Page 29 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution: Obviously M 1 ,M 2 ,M 3 are multiples of 9. Since M has 2000 digits, the sum of
all its digits M 1 ≤ 9 × 2000 = 18000, then M 1 has at most five digits and the first digit is

0 or 1. Thus M 2 ≤ 1+4 × 9 = 37, which implies that M 2 has at most two digits and the

is divisible by
first digit is less than or equal to 3. Thus M 3 ≤ 3+9 = 12. In addition, M 3
9 and M 3 =0, hence M 3 =9.

1 1 2 √
1.77 Let x, y be two distinct positive integers, and + = , evaluate x + y .
x y 5

Solution: Set 1 = 2a , 1 = 2b where a, b are positive integers and coprime, and let
x 5(a + b) y 5(a + b)
5(a + b) b b 1
a>b . Then 2x = = 5 +5 × . Since x is a positive integer, then = , thus
a a a 5
2x =6 ⇒ x =3. On the other hand, 2y = 5(a + b) = 5 +5 × a . Since y is a positive integer,
b b
a √ √ √
then =5, thus y = 15. Therefore, x + y = 18 = 3 2.
b
1.78 The positive integers a, b, c satisfy a +3b +3c + 13 < 2ab +4b + 12c , find
2
2
2
the value of a + b + c .
2
2
2
2
2
2
Solution: a +3b +3c + 13 < 2ab +4b + 12c ⇒ a +3b +3c + 13 − 2ab − 4b − 12c +1 <
2
2
2
2
2
a +3b +3c 2 2 + 13 < 2ab +4b + 12c 2⇒ a +3b +3c + 13 − 2ab − 4b − 12c +1 <
2
1 ⇒ (a − b) + 2(b − 1) + 3(c − 2) < 1
1 ⇒ (a − b) + 2(b − 1) + 3(c − 2) < 1. (a − b) ≥ 0, (b − 1) ≥ 0, (c − 2) ≥ 0, and a, b, c
2
2
2
2
2
2
are positive integers, thus a = b =1,c =2, hence a + b + c =4.
1.79 Find the minimum positive integer n that is a multiple of 75 and has 75
positive integer factors (including 1 and itself).
Solution: n = 75k =3 × 5 k where k is a positive integer. To minimize n, let n =2 · 3 · 5
β
2
α
γ
(γ ≥ 2,β ≥ 1), and (α + 1)(β + 1)(γ + 1) = 75, from which α +1,β +1,γ +1 are all odd numbers,
2
thus α,β,γ are all even numbers. Then γ =2, and (α + 1)(β + 1) =25 =5 =1 × 25.
a
1) If α +1 = 5,β +1 = 5, then α =4,β =4, thus n =2 · 3 · 5 .
4
4
2
2 · 3
00
2 2
2) If α +1 = 1,β + 1 = 25, then α =0,β = 24, thus n =2 · 3 4 · 5 · 5 .
2 24

2 · 3 · 5 = 32400
4 4
4 4
2 2
According to (1)(2), the minimum positive integer n =2 · 3 · 5 = 32400
1.80 Given the sets M = {x, xy, lg(xy)},N = {0, |x|,y}, and M = N, evaluate
1 1 1 1
2
3
(x + ) + (x + ) +(x + )+ ··· +(x 2001 + ).
y y 2 y 3 y 2001
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