Page 26 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.68 Given 0 <a < 1 and [a + 1 ] +[a + 2 ]+ ··· +[a + 39 ] =6, evaluate [50a].
50 50 50
Here [∗] means the integer part of ∗ .

Solution: 0 <a + 1 <a + 2 < ··· <a + 39 < 2, thus [a + 1 ], [a + 2 ], ··· , [a + 39 ] are
50 50 50 50 50 50
equal to 0 or 1. The condition [a + 1 ] +[a + 2 ]+ ··· +[a + 39 ] =6 implies that six of
50 50 50

[a + 1 ], [a + 2 ], ··· , [a + 39 ] are equal to 1. Hence, [a + 1 ]=[a + 2 ]= ··· =[a + 33 ]=0
50 50 50 50 50 50
and [a + 34 ]=[a + 35 ]= ··· =[a + 39 ]=1. Then 0 <a + 33 < 1 and 1 ≤ a + 34 < 2,
50 50 50 50 50
which lead to 16 ≤ 50a< 17 ⇒ [50a] = 16.

1.69 Factoring x + y + z − 3xyz .
3
3
3
Solution:

3
2
3
3
2
3
3
3
2
2
x + y + z − 3xyz = x +3x y +3xy + y + z − 3x y − 3xy − 3xyz =(x +
2
2
3
3
y) + z − 3xy(x + y + z) =(x + y + z)[(x + y) − (x + y)z + z ] − 3xy(x + y + z)=
2
2
2
(x + y + z)(x + y + z − xy − yz − zx) .
1.70 a, b, c are prime numbers, c is a one-digit number, and ab + c = 1993, evaluate
a + b + c .

Solution: The right hand side of ab + c = 1993 is an odd number, thus one of ab and c
is an even number. If c is an even prime number which has to be 2, then
ab = 1993 − 2 = 1991 = 11 × 181, then one of a, b is 11, and the other one is 181. If ab
is an even number, let b be the even prime number 2, then 2a + c = 1993. Since c is a
prime number, then c =3, 5, or 7, and a = 995, 944, or 993, all of which are not prime
numbers. Hence a + b + c = 11 + 181 + 2 = 194.

1.71 Find the minimum positive fraction such that it is an integer when divided by

54/175 or multiplied by 55/36.

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