Page 35 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.93 In the Cartesian plane XOY , all coordinates of the points A(x 1 ,y 1 ) and

B(x 2 ,y 2 ) are one-digit positive integers. The angle between OA and the positive part of

0
x axis is greater than 45 , and the angle between OB and the positive part of x axis is
0
less than 45 . Denote B =(x 2 , 0),A = (0,y 1 ). The area of OB B is 33.5 larger than

the area of OA A . Find the four-digit number x 1 x 2 y 2 y 1 where x 1 ,x 2 ,y 2 ,y 1 are the

four digits.
1 1
+
1
1
OB B =
Solution: S min S max S OA A + 33.5 ⇒ x 2 y 2 = x 1 y 1 + 33.5 ⇒ x 2 y 2 = x 1 y 1 + 67 . Since

2
2
x 1 y 1 > 0, then x 2 y 2 > 67. In addition, x 2 ,y 2 are one-digit positive integers, then x 2 y 2 = 72
or 81. ∠BOB < 45 , then the point B is below the diagonal line y = x , then x 2 >y 2 ,
0

thus x 2 y 2 = 81, then x 2 y 2 = 72, which implies x 2 =9,y 2 =8. Hence, x 1 y 1 =5. Since
0
∠AOB > 45 , then the point A is above the diagonal line y = x , then x 1 <y 1 . Since

x 1 ,y 1 are one-digit positive integers, then x 1 =1,y 1 =5. Therefore the four-digit number
x 1 x 2 y 2 y 1 = 1985.

1.94 Given a positive integer n> 30 and 2002n is divisible by 4n − 1, find the value
of n.

2002n
Solution: Let 2002n = k , then = k = 500 + 2(n+250) . Since 4n − 1 is an odd number, then
4n−1
4n − 1 4n − 1 n + 250
2(n + 250) is divisible by 4n − 1 . Let = p ( p is a positive integer), then
4n − 1
4n + 1000 1001
4p = = 1+ , thus 1001 is divisible by 4n − 1 . Since n ≥ 30 and
4n − 1 4n − 1
1001 = 7 × 11 × 13, then we should have 4n − 1 = 143, which implies n = 36, p =2.

1.95 How many integers satisfying the inequality |x − 2000| + |x|≤ 9999?

Solution: If x ≥ 2000, then the inequality becomes (x − 2000) + x ≤ 9999 ⇔ 2000 ≤ x ≤ 5999.5 .

There are 4000 integers satisfying the inequality. If 0 ≤ x< 2000, then the inequality

becomes (2000 − x)+ x ≤ 9999 ⇔ 2000 ≤ 9999 that is always true, then there are 2000

integers satisfying the inequality. If x< 0 , then the inequality becomes

(2000 − x) +(−x) ≤ 9999 ⇔−3999.5 ≤ x< 0 . There are additionally 3999 integers satisfying

the inequality. Hence, totally there are 4000 + 2000 + 3999 = 9999 integers satisfying the

inequality.

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