assignment5_sol - MATH 135 Fall 2005 Assignment 5 Solutions...

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Unformatted text preview: MATH 135 Fall 2005 Assignment 5 Solutions Hand-In Problems Exercise 3-2: Which of the following integers are congruent modulo 6?- 147 ,- 91 ,- 22 ,- 14 ,- 2 , 2 , 4 , 5 , 21 , 185 Solution: Look at the quotients and remainders on division by 6: x- 147- 91- 22- 14- 2 2 4 5 21 185 q- 25- 16- 4- 3- 1 3 30 r 3 5 2 4 4 2 4 5 3 5 Then the following sets of numbers give members of the same equivalence class under congruence modulo 6: { 2 ,- 22 } , { 21 ,- 147 } , { 4 ,- 2 ,- 14 } , { 185 , 5 ,- 91 } since integers which give the same reminder when divided by 6 are congruent modulo 6. (We could also approach this problem by looking at the difference between each pair and checking if this difference is divisible by 6 or not. This would require more work, though.) Exercise 3-4: Let N = 3 729 . What is the last digit in the decimal representation of N ? What are the last digits in the base 9 and base 8 representations of N ? Solution: Since the last digit of a number written in base b is the remainder when the number is divided by b , then we must calculate N modulo b for each of b = 10, b = 9, and b = 8. Now 3 2 ≡ - 1 (mod 10) and 3 729 = 3 364(2)+1 = (3 2 ) 364 · 3 and so 3 729 ≡ 3 · (3 2 ) 364 (mod 10) ≡ 3(- 1) 364 (mod 10) ≡ 3 (mod 10) so the last digit in the decimal expansion of N is 3. Similarly, the last digit in the base 9 representation is the remainder upon division by 9. However 3 2 ≡ 0 (mod 9), so 3 729 = 3 2 · 3 727 ≡ · 3 727 (mod 9) ≡ (mod 9) and the last digit in the base 9 representation of N is 0. Also, the last digit in the base 8 representation is the remainder upon division by 8. Now 3 2 ≡ 1 (mod 8), so 3 729 = 3 364(2)+1 ≡ 3 · (3 2 ) 364 (mod 8) ≡ 3(1) 364 (mod 8) ≡ 3 (mod 8) and the last digit in the base 8 representation of N is 3. Exercise 3-6: Is 6 17 + 17 6 divisible by 3 or 7? Solution: We consider 6 17 + 17 6 modulo 3 and modulo 7. For divisibility by 3, we see that 6 ≡ 0 (mod 3) and 17 ≡ - 1 (mod 3) . Thus 6 17 + 17 6 ≡ 17 + (- 1) 6 (mod 3) ≡ 1 (mod 3) Therefore 3 does not divide 6 17 + 17 6 , since 6 17 + 17 6 6≡ 0 (mod 3). For divisibility by 7, we have 6 ≡ - 1 (mod 7) and 17 ≡ 3 (mod 7) . Therefore 6 17 + 17 6 ≡ (- 1) 17 + 3 6 (mod 7) ≡ (- 1) + ( 3 2 ) 3 (mod 7) ≡ (- 1) + 9 3 (mod 7) ≡ (- 1) + 2 3 (mod 7) ≡ - 1 + 8 (mod 7) ≡ (mod 7) . Hence 6 17 + 17 6 is divisible by 7....
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This homework help was uploaded on 04/18/2008 for the course LINEAR ALG MATH135 taught by Professor Vanderburgh during the Spring '08 term at Waterloo.

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assignment5_sol - MATH 135 Fall 2005 Assignment 5 Solutions...

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