midterm-solutions

# 4 show that the cardinality of the set of all pairs

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4. Show that the cardinality of the set of all pairs of natural numbers is the same as the cardinality of the set of all natural numbers. Solution. This is Cantor’s other diagonal proof. We write the pairs of natural numbers in a grid (1 , 1) (1 , 2) (1 , 3) (1 , 4) ··· (2 , 1) (2 , 2) (2 , 3) (2 , 4) ··· (3 , 1) (3 , 2) (3 , 3) (3 , 4) ··· (4 , 1) (4 , 2) (4 , 3) (4 , 4) ··· . . . . . . . . . . . . . . . and proceed down the “anti-diagonals” (which run from northeast to southwest) in turn, giving the list (1 , 1) , (1 , 2) , (2 , 1) , (1 , 3) , (2 , 2) , (3 , 1) , (1 , 4) , (2 , 3) , (3 , 2) , (4 , 1) ... 5(a). Compute 3 43 (mod 8) using Euler’s theorem. Solution. First, φ (8) = 4. (This is the number of numbers less than 4 and having no factor in common with it, which are 1 , 3 , 5 , 7.) Euler’s theorem states that a φ ( n ) 1 (mod n ) whenever a and n have no common factor. Since 3 and 8 have no common factor, this gives us 3 4 1 (mod 8). (You could check this explicitly, since 3 4 = 81.) Thus 3 43 = (3 4 ) 10 (3 3 ) 3 3 (mod 8) and 3 3 = 27; the answer is the remainder when 27 is divided by 8, which is 3. 5(b). Why can Fermat’s little theorem not be used in part (a)? Solution. Fermat’s little theorem only applies when the modulus is prime. 8 is not prime. 6. Bank identiﬁcation numbers consist of nine digits n 1 ,n 2 ,...,n 9 , such that 7 n 1 + 3 n 2 + 9 n 3 + 7 n 4 + 3 n 5 + 9 n 6 + 7 n 7 + 3 n 8 + 9 n 9 0 (mod 10) . Consider the bank identiﬁcation number 211 X 72946 where X is a missing digit. What is X ? Solution. Putting the claimed bank identiﬁcation number into the formula, we get 7 · 2 + 3 · 1 + 9 · 1 + 7 · x + 3 · 7 + 9 · 2 + 7 · 9 + 3 · 4 + 9 · 6 0 (mod 10) . Doing the arithmetic gives 194 + 7

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