Download Solutions to Practice Problems, Math 312 1 Find all prime numbers

Solutions to Practice Problems, Math 312
1 Find all prime numbers of the form 24k+2 + 1.
Solution Since 24k+2 +1 ≡ 0 modulo 5, if k ≥ 1, it follows that 24k+2 +1
is composite and hence the only prime is with k = 0 (i.e. 5).
2 A child has six dollars and fifty cents in dimes and quarters. Determine the number of possible combinations of coins she can have.
Solution We solve
10x + 25y = 650
or
2x + 5y = 130.
Since one solution is, by inspection, x0 = 0, y0 = 26, all integral
solutions are given by x = 5k, y = 26 − 2k for k ∈ Z. The positive
solutions correspond to 0 ≤ k ≤ 13.
3 Find a number n which, when divided by 29 leaves a remainder of
17; and when divided by 78 leaves a remainder of 37.
Solution We solve
n ≡ 17 mod 29,
n ≡ 37 mod 78.
Work leads to n ≡ −41 mod 2262 and so a positive solution is n =
2221.
4 Is the sum of three consecutive cubes always divisible by 9?
Solution We have
n3 + (n + 1)3 + (n + 2)3 = 3n3 + 9n2 + 15n + 9 ≡ 3(n3 − n) mod 9
and n3 − n = (n − 1)n(n + 1) is a multiple of 3 and so the product of
three consecutive cubes is, in fact, divisible by 9.
5 Compute 999 modulo 100.
Solution Since φ(100) = 40, we have
999 ≡ 919 ≡ 338 mod 100
and 340 is congruent to 1 mod 100, so we have to solve 9x ≡ 1 modulo
100, i.e. x ≡ −11 mod 100. It follows that
999 ≡ −11 ≡ 89 mod 100.
1
2
6 Determine the inverse of 67 modulo 119.
Solution We want to solve 67x ≡ 1 mod 119. This implies that 67x ≡
1 mod 7 and 67x ≡ 1 mod 17 and so 4x ≡ 1 mod 7 or x ≡ 2 mod 7,
and −x ≡ 1 mod 17 or x ≡ −1 mod 17. By inspection, we have x ≡
16 mod 119.
7 Let p, q and r be distinct primes. Show that there exists n such
that p|n, q|n + 1 and r|n + 2.
Solution This amounts to solving n ≡ 0 mod p, n ≡ −1 mod q and
n ≡ −2 mod r which, since p, q and r are distinct primes, has a common solution by the Chinese Remainder Theorem.
8 If p is a prime and a is an integer, coprime to p, define the order
of a modulo p to be the smallest positive integer k such that ak ≡ 1
modulo p. Prove that k divides p − 1.
Solution Use the division algorithm to write p − 1 = bk + c where
0 ≤ c < k. Then
1 ≡ ap−1 ≡ abk+c ≡ (ak )b · ac ≡ ac mod p.
Since k is the smallest positive integer for which ak ≡ 1 modulo p, we
thus have c = 0 and so k divides p − 1.
9 Suppose that in the course of applying Pollard rho to try to factor
n, we find that xi ≡ f (xi ) modulo n for all i ≥ 1. What conclusion do
we reach?
Solution If xi ≡ f (xi ) modulo n for all i ≥ 1, then x2k − xk = 0 and
so we gain no information on the factorization of n.