Solutions to Math 332 Homework 4. 1. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a complete
residue system modulo 11. Since 1 ≡ 12 (mod 11), 3 ≡ 14. (mod 11),... , 9 ...
Solutions to Math 332 Homework 4
1. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a complete residue system modulo 11. Since 1 ≡ 12 (mod 11), 3 ≡ 14 (mod 11), . . . , 9 ≡ 20 (mod 11), a complete residue system consisting entirely of even integers is {0, 12, 2, 14, 4, 16, 6, 18, 8, 20, 10}. Likewise, since 0 ≡ 11 (mod 11), 2 ≡ 13 (mod 11), . . . , 21 ≡ 10 (mod 11), a complete residue system consisting entirely of odd integers is {11, 1, 13, 3, 15, 5, 17, 7, 19, 9, 21}. 3. Since 189 = 33 × 7 divides 9!, we have n! ≡ 0 (mod 189) for every n ≥ 9. Hence 1! + 2! + · · · + 500! ≡ 1! + 2! + · · · + 8! (mod 189) ≡ 117 (mod 189). 8. 33 = 27 ≡ 1 (mod 26), so 3999 ≡ 1 (mod 26), ie. 3999 − 1 = 26k for some integer k. Note that ( 0 (mod 26) if k is even; 3999 − 1 = 13k ≡ 2 13 (mod 26) if k is odd. Since 3 ≡ −1 (mod 4), so 3999 ≡ −1 (mod 4) and so 3999 − 1 ≡ −2 6≡ 0 (mod 4), ie. 4 - 3999 − 1. Hence 2 - (3999 − 1)/2, ie. (3999 − 1)/2 is odd. Alternatively, one may factorize 3999 − 1 27333 − 1 = 2 2 (27 − 1)(27332 + 27331 + · · · + 27 + 1) = 2 = 13 × (27332 + 27331 + · · · + 27 + 1) ≡ 13 × (1 + 1 + · · · + 1 + 1) ≡ 13 × 333 (mod 26) ≡ 13 (mod 26).
(mod 26)
9(a). 35 = 243 ≡ 16 (mod 227) ⇒ 310 ≡ 162 ≡ 29 (mod 227) ⇒ 320 ≡ 292 ≡ 160 (mod 227). 330 = 310 × 320 ≡ 29 × 160 ≡ 100 (mod 227) ⇒ 332 = 330 × 32 ≡ 100 × 9 ≡ 219 (mod 227) ⇒ 332 + 8 ≡ 219 + 8 ≡ 0 (mod 227). 9(b). Note that 117 = 32 × 13. 52 ≡ −1 (mod 13) ⇒ 552 ≡ (−1)26 ≡ 1 (mod 13) ⇒ 553 ≡ 5 (mod 13). Hence 553 − 1 ≡ 4 6≡ 0 (mod 13), ie. 13 - 553 − 1. So 117 - 553 − 1. 10. 169 = 132 and 323 = 17×19. 13 ≡ 1 (mod 12) ⇒ 169 = 132 ≡ 1 (mod 12). 17 ≡ 5 (mod 12) and 19 ≡ −5 (mod 12) ⇒ 323 ≡ −52 ≡ −1 (mod 12). So 169323 ≡ 1 (mod 12) and 323169 ≡ −1 (mod 12). Therefore 169323 + 323169 ≡ 1 − 1 ≡ 0 (mod 12). 13. Since 32 ≡ 24 (mod 7) and 35 ≡ 22 × 3 ≡ −2 (mod 7), we have that 32n+5 = 32n × 35 ≡ 24n × (−2) = −24n+1 (mod 7). Hence 7 | 32n+5 + 24n+1 . 15. 13 ≡ 1 (mod 6) ⇒ 13n2 ≡ n2 (mod 6) ⇒ n(13n2 − 1) ≡ n(n2 − 1) = n(n − 1)(n + 1) ≡ 0 (mod 6) where the last equivalence follows from 2-17. 17. p > 5 and p prime implies that p 6≡ 0, 2, 4, 6, 8 (mod 10) and p 6≡ 5 (mod 10). Hence we are left with p ≡ 1, 3, 7, 9 (mod 10). After squaring, we get p2 ≡ 1, −1, −1, 1 (mod 10). So either p2 − 1 ≡ 0 (mod 10) or p2 + 1 ≡ 0 (mod 10), ie. there are no prime p > 5 such that 10 - p2 − 1 and 10 - p2 + 1.
21. Note that 264 = 23 × 3 × 11. Using (2.5), divisibility by 23 implies 23 | 74z, so z = 4; divisibility by 3 and 11 implies 5+3+x+y+7+4+z ≡0
(mod 3)
and
z−4+7−y+0−x+3−5≡0
(mod 11).
After substituting z = 4 and simplifying, we get x+y ≡1
(mod 3)
and
x+y ≡5
(mod 11).
Since x, y ∈ {0, 1, . . . , 9}, it follows that x + y ∈ {0, 1, . . . , 18}. The only possible value of x + y that satisfies both congruences above is 16. So x + y = 16, implying that x = 7, y = 9; x = y = 8 or x = 9, y = 7.
