ANSWERS OR HINTS TO SELECTED EXERCISES

CHAPTER 1

  • 3. pg203-01
  • 4a. images
  • 9. Exactly one of n and n + 1 is even. Divide that factor by 2 in the denominator of images .
  • 17. images .

CHAPTER 2

  • 2. 6k, 6k + 2, 6k + 3, 6k + 4 are composite.
  • 4. images .
  • 6. Assume that the sum, f(n), of the first n fourth powers is an5 + bn4 + cn3 + dn2 + en + f. Then find six equations by plugging in the values 1, 2, 3, 4, 5, and 6 for n.
  • 13. 21 and 55.
  • 14. Use the fact that the sum of two odd numbers is even and the sum of an odd number and an even number is odd. Then realize that the first two Fibonacci numbers are odd.

CHAPTER 3

  • 5. For any positive integer m, we have m! = m(m − 1)(m − 2)…(3)(2)(1) = m(m − 1)!. Now, let m = n! in which case m − 1 = n! − 1. Then (n!)! = n!(n! − 1)!.
  • 11. pg204-01 pg204-02 images (the sum of the first n − 1 oblong numbers). Then use the fact that the sum of the first n oblong numbers is images . Finally, images .

CHAPTER 4

  • 1. gcd(80,540) = gcd(60,80) = 20.
  • 2. n3 − n = (n − 1)n(n + 1). One of any three consecutive numbers is divisible by 3, and one of any two consecutive numbers is divisible by 2.
  • 4. If m and n are odd, m2 + n2 = (2a + 1)2 + (2b + 1)2 = 4a2 + 4a + 1 + 4b2 + 4b + 1 = 4k + 2, where k = a2 + a + b2 + b.
  • 8. 3n + 1 = (4 − 1)n + 1. Now, use the binomial theorem to evaluate (4 − 1)n and note that when n is odd, it ends in −1, which is canceled by the 1 in (4 − 1)n + 1. Then realize that each remaining term is divisible by 4.
  • 14. First write x as n + y, where n is the integer ⌊x⌋ and y satisfies 0 ≤ y < 1. (For example, 6.7 = 6 + .7) Then observe that 10x = 10n + 10y. Now, show that images . Finally, figure out how big ⌊10y⌋ can get.
  • 18. Find the exponents of the primes 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. The exponent 2, for example, is images .
  • 23. The difference of two odd numbers is even.

CHAPTER 5

  • 1. (3k − 1)3 = 27k3 − 27k2 + 9k − 1 = −1 (mod 9), (3k)3 = 27k3 = 0 (mod 9), and (3k + 1)3 = 27k3 + 27k2 + 9k + 1 = 1 (mod 9). This proves that n3 = 0 or ±1 (mod 9) and that the observed pattern persists.
  • 3. It suffices to verify the statement for n = 0, 1, 2, 3, and 4. Alternatively, we can use Fermat’s little theorem with p = 5.
  • 11. 32 = 42 (mod 7), while 3 ≠ 4 (mod 7).
  • 12. x = 1, 3, 5, and 9.
  • 18. 2100 = (23)33 × 2 = 2 (mod 7).

CHAPTER 6

  • 2. τ(n) = 6 = 2 × 3 implies that either n = pq2 or n = r5, where p, q, and r are primes such that p ≠ q. Now, p = 3 and q = 2 yields n = 12, p = 2 and q = 3 yields n = 18, while r = 2 yields n = 32. The minimum n is 12.
  • 7. By definition, σ(p3) = p3 + p2 + p + 1. Using formula (6.2), we have images .
  • 9. f(n) = f(1n) = f(1)f(n), so f(1) = 1.
  • 14. Let images . Then F is multiplicative. Now, images .

CHAPTER 7

  • 1. images . images .
  • 2a. images .
  • 8. images .
  • 9. 34 = 1 (mod 10).

CHAPTER 8

  • 7. 4 = 2 × 2 and images .
  • 8. (2k)2 − k2 = 3k2.
  • 9. If a = 1, the sum is triangular. If not, the sum is ta + k − 1 − ta − 1.
  • 14. images .
  • 15. Let images . Then multiplying both sides by m yields images . Note that d = m, while the length of the partition is unchanged.

CHAPTER 9

  • 1. p = c − 13 = c + 13 (mod 26).
  • 3. Given c = 3p (mod 26), we have 9c = 27p = p (mod 26), that is, p = 9c (mod 26).
  • 5. (26 − p)2 = p2 (mod 26).
  • 7. Consider p = 1 and p = 3.
  • 10. Use induction to show that images , for each k = 2, 3, …, r. The base case is obvious since n2 > 2n1 > n1. Now, assume that images . Then images .
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