By Ivan Moscovich

E-book by way of Ivan Moscovich

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Extra resources for Fiendishly Difficult Visual Perception Puzzles

Sample 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.