2 questions. (1 Viewer)

[ssj_goku]

Here are some random q's. Not the hardest questions ever made, but I though they were kind of cute.

1. Calculate how many factors are present in the number 6469693230.

2. In a group of 30 people, calculate the probability that there will only be one common birthday (only 2 people share a birthday).

 

[underthesun]

Whoa! They better not turn up in the hSC, even though it'd come with hinting lead-up questions beforehand anyways..

 

[freaking_out]

Originally posted by ssj_goku


2. In a group of 30 people, calculate the probability that there will only be one common birthday (only 2 people share a birthday).

yeah, i've heard of this one...i think its something like 50% or more.

 

[freaking_out]

actually look at this link:

http://science.howstuffworks.com/question261.htm

 

[ssj_goku]

Good pick up freaking out. I was reading this book and it mentioned the birthday paradox. We had just started year 10 prob and so I wanted to know how the heck you'd go about doing it. But fortunately the paradox it mentioned was the same one in the article (P of at least one common birthday) and could be done with year 10 lvl prob. The prob I gave is solved in a similar way, you just need the senior extension to do it.

 

[turtle_2468]

ssj_goku, u a terry pratchett fan?
6469693230... I like this one..

 

[Lazarus]

Originally posted by ssj_goku
1. Calculate how many factors are present in the number 6469693230.

Mmm discrete maths... I should know how to do that. *frets about lack of study for upcoming exam*

 

[deyveed]

Originally posted by ssj_goku

1. Calculate how many factors are present in the number 6469693230.

How do you do this one?
I can only get 1,2,3,5,10 and 6469693230

 

[MyLuv]

Originally posted by deyveed
I can only get 1,2,3,5,10 and 6469693230

---> from here U can find more factors by multiply them together like 2*3 or 2*10 or 2*3*5 but there're many missing.
The only way I can think of is fractorise it to product of prime number like 2^n*3^m*5^k.... but it take forever

 

[freaking_out]

Originally posted by MyLuv
---> from here U can find more factors by multiply them together like 2*3 or 2*10 or 2*3*5 but there're many missing.
The only way I can think of is fractorise it to product of prime number like 2^n*3^m*5^k.... but it take forever

this is a question for computers, not humans.

 

[underthesun]

Originally posted by freaking_out
this is a question for computers, not humans.

yes but unless you haven't realised, many people in this forum are hybrids between humans and computers, shown from their superior mathematical skills.

some even finish the exam in 2 hours time...

 

[Hotdog1]

Any chance of a solution of that first problem? It resembles more like those questions you get in competitions rather than the HSC.

 

[ssj_goku]

Originally posted by deyveed
How do you do this one?
I can only get 1,2,3,5,10 and 6469693230

7 and 11 are also factors. Think about it. Btw turtle, I am a bit of a terry pratchet fan.

 

[Affinity]

underthesun, freaking out:
...

Freeware from Shamus Software, Dublin, Ireland
Full C source code and MIRACL multiprecision library available
Email to mscott@indigo.ie for details
Web page http://indigo.ie/~mscott
Source code from ftp://ftp.compapp.dcu.ie/pub/crypto/miracl.zip

C:\a>factor 6469693230
first trying brute force division by small primes
PRIME FACTOR 2
...

 

[ssj_goku]

No one seems to have done q1. Don't worry, I purposely wrote the q to avoid a lot of computation. Its actually the product of all the primes up to and including 29.

I'm not bad, I'm just drawn that way.

 

[Affinity]

1) there are 1024 factors.. (10 prime factors, each to the power of 1, so it's 2*2*2*2*2*2*2*2*2*2)

 

[ssj_goku]

Nice work Affinity.

 

[MyLuv]

Originally posted by ssj_goku
. Its actually the product of all the primes up to and including 29.

Ahh..just a question,how to get this???

 

[Affinity]

divide

alternative.. you can use that factoring program I used if you are lazy, check my other post