SOLUTIONS MANUAL for Discrete Mathematics with Ducks 2nd Edition by Sarah Marie Belcastro

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SOLUTIONS MANUAL Discrete Mathematics with Ducks 2nd Edition by Sarah Marie Belcastro Chapter 1

Counting and Proofs 1.2 Try This! Let’s Count 1. 2 + 6 + 1 + 1 = 10.

2. 3 · 5 = 15.

3. Six as there are five different ice-cream containers.

4. 5 · 5 · 5 = 125 orders because one might want all flavors the same.

5. This one is a pain, intentionally. First, 51 + 2 52 + 53 = 5+ 20+ 10 = 35 (the 2 52 is because it matters which flavor is ordered with 1 quart and which is ordered with 2 quarts), then just 5 4 4 4 4 = 10, then + + = 4 + 6 + 4 = 14 and = 4. Of course, students do not yet 3 1 2 3 3 know this notation, so they will likely write out solutions exhaustively and at length.

6. 3 + 2 + 1 = 6.

7. Three, as there are only two colors. To assure 7 monochromatic armbands, 13 armbands must be drawn (worst case: b-g-b-g-b-g-b-g-b-g-b-g-b/g).

8. 5 · 3 · 6 = 90.

9.

(a) n − 1.

(b) n − 1, then n − 2. (c) n − i.

(d) (n − 1) + (n − 2) + . . . 1 = n(n − 1)/2.

10.

(a) 3 · k.

(b) k + 1 as there are k containers.

11.

(a) The sum of their sizes. (b) The sum of their sizes! (c) Yes, the WEBS problem and the Frisbee tournament problems. (d) m + m + · · · + m = n · m.

12. m · k; yes, related to the ice-cream and snack-box problems.

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1.9. Problems That Use Counting or Proofs

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1.9 Problems That Use Counting or Proofs 1. 4 · 7 = 28.

2. 28 · 3 · 2 = 168.

3. An even number n can be written as 2k for some integer k. So, the sum of two even numbers is n1 + n2 = 2k1 + 2k2 = 2(k1 + k2 ) = 2k3 , which is even.

4. An odd number n can be written as 2k + 1 for some integer k. So, the sum of two odd numbers is n1 + n2 = 2k1 + 1 + 2k2 + 1 = 2k1 + 2k2 + 2 = 2(k1 + k2 + 1) = 2k3 , which is even.

5. 28 by the product principle using slots.

6. 5 · 4 + 3 · 3 = 29.

7. A binary number n that ends in 0 is the sum of some powers of 2, e.g., n = 2a + 2b + · · · + 2c where the lowest possible power of 2 is 21 = 2. Then we may write n = 2(2a−1 + 2b−1 + · · · + 2c−1 ) and so n is even.

8. An odd number n can be written as 2k + 1 for some integer k. To write n as a binary number, we must write it as a sum of powers of 2. This sum must have lowest power 0 (as in 20 ) or else every term of the sum will be divisible by 2. That means that the binary representation will end in a 1 = 20 .

9. There are 11 · 3 · 5 · k = 165k choices and 1,095 days, so k ≥ 6.63. But, of course, k must be an integer . . . so there are at least seven possible drinks.

10. Two consecutive perfect squares are n2 and (n + 1)2 . n2 − (n + 1)2 = n2 − n2 − 2n − 1 = −2n − 1 = −2n + 1 − 2 = 2(−n − 1) + 1, which is odd.

11. 11 = 10 + 1; and, 3 because there are only two different first digits (1, 2).

12. Counterexample: 32 + 42 = 25 is odd.

13. 15 · 8 = 120.

14. 15 · 6 = 90.

15. 5 + 6 + 3 + 4 = 18.

16. Nope. 2 + 3 = 5 is odd. (However, if we exclude 2, then the remaining primes are odd and so the sum of any two of them will be even.)

17. There are 1,440 minutes per day and so (1,440) · 23 = 960 90-second intervals. 1,042 > 960 so two flights must take off within the same interval.

18. Write n = 2k for some integer k. Then (−1)n = (−1)2k = ((−1)2 )k = 1k = 1.

19. 107 .

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1. Counting and Proofs

20. 94 = 6561.

21. An odd number 2k + 1 times an odd number 2r + 1 is (2r + 1)(2k + 1) = 4kr + 2k + 2r + 1 = 2(2kr + k + r) + 1 and thus is odd.

22.

(a) Check 2,000,000 versus 220 = 1, 048, 576: not enough subsets. (b) By trial and error, see that 2,100,000 > 221 = 2,097,152 but 2,200,000 < 222 = 4,194,304.

23. For even n = 2k, 3(2k)3 + (2k) + 5 = 2(3 · 4k3 ) + 2k + 2 · 2 + 1 = 2(3 · 4k3 + k + 2) + 1, which is odd.

For odd n = 2k + 1, 3(2k + 1)3 + (2k + 1) + 5 = odd + odd + odd = odd.

24. 99 · 2 + 1 = 199; 99 · 3 + 1 = 298; 99 · k + 1.

25. Green and brown socks: three for the first pair, then another two for the next pair, and so forth and so on, so 50 · 2 + 1 = 101.

Three colors of socks: four for the first pair, then another two for the second pair (using the leftover two socks), and so forth and so on: 50 · 2 + 2 = 102.

k colors of socks: k + 1 for the first pair, then another two for the second pair (using the leftover k − 1 socks), and so forth and so on: 50 · 2 + k − 1 = 99 + k.

26. All the usual primes are still prime, and 1 is also prime.

If 1 were prime, we would no longer have unique factorization of natural numbers—for example, 6 = 2 · 3 = 2 · 1 · 1 · 1 · 3 = 1 · 2 · 3 . . . .

27. There are no prime numbers with this definition, because 1 is a positive divisor of every integer, and 1 is not included in this definition!

It would be silly to have a class of numbers with no members.

28. There are 3 · 2 = 6 regular options plus 2 · 1 Fancy Tips, so 6 + 2 = 8.

29. By the pigeonhole principle, if each bin had only 13 skeins then you’d only have 104 miniskeins.

30.

(a) Note that the numbers 1–6, paired with the numbers 12–7, form six sums of 13. If no pair of the 7 numbers sums to 13, then all the numbers must be in one of these two subsets—but that’s only 6 numbers, so the seventh number must come from the other set and pair to make a sum of 13. (b) Similarly, the numbers 12–7, paired with the numbers 6–1, form differences of 6. The same argument applies. (c) Now we look at the numbers 1–3 with 4–6, and 7–9 with 10–12 to get six pairs with differences of 3. The same argument as above applies.

31. We have 7 numbers in the range 1–20. There are 27 = 128 possible subsets, and the highest possible sum is 14 + 15 + 16 + 17 + 18 + 19 + 20 = 119. So there are at most 119 different subset sums, and more subsets than that—thus, by the pigeonhole principle, there are two subsets that have the same sum.

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1.9. Problems That Use Counting or Proofs

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32. We can subdivide the square mile into four half-mile squares. By the pigeonhole principle, we know that two of the five Dunkin Donutses must be in the same half-mile square. The furthest these could be apart is the diagonal of the square, or √12 ≈ .71 < .75 miles apart.

However, we could have four Dunkin Donutses at the corners of our square mile, and one in the center, and no two of these would be within half a mile of each other.

33. Well, only one at a time, but there are 2 + 2 · 6 + 2 + 1 + 3 = 20 shelves on which it might be found. And yes, the author does own this book.

34. 12 times 2 is 24.

35. If a(ab + 1) is odd, then both a and ab + 1 must be odd. Therefore a is odd, so a = 2k + 1.

Now (2k + 1)b + 1 = 2kb + b + 1 is odd, which means that b + 1 is odd (because 2kb is even) and therefore b is even.

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Chapter 2

Sets and Logic 2.4 Try This! Problems on Sets and Logic 1. 4; {0, / {0}, {cat}, {{dog}}, {{2.1, 6}}, {0, cat}, {0, {dog}}, {0, {2.1, 6}}, {cat, {dog}}, {cat, {2.1, 6}}, {{dog}, {2.1, 6}}, {0, cat, {dog}}, {0, cat, {2.1, 6}}, {0, {dog}, {2.1, 6}}, {cat, {dog}, {2.1, 6}}, {0, cat, {dog}, {2.1, 6}}}. (24 = 16.)

2. First logic-language it. For all times t, and for all people p, p can be fooled at time t. (There are certainly other possible interpretations.) The negation will be, There exists a time t and a person p such that p cannot be fooled at time t.

3. For example, −16, 8, 12, . . . . These are the even integers.

4.

(a) A ∪ B = A ∩ B; A ∩ B = A ∪ B.

(b)

P T T F F

Q T F T F

P∨Q T T T F

¬(P ∨ Q) F F F T

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

(¬P) ∧ (¬Q) F F F T

P T T F F

Q T F T F

P∧Q T F F F

¬(P ∧ Q) F T T T

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

(¬P) ∨ (¬Q) F T T T

(c)

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