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2004 AMC12B 真题及答案详细解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月5日下午4:55

2004 AMC 12B 真题

Problem 1

At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?

$(mathrm {A}) 3qquad (mathrm {B}) 6 qquad (mathrm {C}) 9 qquad (mathrm {D}) 12 qquad (mathrm {E}) 15$

Problem 2

In the expression $ccdot a^b-d$ , the values of $a$ , $b$ , $c$ , and $d$ are 0, 1, 2, and 3, although not necessarily in that order. What is the maximum possible value of the result?

$(mathrm {A}) 5qquad (mathrm {B}) 6 qquad (mathrm {C}) 8 qquad (mathrm {D}) 9 qquad (mathrm {E}) 10$

Problem 3

If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$ ?

$(mathrm {A}) 8qquad (mathrm {B}) 9 qquad (mathrm {C}) 10 qquad (mathrm {D}) 11 qquad (mathrm {E}) 12$

Problem 4

An integer $x$ , with $10leq xleq 99$ , is to be chosen. If all choices are equally likely, what is the probability that at least one digit of $x$ is a 7?

$(mathrm {A}) dfrac{1}{9} qquad (mathrm {B}) dfrac{1}{5} qquad (mathrm {C}) dfrac{19}{90} qquad (mathrm {D}) dfrac{2}{9} qquad (mathrm {E}) dfrac{1}{3}$

Problem 5

On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving 10 Canadian dollars for every 7 U.S. dollars. After spending 60 Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$ ?

$(mathrm {A}) 5qquad (mathrm {B}) 6 qquad (mathrm {C}) 7 qquad (mathrm {D}) 8 qquad (mathrm {E}) 9$

Problem 6

Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$(mathrm {A}) 13qquad (mathrm {B}) 14 qquad (mathrm {C}) 15 qquad (mathrm {D}) 16 qquad (mathrm {E}) 17$

Problem 7

A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?

$(mathrm {A}) 200+25pi quad (mathrm {B}) 100+75pi quad (mathrm {C}) 75+100pi quad (mathrm {D}) 100+100pi quad (mathrm {E}) 100+125pi$

Problem 8

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains 100 cans, how many rows does it contain?

$(mathrm {A}) 5 qquad (mathrm {B}) 8 qquad (mathrm {C}) 9 qquad (mathrm {D}) 10 qquad (mathrm {E}) 11$

Problem 9

The point $(-3,2)$ is rotated $90^circ$ clockwise around the origin to point $B$ . Point $B$ is then reflected over the line $x=y$ to point $C$ . What are the coordinates of $C$ ?

$mathrm{(A)} (-3,-2) qquad mathrm{(B)} (-2,-3) qquad mathrm{(C)} (2,-3) qquad mathrm{(D)} (2,3) qquad mathrm{(E)} (3,2)$

Problem 10

An annulus is the region between two concentric circles. The concentric circles in the ﬁgure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ , $d=YZ$ , and $e=XY$ . What is the area of the annulus?

$[asy] import graph; unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]$

$mathrm{(A) } pi a^2 qquad mathrm{(B) } pi b^2 qquad mathrm{(C) } pi c^2 qquad mathrm{(D) } pi d^2 qquad mathrm{(E) } pi e^2$

Problem 11

All the students in an algebra class took a $100$ -point test. Five students scored $100$ , each student scored at least $60$ , and the mean score was $76$ . What is the smallest possible number of students in the class?

$mathrm{(A)} 10 qquad mathrm{(B)} 11 qquad mathrm{(C)} 12 qquad mathrm{(D)} 13 qquad mathrm{(E)} 14$

Problem 12

In the sequence $2001$ , $2002$ , $2003$ , $ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^textrm{th}$ term in this sequence?

$mathrm{(A) } -2004 qquad mathrm{(B) } -2 qquad mathrm{(C) } 0 qquad mathrm{(D) } 4003 qquad mathrm{(E) } 6007$

Problem 13

If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$ ?

$mathrm{(A)} -2 qquadmathrm{(B)} -1 qquadmathrm{(C)} 0 qquadmathrm{(D)} 1 qquadmathrm{(E)} 2$

Problem 14

In $triangle ABC$ , $AB=13$ , $AC=5$ , and $BC=12$ . Points $M$ and $N$ lie on $AC$ and $BC$ , respectively, with $CM=CN=4$ . Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$ . What is the area of pentagon $CMJKN$ ?

$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); [/asy]$

$mathrm{(A)} 15 qquad mathrm{(B)} frac{81}{5} qquad mathrm{(C)} frac{205}{12} qquad mathrm{(D)} frac{240}{13} qquad mathrm{(E)} 20$

Problem 15

The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

$mathrm{(A) } 9 qquad mathrm{(B) } 18 qquad mathrm{(C) } 27 qquad mathrm{(D) } 36qquad mathrm{(E) } 45$

Problem 16

A function $f$ is defined by $f(z) = ioverline{z}$ , where $i=sqrt{-1}$ and $overline{z}$ is the complex conjugate of $z$ . How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$ ?

$mathrm{(A)} 0 qquadmathrm{(B)} 1 qquadmathrm{(C)} 2 qquadmathrm{(D)} 4 qquadmathrm{(E)} 8$

Problem 17

For some real numbers $a$ and $b$ , the equation $[8x^3 + 4ax^2 + 2bx + a = 0]$ has three distinct positive roots. If the sum of the base- $2$ logarithms of the roots is $5$ , what is the value of $a$ ?

$mathrm{(A)} -256 qquadmathrm{(B)} -64 qquadmathrm{(C)} -8 qquadmathrm{(D)} 64 qquadmathrm{(E)} 256$

Problem 18

Points $A$ and $B$ are on the parabola $y=4x^2+7x-1$ , and the origin is the midpoint of $AB$ . What is the length of $AB$ ?

$mathrm{(A)} 2sqrt5 qquad mathrm{(B)} 5+frac{sqrt2}{2} qquad mathrm{(C)} 5+sqrt2 qquad mathrm{(D)} 7 qquad mathrm{(E)} 5sqrt2$

Problem 19

A truncated cone has horizontal bases with radii $18$ and $2$ . A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

$mathrm{(A)} 6 qquadmathrm{(B)} 4sqrt{5} qquadmathrm{(C)} 9 qquadmathrm{(D)} 10 qquadmathrm{(E)} 6sqrt{3}$

Problem 20

Each face of a cube is painted either red or blue, each with probability $1/2$ . The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

$textbf{(A)} frac14 qquad textbf{(B)} frac {5}{16} qquad textbf{(C)} frac38 qquad textbf{(D)} frac {7}{16} qquad textbf{(E)} frac12$

Problem 21

The graph of $2x^2 + xy + 3y^2 - 11x - 20y + 40 = 0$ is an ellipse in the first quadrant of the $xy$ -plane. Let $a$ and $b$ be the maximum and minimum values of $frac yx$ over all points $(x,y)$ on the ellipse. What is the value of $a+b$ ?

$mathrm{(A)} 3 qquadmathrm{(B)} sqrt{10} qquadmathrm{(C)} frac 72 qquadmathrm{(D)} frac 92 qquadmathrm{(E)} 2sqrt{14}$

Problem 22

The square

$begin{tabular}{|c|c|c|} hline 50 & textit{b} & textit{c} \ hline textit{d} & textit{e} & textit{f} \ hline textit{g} & textit{h} & 2 \ hline end{tabular}$ is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$ ?

$textbf{(A)} 10 qquad textbf{(B)} 25 qquad textbf{(C)} 35 qquad textbf{(D)} 62 qquad textbf{(E)} 136$

Problem 23

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?

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

Problem 24

In $triangle ABC$ , $AB = BC$ , and $overline{BD}$ is an altitude. Point $E$ is on the extension of $overline{AC}$ such that $BE = 10$ . The values of $tan angle CBE$ , $tan angle DBE$ , and $tan angle ABE$ form a geometric progression, and the values of $cot angle DBE,$ $cot angle CBE,$ $cot angle DBC$ form an arithmetic progression. What is the area of $triangle ABC$ ?

$[asy] size(120); defaultpen(0.7); pair A = (0,0), D = (5*2^.5/3,0), C = (10*2^.5/3,0), B = (5*2^.5/3,5*2^.5), E = (13*2^.5/3,0); draw(A--D--C--E--B--C--D--B--cycle); label("(A)",A,S); label("(B)",B,N); label("(C)",C,S); label("(D)",D,S); label("(E)",E,S); [/asy]$ $mathrm{(A)} 16 qquadmathrm{(B)} frac {50}3 qquadmathrm{(C)} 10sqrt{3} qquadmathrm{(D)} 8sqrt{5} qquadmathrm{(E)} 18$

Problem 25

Given that $2^{2004}$ is a $604$ -digit number whose first digit is $1$ , how many elements of the set $S = {2^0,2^1,2^2,ldots ,2^{2003}}$ have a first digit of $4$ ?

$mathrm{(A)} 194 qquadmathrm{(B)} 195 qquadmathrm{(C)} 196 qquadmathrm{(D)} 197 qquadmathrm{(E)} 198$

Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $frac{1}{2} cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 Rightarrow mathrm{(A)}$ .
Because there are five days, or four transformations between days (day 1 $rightarrow$ day 2 $rightarrow$ day 3 $rightarrow$ day 4 $rightarrow$ day 5), she makes $48 cdot frac{1}{2^4} = boxed{mathrm{(A)} 3}$
If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ .
If $b=0$ , the expression evaluates to $c-dleq 2$ .
Case $d=0$ remains. In that case, we want to maximize $ccdot a^b$ where ${a,b,c}={1,2,3}$ . Trying out the six possibilities we get that the best one is $(a,b,c)=(3,2,1)$ , where $ccdot a^b=1cdot 3^2=boxed{mathrm{(D)} 9}$ .
$1296 = 2^4 3^4$ and $4+4=boxed{8} Longrightarrow mathrm{(A)}$ .
The digit $7$ can be either the tens digit ( $70, 71, dots, 79$ : $10$ possibilities), or the ones digit ( $17, 27, dots, 97$ : $9$ possibilities), but we counted the number $77$ twice. This means that out of the $90$ two-digit numbers, $10+9-1=18$ have at least one digit equal to $7$ . Therefore the probability is $dfrac{18}{90} = boxed{dfrac{1}{5}} Longrightarrow mathrm{(B)}$ .By complementary counting, we count the numbers that do not contain a $7$ , then subtract from the total. There is a $frac{8}{9}cdotfrac{9}{10}$ probability of choosing a number that does NOT contain a $7$ . Subtract this from $1$ and simplify yields $1 - frac{8}{9}cdotfrac{9}{10} = frac{90}{90} - frac{72}{90} = frac{18}{90} = frac{1}{5}$ .
Solution 1

Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=frac{7}{10}cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $boxed{mathrm{(A)} 5}$ .

Solution 2

Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $frac{60}{3}=20$ times. Thus $d=7cdot20=140$ , and the sum of the digits is $boxed{mathrm{(A)} 5}$ .
The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $sqrt{8^2 + 10^2}approx12.8$ , and thus the answer is $boxed{mathrm{(A)} 13}$ .Without a calculator one can note that $8^2+10^2=164<169=13^2Rightarrowboxed{mathrm{(A)} 13}$ .
The area of the circle is $S_{bigcirc}=100pi$ ; the area of the square is $S_{square}=100$ .Exactly $frac{1}{4}$ of the circle lies inside the square. Thus the total area is $dfrac34 S_{bigcirc}+S_{square}=boxed{mathrm{(B) }100+75pi}$ . $[asy] Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); [/asy]$
The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=boxed{mathrm{(D) }10}$ .
The entire situation is in the picture below. The correct answer is $boxed{mathrm{(E)} (3,2)}$ .
The area of the large circle is $pi b^2$ , the area of the small one is $pi c^2$ , hence the shaded area is $pi(b^2-c^2)$ .From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $boxed{mathrm{(A) }pi a^2}$
Let the number of students be $ngeq 5$ . Then the sum of their scores is at least $5cdot 100 + (n-5)cdot 60$ . At the same time, we need to achieve the mean $76$ , which is equivalent to achieving the sum $76n$ .Hence we get a necessary condition on $n$ : we must have $5cdot 100 + (n-5)cdot 60 leq 76n$ . This can be simplified to $200 leq 16n$ . The smallest integer $n$ for which this is true is $n=13$ .To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13cdot 76 = 988$ points to divide among them. The five $100$ s make $500$ , hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.Hence the smallest possible number of students is $boxed{mathrm{(D)} 13}$ .
Solution 1

We already know that $a_1=2001$ , $a_2=2002$ , $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ , $a_6=2003+2000-2005=1998$ , $a_7=2000+2005-1998=2007$ , and so on.

We can now discover the following pattern: $a_{2k+1} = 1999+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2cdot 1002} = 2004 - 2cdot 1002 = boxed{0}$ .

Solution 2

Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$ .

Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= cdots$

From the values given in the problem statement we see that $a_3=a_1+2$ .

From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$ .

From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$ .

Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = cdots = a_2 - 2002 = boxed{0}$ .
By the definition of an inverse function, $x = f(f^{-1}(x)) = a(bx+a)+b = abx + a^2 + b$ . By comparing coefficients, we have $ab = 1 Longrightarrow b = frac 1a$ and $a^2 + b = a^2 + frac{1}{a} = 0$ . Simplifying, $[a^3 + 1 = 0]$ and $a = b = -1$ . Thus $a+b = -2 Rightarrow mathrm{(A)}$ .
Solution 1

The triangle $ABC$ is clearly a right triangle, its area is $frac{5cdot 12}2 = 30$ . If we knew the areas of triangles $AMJ$ and $BNK$ , we could subtract them to get the area of the pentagon.

Draw the height $CL$ from $C$ onto $AB$ . As $AB=13$ and the area is $30$ , we get $CL=frac{60}{13}$ . The situation is shown in the picture below:

$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( C--L, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE); [/asy]$

Now note that the triangles $ABC$ , $AMJ$ , $ACL$ , $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$ , their areas have ratio $k^2$ . We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$ .

We have $frac{CL}{BC} = frac{60/13}{12} = frac 5{13}$ , hence $[ACL] = frac{ 25[ABC] }{169} = frac{750}{169}$ .

Also, $frac{CL}{AC} = frac{60/13}5 = frac{12}{13}$ , hence $[CBL] = frac{ 144[ABC] }{169} = frac{4320}{169}$ .

Now for the smaller triangles:

We know that $frac{AM}{AC}=frac 15$ , hence $[AMJ] = frac{[ACL]}{25} = frac{30}{169}$ .

Similarly, $frac{BN}{BC}=frac 8{12} = frac 23$ , hence $[NBK] = frac{4[CBL]}9 = frac{1920}{169}$ .

Finally, the area of the pentagon is $30 - frac{30}{169} - frac{1920}{169} = boxed{frac{240}{13}}$ .

Solution 2

Split the pentagon along a different diagonal as follows:

$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( M--N, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE); [/asy]$
The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles $ABC$ , $AMJ$ , and $NBK$ are all similar.

Since $BN=12-4=8$ , $NK=frac{5}{13}(8)=frac{40}{13}$ and $BK=frac{12}{13}(8)=frac{96}{13}$ . Since $AM=5-4=1$ , $JM=frac{12}{13}$ and $AJ=frac{5}{13}$ .

The trapezoid's height is therefore $13-frac{5}{13}-frac{96}{13}=frac{68}{13}$ , and its area is $frac{1}{2}left(frac{68}{13}right)left(frac{12}{13}+frac{40}{13}right)=frac{34}{13}(4)=frac{136}{13}$ .

Triangle $MCN$ has area $frac{1}{2}(4)(4)=8$ , and the total area is $frac{104+136}{13}=boxed{frac{240}{13}}$ .

Solution 3

Because triangle ABC, triangle NBK, and triangle AMJ are similar right triangles whose hypotenuses are in the ratio 13 : 8 : 1, their areas are in the ratio 169 : 64 : 1. The area of triangle ABC is 1/2 (12)(5) = 30, so the areas of triangle NBK and triangle AMJ are (64/169) (30) and (1/169)(30), respectively. Thus the area of pentagon CMJKN is (1 − 64/169 - 1/169)(30)= $boxed{mathbf{(D)}240/13}$
Solution 1

If Jack's current age is $overline{ab}=10a+b$ , then Bill's current age is $overline{ba}=10b+a$ .

In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$ .

We are given that $10a+b+5=2(10b+a+5)$ . Thus $8a=19b+5$ .

For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $frac{19b+5}8$ is not an integer, and for $bgeq 4$ it is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=boxed{mathrm{(B) }18}$ .

Solution 2

Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.

The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$ .

Thus Bill's age is in the set ${13,22,31,40,49,58,67,76,85,94}$ .

As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options ${13,49,58,67}$ .

Checking each of them, we see that only $13$ works, and gives the solution $31-13=boxed{mathrm{(B) }18}$ .
Let $z = a+bi$ , so $overline{z} = a-bi$ . By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$ , which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 Rightarrow mathrm{(C)}$ intersections.
Let the three roots be $x_1,x_2,x_3$ . $[log_2 x_1 + log_2 x_2 + log_2 x_3 = log_2 x_1x_2x_3= 5 Longrightarrow x_1x_2x_3 = 32]$ By Vieta’s formulas, $[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a]$ gives us that $a = -8x_1x_2x_3 = -256 Rightarrow mathrm{(A)}$ .
Let the coordinates of $A$ be $(x_A,y_A)$ . As $A$ lies on the parabola, we have $y_A=4x_A^2+7x_A-1$ . As the origin is the midpoint of $AB$ , the coordinates of $B$ are $(-x_A,-y_A)$ . We need to choose $x_A$ so that $B$ will lie on the parabola as well. In other words, we need $-y_A = 4(-x_A)^2 + 7(-x_A) - 1$ .Substituting for $y_A$ , we get: $-4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1$ .This simplifies to $8x_A^2 - 2 = 0$ , which solves to $x_A = pm 1/2$ . Both roots lead to the same pair of points: $(1/2,7/2)$ and $(-1/2,-7/2)$ . Their distance is $sqrt{ 1^2 + 7^2 } = sqrt{50} = boxed{5sqrt2}$ .

Alternate Solution

Let the coordinates of $A$ and $B$ be $(x_A, y_A)$ and $(x_B, y_B)$ , respectively. Since the median of the points lies on the origin, $x_A + x_B = y_A + y_B = 0$ and expanding $y_A + y_B$ , we find: $[4x_A^2 + 7x_A - 1 + 4x_B^2 + 7x_B - 1 = 0]$ $[4(x_A^2 + x_B^2) + 7(x_A + x_B) = 2]$ $[x_A^2 + x_B^2 = frac{1}{2}.]$

It also follows that $(x_A + x_B)^2 = 0$ . Expanding this, we find: $[x_A^2 + 2x_A x_B + x_B^2 = 0]$ $[frac{1}{2} + 2x_A x_B = 0]$ $[x_A x_B = -frac{1}{4}.]$

To find the distance between the points, $sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}$ must be found. Expanding $y_A - y_B$ : $[y_A - y_B = 4x_A^2 + 7x_A - 1 - 4x_B^2 - 7x_B + 1]$ $[= 4(x_A^2 - x_B^2) + 7(x_A - x_B)]$ $[= 4(x_A + x_B)(x_A - x_B) + 7(x_A - x_B)]$ $[= 7(x_A - x_B)]$ we find the distance to be $sqrt{50(x_A - x_B)^2}$ . Expanding this yields $5sqrt{2(x_A^2 + x_B^2 - 2x_A x_B)} = boxed{5sqrt{2}}$ .
Solution #1

Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:

$[asy] import olympiad; size(220); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2; draw(A--B--C--D--cycle); draw(circumcircle(E,F,G)); dot(E); dot(F); dot(G); label("(A)",A,S); label("(B)",B,S); label("(C)",C,NE); label("(D)",D,NW); label("(E)",E,S); label("(F)",F,NE); label("(G)",G,N); [/asy]$ Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$ , so $BC = 20$ . We draw $H$ such that it is the foot of the altitude $overline{HD}$ to $overline{AB}$ :

$[asy] import olympiad; size(220); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2, H=(16,0); pair P=(D+G)/2, Q=(D+H)/2, R=(B+E)/2, T=(A+H)/2, O=(E+G)/2; draw(A--B--C--D--cycle); draw(G--E--H--D); draw(circumcircle(E,F,G)); dot(E);dot(F);dot(G);dot(H);dot(O); label("(A)",A,S); label("(B)",B,S); label("(C)",C,NE); label("(D)",D,NW); label("(E)",E,S); label("(F)",F,NE); label("(G)",G,N); label("(H)",H,S); label("(O)",O,NE); label("(2)",P,N); label("(12)",Q,W); label("(18)",R,S); label("(16)",T,S); label("(20)",(A+D)/2,NW); label("(r)",(O+E)/2,SE); [/asy]$ By the Pythagorean Theorem, $[r = frac{DH}2 = frac{sqrt{20^2 - 16^2}}2 = boxed{6} Rightarrow mathrm{(A)}.]$

Solution #2

Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $overline{AD}$ and $overline{BC}$ from the solution above and label the point at where they meet $H$ . Because $frac{overline{GC}}{overline{BE}}$ = $frac{1}{9}$ , $frac{overline{HG}}{overline{HE}}$ = $frac{1}{9}$ . Let $overline{HG} = x$ and $overline{GE} = 8x$ .

Because these are radii, $overline{GO} = overline{OE} = overline{OF} = 4x$ . $overline{OF} perp overline{BH}$ so $overline{OF}^2 + overline{FH}^2 = overline{OH}^2$ . Plugging in, we get $4x^2 + overline{FH}^2 = 5x^2$ so $overline{FH} = 3x$ .Triangles $OFH$ and $BEH$ are similar so $frac{overline{OF}}{overline{BE}} = frac{overline{FH}}{overline{EH}}$ which gives us $frac{4x}{18} = frac{3x}{9x}$ . Solving for x, we get $[x = 1.5]$ and $[4x = boxed{6} Rightarrow mathrm{(A)}.]$
There are $2^6$ possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if $5$ or $6$ of the faces are colored the same, which for each color can happen in $6 + 1 = 7$ ways. If $4$ of the faces are colored the same, there are $3$ possible cubes (corresponding to the $3$ possible ways to pick pairs of opposite faces for the other color). If $3$ of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of $2(7 + 3) = 20$ ways for this to occur, and the desired probability is $frac{20}{2^6} = frac{5}{16} mathbf{(B)}$ .
$frac yx$ represents the slope of a line passing through the origin. It follows that since a line $y = mx$ intersects the ellipse at either $0, 1,$ or $2$ points, the minimum and maximum are given when the line $y = mx$ is a tangent, with only one point of intersection. Substituting, $[2x^2 + x(mx) + 3(mx)^2 - 11x - 20(mx) + 40 = 0]$ Rearranging by the degree of $x$ , $[(3m^2 + m + 2)x^2 - (20m + 11)x + 40 = 0]$ Since the line $y=mx$ , we want the discriminant, $[(20m+11)^2 - 4cdot 40 cdot (3m^2 + m + 2) = -80m^2 + 280m - 199]$ to be equal to $0$ . We want $a+b$ , which is the sum of the roots of the above quadratic. By Vieta’s formulas, that is $frac{280}{80} = frac{7}{2} Rightarrow mathrm{(C)}$ .
Solution A

If the power of a prime $p^n$ other than $2,5$ divides $g$ , then from $50 cdot 2 e = 50dg$ it follows that $p^n|e$ , but then considering the product of the diagonals, $p^{2n} |gec$ but $p^{2n} nmid 100e$ , contradiction. So the only prime factors of $g$ are $2$ and $5$ .

It suffices now to consider the two magic squares comprised of the powers of $2$ and $5$ of the corresponding terms. These satisfy the normal requirement that the sums of rows, columns, and diagonals are the same, owing to our rules of exponents; additionally, all terms are non-negative.

The powers of $2$ :

$begin{tabular}{|c|c|c|} hline 1 & textit{b} & textit{c} \ hline textit{d} & textit{e} & textit{f} \ hline textit{g} & textit{h} & 1 \ hline end{tabular}$ So $1 + 1 + e = g + e + c Longrightarrow g = 2 - c$ , so $g = 0,1,2$ . Indeed, we have the magic squares

$begin{tabular}{|c|c|c|} hline 1 & 0 & 2 \ hline 2 & 1 & 0 \ hline 0 & 2 & 1 \ hline end{tabular}, quad begin{tabular}{|c|c|c|} hline 1 & 1 & 1 \ hline 1 & 1 & 1 \ hline 1 & 1 & 1 \ hline end{tabular}, quad begin{tabular}{|c|c|c|} hline 1 & 2 & 0 \ hline 0 & 1 & 2 \ hline 2 & 0 & 1 \ hline end{tabular},$ The powers of $5$ :

$begin{tabular}{|c|c|c|} hline 2 & textit{b} & textit{c} \ hline textit{d} & textit{e} & textit{f} \ hline textit{g} & textit{h} & 0 \ hline end{tabular}$ Again, we get $2 + e = g + e + c Longrightarrow g = 0,1,2$ . However, if we let $g = 2, c = 0$ , then $e = d + e + f Longrightarrow d = f = 0$ , which obviously gives us a contradiction, and similarly for $g = 0, c = 2$ . For $g = 1$ , we get

$begin{tabular}{|c|c|c|} hline 2 & 0 & 1 \ hline 0 & 1 & 2 \ hline 1 & 2 & 0 \ hline end{tabular}$ In conclusion, $g$ can be $2^0 cdot 5^1, 2^1 cdot 5^1, 2^2 cdot 5^1$ , and their sum is $boxed{mathbf{(C)}35}$ .

Solution B

All the unknown entries can be expressed in terms of $b$ . Since $100e = beh = ceg = def$ , it follows that $h = frac{100}{b}, g = frac{100}{c}$ , and $f = frac{100}{d}$ . Comparing rows $1$ and $3$ then gives $50bc = 2 cdot frac{100}{b} cdot frac{100}{c}$ , from which $c = frac{20}{b}$ . Comparing columns $1$ and $3$ gives $50d cdot frac{100}{c}= 2c cdot frac{100}{d}$ , from which $d = frac{c}{5} = frac{4}{b}$ . Finally, $f = 25b, g = 5b$ , and $e = 10$ . All the entries are positive integers if and only if $b = 1, 2,$ or $4$ . The corresponding values for $g$ are $5, 10,$ and $20$ , and their sum is $boxed{mathbf{(C)}35}$ .

Solution C

We know because this is a multiplicative magic square that each of the following are equal to each other: $100e=ceg=50dg=beh=2cf=50bc=def=2gh$

From this we know that $50dg=2hg$ , thus $h=25d$ . Thus $beh=be(25d)$ and $be(25d)=100e$ . Thus $b=frac{4}{d}$ From this we know that $50bc=(50)(frac{4}{d})(c)=50dg$ . Thus $c=frac{d^2g}{4}$ . Now we know from the very beginning that $100e=ceg$ or $100=cg$ or $100=frac{d^2g}{4}(g)$ or $frac{d^2g^2}{4}$ . Rearranging the equation $100=frac{d^2g^2}{4}$ we have $(d^2)(g^2)=400$ or $dg=20$ due to $d$ and $g$ both being positive. Now that $dg=20$ we find all pairs of positive integers that multiply to $20$ . There is $(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)$ . Now we know that $b=frac{4}{d}$ and b has to be a positive integer. Thus $d$ can only be $1$ , $2$ , or $4$ . Thus $g$ can only be $20$ , $10$ ,or $5$ . Thus sum of $20+10+5$ = $35$ . The answer is $boxed{mathbf{(C)}35}$ .
Letting the roots be $r$ , $s$ , and $t$ , where $t = r+s$ , we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$ , and so $t = 1002$ . Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$ . Letting $x = 0$ gives that $1002 mid n$ because $x - 1002 mid x^3 - 2004x^2 + mx + n$ . Letting $n = 1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$ , we see that $b$ is the sum of the roots of $x^2 - bx + a$ , $r$ and $s$ , and so $b = 1002$ . Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$ , and we wish to find the number of possible values of $a$ . By the quadratic formula, we see that $[frac{1002 pm sqrt{1002^2 - 4a}}{2} = 501 pm sqrt{501^2 - a}]$ are the two values of noninteger positive real numbers $r$ and $s$ , neither of which is equal to $1002$ . This information gives us that $0 < 501^2 - a < 501^2$ , and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = 250,!500 mathrm{(C)}$ possible values of $n = 1002a$ .
Let $alpha = DBC$ . Then the first condition tells us that $[tan^2 DBE = tan(DBE - alpha)tan(DBE + alpha) = frac {tan^2 DBE - tan^2 alpha}{1 - tan ^2 DBE tan^2 alpha},]$ and multiplying out gives us $(tan^4 DBE - 1) tan^2 alpha = 0$ . Since $tanalpha neq 0$ , we have $tan^4 DBE = 1 Longrightarrow angle DBE = 45^{circ}$ .The second condition tells us that $2cot (45 - alpha) = 1 + cot alpha$ . Expanding, we have $1 + cot alpha = 2left[frac {cot alpha + 1}{cot alpha - 1}right] Longrightarrow (cot alpha - 3)(cot alpha + 1) = 0$ . Evidently $cot alpha neq - 1$ , so we get $cot alpha = 3$ .Now $BD = 5sqrt {2}$ and $AC = frac {2BD} {cot alpha} = frac {10sqrt {2}}{3}$ . Thus, $[ABC] = frac {1}{2} cdot 5sqrt {2} cdot frac {10sqrt {2}}{3} = frac {50}{3} mathrm{(B)}$ .
Solution 1

Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$ . Thus $S$ contains $603$ elements with a first digit of $1$ . For each number in the form of $2^k$ such that its first digit is $1$ , then $2^{k+1}$ must either have a first digit of $2$ or $3$ , and $2^{k+2}$ must have a first digit of $4,5,6,7$ . Thus there are also $603$ numbers with first digit ${2,3}$ and $603$ numbers with first digit ${4,5,6,7}$ . By using complementary counting, there are $2004 - 3 times 603 = 195$ elements of $S$ with a first digit of ${8,9}$ . Now, $2^k$ has a first digit of ${8,9}$ if and only if the first digit of $2^{k-1}$ is $4$ , so there are $boxed{195} Rightarrow mathrm{(B)}$ elements of $S$ with a first digit of $4$ .

Solution 2

We can make the following chart for the possible loops of leading digits: $[1 rightarrow 2 rightarrow 4 rightarrow 8 rightarrow 1]$ $[1 rightarrow 2 rightarrow 4 rightarrow 9 rightarrow 1]$ $[1 rightarrow 2 rightarrow 5 rightarrow 1]$ $[1 rightarrow 3 rightarrow 6 rightarrow 1]$ $[1 rightarrow 3 rightarrow 7 rightarrow 1]$

Thus each loop from $1 rightarrow 1$ can either have $3$ or $4$ numbers. Let there be $x$ of the sequences of $3$ numbers, and let there be $y$ of the sequences of $4$ numbers. We note that a $4$ appears only in the loops of $4$ , and also we are given that $2^{2004}$ has $604$ digits. $[3x+4y=2004]$ $[x+y=603]$ Solving gives $x = 408$ and $y = 195$ , thus the answer is $(B)$ .