Page 37 - Elementary Algebra Exercise Book I
P. 37
ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers
Solution: The three equalities imply that we can treat a, b, c as three distinct roots of the equation
1 1 1 1
2
3
+ + = , which is equivalent to 2t +(x + y + z)t − xyz =0 . Vieta’s
x + t y + t z + t t
formulas lead to ab + bc + ca =0, thus 1 + 1 + 1 = ab + bc + ca =0.
a b c abc
1.100 If m is a natural number, S m represents the sum of all digits of m, and the largest
common divisor of S m and S m+1 is a prime greater than 2, find the minimum value of m.
Solution: (S m ,S m+1 ) > 2 ⇒ S m+1 − S m �=1. Assume m has 9’s as the last n digits (n ≥ 0),
then S m+1 = S m − 9n +1 . Let (S m ,S m+1 ) = d , then d =(S m , 9n − 1) , d|9n − 1, thus
n =0, 1 (since d> 2). If n =2, then d|17, d = 17, S m has the minimum value 34 (since
S m ≥ 18) and m has the minimum value 8899. If n =3, then d|26, d = 13, S m has the
minimum value 39 (since S m ≥ 27) and m has the minimum value 48999. If n ≥ 4, then
m ≥ 9999 when d exists. Hence, m has the minimum value 8899.
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