Official TSR Mathematical Society

I'll post a few slightly more accessible problems with brief hints in spoiler boxes underneath. I didn't make these problems up myself and while they don't have any far reaching consequences they are perhaps fairly interesting.

1: Let [x] denote the largest integer not exceeding x (for example, [2.5]=2, [-1.1]=-2 and [

\pi

]=3).
Show that the number of zeros at the end of the integer n! is:

\left[ \frac{n}{5}\right]+\left[ \frac{n}{25}\right]+\left[ \frac{n}{125}\right]+\left[ \frac{n}{625}\right]+\ldots

Spoiler

2: Does there exist a UNIQUE positive integer t such that t, t+2 and t+4 are all primes?

Spoiler

Reply 221

RichE

15

These were pointed out to me recently by a friend

http://www.maths.usyd.edu.au/u/SUMS/sums2005.pdf

They're probably about BM02 level.

Have done 6 and 7 - working my way through the others (I hope).

Reply 222

evariste_3086

1

RichE

These were pointed out to me recently by a friend

http://www.maths.usyd.edu.au/u/SUMS/sums2005.pdf

They're probably about BM02 level.

Have done 6 and 7 - working my way through the others (I hope).

Have any of you done BMO and/or IMO ? How did you get on?
I never had chance to try and gain entry for these when i was at school. My teacher had trouble going beyond O level! I feel i have missed out on many skills by not encountering these maths challenges before.

Reply 223

J.F.N

1

RichE

These were pointed out to me recently by a friend

http://www.maths.usyd.edu.au/u/SUMS/sums2005.pdf

They're probably about BM02 level.

Have done 6 and 7 - working my way through the others (I hope).

Nice questions, but the prize is a bit shocking, don't you think? The Putnam, the Undergrad's math competition here in America, pays the highest scorer $2,500, a $12,000 scholarship, and graduate school tuition. Other high scorers also get substantial prizes. And never mind that they become mini-celebrities in Undergrad math communities.

Reply 224

RichE

15

J.F.N

Nice questions, but the prize is a bit shocking, don't you think? The Putnam, the Undergrad's math competition here in America, pays the highest scorer $2,500, a $12,000 scholarship, and graduate school tuition. Other high scorers also get substantial prizes. And never mind that they become mini-celebrities in Undergrad math communities.

But these questions don't seem that hard. It sounds like something a maths society set up for a bit of fun. Certainly not as respected as the Putnam.

Reply 225

KAISER_MOLE

7

I can't open the questions, my adobe must be messed up >.<

and for Gaz question 2.

Spoiler

And my thoughts on question 1.

Spoiler

Reply 226

Gaz031

15

Kaiser,
You're pretty much right with Q2. :smile:

You can also prove that either t, t+2 or t+4 are composite for all t>3 by letting t=3k, 3k+1 or 3k+2 and showing that for each possible form for t, at least one of t, t+2 or t+4 is divisible by 3.

Can I join?

will like 2 join 2

Aha, Im now a member. Thanks Euler!

Im sure you all know this, but just something to bring up about primes. If only mathematicians can reason why primes come up in pairs

p, p+2

. There has to be a link between that and the Gaussian approximation of density of prime numbers among the integers

\frac {1}{\log n}\

I refer you to the "the theory of numbers" chapter in "What is Mathematics" pp 28-29.

Just a few thoughts as my first post as an inaugarated member :smile:

Reply 230

J.F.N

1

HTale

Aha, Im now a member. Thanks Euler!

Im sure you all know this, but just something to bring up about primes. If only mathematicians can reason why primes come up in pairs

p, p+2

. There has to be a link between that and the Gaussian approximation of density of prime numbers among the integers

\frac {1}{\log n}\

I refer you to the "the theory of numbers" chapter in "What is Mathematics" pp 28-29.

Just a few thoughts as my first post as an inaugarated member :smile:

The best progress on the problem was made by J.R. Chen, who has proven that there are infinitely many primes p such that p+2 is either a prime or has two prime factors. But in the theory of numbers there will always be more unsolved problems than solved ones. For example, it is conjectured that there are infinitely many primes p such that p+2, p+6 are also primes.

Advancing even further, let N be any given number. Can we always find a prime p such that n²-n+p is always prime when 0<=n<=N? (for reference, call this conjecture [1]). (The famous example of this is Euler's n²-n+17, which is always prime when 0<=n<=16). Conjecture [1] is also an unsolved problem, and if it is solved then the twin prime problem can be settled. This is because in order for the polynomial n²-n+p to (successively) take prime values, n must be restricted to the integers from 0 to p-1. We now construct a sequence of polynomials n²-n+p_i with the following property: when 0<=n<=p_(i-1), the number n²-n+p_i is always prime. If [1] is solved, then this construction is certainly possible. Now taking n=1 and 2 will give p_i and p_(i+2) both primes, settling the twin prime problem. In a similar way, taking n=1,2,3 will give p_i,p_(i+2),p_(i+6) all primes, settling the conjecture in the end of the first paragraph.

Reply 231

Starsailor

1

I'm doing AS Maths, can I join please?

Michael

Reply 232

HTale

14

J.F.N

The best progress on the problem was made by J.R. Chen, who has proven that there are infinitely many primes p such that p+2 is either a prime or has two prime factors. But in the theory of numbers there will always be more unsolved problems than solved ones. For example, it is conjectured that there are infinitely many primes p such that p+2, p+6 are also primes.

Advancing even further, let N be any given number. Can we always find a prime p such that n²-n+p is always prime when 0<=n<=N? (for reference, call this conjecture [1]). (The famous example of this is Euler's n²-n+17, which is always prime when 0<=n<=16). Conjecture [1] is also an unsolved problem, and if it is solved then the twin prime problem can be settled. This is because in order for the polynomial n²-n+p to (successively) take prime values, n must be restricted to the integers from 0 to p-1. We now construct a sequence of polynomials n²-n+p_i with the following property: when 0<=n<=p_(i-1), the number n²-n+p_i is always prime. If [1] is solved, then this construction is certainly possible. Now taking n=1 and 2 will give p_i and p_(i+2) both primes, settling the twin prime problem. In a similar way, taking n=1,2,3 will give p_i,p_(i+2),p_(i+6) all primes, settling the conjecture in the end of the first paragraph.

Is this the same Chen that showed that every even number is the difference between a prime and a

P_2

?.

Reply 233

J.F.N

1

HTale

Is this the same Chen that showed that every even number is the difference between a prime and a

P_2

?.

Yes.

Reply 234

Neapolitan

2

Can I join, please? :smile:

Reply 235

dvs

10

J.F.N

Nice questions, but the prize is a bit shocking, don't you think? The Putnam, the Undergrad's math competition here in America, pays the highest scorer $2,500, a $12,000 scholarship, and graduate school tuition. Other high scorers also get substantial prizes. And never mind that they become mini-celebrities in Undergrad math communities.

Will you be taking the Putnam this year?

Reply 236

AlphaNumeric

10

Do you have to PM someone to join or just post here? If so, do you have to solve one of those questions or not?
I assume its relevant what course I'm doing (everyone else is saying there), I'm a 4th year maths student at Cambridge.

Reply 237

cktlee1

can someone enlighten me as to why "falsity implies anything"?....i'm looking at the table :

p q p=>q
T T T
T F F
F T T
F F T

where p and q are different statements, T is True, F is False, and p=>q means "p implies q "

please give me reasons rather than yet another definition of what "implies" means.... (ie. defined as "¬p or q" )

Reply 238

AlphaNumeric

10

Its a convention. When considering "p => q", that means if p is true, then it implies q is true. However, if p is not true, you cannot say anything about "p => q", because you need p to be true before you can compare it to the state of q. The convention then is to say that since "p" is false, you can say that "p => q" is true, no matter the state of "q".

You are free to make such a definition, because nothing contradicts it. If further down the line of constructing your logic you find that the choice of convention is important, then something would have been done about it, but that doesn't happen so the choice of convention is allowable.

I'm not a "logic-stian" unfortunately, I'm getting this from my 1st year course which had something on this in them, but thats what I think they are saying.

Reply 239

Cexy

There's a little bit more to it than just convention. This may get a little tricky, but hopefully if you work at it for a bit, you'll be able to follow it.

Consider the statements P = (pigs can fly) and Q = (the queen is rich)

Consider the statement "Q and ¬Q => P". Here the premise, that the queen both is and isn't rich, is clearly false. The conclusion may or may not be true - we don't know as there could be flying pigs that we've never seen. Let's assume that the statement is false, and we'll see why we have to count this statement as true.

Look at "Q => Q or P". This is definitely true, since if the queen is rich, then at least one of "the queen is rich" or "pigs can fly" is true. -(1)

What about "(Q or P) and ¬Q => P". This is true as well, since if either the queen is rich and pigs can fly, but the queen isn't rich, then pigs can fly. -(2)

But now look what happens if we combine the last two statements:

"Q (=> Q or P) and ¬Q => P".

Here we assume that "the queen is rich", and "the queen is not rich". Since the queen is rich, then "either the queen is rich or pigs can fly", from statement (1). So "either the queen is rich or pigs can fly", and "the queen is not rich". Therefore "pigs can fly", according to statement 2.

But this is just the statement we started out with - that "the queen is rich" and "the queen isn't rich", so "pigs can fly". But we assumed this was false - so we've combined two true statements and ended up with a false one.

This is clearly nonsense, so our only option is to assume that the original statement is true - i.e. the statement "something false => something true or false" has to count as true.

Hopefully, that explains why a statement is only false when it has the form "something true => something false".

:smile:

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