# Brain Teasers and Math Problems

1. Apr 11, 2006

### siddharth

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

Last edited by a moderator: May 2, 2017
2. Apr 11, 2006

### davee123

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

3. Apr 11, 2006

### NateTG

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.

4. Apr 11, 2006

### davee123

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

5. Apr 11, 2006

### Jimmy Snyder

I got a few of the quickies.

4. $-1 = e^{i\pi}$
5. $\left(\frac{499}{500}\right)^{500}$
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.

Last edited: Apr 11, 2006
6. Apr 11, 2006

### Euclid

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

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

7. Apr 12, 2006

### NateTG

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

I'd prefer something like:
Consider the real solution to:
$$x^x=2$$

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

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

8. Apr 12, 2006

### 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.

9. Apr 12, 2006

### Jimmy Snyder

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

10. Apr 12, 2006

### siddharth

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

11. Apr 12, 2006

### 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

Last edited: Apr 12, 2006
12. Apr 12, 2006

### Jimmy Snyder

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

13. Apr 12, 2006

### Jimmy Snyder

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?

14. Apr 12, 2006

### NateTG

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

15. Apr 12, 2006

### Jimmy Snyder

I don't believe so.

16. Apr 12, 2006

### siddharth

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

17. Apr 12, 2006

### arunbg

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
$$^{n+r-1}C_{r-1}$$

For this case,the no of favourable outcomes (no error in page 29) from the above formula will be $^{998}C_{498}$ (distributing 500 errors in 499 pages leaving page 29 vacant).
No of possible outcomes will be $^{999}C_{499}$(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

18. Apr 12, 2006

### 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
$$\left(\frac{499}{500}\right)^{500}$$

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.

19. Apr 12, 2006

### Jimmy Snyder

For this you receive an AAA+.

Edit: Sorry, make that an A+.

20. Apr 12, 2006

### Jimmy Snyder

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

Using your method, I get)
$$\frac{^{n+r-1}C_{r-1}}{^{n+r}C_{r}}$$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.

Last edited: Apr 12, 2006
21. Apr 13, 2006

### arunbg

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.

22. Apr 13, 2006

### Jimmy Snyder

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

23. Apr 13, 2006

### 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.

24. Apr 13, 2006

### Jimmy Snyder

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

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.

25. Apr 13, 2006

### 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.