By Kenneth H. Rosen
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Additional info for Elementary Number Theory and Its Applications, 5th edition, Instructor's Solutions Manual
When n = 1 we have p1 = 2 ≤ 22 = 2. Now assume that pk ≤ 22 for k = 1, 2, . . , n − 1. Then by Euclid’s proof, a prime q other than p1 , p2 , . . , pn divides Qn . Then pn < 0 1 n−1 0 1 n−1 n−1 q ≤ Qn = p1 p2 · · · pn + 1 ≤ 22 22 · · · 22 = 22 +2 +···+2 = 22 −1 + 1. Since the inequality is n−1 n−1 strict and we are dealing with integers we have pn ≤ 22 −1 ≤ 22 , which completes the induction step. b. , the (n + 1)st prime is less than or equal to 22 , and since a power of 2 can not be prime n itself when n > 0, we must have at least n + 1 primes strictly less than 22 .
Suppose that f is O(g) where f (n) and g(n) are positive integers for every integer n. Then there is an integer C such that f (n) < Cg(n) for all x ∈ S. Then f k (n) < C k g k (n) for all x ∈ S. Hence f k is O(g k ). 10. Suppose f (n) = O(log2 n). Then f (n) ≤ k log2 n = k log2 r logr n = k logr n. Conversely, if f (n) ≤ k logr n = k(log2 n)/(log2 r) = k log2 n, and so f (n) = O(log2 n). 11. The number of digits in the base b expansion of n is 1 + k where k is the largest integer such that bk ≤ n < bk+1 since there is a digit for each of the powers of b0 , b1 , .
As, b − c − 1 < a − c, a − c > b − c − 1 < b, and c + 9 − b ≥ 10 + c − a, we know that a and b are a − c and c + 9 − b, perhaps not respectively. If a = a − c, then c = 0. But then b = 9 − b, which is impossible. If a = c + 9 − b, then b = a − c and a = c + 9 − a − c, from which it follows that 9 is even. So we conclude that c = d. Suppose that a = b > c > d. Then (a, a, c, d) − (d, c, a, a) = (a − d, a − c − 1, c − a + 9, 10 + d − a). From the inequalities a ≥ a − d ≥ a − c > a − c − 1 and c − a + 9 ≥ d + 1 − a + 9 = 10 + d − a we may conclude that c and d are a − c − 1 and 10 + d − a, perhaps not respectively.
Elementary Number Theory and Its Applications, 5th edition, Instructor's Solutions Manual by Kenneth H. Rosen