answer_4

# answer_4 - 2.2.1(b 187 = 11 ย 17 0(d-24 =-6 ย 4 0(f(9 ๎ฏ...

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Unformatted text preview: 2.2.1.(b) 187 = 11 ยท 17 + 0 (d) -24 = (-6) ยท 4 + 0 (f) (9 ๎ฏ ๎ต +5) = ๎๎ถ๎ฏ ๎ต ๎ ๎ด๎ ยท 3 + 2 2.2.2.(b) 0 (d) 0 (f) 2 2.2.3.(h) 0 2.2.4.(c) a = 1, b = 2, c = 3 2.2.14(b) Case 1: It might be that n = 6q for some integer q. But we know this case does not happen for this n, since ๎ฒ ๎ต ๎ ๎ถ๎น๎ต ๎ต ๎ ๎ด๎ต ยท ๎๎ถ๎ต ๎ต ๎ Case 2: It might be that n = 6q+1 for some integer q. In this case, ๎ฒ ๎ต = ๎ถ๎น๎ต ๎ต ๎ ๎ด๎ต๎ต ๎ ๎ด = 12 ยท ๎๎ถ๎ต ๎ต ๎ ๎ต๎ +1 Case 3: It might be that n = 6q+2 for some integer q. This case does not happen for this n, since ๎ฒ ๎ต ๎ ๎ด๎ต ยท ๎๎ถ๎ต ๎ต ๎ ๎ต๎ ๎ ๎ท Case 4: It might be that n = 6q+3 for some integer q. This case does not happen for this n, since ๎ฒ ๎ต ๎ ๎ด๎ต ยท ๎๎ถ๎ต ๎ต ๎ ๎ถ๎ต๎ ๎ ๎ผ Case 5: It might be that n = 6q+4 for some integer q. This case does not happen for this n, since ๎ฒ ๎ต ๎ ๎ด๎ต ยท ๎๎ถ๎ต ๎ต ๎ ๎ท๎ต๎ ๎ ๎ด๎น ๎ ๎ด๎ต ยท ๎๎ถ๎ต ๎ต ๎ ๎ท๎ต ๎ ๎ด๎ ๎ ๎ท Case 6: It might be that n = 6q+5 for some integer q. In this case, ๎ฒ ๎ต ๎ ๎ถ๎น๎ต ๎ต ๎ ๎น๎ฝ๎ต ๎ ๎ต๎ธ = 12 ยท ๎๎ถ๎ต ๎ต ๎ ๎ธ๎ต ๎ ๎ท๎ +1 Therefore ๎ฒ ๎ต is of the form 12 ยท (an integer) + 1 in all possible cases. 2.2.15. By the Division Theorem, this means that when n is divided by 3 or 4, there is no remainder. Also, when n is divided by 12, it leaves a remainder of 0,1,2,....,11. That is, one of the following cases must be true: Case 1: It might be that n=12q for some integer q. In this case, n=3(4q) and n=4(3q) can be happen Case 2: It might be that n=12q +1 for some integer q. In this case, n=3(4q)+1 and n=4(3q)+1 which means that when n is divided by 3, the remainder is 1. This case is impossible since we know that the given value of n is divisible by 3 Case 3: It might be that n=12q +2 for some integer q. In this case, n=3(4q)+2 and n=4(3q)+2 which means that when n is divided by 3, the remainder is 2. This case is impossible since we know that the given value of n is divisible by 3 Case 4: It might be that n=12q +3 for some integer q. In this case, n=3(4q+1) and n=4(3q)+3 which means that when n is divided by 4, the remainder is 3. This case is impossible since we know that the given value of n is divisible by 4 Case 5: It might be that n=12q +4 for some integer q. In this case, n=3(4q+1)+1and n=4(3q+1) which means that when n is divided by 3, the remainder is 1. This case is impossible since we know that the given value of n is divisible by 3 Case 6: It might be that n=12q +5 for some integer q. In this case, n=3(4q+1)+2and n=4(3q+1)+1 which means that when n is divided by 3, the remainder is 2. This case is impossible since we know that the given value of n is divisible by 3 Case 7: It might be that n=12q +6 for some integer q. In this case, n=3(4q+2)and n=4(3q+1)+2 which means that when n is divided by 4, the remainder is 2. This case is impossible since we know that the given value of n is divisible by 4.impossible since we know that the given value of n is divisible by 4....
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