Recent content by miren324

I have a homework problem and I use a rule to solve it that seems to be true, at least for small numbers, but I cannot seem to find a clearly stated theorem assuring me that it is true. Here's the problem with my solution: Find the remainder of the division of 2^(36!) by 37. Proof: By...

So, by Wilson's Theorem I have 210! \equiv -1 (mod 211). Now I divide by 131*132*133*134*135\equiv 208 (mod 211). Now divide 210 by 208 (mod 211). 210/208 = x (mod 211) 210 = 208x (mod 211) 210 = 208x + 211y 211 = 210(1)+3 so 3=211-208(1) 208 = 3(69) + 1 so 1 = 208-3(69) 3 = 1(3) so...

Thanks a lot. Found the proof. You may have just saved my MAT445 grade. Haha.

Just to make sure I'm clear, since 75!=((-1)^75)(-1)(-2)...(-75), then 75!130!=(-1)(-1)(-2)...(-75)(1)(2)...(130). Now, 75!130! is the product of all (but 5) nonzero elements of Z/211Z because in mod 211 -1 is 210, -2 is 209, and -75 is 136, so all we are missing in our product are 131, 132...

Prove that the last three digits of n^100 can be only: 000, 001, 376, or 625. I can easily show that the last digit is either 0, 1, 6 or 5 because n^100=((n^25)^2)^2, so if our last three digits are 100a+10b+c, with a, b, c belonging to {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, any digit for c...

I'm struggling with how to even begin with this problem. Find the remainder of the division of 75!*130! by 211. 211 is prime, so I know the remainder is not 0. I'm not sure where to start though. Thanks!

Not sure exactly where you're headed with that. For all I know the two separate entries for x^2 could be on exact opposite sides, meaning that the smallest sub-tree has 2 as it's root.

Here's the question. Starting with an integer a≥2, we write on its left, below it, the number a+1, and on its right, below it, the number a^2, and obtain four numbers, to which we continue the process. We thus obtain a binary tree, whose root is a. Prove that the numbers in every line of the...

@Mandeep Deka: I have that book. Do you know where in the book this problem is? I don't remember seeing it, but of course I didn't look at every problem in the whole book.

I need help. I'm trying to prove that if (2^n)+1 is prime, then there exists an integer k>=0 such that n=2^k. If n is odd, then (2^n)+1=(2^(2k+1))-(-1)^(2k+1)=(2+1)(stuff...)=(3)(stuff) so it's not prime, a contradiction. So I've knocked out half of the possible n's. What I'm struggling...

Here's the problem: Let f(x)={x, x in Q; 0, x in R\Q. Show f is continuous at c if and only if c = 0. Hint: You may want to use the following theorem: Let A and B be two disjoint subsets of R and f1:A\rightarrowR and f2:B\rightarrowR. Define f:A\cupB\rightarrowR by f(x)={f1(x), x in A...

It just makes sense that it would be true. I know that if x does not equal c, the limit as x approaches c of f(x) is L, and L is "near" f(c-.00001) and f(c+.00001) and f(c-.000000001), etc. But I only know that from experience with linear equations. I'm not sure how to translate this into a...

Here's the problem: Let f:D-R and c in R be and a accumulation point of D, which is a subset of R. Suppose that a<=f(x)<=b for all x in D, x not equal to c, and suppose that limx\rightarrowc f(x) = L. Prove that a<=L<=b. I'm having trouble here. I've tried to prove by contradiction, by...

Homework Statement Let (sn) be a sequence in R that is bounded but diverges. Show that (sn) has (at least) two convergent subsequences, the limits of which are different. Homework Equations The Attempt at a Solution I know that a convergent subsequence exists by...