Yes we are up to that part but I'm not sure how to even start it. Do we need to say for all \epsilon> there exists a n,m /geqN such that abs(Snk-Smk)<E?
to use the chinese remainder theorem wouldn't you have to rewrite the equation s for the x's as
x==-3(mod5)
x==-2(mod6)
x==-3(mod7)
so then we can rewrite it using the q's. and once we figure out what the general form of it the x would be the remainder of it
Thank you for answering my question. I just need some clarification though. When you have the
x + 3 == 0 (mod 6),
y + 1 == 0 (mod 6).
do you solve the congruence x + 3 == 0(mod6) and then you solve the second y+1 == 0(mod6) and once you do you subsitute the remainder into the n = 2^x *...
fi have no idea what to do and i tried posting it on another forum and nobody replied so please help me! thank you so much!
find a positive integer n so that 40n is a fifth power (of an integer) 500n is a sixth power, and 200n is a seventh power, or explain why it is impossible to do so...
a = 238000 = 2^4 x 5^3 x 7 x 17 and b = 299880 = 2^3 x 3^2 x 5 x 7^2 x 17
is there an integer n so that a divides b^n if so what is the smallest possibility for n
I have 2 questions.
1)what is the remainder with 100! is divided by 103? explain your answer
2)a = 238000 = 2^4 x 5^3 x 7 x 17 and b=299880 = 2^3 x 3^2 x 5 7^2 17. Is there an integer so that a divides b^n? if so what is the smallest possibility for n?
the first one i have no...