Mod Questions

Does anybody know any good math problems with mod. I don't know much about it and i'd like to practice a bit:S

Try to find primes...;)

Huh? Try to find primes? I asked for mod problems? What do you mean?

If you are thinking of MODULUS, finding primes is a perfect way to practice that....

if a^3+b^3=0 (mod 11)
prove a+b=0 (mod 11)


prove a^p=a (mod p) for some prime p

find the smallest positive k such that
k is divisible by 5
k+1 is divisible by 7
k+2 is divisible 9
k+3 is divisible by 11

:p

here is interesting one:

x^2+y^2=13

and use the fact that i=5 (mod 13)


:p :p

prove (a+b)^p=a^p+b^p (mod p) for some prime p.

(a+b)^p
= a^p + pC1*a^p-1b+pC2*a^p-2*b²...+ pC(p-2)*a²*b^p+2+ pC(p-1)*a*b^p-1 + b^p

Now, all of the pCx have a factor of p (cos p is prime, and so p!/(something-less-than-p)! will always have one factor of p.)
=>all of these terms == 0 (mod p)
=> (a+b)^p == a^p + 0 + 0 +...+ 0 + 0 + b^p
== a^p + b^p

a cool MOD q coming straight from the origional IMO:
"Proove that 2^x + 1 is never a multiple of seven."
That's right, it was an IMO question.

I still think that finding primes is the easiest way of learning...everyone knows how to find a prime, and if you don't it will take you 1 sec to learn it...

sql_lall why go through all the trouble? fermats little theorem would do in one line

now now, it isnt good to use "IMO" to scare people away...
:D:D:D
thats pretty easy for an imo question. what year?

anywayz? assume 2^3-1 (thats right, thats 7) divides into 2^x+1

then use long division (easiest way to explain it, but on a contest i would use a=pq+r), then we would get a remainder of 0 right? if you try it out with a few terms you see the remainder would always be in the form of 2^p+1

so 2^p+1=2^3-1

but no integer p works. therefore 2^x+1 is never divisible by 2^3-1