Prime Number and Mod
Section 4.1
Divisibility and Modular Arithmetic
87
CHAPTER 4
Number Theory and Cryptography
SECTION 4.1
Divisibility and Modular Arithmetic
2. a) 1 | a since a = 1 · a.
b) a | 0 since 0 = a · 0.
4. Suppose a | b , so that b = at for some t , and b | c, so that c = bs for some s. Then substituting the first equation into the second, we obtain c = (at)s = a(ts). This means that a | c, as desired.
6. Under the hypotheses, we have c = as and d = bt for some s and t . Multiplying we obtain cd = ab(st), which means that ab | cd , as desired.
8. The simplest counterexample is provided by a = 4 and b = c = 2.
10. In each case we can carry out the arithmetic on a calculator.
a) Since 8 · 5 = 40 and 44 − 40 = 4 , we have quotient 44 div 8 = 5 and remainder 44 mod 8 = 4.
b) Since 21 · 37 = 777, we have quotient 777 div 21 = 37 and remainder 777 mod 21 = 0 .
c) As above, we can compute 123 div 19 = 6 and 123 mod 19 = 9. However, since the dividend is negative and the remainder is nonzero, the quotient is −(6 + 1) = −7 and the remainder is 19 − 9 = 10. To check that
−123 div 19 = −7 and −123 mod 19 = 10 , we note that −123 = (−7)(19) + 10.
d) Since 1 div 23 = 0 and 1 mod 23 = 1 , we have −1 div 23 = −1 and −1 mod 23 = 22 .
e) Since 2002 div 87 = 23 and 2002 mod 87 = 1, we have −2002 div 87 = −24 and 2002 mod 87 = 86 .
f) Clearly 0 div 17 = 0 and 0 mod 17 = 0.
g) We have 1234567 div 1001 = 1233 and 1234567 mod 1001 = 334.
h) Since 100 div 101 = 0 and 100 mod 101 = 100, we have −100 div 101 = −1 and −100 mod 101 = 1 .
12. a) Because 100 mod 24 = 4 , the clock reads the same as 4 hours after 2:00, namely 6:00.
b) Essentially we are asked to compute 12 − 45 mod 24 = −33 mod 24 = −33 + 48 mod 24 = 15 . The clock reads 15:00.
c) Because 168 ≡ 0 (mod 24), the clock read 19:00.
14. This problem is equivalent to asking for the right-hand side mod 19 . So we just do the arithmetic and compute the remainder upon division by 19.
a) 13 · 11 =
Divisibility and Modular Arithmetic
87
CHAPTER 4
Number Theory and Cryptography
SECTION 4.1
Divisibility and Modular Arithmetic
2. a) 1 | a since a = 1 · a.
b) a | 0 since 0 = a · 0.
4. Suppose a | b , so that b = at for some t , and b | c, so that c = bs for some s. Then substituting the first equation into the second, we obtain c = (at)s = a(ts). This means that a | c, as desired.
6. Under the hypotheses, we have c = as and d = bt for some s and t . Multiplying we obtain cd = ab(st), which means that ab | cd , as desired.
8. The simplest counterexample is provided by a = 4 and b = c = 2.
10. In each case we can carry out the arithmetic on a calculator.
a) Since 8 · 5 = 40 and 44 − 40 = 4 , we have quotient 44 div 8 = 5 and remainder 44 mod 8 = 4.
b) Since 21 · 37 = 777, we have quotient 777 div 21 = 37 and remainder 777 mod 21 = 0 .
c) As above, we can compute 123 div 19 = 6 and 123 mod 19 = 9. However, since the dividend is negative and the remainder is nonzero, the quotient is −(6 + 1) = −7 and the remainder is 19 − 9 = 10. To check that
−123 div 19 = −7 and −123 mod 19 = 10 , we note that −123 = (−7)(19) + 10.
d) Since 1 div 23 = 0 and 1 mod 23 = 1 , we have −1 div 23 = −1 and −1 mod 23 = 22 .
e) Since 2002 div 87 = 23 and 2002 mod 87 = 1, we have −2002 div 87 = −24 and 2002 mod 87 = 86 .
f) Clearly 0 div 17 = 0 and 0 mod 17 = 0.
g) We have 1234567 div 1001 = 1233 and 1234567 mod 1001 = 334.
h) Since 100 div 101 = 0 and 100 mod 101 = 100, we have −100 div 101 = −1 and −100 mod 101 = 1 .
12. a) Because 100 mod 24 = 4 , the clock reads the same as 4 hours after 2:00, namely 6:00.
b) Essentially we are asked to compute 12 − 45 mod 24 = −33 mod 24 = −33 + 48 mod 24 = 15 . The clock reads 15:00.
c) Because 168 ≡ 0 (mod 24), the clock read 19:00.
14. This problem is equivalent to asking for the right-hand side mod 19 . So we just do the arithmetic and compute the remainder upon division by 19.
a) 13 · 11 =