Page 12 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution: Let T = t 1 + t 2 + ··· +(2+0+0+8) +(2+0+0+9), and then reverse the
order of right hand side to obtain T = (2+0+0+9) +(2+0+0+8) + ··· +2+1.
Add up these two equalities to obtain

2T = [1+(2+0+0+9)]+[2+(2+0+0+8)]+·· ·+[(2+0+0+8)+2]+[(2+0+0+9)+1] =2T = [1+(2+0+0+9)]+[2+(2+0+0+8)]+·· ·+[(2+0+0+8)+2]+[(2+0+0+9)+1] = =

2T = [1+(2+0+0+9)]+[2+(2+0+0+8)]+·· ·+[(2+0+0+8)+2]+[(2+0+0+9)+1]
12 × 2009 ⇒ T = 12 × 2009/2 = 12054.
12 × 2009 ⇒ T = 12 × 2009/2 = 1205412 × 2009 ⇒ T = 12 × 2009/2 = 12054
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1.22 Let a be a positive integer, b and c are prime numbers, and they satisfy a = bc, + 1 b = ,
c
a
find the value of a.
1 1 1 1 1 1 b − c
Solution: + = ⇒ = − = . Since a = bc , we have b − c =1, thus c and
a b c a c b bc
b are two consecutive prime numbers, which has the only choice c =2,b =3, thus a =6.

1.23 Let x, y are two natural numbers and they satisfy x > y, x + y = 667. Let p be
the least common multiple of x and y , let d be the greatest common divisor of x and

y , and p = 120d . Find the maximum value of x − y .

Solution: Let x = md, y = nd , then m, n should be coprime since d is the greatest common
divisor. x> y implies m>n . p = mnd = 120d ⇒ mn = 120 . In addition,

(m + n)d = 667 = 23 × 29 = 1 × 667(m + n)d = 667 = 23 × 29 = 1 × 667. Since 23 and 29 are coprime, there are only three

possibilities: (1) m + n = 23,d = 29; (2) m + n = 29,d = 23; (3) m + n = 667,d =1. Together

with mn = 120, we have (1) m = 15,n =8 ; (2) m = 24,n =5 ; (3) no solution. Thus
(m − n) max = 24 − 5 = 19 which leads to (x − y) max = (24 − 5) × 23 = 437.

1.24 If x,y,z satisfy 3x +7y + z =5 (i), 4x + 10y + z = 39 (ii), find the value of
x + y + z
.
x +3y

Solution: (i) × 3 − (ii) × 2 ⇒ x + y + z = −63. (ii) − (i) ⇒ x +3y = 34.

x + y + z z
x + y + 63 63
Hence, = .
= − −
x +3y 34 34
x +3y
√ √ 3 3 1
1.25 Given a = 3 4+ 3 2+1, find the value of + + .
a a 2 a 3

3
2
√ √ √ √ 1 √ 3 1 2 a +3a +3a +1 − a 3 a +1 3 3 1 3
3

3
3 2
3
3
3
3
1
a +3a
3a
Solution: ( 2 − 1)a =( 2 − 1)( 4+ 3 2+1) = 2 − 1= 1 ⇒ = 3 3 2 − 1, thus + +3a +1 3a +3a +1 2 = +3a +1 − a 3 a +1 = −1= 1 1+ −
+
3
=
+
a
3
a a a 2 + a 3 = a a 2 a 3 a = a 3 a 3 a = a −1= 1+ a − a
2
3
2
3 3 1 3a +3a +1 a +3a +3a +1 − a 3 a +1 3 1 3 − 1= 2 − 1= 1 . 1= 2 − 1= 1
+ + = = = −1= 1+
a a 2 a 3 a 3 a 3 a a
1= 2 − 1= 1 x x 2
1.26 a =0 is a real number, and = a , express x + x +1 as a function of a .
4
2
2
x + x +1
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