Section 2.5

• 6. x = 9.
• 10. If x is an inverse of a modulo m, then ax - 1 = tm. If a and m have a common divisor greater than 1, then 1 must also have this same common divisor , since 1 = ax - tm. Of course, this isn't true, so x doesn't exist.
• 12. 12
• 14. I'll use "cong" to mean "is congruent to". a) 2(6) cong 1 (mod 11), 3(4) cong 1 (mod 11), 5(9) cong 1 (mod 11), 7(8) cong 1 (mod 11). b) 10! = 1 *(2*6)*(3*4)*(5*9)*(7*8)*10 cong 1*1*1*1*10 = 10 cong -1 (mod 11).
• 16.
• 18. Just a restatement of the Chinese Remainder Theorem.
• 20. Suppose that there are two solutions to the set of congruences, i.e., suppose x cong ai (mod mi) and y cong ai (mod mi) for each i. We want to show that these solutions are the same modulo m. The assumption implies that x cong y (mod mi) for each i. From Exercise 19, x cong y (mod m).
• 22. Solve x cong 0 (mod 5) and x cong 1 (mod 3). The Chinese Remainder Theorem says there is a unique answer modulo 15. Just look at ultiples of 5 and divide each by 3. You find that x = 10 satisfies the conditions.
• 26. 4 is represented by (1,4) and 7 by (1,2). Adding coordinate-wise, we see that the sum is represented by (1+1,4+2) = (2,6) = (2,1). We are working modulo 5 in the second coordinate. (2,1) represents 11 in the table formed in Exercise 25. So we conclude 4 + 7 = 11.
• 28. Divide 2^a-1 by 2^b-1. First we get that 2^a-1 = (2^b-1)2^(a-b) + 2^(a-b) - 1. This gives us a remainder of 2^(a-b)-1. Do this again, and the remainder is 2^(a-2b) - 1. Continue until the final remainder is 2^(a-kb)-1, but a-kb is just a mod b, so we get 2^(a mod b) - 1.
• 32. 2299 1317 2117. Maple really helps with this one.