Elementary Algebra Exercise Book I by Bookboon.com

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When z = 1 , then x + y = 1 , then (x, y) = (0, 1) or (1, 0) . When z = 0 , then x + y = 3 , then there are four possibilities of (x, y) . Hence, a3 = 2 + 4 = 6 . When n = 2001 , we have x + y + 2z = 2001 , thus 0 ≤ z ≤ 1000 . When z = 0 , then x + y = 2001 , then there are 2002 possibilities of (x, y) . When z = 1 , then x + y = 1999 , then there are 2000 possibilities of (x, y) . . . . When z = 1000 , then x + y = 1 , then there are two possibilities of (x, y) . As a conclusion, a2001 = 2002 + 2000 + 1998 + · · · + 4 + 2 = 1003002 .

108 Let a, b, c are positive integers and follow a geometric sequence, and b − a is a perfect square, log6 a + log6 b + log6 c = 6 , find the value of a + b + c . Solution: log6 a + log6 b + log6 c = 6 ⇒ log6 abc = 6 ⇒ abc = 66 . In addition, b2 = ac , then b = 62 = 36, ac = 362 . In order to make 36 − a a perfect square, a can only be 11,27,32,35, and a is a divisor of 362 , thus a = 27 , then c = 48 . Therefore, a + b + c = 27 + 36 + 48 = 111 . 109 Real numbers The real numbers a, b, c, d satisfy a + b = c + d, a3 + b3 = c3 + d3 , show a2011 + b2011 = c2011 + d2011 .

27 Solution: When n = 3 , we have x + y + 2z = 3 . Since x ≥ 0, y ≥ 0, z ≥ 0 , we have 0 ≤ z ≤ 1 . When z = 1 , then x + y = 1 , then (x, y) = (0, 1) or (1, 0) . When z = 0 , then x + y = 3 , then there are four possibilities of (x, y) . Hence, a3 = 2 + 4 = 6 . When n = 2001 , we have x + y + 2z = 2001 , thus 0 ≤ z ≤ 1000 . When z = 0 , then x + y = 2001 , then there are 2002 possibilities of (x, y) . When z = 1 , then x + y = 1999 , then there are 2000 possibilities of (x, y) . . . . When z = 1000 , then x + y = 1 , then there are two possibilities of (x, y) .

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