that's 1 number 2, and 1 number 1, so we write [1, 2, 1, 1],...
So a few next sets are:
[1, 1, 1, 2, 2, 1]
[3, 1, 2, 2, 1, 1]
[1, 3, 1, 1, 2, 2, 2, 1]
[1, 1, 1, 3, 2, 1, 3, 2, 1, 1]
...
For # 15,
1023 players (yes, not 1024) participate in a tournament in which each game produces a decisive winner. Players are eliminated by knock-out with byes being given when odd number of players occur at any given round. How many matches need to be played to find a winner ?
I think it's 1022, since 1022 playes must be knocked-out before the winner is found.

 

[Jimmy Snyder]

The answer to quickie #11 is yes
The answer to quickie #12 is use base 15

 

[siddharth]

jimmy,
how is the fifth one [itex]\left(\frac{499}{500}\right)^{500}[/itex]?
Isn't that very high?

 

[arunbg]

Answer to quickie no 5: [color="#black"]Probability of 0 typos = 499/999[/COLOR]
Answer to quickie no 13:[color="#black"]2.3 , 2ln3[/COLOR]
Didn't find time 4 the rest.
Nice collection anyways.

--Arun

 

[Jimmy Snyder]

Answer to quickie #3 is 1001! + 2, 1001! + 3, ... 1001! + 1001

 

[Jimmy Snyder]

Isn't that very high?

It's about 0.3675
Suppose there were only 1 error in the book. Then what would the answer be? How about 2 errors? Since the errors are randomly distributed, the probablility of both errors not being on page 29 is the product of the probablilities that either one is not on page 29. How about 500 errors?

 

[NateTG]

It's about 0.3675
Suppose there were only 1 error in the book. Then what would the answer be? How about 2 errors? Since the errors are randomly distributed, the probablility of both errors not being on page 29 is the product of the probablilities that either one is not on page 29. How about 500 errors?

You're making some assumptions about the distribution of errors that aren't necessarily true.

 

[Jimmy Snyder]

You're making some assumptions about the distribution of errors that aren't necessarily true.

I don't believe so.

 

[siddharth]

Answer to quickie #8 is staring right out of the keyboard

 

[arunbg]

You're making some assumptions about the distribution of errors that aren't necessarily true.

What jimmy is trying to ask is how the probability will be affected if the no of errors were different from the number of pages.
However I feel his answer is wrong.
Let me explain how I arrived at my answer( which can be extended to general cases)
The general formula for distributing n things among r different people is given by
[tex]^{n+r-1}C_{r-1}[/tex]

For this case,the no of favourable outcomes (no error in page 29) from the above formula will be [itex]^{998}C_{498}[/itex] (distributing 500 errors in 499 pages leaving page 29 vacant).
No of possible outcomes will be [itex]^{999}C_{499}[/itex](distributing 500 errors in 500 pages)
So the probability will be their ratio which turns out to be 499/999
(slightly less than half).
The same procedure may be applied for other variations od the problem.

- Arun

 

[NateTG]

I agree that if one assumes that the typos are independantly assigned with an even random distribution, then the probability that page 29 has no typos is going to be
[tex]\left(\frac{499}{500}\right)^{500}[/tex]

However, consider, for example, a book that has 1 character per page, 500 pages, and 500 typos. The probability that a typo occurs on page 29 is then 1.

 

[Jimmy Snyder]

However, consider, for example, a book that has 1 character per page, 500 pages, and 500 typos. The probability that a typo occurs on page 29 is then 1.

For this you receive an AAA+.

Edit: Sorry, make that an A+.

 

[Jimmy Snyder]

What jimmy is trying to ask is how the probability will be affected if the no of errors were different from the number of pages.
However I feel his answer is wrong.
Let me explain how I arrived at my answer( which can be extended to general cases)
The general formula for distributing n things among r different people is given by
[tex]^{n+r-1}C_{r-1}[/tex]

For this case,the no of favourable outcomes (no error in page 29) from the above formula will be [itex]^{998}C_{498}[/itex] (distributing 500 errors in 499 pages leaving page 29 vacant).
No of possible outcomes will be [itex]^{999}C_{499}[/itex](distributing 500 errors in 500 pages)
So the probability will be their ratio which turns out to be 499/999
(slightly less than half).
The same procedure may be applied for other variations od the problem.

- Arun

Try this out with 2 errors distributed between 2 pages. I get 1/4 as the probability that page 1 is free of errors in two different ways:

method 1)
error 1 on page 1, error 2 on page 1: page 1 is not error free
error 1 on page 1, error 2 on page 2: page 1 is not error free
error 1 on page 2, error 2 on page 1: page 1 is not error free
error 1 on page 2, error 2 on page 2: page 1 is error free
probability that page 1 is error free: 1/4

method 2)
[tex]\left(\frac{n-1}{n}\right)^n[/tex] where n = 2
probability that page 1 is error free: 1/4

Using your method, I get)
[tex]\frac{^{n+r-1}C_{r-1}}{^{n+r}C_{r}}[/tex]where n = 2 and r = 1
probability that page 1 is error free: 1/3

I think your method is wrong. I think it counts both instances of 1 error per page as if they were a single instance.

 

[arunbg]

error 1 on page 1, error 2 on page 1: page 1 is not error free
error 1 on page 1, error 2 on page 2: page 1 is not error free
error 1 on page 2, error 2 on page 1: page 1 is not error free
error 1 on page 2, error 2 on page 2: page 1 is error free

How can a typo on one page be exchanged with the typo on another page?
Also as per your logic you also have to take into account the no of characters in each page since these would have to be treated as different errors.
The right approach (as per question) is to simply find the distribution or no of errors on the pages and not name them individually.

Please enlighten me on how you obtained your formula for clearer understanding.

 

[Jimmy Snyder]

How can a typo on one page be exchanged with the typo on another page?

You have to count it twice because it is twice as likely to happen than that both typos should end up on page two.

 

[arunbg]

Firstly typos cannot be named typo1, typo2 etc. This is because a typo in one page cannot be made a typo of another page in some other case, unless the pages are all same( in which case the book is useless).
ONLY the total no of typos is conserved.So interchanging the typos to get different outcomes is wrong.
Also if we consider the typos to be unique (typo1, typo2,...), then we'll have to examine Nate's argument regarding the no of characters in a page and typo distribution treating each location of the "same typo" as different.Unnecessary complications.

 

[Jimmy Snyder]

Arun, let's flip a coin twice and record the number of heads. The possible outcomes are:

0 heads
1 heads
2 heads.

Would you argue that each of these three outcomes is equally likely? I assume you would not, but I will be interested to hear it directly from you.

 

[NateTG]

Did I mention something about assumptions?

Jimmy is assuming that the placement of typos is individually equi-probable.

With 2 pages, and 2 typos, Jimmy's predictions would be:
2,0 (25%)
1,1 (75%)
0,2 (25%)

Arunbg is assuming that the arrangements of typos are individually equi-probable.
For 2 pages and 2 typos, Arunbg's predictions would be:
2,0 (33.333...%)
1,1 (33.333...%)
0,2 (33.333...%)

Which model should be selected depends on what sort of thing we're looking at.

My argument is only intended to demonstrate that there are unstated but singificant qualities. And, it's deliberately extreme in order to illustrate said pathology. It's quite clear that if there are 500 pages, 500 typos, and each page can only hold one typo, that, indeed, the number of typos on each page will be 1.

 

[hellraiser]

There are two methods being argued
one is n/(2n-1) (assuming n and r are equal)
and {(n-1)/n}^n
for the simple case of n=1 there is no chance that the error can be located on anothere page. so for n = 1 answer should be 0. Hence put n = 500 the answer must be (499/500)^500

another method to see this might be:
If there is one typo then the probability that it is in one of pages other than 29 is 499/500. Multiplying 500 times for 500 typos (499/500)^500.

 

[Jimmy Snyder]

Jimmy is assuming that the placement of typos is individually equi-probable.

Another way of saying this is "Jimmy is assuming that the distribution of typos is random". This assumption is acceptable because it is the distribution stated in the puzzle. I intend to deliberately place a typo on page 1 if the coin comes up heads, and on page 2 if it comes up tails. That is what a random distribution of typos means. I don't know the name of the distribution Arun is dealing with, but it is not the one stated in the puzzle.

As a side issue, not related to the matter above:
You are right that I assumed that there can be more than one typo per page and this is not stated in the puzzle. My answer is wrong for the case when no more than one typo per page is allowed, and you have provided us with the correct answer.

 

[hellraiser]

This question came in the JEE exam this Sunday.
What is the probability that a nucleus decays during two half lives?

I marked 0.75 ---
1/2 if it decays in 1st life itself
1/2 * 1/2 if it does not decay in first half life and then decays in next one.
But I don't undestand why FIITJEE has given answer as 1/2 (as one of my friend told me. I have not mustered enuf courage to see the solns. :) )

 

[NateTG]

Another way of saying this is "Jimmy is assuming that the distribution of typos is random".

Well, yes, but the same is true of Arunbg's notion interpretation

It's a bit like 'I have two cards. One of them is not an ace. What is the probability that one of my cards is an ace?' is ambiguous. (You could reasonably answer 1 in 17, or 5 in 33.)

 

[Jimmy Snyder]

You could reasonably answer 1 in 17, or 5 in 33.

No one would consider the toss of an unfair coin to be random. If they did then you would have to consider the steady stream of heads obtained by tossing a two-headed coin to be random. A coin that comes up two heads in a row one third of the time is an unfair coin. Given the wording of the puzzle, this so-called solution is just plain wrong.

 

[arunbg]

Firstly you are not the one "distributing the typos".
If you open a book filled with typos at random, would you argue that it is more probable to find 1 typo in one page than say all typos in one page?
I don't think so.Unless more information is given regarding the actual distribution of typos, you just don't know.

This question came in the JEE exam this Sunday.
What is the probability that a nucleus decays during two half lives?

Whoa! I marked 3/4 too. I used the same line of reasoning as you did.
Hope FIITJEE is wrong for once.
Doesn't make a difference though, I am not going to get through anyway.

 

[Jimmy Snyder]

Firstly you are not the one "distributing the typos".

The issue is not who distributes them, but rather whether they are randomly distributed. I claim that by the procedure I laid out, they would be. Good luck on the test. I would have answered as you did even though the question doesn't say whether the half-lives are for the nucleus or for something else altogether.

Nate, I see no ambiguity in this puzzle. You could just as easily argue to me that the words book and page are ambiguous because both can also be verbs.

A book of 500 pages has 500 typos randomly distributed. What is the probability that page 29 has no typos ?

You could say that my hidden assumption is that the puzzler is using the English language in the same way that the rest of us do.

 

[Promethium147]

The four hackers made their way to the cooling control center, which has four consoles, each controlling a separate plasma vent. I've set a different security protocol on each plasma vent, each wildly different from the other. Unfortunately, the humans knew this, and so each elite hacker is an expert in bypassing each type of security, but only the hacker who's an expert at that particular security will be able to bypass it, and no one else.

Knowing this vulnerability, and as a last ditch effort to save my station, I flooded the control room with intense radiation. Since the control room is only big enough to allow one person to work on it, the four hackers are forced to go only one at a time. Each hacker must enter the control room via a series of tight and lengthy corridors (which are designed to stop radiation from leaking into the main station), and pick one of the four consoles to check. If he is capable of disarming that particular security, he will do it, and disable that particular vent, at which he will die knowing his task is done. (He will not have enough time to go back and inform his comerades about the result) If he is not, then he has time to check a second console before the radiation turns him into green goo.

All four vents must be shut down in order for the station to be destroyed.

The four hackers, of course, knew this all beforehand, and so started to devised a plan which will maximize their probability of success. They thought as follows: if all of them pick consoles 1 and 2, the probability of success is 0, because there will always be two failures. If all of them pick 2 random consoles, then the probability of success will be 1/2 * 1/2 * 1/2 * 1/2 = 1/16, or 6.25%. At last, a few minutes later, they came up with a plan that will allow them to have a probability of success of over 40%. To make things simpler, they designated the four consoles to be A B C and D, and the different security protocols to be 1 2 3 and 4.

What plan is this?

This is supposedly possible but I haven't figured anything out yet. See if you can get it.

EDIT: This comes from a story, but the rest is not necessary to solve.

 

[davee123]

This is supposedly possible but I haven't figured anything out yet. See if you can get it.

EDIT: This comes from a story, but the rest is not necessary to solve.

It seems that it's impossible. It nearly expressly states that no hacker can tell the others of his success. But even if he *could*, the maximum success rate that I can muster is 33%.

Effectively, the first hacker's decision is random. He has a 50% chance of success at shutting down 1 vent. If he fails, they all fail. Assuming the first hacker's success, if the next hacker can *tell* which console was successfully hacked, he has a 66% chance of successfully hacking, since he may try 2 of the remaining 3 consoles. Again, if he fails, so does the mission. The chance that both the first and second hackers succeed is now 33%. If the first two succeed, and the last two can again tell which consoles were successfully hacked, they each have a 100% chance at success, so they stay at 33% for overall success.

So, I'm pretty sure that you can't get more than 33%. And maybe there's a sneaky way of getting 33% without actually knowing the successes of your preceeding hackers, but I don't think you can get more than 40% success.

DaveE

 

[Jimmy Snyder]

I came to the same conclusion as davee123. I looked into the possibility of using some variation of the 'Monty Hall' problem to improve chances, but with no success. The conditions necessary for that problem do not prevail in this one.

If you consider that the first person has a 50% chance of success, then in order to achieve a better than 40% overall chance of success, the second person would need a better than 80% chance of succeeding.

I did assume that the second and subsequent hackers could identify the hacked consoles by some method, perhaps by the puddle of green goo next to it (if there is goo by the console then either the console is hacked or the jig is up). If there is no way to identify the hacked consoles, then the best chance for success is one in sixteen. To achieve this, the hackers simply arrange ahead of time to evenly distribute the attacks among the consoles.