已知函数y=x2-1840x+2003与x轴的交点为(m,0),(n,0),则(m2-1841m+2003)(n2-1841n+2003)的值为______.
∵函数y=x2-1840x+2003与x轴的交点为(m,0),(n,0),
∴m,n是方程x2-1840x+2003=0的两个根,即m2-1840m+2003=0,n2-1840n+2003=0,
∴m+n=1840,mn=2003,
(m2-1841m+2003)(n2-1841n+2003)
=(m2-1840m+2003+m)(n2-1840n+2003+n)
=mn
=2003.