SOLUTIONS TO HOMEWORK 2
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
Three Transcendental Numbers From the Last Non
... trivial result from Diophantine approximation [7], [11]: if we have α rational and > 0 then there exist only finitely many many rational numbers p/q ...
... trivial result from Diophantine approximation [7], [11]: if we have α rational and > 0 then there exist only finitely many many rational numbers p/q ...
A New Connection Between the Triangles of Stirling and Pascal
... But the sequence of the number of elements in each white triangle began with 1 and this isn’t a perfect number! That’s true, Pascal’s triangle doesn’t always yield perfect numbers in this manner, but every even perfect number does appear somewhere in this sequence. This is because the number of elem ...
... But the sequence of the number of elements in each white triangle began with 1 and this isn’t a perfect number! That’s true, Pascal’s triangle doesn’t always yield perfect numbers in this manner, but every even perfect number does appear somewhere in this sequence. This is because the number of elem ...
Integer Factoring
... All run times involving L n are for n → ∞. This implies that the o(1)’s that might occur in run time expressions go to zero. In practice, however, the o(1)’s are not zero. Therefore we cannot encourage the practice of evaluating the run time expression for any of the factoring methods presented here ...
... All run times involving L n are for n → ∞. This implies that the o(1)’s that might occur in run time expressions go to zero. In practice, however, the o(1)’s are not zero. Therefore we cannot encourage the practice of evaluating the run time expression for any of the factoring methods presented here ...
What is the sum of the first 100 positive integers?
... First, a list consisting of only one item cannot be re-arranged, so D1 = 0. Second, D2 = 1 since the only way to get two items completely out of order is to transpose them. Now consider the number Dn of derangements of the set of the first n positive integers: {1, 2, 3, . . . , n}. The number 1 can ...
... First, a list consisting of only one item cannot be re-arranged, so D1 = 0. Second, D2 = 1 since the only way to get two items completely out of order is to transpose them. Now consider the number Dn of derangements of the set of the first n positive integers: {1, 2, 3, . . . , n}. The number 1 can ...
x - Dr Frost Maths
... Explain why k and k+1 are coprime for any positive integer k. Answer: Suppose k had some factor q. Then k+1 must have a remainder of 1 when divided? by q, so is not divisible by q. The same reasoning underpins Euclid’s proof that there are infinitely many primes. Suppose we have a list of all known ...
... Explain why k and k+1 are coprime for any positive integer k. Answer: Suppose k had some factor q. Then k+1 must have a remainder of 1 when divided? by q, so is not divisible by q. The same reasoning underpins Euclid’s proof that there are infinitely many primes. Suppose we have a list of all known ...
Variant of a theorem of Erdős on the sum-of-proper
... it is possible to show this set has lower density larger than 10−8 . To do this one can exploit the fact that for n squarefree we have s(n) = s∗ (n). Thus, in trying to find values for s or s∗ with 510 as a unitary divisor and in the residue class c (mod d), for both problems we are restricted to se ...
... it is possible to show this set has lower density larger than 10−8 . To do this one can exploit the fact that for n squarefree we have s(n) = s∗ (n). Thus, in trying to find values for s or s∗ with 510 as a unitary divisor and in the residue class c (mod d), for both problems we are restricted to se ...