By Gloria Rand

A boy and his father prefer to hike within the historical wooded area close to their domestic. yet at some point they notice blue marks on some of the trees--the marks of loggers. The boy makes a decision they have to do whatever to attempt to save lots of the woodland. A crusade is introduced and the struggle is on.Gloria and Ted Rand have been encouraged to create this e-book after listening to real-life tales from their son, Martin, who's an energetic conservationist in Washington nation. jointly, this writer and illustrator group has captured the quiet majesty of our nation's historical forests. Bordering the paintings are images of local crops and animals; a quick nature advisor on the finish of the booklet offers younger naturalists with pointers on determining bushes and animal tracks.

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**Example text**

Here we say that, for a given positive integer n, two integers, a and b, are equal modulo n (more often we use the word congruent) if a – b = kn for some integer k. We write this as a ≡ b (mod n). For example, 20 ≡ 2 (mod 6) as 20 – 2 = 18 = 3 × 6. The arithmetic of the integers modulo n follows similar but not identical rules to that of ordinary algebra. We will not call upon the idea too much, except the notation crops up in the explanation of Problem 8 on coding theory. Combinations We will assume familiarity with expansions such as (a + b)2 = a2 + 2ab + b2 and the diﬀerence of two squares: a2 – b2 = (a – b)(a + b).

A famous inﬁnite series is the following: ∞ S= n=1 1 . n2 It comprises an inﬁnite sum of positive terms so it either diverges to +∞, as does for example 1 the sum of reciprocals, ∞ n=1 n , or converges to a limit. To prove that it converges to some limit, we only need to prove that the sum is bounded above by some positive number. This can be done in a number of ways, one of the quickest being the observation that the sum from n = 2 to ∞ is less than the area under the curve y = x12 from x = 1 to ∞.

1 so that (3) applies in all cases. Two consequences of these formulae are n n = k n–k 24 PROFESSOR HIGGINS’ S PROBLEM COLLECTION which follows from (3) and n 2n = k=0 n . k ( 4) To see the latter, put a = b = 1 in (2). Moreover, 2n is the total number of subsets of a set of size n, as the right-hand side of (4) sums the number of subsets of size k of a set of size n as k ranges from 0 up to n. Complex numbers Although not required in the previous problem set, so-called imaginary and complex numbers do arise later and, being numbers, we use this opportunity to outline their properties.