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Unformatted text preview: Math 453 Problem Solutions, Fall 2006 HW# 1 1.3 #12. Show that ∑ n j =1 j · j ! = ( n + 1)! 1 for every positive integer n . When n = 1, both sides of the equation are 1, and this suffices for the base step. Now suppose that the equation is true for n = k , that is, ∑ k j =1 j · j ! = ( k + 1)! 1. Adding ( k + 1) · ( k + 1)! to both sides gives k +1 X j =1 j · j ! = ( k + 1)! 1 + ( k + 1) · ( k + 1)! = ( k + 2)( k + 1)! 1 = ( k + 2)! 1 . This shows that the equation is true when n = k + 1. By induction, the equation is true for every positive integer n . 1.3 #20. Prove that 2 n < n ! for n ≥ 4. When n = 4, we check that 2 4 < 4! (i.e., 16 < 24), and this completes the base step. Assume that 2 k < k ! for some k ≥ 4. Since 2 < k + 1, 2 k +1 = 2 · 2 k < 2 · k ! < ( k + 1) · k ! = ( k + 1)! . By induction, 2 n < n ! for n ≥ 4. 1.5#8. Are there integers a,b,c so that a  bc but a b and a c ? Yes, for example a = 6, b = 2, c = 3. 1.5# 24. How many positive integers ≤ 1000 are not divisible by 3 or by 5? . Of the 1000 positive integers ≤ 1000, [1000 / 3] = 333 are divisible by 3 and [1000 / 5] = 200 are divisible by 5. However, [1000 / 15] = 66 numbers are divisible by both 3 and 5. Hence the answer is 1000 [1000 / 3] [1000 / 5] + [1000 / 15] = 1000 333 200 + 66 = 533 . 1.5# 28. Prove that if a is an integer, then 3  ( a 3 a ) . Let a be an integer. By the division algorithm, there are integers q and r with 0 ≤ r ≤ 2 and a = 3 q + r . If r = 0, then a 3 a = ( a 1) a ( a + 1) = (3 q 1)3 q (3 q + 1) = 3[ q (3 q 1)(3 q + 1)] , so 3  ( a 3 a ). If r = 1, then a 3 a = 3 q (3 q + 1)(3 q + 2) , so 3  ( a 3 a ). If r = 2, then a 3 a = (3 q + 1)(3 q + 2)(3 q + 3) = 3[(3 q + 1)(3 q + 2)( q + 1)] , so 3  ( a 3 a ). Notice that a 3 a is always the product of three consecutive integers. One of these three integers is a multiple of 3. 1 Math 453 Problem Solutions, Fall 2006 HW# 2 3.3#6. If a is a positive integer, what is ( a,a + 2) ? Suppose d  a and d  ( a + 2). Then d  ( a + 2 a ), so d  2. Therefore, ( a,a + 2) is 1 or 2. If a is odd, then 2 a , so ( a,a + 2) = 1. If a is even, then 2  a and 2  ( a + 2), so ( a,a + 2) = 2. 3.3# 14. (a) Show that ( a,b ) = ( a,c ) = 1 implies ( a,bc ) = 1. From a theorem proved in class, there are integers x,y such that ax + by = 1 and there are integers u,v so that au + cv = 1. Therefore, (1) 1 = ( ax + by )( au + cv ) = a ( axu + xcv + uby ) + ( bc )( yv ) . Let d = ( a,bc ). Clearly d divides the right side of the equation (1), so d  1 and therefore d = 1. (b) Use induction to show that for any integer n ≥ 2 and integers a 1 ,... ,a n ,b , if ( a 1 ,b ) = ( a 2 ,b ) = ··· = ( a n ,b ) = 1 , then ( a 1 a 2 ··· a n ,b ) = 1 . ( Extra credit ). The n = 2 case was proved in part (a) and this constitutes the base step of the induction. Now suppose k ≥ 2 and we know that ( a 1 ,b ) = ··· = ( a k ,b ) = 1 implies ( a 1 ··· a k ,b ) = 1 for any integers a 1 ,... ,a,....
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This note was uploaded on 11/03/2010 for the course MATH 26233820 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.
 Fall '10
 GIESEKER,D.
 Math, Algebra

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