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Unformatted text preview: PARTIAL SOLUTION SET 4—DUE 2/19/2008 In the interest of getting this out before the midterm, this is current a par- tial solution set only. Please report any errors in this document to Ian Sammis ( email@example.com ). For more practice try odd problems. Problem 1 (#3.5.32) . Show that if a , b , and m are integers such that m ≥ 2 and a ≡ b (mod m ) , then gcd ( a,m ) = gcd ( b,m ) . Proof. Let d = gcd( a,m ). Then d | a and d | m , so for some k,ℓ ∈ Z , a = kd and m = ℓd . Since a ≡ b (mod m ), for some r ∈ Z , a- b = rm . Substituting in, kd- b = rℓd , so b = d ( k- rℓ ). Thus d | b . Since the gcd of a and m divides b , it’s a common divisor of b and m , so gcd( a,m ) ≤ gcd( b,m ). By an analogous argument, the greatest common divisor of b and m is a common divisor of a and m , so gcd( a,m ) ≥ gcd( b,m ). Thus gcd( a,m ) = gcd( b,m ), as desired. Problem 2 (#3.6.30) . Show that a positive integer is divisible by 11 if and only if the difference between the sum of the decimal digits in even-numbered positions...
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This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.
- Spring '11