### (2n choose n) is never a perfect power

**Source:**Cute problem sent by Sudeep Kanath

**Problem:**Prove that (2n choose n) is never a perfect power

Update ( 21 June 2014 ):

**Solution:**Posted by Sandeep, Dinesh Krithivasan and Vishal Khatri in comments.

- Get link
- Other Apps

Update ( 21 June 2014 ):

- Get link
- Other Apps

Most of my friends already read an article that I wrote more than an year back - "Speak Up"

Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it!

Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis.

@Consultants, correct me if I am wrong in my estimates. :P

Why is there a need to be selective?

From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once every week right …

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

1) Calculate P{X>Y}

2) What's the expected value of X

3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

Solution:

1)

(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

1) Calculate P{X>Y}

2) What's the expected value of X

3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

Solution:

1)

(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

Sent to me by Gaurav Sinha

Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions. In total, Siddhant scores 25 out of 34.

Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34.

Note that

a) Vaibhav scores more than Siddhant

b) Siddhant score better than Vaibhav in both individual topics - 5/6 > 20/25 and 20/28 > 6/9

How is it possible?

There is always atleast one prime between n and 2n. (Bertrand's postulate). These primes occur only once in factorization of 2nCn. So, 2nCn can never be a perfect power.

ReplyDeleteThanks

DeleteDirect consequence of Bertrand's postulate. After canceling out one of the n! in the denominator, the numerator will be the product of (n+1)(n+2)... up to 2n. By Bertrand's postulate, there is a prime in this group of numbers, say p. Then, (2n choose n) is divisible by p but not any powers of p and so cannot be a perfect power.

ReplyDeleteThis is probably nuking a mosquito though - there ought to be a simple proof.

Most proofs of Bertrand's postulate (at least the ones on wiki) start by studying the prime factorization of 2nCn. So, this reasoning is likely to be circular.

DeleteThanks.

DeleteIt will always be some power of 2 multiplied with some odd number. So, it can't be a perfect power.

ReplyDelete?? Wrong solution!

DeleteJust saw your comment on this..I agree, I have written it wrong... just have to work out again to what i wanted to write. Thanks for bringing it to notice :-)

DeleteI meant to say that it will always be a power of 2 multiplied with some odd numbers of which at least one would a prime. And after reading the comments above and below, I noticed that the prime would be lying in n to 2n.

DeleteSo, I wrote it wrong, Thank you for pointing out :-)

There is always a prime between n and 2n. Hence!

ReplyDeleteThanks

DeleteBertrand's postulate it is !!!

ReplyDeletehttp://en.wikipedia.org/wiki/Bertrand%27s_postulate