Difference between revisions of "2004 AMC 12B Problems/Problem 23"

Revision as of 18:59, 3 July 2013

Problem

The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $n$ are possible?

$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$

Solution

Let the roots be $r,s,r + s$ , and let $t = rs$ . Then

$(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$

and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$ . Then our polynomial looks like $x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0$ and we need the number of possible products $t = rs = r(1002 - r)$ .

Since $r > 0$ and $t > 0$ , it follows that $0 < t = r(1002-r) < 501^2 = 251001$ , with the endpoints not achievable because the roots must be distinct. Because $r$ cannot be an integer, there are $251000 - 500 = 250,\!500\ \mathrm{(C)}$ possible values of $n = -1002t$ .

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 21:05, 6 January 2009 (view source) Yongyi781 (talk \| contribs) (Fixed spacing) ← Older edit		Revision as of 18:59, 3 July 2013 (view source) Nathan wailes (talk \| contribs) (→‎See also) Newer edit →
Line 23:		Line 23:

	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
		+	{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2004 AMC 12B Problems/Problem 23"

Revision as of 18:59, 3 July 2013

Problem

Solution

See also