Problem: Ο β 1 Ο ξ = 1 13 1 3
β k = 0 12 ( 2 β 2 Ο k + Ο 2 k ) k = 0 β 1 2 β ( 2 β 2 Ο k + Ο 2 k )
is divided by 1000 1 0 0 0
Solution:
Suppose the roots of p ( x ) = x 2 β 2 x + 2 p ( x ) = x 2 β 2 x + 2 r 1 r 1 β r 2 r 2 β
β k = 0 12 p ( Ο k ) = β k = 0 12 ( Ο k β r 1 ) β k = 0 12 ( Ο k β r 2 ) = ( r 1 13 β 1 ) ( r 2 13 β 1 ) . k = 0 β 1 2 β p ( Ο k ) = k = 0 β 1 2 β ( Ο k β r 1 β ) k = 0 β 1 2 β ( Ο k β r 2 β ) = ( r 1 1 3 β β 1 ) ( r 2 1 3 β β 1 ) .
By Vieta's Theorem, r 1 + r 2 = 2 r 1 β + r 2 β = 2 r 1 r 2 = 2 r 1 β r 2 β = 2 r r r 1 r 1 β r 2 r 2 β r 2 β 2 r + 2 = 0 r 2 β 2 r + 2 = 0 r 4 + 4 = ( r 2 β 2 r + 2 ) ( r 2 + 2 r + 2 ) = 0 r 4 + 4 = ( r 2 β 2 r + 2 ) ( r 2 + 2 r + 2 ) = 0 r 4 = β 4 , r 12 = β 64 , r 4 = β 4 , r 1 2 = β 6 4 , r 13 = β 1 β 64 r r 1 3 = β 1 β 6 4 r
( 1 + 64 r 1 ) ( 1 + 64 r 2 ) = 1 + 64 ( r 1 + r 2 ) + 4096 r 1 r 2 = 1 + 64 β
2 + 4096 β
2 = 8321. ( 1 + 6 4 r 1 β ) ( 1 + 6 4 r 2 β ) = 1 + 6 4 ( r 1 β + r 2 β ) + 4 0 9 6 r 1 β r 2 β = 1 + 6 4 β
2 + 4 0 9 6 β
2 = 8 3 2 1 .
The requested remainder is 321 3 2 1 β
OR OR
As in the first solution, the product can be expressed as ( r 1 13 β 1 ) ( r 2 13 β 1 ) ( r 1 1 3 β β 1 ) ( r 2 1 3 β β 1 ) r 1 r 1 β r 2 r 2 β p ( x ) = x 2 β 2 x + 2 p ( x ) = x 2 β 2 x + 2 1 Β± i 1 Β± i
1 Β± i = 2 ( cos β‘ ( Ο 4 ) + i sin β‘ ( Β± Ο 4 ) ) . 1 Β± i = 2 β ( cos ( 4 Ο β ) + i sin ( Β± 4 Ο β ) ) .
By de Moivre's Theorem
( 1 + i ) 13 = ( 2 ) 13 ( cos β‘ 13 Ο 4 + i sin β‘ 13 Ο 4 ) = 64 2 ( β 1 2 β i 2 ) = β 64 β 64 i . ( 1 + i ) 1 3 = ( 2 β ) 1 3 ( cos 4 1 3 Ο β + i sin 4 1 3 Ο β ) = 6 4 2 β ( β 2 β 1 β β 2 β i β ) = β 6 4 β 6 4 i .
Taking the conjugate of both sides gives ( 1 β i ) 13 = β 64 + 64 i ( 1 β i ) 1 3 = β 6 4 + 6 4 i
( r 1 13 β 1 ) ( r 2 13 β 1 ) = ( β 65 β 64 i ) ( β 65 + 64 i ) = 6 5 2 + 6 4 2 = 8321 , ( r 1 1 3 β β 1 ) ( r 2 1 3 β β 1 ) = ( β 6 5 β 6 4 i ) ( β 6 5 + 6 4 i ) = 6 5 2 + 6 4 2 = 8 3 2 1 ,
as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions