2004 AMC 12B Solutions

Solutions 2004 55 th AMC 12 B 723. (C) Let a denote the zero that is an integer. Because the coefficient of x 3 is 1,there can be no other rational zeros, so the two other zeros must be a 2 ± r forsome irrational number r. The polynomial is then(x − a)[x −( a)] [ ( a)]2 + r x −2 − r ( ) 54 a2 − r 2 x − a= x 3 − 2ax 2 +Therefore a = 1002 and the polynomial is( 14 a2 − r 2 ).x 3 − 2004x 2 + (5(501) 2 − r 2 )x − 1002((501) 2 − r 2 ).All coefficients are integers if and only if r 2 is an integer, and the zeros arepositive and distinct if and only if 1 ≤ r 2 ≤ 501 2 − 1 = 251,000. Because rcannot be an integer, there are 251,000 − 500 = 250,500 possible values of n.24. (B) Let ̸ DBE = α and ̸ DBC = β. Then ̸ CBE = α−β and ̸ ABE = α+β,so tan(α − β) tan(α + β) = tan 2 α. Thusfrom which it follows thattan α − tan β1 + tan α tan β · tan α + tan β1 − tan α tan β = tan2 α,tan 2 α − tan 2 β = tan 2 α(1 − tan 2 α tan 2 β).Upon simplifying, tan 2 β(tan 4 α − 1) = 0, so tan α = 1 and α = π 4. Let DC = aand BD = b. Then cot ̸ DBC = b a . Because ̸ CBE = π 4 − β and ̸ ABE =π4 + β, it follows that cot ̸ CBE = tan ̸ ABE = tan ( π4 + β) = 1+ a b1−= b+aab b−a .Thus the numbers 1, b+ab−a , and b a form an arithmetic progression, so b a = b+3ab−a .Setting b = ka yields k 2 − 2k − 3 = 0, and the only positive solution is k = 3.Hence b = BE √2= 5 √ 2, a = 5√ 23 , and the area of △ABC is ab = 50 3 .25. (B) The smallest power of 2 with a given number of digits has a first digit of 1,and there are elements of S with n digits for each positive integer n ≤ 603, sothere are 603 elements of S whose first digit is 1. Furthermore, if the first digitof 2 k is 1, then the first digit of 2 k+1 is either 2 or 3, and the first digit of 2 k+2is either 4,5,6, or 7. Therefore there are 603 elements of S whose first digit is 2or 3, 603 elements whose first digit is 4,5,6, or 7, and 2004 − 3(603) = 195 whosefirst digit is 8 or 9. Finally, note that the first digit of 2 k is 8 or 9 if and only ifthe first digit of 2 k−1 is 4, so there are 195 elements of S whose first digit is 4.