2010 AMC 10A Problem 21

Question

The polynomial

x^3-ax^2+bx-2010

has three positive integer zeros. What is the smallest possible value of

a?

78

88

98

108

118

The roots multiply to

2010

and sum to

a

Put

67

alone, then split

30

into two factors with smallest sum

Let the roots be positive integers

r\le s\le t.

By Vieta’s formulas,

rst=2010

and

a=r+s+t.

Since

2010=2\cdot3\cdot5\cdot67,

one root must be divisible by

67.

If that root is larger than

67,

then it is at least

134,

which is already worse than the construction below.

Thus take

t=67

and minimize

r+s

with

rs=30.

The factor pair with smallest sum is

5

and

6,

so

a=5+6+67=78.

Thus, A is the correct answer.

Problem 21 in other years