# An Elementary Swahili Newspaper Reader by Agnes Musyoki

By Agnes Musyoki

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**Additional resources for An Elementary Swahili Newspaper Reader **

**Example text**

The binomial theorem shows that this is nbn−1 + k · c. Thus (((b + c)n − bn )/c, c) divides (nbn−1 , c). Rewriting (an − bn )/(a − b) as (an − (a − c)n )/c shows that (((b + c)n − bn )/c, c) divides (nan−1 , c). Therefore (((b + c)n − bn )/c, c) divides (n(a, b)n−1 , c). These expansions also make it clear that (n(a, b)n−1 , c) is a divisor of (an − bn )/(a − b). If a and b are relatively prime then (a, b) = 1. Apply part (a). Since (a/b) + (c/d) = (ad + bc)/bd is an integer, bd | ad + bc.

Consider the function f (k) = (n + 1 − k)k, whose graph is a concave-down parabola with k-intercepts at k = 0 and k = n + 1. Since f (1) = f (n) = n, it is clear that n n f (k) ≥ n for k = 1, 2, 3, . . , n. )2 = k=1 k(n + 1 − k) ≥ k=1 n, by the inequality above. This last is equal to nn . )2 . )). 8. There exist by hypothesis k1 and k2 such that f1 ≤ k1 O(g1 ) and f2 ≤ k2 O(g2 ). Let k = max{c1 k1 , c2 k2 }. Then c1 f1 + c2 f2 ≤ c1 k1 O(g1 ) + c2 k2 O(g2 ) ≤ k(O(g1 ) + O(g2 )) = kO(g1 + g2 ). 9.

Let n ≥ 128 and r suppose there is no prime between n and 2n. Let 2n = p≤2n p be the prime factorization for n 2n 2n r = p≤n p . If p is a prime in the range n . But there are no primes between n and 2n, so n 2n 2n 2n/3 < p ≤ n then p divides n! exactly once and (2n)! exactly twice. Thus p = n . Therefore n √ r r √ √ √ p p ≤ 2n p since if p is in the range 2n < p ≤ 2n/3, then p p≤2n/3 p≤ 2n 2n