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2006 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6日上午9:29

2006 AMC 12 B 真题

答案详细解析请参考文末

Problem 1

What is $( - 1)^1 + ( - 1)^2 + cdots + ( - 1)^{2006}$ ?

$text {(A) } - 2006 qquad text {(B) } - 1 qquad text {(C) } 0 qquad text {(D) } 1 qquad text {(E) } 2006$

Problem 2

For real numbers $x$ and $y$ , define $xspadesuit y = (x + y)(x - y)$ . What is $3spadesuit(4spadesuit 5)$ ?

$text {(A) } - 72 qquad text {(B) } - 27 qquad text {(C) } - 24 qquad text {(D) } 24 qquad text {(E) } 72$

Problem 3

A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?

$text {(A) } 10 qquad text {(B) } 14 qquad text {(C) } 17 qquad text {(D) } 20 qquad text {(E) } 24$

Problem 4

Mary is about to pay for five items at the grocery store. The prices of the items are $textdollar7.99$ , $textdollar4.99$ , $textdollar2.99$ , $textdollar1.99$ , and $textdollar0.99$ . Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the $textdollar20.00$ that she will receive in change?

$text {(A) } 5 qquad text {(B) } 10 qquad text {(C) } 15 qquad text {(D) } 20 qquad text {(E) } 25$

Problem 5

John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?

$text {(A) } 30 qquad text {(B) } 50 qquad text {(C) } 60 qquad text {(D) } 90 qquad text {(E) } 120$

Problem 6

Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?

$text {(A) } 129 qquad text {(B) } 137 qquad text {(C) } 174 qquad text {(D) } 223 qquad text {(E) } 411$

Problem 7

Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?

$text {(A) } 4 qquad text {(B) } 12 qquad text {(C) } 16 qquad text {(D) } 24 qquad text {(E) } 48$

Problem 8

The lines $x = frac 14y + a$ and $y = frac 14x + b$ intersect at the point $(1,2)$ . What is $a + b$ ?

$text {(A) } 0 qquad text {(B) } frac 34 qquad text {(C) } 1 qquad text {(D) } 2 qquad text {(E) } frac 94$

Problem 9

How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?

$text {(A) } 21 qquad text {(B) } 34 qquad text {(C) } 51 qquad text {(D) } 72 qquad text {(E) } 150$

Problem 10

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?

$text {(A) } 43 qquad text {(B) } 44 qquad text {(C) } 45 qquad text {(D) } 46 qquad text {(E) } 47$

Problem 11

Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?

$text {(A) } frac 67 qquad text {(B) } frac {13}{14} qquad text {(C) } 1 qquad text {(D) } frac {14}{13} qquad text {(E) } frac 76$

Problem 12

The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $pne 0$ . What is $b$ ?

$text {(A) } -p qquad text {(B) } 0 qquad text {(C) } 2 qquad text {(D) } 4 qquad text {(E) } p$

Problem 13

Rhombus $ABCD$ is similar to rhombus $BFDE$ . The area of rhombus $ABCD$ is 24, and $angle BAD = 60^circ$ . What is the area of rhombus $BFDE$ ?

$[asy] defaultpen(linewidth(0.7)+fontsize(11)); unitsize(2cm); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]$

$textrm{(A) } 6 qquad textrm{(B) } 4sqrt {3} qquad textrm{(C) } 8 qquad textrm{(D) } 9 qquad textrm{(E) } 6sqrt {3}$

Problem 14

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4$ cents per glob and $J$ blobs of jam at $5$ cents per glob. The cost of the peanut butter and jam to make all the sandwiches is $textdollar 2.53$ . Assume that $B$ , $J$ and $N$ are all positive integers with $N>1$ . What is the cost of the jam Elmo uses to make the sandwiches?

$mathrm{(A)} 1.05 qquad mathrm{(B)} 1.25 qquad mathrm{(C)} 1.45 qquad mathrm{(D)} 1.65 qquad mathrm{(E)} 1.85$

Problem 15

Circles with centers $O$ and $P$ have radii 2 and 4, respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O$ , and points $C$ and $D$ are on the circle centered at $P$ , such that $overline{AD}$ and $overline{BC}$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$ ?

$[asy] unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11)); pair A, B, C, D; pair[] O; O[1] = (6,0); O[2] = (12,0); A = (32/6,8*sqrt(2)/6); B = (32/6,-8*sqrt(2)/6); C = 2*B; D = 2*A; draw(Circle(O[1],2)); draw(Circle(O[2],4)); draw((0.7*A)--(1.2*D)); draw((0.7*B)--(1.2*C)); draw(O[1]--O[2]); draw(A--O[1]); draw(B--O[1]); draw(C--O[2]); draw(D--O[2]); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SW); label("$D$", D, NW); dot("$O$", O[1], SE); dot("$P$", O[2], SE); label("$2$", (A + O[1])/2, E); label("$4$", (D + O[2])/2, E);[/asy]$

$textbf{(A) } 18sqrt {3} qquad textbf{(B) } 24sqrt {2} qquad textbf{(C) } 36 qquad textbf{(D) } 24sqrt {3} qquad textbf{(E) } 32sqrt {2}$

Problem 16

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$ , respectively. What is its area?

$mathrm{(A)} 20sqrt {3} qquad mathrm{(B)} 22sqrt {3} qquad mathrm{(C)} 25sqrt {3} qquad mathrm{(D)} 27sqrt {3} qquad mathrm{(E)} 50$

Problem 17

For a particular peculiar pair of dice, the probabilities of rolling $1$ , $2$ , $3$ , $4$ , $5$ and $6$ on each die are in the ratio $1:2:3:4:5:6$ . What is the probability of rolling a total of $7$ on the two dice?

$mathrm{(A)} frac 4{63} qquad mathrm{(B)} frac 18 qquad mathrm{(C)} frac 8{63} qquad mathrm{(D)} frac 16 qquad mathrm{(E)} frac 27$

Problem 18

An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

$mathrm{(A)} 120 qquad mathrm{(B)} 121 qquad mathrm{(C)} 221 qquad mathrm{(D)} 230 qquad mathrm{(E)} 231$

Problem 19

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?

$mathrm{(A)} 4 qquad mathrm{(B)} 5 qquad mathrm{(C)} 6 qquad mathrm{(D)} 7 qquad mathrm{(E)} 8$

Problem 20

Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $lfloorlog_{10}4xrfloor - lfloorlog_{10}xrfloor = 0$ ? Here $lfloor xrfloor$ denotes the greatest integer that is less than or equal to $x$ .

$mathrm{(A)} frac 18 qquad mathrm{(B)} frac 3{20} qquad mathrm{(C)} frac 16 qquad mathrm{(D)} frac 15 qquad mathrm{(E)} frac 14$

Problem 21

Rectangle $ABCD$ has area $2006$ . An ellipse with area $2006pi$ passes through $A$ and $C$ and has foci at $B$ and $D$ . What is the perimeter of the rectangle? (The area of an ellipse is $abpi$ where $2a$ and $2b$ are the lengths of the axes.)

$mathrm{(A)} frac {16sqrt {2006}}{pi} qquad mathrm{(B)} frac {1003}4 qquad mathrm{(C)} 8sqrt {1003} qquad mathrm{(D)} 6sqrt {2006} qquad mathrm{(E)} frac {32sqrt {1003}}pi$

Problem 22

Suppose $a$ , $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=mcdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$ ?

$mathrm{(A)} 489 qquad mathrm{(B)} 492 qquad mathrm{(C)} 495 qquad mathrm{(D)} 498 qquad mathrm{(E)} 501$

Problem 23

Isosceles $triangle ABC$ has a right angle at $C$ . Point $P$ is inside $triangle ABC$ , such that $PA=11$ , $PB=7$ , and $PC=6$ . Legs $overline{AC}$ and $overline{BC}$ have length $s=sqrt{a+bsqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$ ?

$[asy] pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]$

$mathrm{(A)} 85 qquad mathrm{(B)} 91 qquad mathrm{(C)} 108 qquad mathrm{(D)} 121 qquad mathrm{(E)} 127$

Problem 24

Let $S$ be the set of all points $(x,y)$ in the coordinate plane such that $0leq xleq fracpi 2$ and $0leq yleq fracpi 2$ . What is the area of the subset of $S$ for which $sin^2 x - sin xsin y + sin^2 yle frac 34$ ?

$mathrm{(A)} frac {pi^2}9 qquad mathrm{(B)} frac {pi^2}8 qquad mathrm{(C)} frac {pi^2}6 qquad mathrm{(D)} frac {3pi^2}{16} qquad mathrm{(E)} frac {2pi^2}9$

Problem 25

A sequence $a_1,a_2,dots$ of non-negative integers is defined by the rule $a_{n+2}=|a_{n+1}-a_n|$ for $ngeq 1$ . If $a_1=999$ , $a_2<999$ and $a_{2006}=1$ , how many different values of $a_2$ are possible?

$mathrm{(A)} 165 qquad mathrm{(B)} 324 qquad mathrm{(C)} 495 qquad mathrm{(D)} 499 qquad mathrm{(E)} 660$

2006 AMC12 B 真题答案详细解析

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$(-1)^n=1$ if n is even and $-1$ if n is odd. So we have
$-1+1-1+1-cdots-1+1=0+0+cdots+0+0=0 Rightarrow text{(C)}$

$4spadesuit 5=-9$ $3spadesuit -9=-72 Rightarrow text{(A)}$

Solution 1

If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $Rightarrow boxed{text{(A)}}$ .

Solution 2

Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 rightarrow 2x+14=34 rightarrow 2x=20 rightarrow x = 10$ ,

and the answer is $boxed{text{(A)}}$ .

The total price of the items is $(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95$ $20-18.95=1.05$ $frac{1.05}{20}=.0525 Rightarrow text{(A)}$

Alternative Solution

We can round the prices to $8$ , $5$ , $3$ , $2$ , and $1$ .

So $20 - (8+5+3+2+1) = 1$

We can make an equation: $20* frac{x}{100}=1$

If we simplify the equation to "x", we get $x=5$

The speed that Bob is catching up to John is $5-3=2$ miles per hour. Since Bob is one mile behind John, it will take $frac{1}{2} Rightarrow text{(A)}$ of an hour to catch up to John.

Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ by $3$ to get $137impliesboxed{(text{B})}$ .

First, we seat the children.The first child can be seated in $3$ spaces.The second child can be seated in $2$ spaces.Now there are $2 times 1$ ways to seat the adults. $3 times 2 times 2 = 12 Rightarrow text{(B)}$ Alternative solution:If there was no restriction, there would be 4!=24 ways to sit. However, only 2/4 of the people can sit in the driver's seat, so our answer is $frac{2}{4}cdot 24= 12 Rightarrow text{(B)}$

Solution 1

$4x-4a=y$

$4x-4a=frac{1}{4}x+b$

$4cdot1-4a=frac{1}{4}cdot1+b=2$

$a=frac{1}{2}$

$b=frac{7}{4}$

$a+b=frac{9}{4} Rightarrow text{(E)}$

Solution 2

Add both equations:

$begin{cases}x=frac{1}{4}y+a y=frac{1}{4}x+b end{cases}$

Simplify:

$frac{4}{4}(x+y)=frac{1}{4}(x+y)+(a+b)$

Isolate our solution:

$frac{3}{4}(x+y)=a+b$

Substitute the point of intersection $[x=1, y=2]$

$a+b=frac{3}{4}cdot3=frac{9}{4} Rightarrow text{(E)}$

Solution 3

Plugging in $(1,2)$ into the first equation, and solving for $a$ we get $a$ as $frac{1}{2}$ .

Doing the same for the second equation for the second equation, we get $b$ as $frac{7}{4}$

Adding $a+b = frac{1}{2} + frac{7}{4} = frac{9}{4} Rightarrow text{(E)}$

Solution 1

Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).

If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,

and so on.

So, the answer is $3(1+2)+2(3+4)+1(5+6)=boxed{34} Rightarrow B$ .

Solution 2

The last digit is 4, 6, or 8.

If the last digit is $x$ , the possibilities for the first two digits correspond to 2-element subsets of ${1,2,dots,x-1}$ .

Thus the answer is ${3choose 2} + {5choose 2} + {7choose 2} = 3 + 10 + 21 = 34$ .

Solution 3

The answer must be half of a triangular number (evens and decreasing/increasing) so $boxed{34}$ or B. -zoevv

If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: $[ x+15>3x Rightarrow 2x<15 Rightarrow x<7.5]$ Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43 Rightarrow boxed{text {(A)}}$ .

Joe has 2 ounces of cream, as stated in the problem.JoAnn had 14 ounces of liquid, and drank $frac{1}{7}$ of it. Therefore, she drank $frac{1}{7}$ of her cream, leaving her $2*frac{6}{7}$ . $frac{2}{2*frac{6}{7}}=frac{7}{6} Rightarrow boxed{text{(E)}}$

Substituting $(0,-p)$ , we find that $y = -p = a(0)^2 + b(0) + c = c$ , so our parabola is $y = ax^2 + bx - p$ .The x-coordinate of the vertex of a parabola is given by $x = p = frac{-b}{2a} Longleftrightarrow a = frac{-b}{2p}$ . Additionally, substituting $(p,p)$ , we find that $y = p = a(p)^2 + b(p) - p Longleftrightarrow ap^2 + (b-2)p = left(frac{-b}{2p}right)p^2 + (b-2)p = pleft(frac b2-2right) = 0$ . Since it is given that $p neq 0$ , then $frac{b}{2} = 2 Longrightarrow b = 4 mathrm{(D)}$ .

Solution 2

A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $pne 0$ , it follows that $a=-2/p$ . Thus $[y=-frac{2}{p}(x^2-2px+p^2)+p=-frac{2}{p}x^2+4x-p,]$ so $boxed{b=4}$ .

Solution 1

The ratio of any length on $ABCD$ to a corresponding length on $BFDE^2$ is equal to the ratio of their areas. Since $angle BAD=60$ , $triangle ADB$ and $triangle DBC$ are equilateral. $DB$ , which is equal to $AB$ , is the diagonal of rhombus $ABCD$ . Therefore, $AC=frac{DB(2)}{2sqrt{3}}=frac{DB}{sqrt{3}}$ . $DB$ and $AC$ are the longer diagonal of rhombuses $BEDF$ and $ABCD$ , respectively. So the ratio of their areas is $(frac{1}{sqrt{3}})^2$ or $frac{1}{3}$ . One-third the area of $ABCD$ is equal to $8$ . So the answer is $boxed{text{C}}$ .

Solution 2

Draw the line $overline{DB}$ to form an equilateral triangle, since $angle BAD=60$ , and line segments $overline{AB}$ and $overline{AD}$ are equal in length. To find the area of the smaller rhombus, we only need to find the value of any arbitrary base, then square the result. To find the value of the base, use the line we just drew and connect it to point $E$ at a right angle along line $overline{DB}$ . Call the connected point $P$ , with triangles $DPE$ and $EPB$ being 30-60-90 triangles, meaning we can find the length of $overline{ED}$ or $overline{EB}$ . The base of $ABCD$ must be $sqrt{24}$ , and half of that length must be $sqrt{6}$ (triangles $DPE$ and $EPB$ are congruent by $SSS$ ). Solving for the third length yields $sqrt{8}$ , which we square to get the answer $boxed{text{C}}$ .

Solution 3

Draw line DB, cutting rhombus BFDE into two triangles which fit nicely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8.

From the given, we know that $253=N(4B+5J)$ (The numbers are in cents)since $253=11cdot23$ , and since $N$ is an integer, then $4B+5J=11$ or $23$ . It is easily deduced that $11$ is impossible to make with $B$ and $J$ integers, so $N=11$ and $4B+5J=23$ . Then, it can be guessed and checked quite simply that if $B=2$ and $J=3$ , then $4B+5J=4(2)+5(3)=23$ . The problem asks for the total cost of jam, or $N(5J)=11(15)=165$ cents, or $1.65impliesmathrm{(D)}$

Draw the altitude from $O$ onto $DP$ and call the point $H$ . Because $angle OAD$ and $angle ADP$ are right angles due to being tangent to the circles, and the altitude creates $angle OHD$ as a right angle. $ADHO$ is a rectangle with $OH$ bisecting $DP$ . The length $OP$ is $4+2$ and $HP$ has a length of $2$ , so by pythagorean's, $OH$ is $sqrt{32}$ . $2 cdot sqrt{32} + frac{1}{2}cdot2cdot sqrt{32} = 3sqrt{32} = 12sqrt{2}$ , which is half the area of the hexagon, so the area of the entire hexagon is $2cdot12sqrt{2} = boxed{(B)24sqrt{2}}$

Solution 2

$ADPO$ and $OPBC$ are congruent right trapezoids with legs $2$ and $4$ and with $OP$ equal to $6$ . Draw an altitude from $O$ to either $DP$ or $CP$ , creating a rectangle with width $2$ and base $x$ , and a right triangle with one leg $2$ , the hypotenuse $6$ , and the other $x$ . Using the Pythagorean theorem, $2006 AMC12B 真题及答案详细解析$ is equal to $4sqrt{2}$ , and $x$ is also equal to the height of the trapezoid. The area of the trapezoid is thus $frac{1}{2}cdot(4+2)cdot4sqrt{2} = 12sqrt{2}$ , and the total area is two trapezoids, or $boxed{24sqrt{2}}$ .

To find the area of the regular hexagon, we only need to calculate the side length. a distance of $sqrt{7^2+1^2} = sqrt{50} = 5sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $frac{5sqrt{2}}{2}cdotfrac{1}{sqrt{3}}cdot2 = frac{5sqrt{6}}{3}$ .The apothem is thus $frac{1}{2}cdotfrac{5sqrt{6}}{3}cdotsqrt{3} = frac{5sqrt{2}}{2}$ , yielding an area of $frac{1}{2}cdot10sqrt{6}cdotfrac{5sqrt{2}}{2}=25sqrt{3}$ .

The probability of getting an $x$ on one of these dice is $frac{x}{21}$ .The probability of getting $1$ on the first and $6$ on the second die is $frac 1{21}cdotfrac 6{21}$ . Similarly we can express the probabilities for the other five ways how we can get a total $7$ . (Note that we only need the first three, the other three are symmetric.)Summing these, the probability of getting a total $7$ is: $[2cdotleft( frac 1{21}cdotfrac 6{21} + frac 2{21}cdotfrac 5{21} + frac 3{21}cdotfrac 4{21} right) = frac{56}{441} = boxed{frac{8}{63}}]$

Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $2006 AMC12B 真题及答案详细解析$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.Once we pick $xin{-10,dots,10}$ , we have $11-|x|$ valid choices for $y$ , giving a total of $boxed{121}$ possible positions.

Solution 2

$10$ moves results in a lot of possible endpoints, so we try small cases first.

If the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to. If the object makes $2$ moves, it can move to $(0, 2)$ , $(1, 1)$ , $(2, 0)$ , $(1, -1)$ , $(0, -2)$ , $(-1, -1)$ , $(-2, 0)$ as well as $(0, 0)$ , for a total of $9$ moves. If the object makes $3$ moves, it can end up at $(0, 3)$ , $(2, 1)$ , $(1, 2)$ , $(3, 0)$ , $(2, -1)$ , $(1, -2)$ , $(0, -3)$ , $(-1, -2)$ , $(-2, -1)$ , $(0, -3)$ etc. for a total of $16$ moves.

At this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $boxed{B}$ .

First, The number of the plate is divisible by $9$ and in the form of $aabb$ , $abba$ or $abab$ .We can conclude straight away that $a+b= 9$ using the $9$ divisibility rule.If $b=1$ , the number is not divisible by $2$ (unless it's $1818$ , which is not divisible by $4$ ), which means there are no $2$ , $4$ , $6$ , or $8$ year olds on the car, but that can't be true, as that would mean there are less than $8$ kids on the car.If $b=2$ , then the only possible number is $7272$ . $7272$ is divisible by $4$ , $6$ , and $8$ , but not by $5$ and $7$ , so that doesn't work.If $b=3$ , then the only number is $6336$ , also not divisible by $5$ or $7$ .If $b=4$ , the only number is $5544$ . It is divisible by $4$ , $6$ , $7$ , and $8$ .Therefore, we conclude that the answer is $mathrm{(B)} 5$

NOTE: Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to $5$ or $8$ .

Alternate Solution

We know that the number of the plate is divisible by $9$ and in the form $aabb$ , $abba$ , or $abab$ for distinct digits $a,b$ .

Using the divisibility rule for $9$ , we can conclude that $a+bequiv 0pmod{9}$ .

We also know that the number of the plate is even, because you can only discard one number from the integers $1$ through $9$ , inclusive ( $8$ children, oldest is $9$ ), and there's always going to be an even number left.

If one of the children was $5$ years old, then the plate number would be divisible by both $5$ and $2$ . Thus, the units digit must be $0$ . Then, the possible form of the plate number would be $aa00$ , $0bb0$ , or $a0a0$ . The first case is not possible because $00$ is not a possible age for the father.

We have the remaining forms $0bb0$ and $a0a0$ .

$textbf{Case 1: 0bb0:}$ We know that $2bequiv 0pmod{9}$ , so $b$ has to be either $0$ or $9$ . $b$ can't be $0$ because $a, b$ are distinct. So, $b=9$ . However, the number $0990$ is not divisible by both $4$ and $7$ , so whichever number you discard, the other one will still be there. This creates a contradiction, so Case $1$ cannot be true.

$textbf{Case 2: a0a0:}$ Similarly to Case $1$ , we can determine that $a$ has to be $9$ . However, the number $9090$ is not divisible by both $4$ and $7$ . Using the same logic in case $1$ , we conclude that Case $2$ cannot be true.

Since we have disproven all of our cases, we know that it is impossible for one of the children to be $mathrm{(B)} 5$ years old.

NOTE: This might look tedious, but it only took me around 30 seconds to do on paper.

Let $k$ be an arbitrary integer. For which $x$ do we have $lfloorlog_{10}4xrfloor = lfloorlog_{10}xrfloor = k$ ?The equation $lfloorlog_{10}xrfloor = k$ can be rewritten as $10^k leq x < 10^{k+1}$ . The second one gives us $10^k leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if and only if $10^k leq x < frac{10^{k+1}}4$ .Hence for each integer $k$ we get an interval of values for which $lfloorlog_{10}4xrfloor - lfloorlog_{10}xrfloor = 0$ . These intervals are obviously pairwise disjoint.For any $kgeq 0$ the corresponding interval is disjoint with $(0,1)$ , so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$ . Hence our answer is the sum of the lengths of the intervals for $k<0$ .For a fixed $k$ the length of the interval $left[ 10^k, frac{10^{k+1}}4 right)$ is $frac 32cdot 10^k$ .This means that our result is $frac 32 left( 10^{-1} + 10^{-2} + cdots right) = frac 32 cdot frac 19 = boxed{frac 16}$ .

Solution 2

The largest value for $x$ is $10^{0}$ . If $x > 10^{-1}$ , then $lfloorlog_{10}4xrfloor$ doesn't fulfill the condition unless $10^{-2} leq x < 0.25 * 10^{-1}$ . The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.

Recognize that this is a geometric sequence. The probability of choosing $x$ such that $lfloorlog_{10}4xrfloor$ and $lfloorlog_{10}xrfloor$ both equal $-1$ is $(9/10)* (15/90) =15/100$ , because there is a 90 percent chance of choosing $x > 10^{-1}$ , and only values of $x$ between $10^{-1}$ and $0.25*10^{0}$ work in this case. Then, for $x$ such that $lfloorlog_{10}4xrfloor$ and $lfloorlog_{10}xrfloor$ both equal $-2$ , you have $(1/10) * ((9/10) *(15/90))$ . This is a geometric series with ratio $1/10$ . Using $a/(1-r)$ for the sum of an infinite geometric sequence, we get $(15/100)/(1-(1/10)) = boxed{frac 16}$ .

$[asy] size(7cm); real l=10, w=7, ang=asin(w/sqrt(l*l+w*w))*180/pi; draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle); draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2))); draw(rotate(ang)*((0,0)--(2l+2w,0)),red); draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red); label("$A$",(-l,w),NW); label("$B$",(-l,-w),SW); label("$C$",(l,-w),SE); label("$D$",(l,w),SE); // Made by chezbgone2 [/asy]$ Let the rectangle have side lengths $l$ and $w$ . Let the axis of the ellipse on which the foci lie have length $2a$ , and let the other axis have length $2b$ . We have $[lw=ab=2006]$ From the definition of an ellipse, $l+w=2aLongrightarrow frac{l+w}{2}=a$ . Also, the diagonal of the rectangle has length $sqrt{l^2+w^2}$ . Comparing the lengths of the axes and the distance from the foci to the center, we have $[a^2=frac{l^2+w^2}{4}+b^2Longrightarrow frac{l^2+2lw+w^2}{4}=frac{l^2+w^2}{4}+b^2Longrightarrow frac{lw}{2}=b^2Longrightarrow b=sqrt{1003}]$ Since $ab=2006$ , we now know $asqrt{1003}=2006Longrightarrow a=2sqrt{1003}$ and because $a=frac{l+w}{2}$ , or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of $boxed{8sqrt{1003}}$ .

Solution 1

The power of $10$ for any factorial is given by the well-known algorithm $[leftlfloor frac n{5}rightrfloor + leftlfloor frac n{25}rightrfloor + leftlfloor frac n{125}rightrfloor + cdots]$ It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$ . As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624cdot 2)=758$ . Applying the algorithm, we see that our answer is $152+152+188= boxed{492}$ .

Solution 2

Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides it. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$ .

Consider $n! = 1cdot 2 cdot dots cdot n$ . Each fifth term is divisible by $5$ , each $25$ -th one by $25$ , and so on. Hence the total power of $5$ that divides $n$ is $leftlfloor frac n{5}rightrfloor + leftlfloor frac n{25}rightrfloor + cdots$ . (For any $n$ only finitely many terms in the sum are non-zero.)

In our case we have $a<2006$ , so the largest power of $5$ that will be less than $a$ is at most $5^4 = 625$ . Therefore the power of $5$ that divides $a!$ is equal to $leftlfloor frac a{5}rightrfloor + leftlfloor frac a{25}rightrfloor + leftlfloor frac a{125}rightrfloor + leftlfloor frac a{625}rightrfloor$ . The same is true for $b$ and $c$ .

Intuition may now try to lure us to split $2006$ into $a+b+c$ as evenly as possible, giving $a=b=669$ and $c=668$ . However, this solution is not optimal.

To see how we can do better, let's rearrange the terms as follows:

$begin{align*} result & = Biglfloor frac a{5}Bigrfloor + Biglfloor frac b{5}Bigrfloor + Biglfloor frac c{5}Bigrfloor & + Biglfloor frac a{25}Bigrfloor + Biglfloor frac b{25}Bigrfloor + Biglfloor frac c{25}Bigrfloor & + Biglfloor frac a{125}Bigrfloor + Biglfloor frac b{125}Bigrfloor + Biglfloor frac c{125}Bigrfloor & + Biglfloor frac a{625}Bigrfloor + Biglfloor frac b{625}Bigrfloor + Biglfloor frac c{625}Bigrfloor end{align*}$

The idea is that the rows of the above equation are roughly equal to $leftlfloor frac n{5}rightrfloor$ , $leftlfloor frac n{25}rightrfloor$ , etc.

More precisely, we can now notice that for any positive integers $a,b,c,k$ we can write $a,b,c$ in the form $a=a_0k + a_1$ , $b=b_0k+b_1$ , $c=c_0k + c_1$ , where all $a_i,b_i,c_i$ are integers and $0leq a_1,b_1,c_1<k$ .

It follows that $[Biglfloor frac a{k}Bigrfloor + Biglfloor frac b{k}Bigrfloor + Biglfloor frac c{k}Bigrfloor = a_0+b_0+c_0]$ and $[Biglfloor frac {a+b+c}kBigrfloor = a_0 + b_0 + c_0 + Biglfloor frac {a_1+b_1+c_1}kBigrfloor leq a_0 + b_0 + c_0 + 2]$

Hence we get that for any positive integers $a,b,c,k$ we have $[Biglfloor frac a{k}Bigrfloor + Biglfloor frac b{k}Bigrfloor + Biglfloor frac c{k}Bigrfloor quad geq quad Biglfloor frac {a+b+c}kBigrfloor - 2]$

Therefore for any $a,b,c$ the result is at least $leftlfloor frac n{5}rightrfloor + leftlfloor frac n{25}rightrfloor + leftlfloor frac n{125}rightrfloor + leftlfloor frac n{625}rightrfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492$ .

If we now show how to pick $a,b,c$ so that we'll get the result $492$ , we will be done.

Consider the row with $625$ in the denominator. We need to achieve sum $1$ in this row, hence we need to make two of the numbers smaller than $625$ . Choosing $a=b=624$ does this, and it will give us the largest possible remainders for $a$ and $b$ in the other three rows, so this is a pretty good candidate. We can compute $c=2006-a-b=758$ and verify that this triple gives the desired result $boxed{492}$ .

$[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("alpha",C,5*dir(80),f); MP("90^circ-alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); [/asy]$
Using the Law of Cosines on $triangle PBC$ , we have: $begin{align*} PB^2&=BC^2+PC^2-2cdot BCcdot PCcdot cos(alpha) Rightarrow 49 = 36 + s^2 - 12scos(alpha) Rightarrow cos(alpha) = dfrac{s^2-13}{12s}. end{align*}$ Using the Law of Cosines on $triangle PAC$ , we have: $begin{align*} PA^2&=AC^2+PC^2-2cdot ACcdot PCcdot cos(90^circ-alpha) Rightarrow 121 = 36 + s^2 - 12ssin(alpha) Rightarrow sin(alpha) = dfrac{s^2-85}{12s}. end{align*}$ Now we use $sin^2(alpha) + cos^2(alpha) = 1$ . $begin{align*} sin^2(alpha)+cos^2(alpha) = 1 &Rightarrow frac{s^4-26s^2+169}{144s^2} + frac{s^4-170s^2+7225}{144s^2} = 1 &Rightarrow 2s^4-340s^2+7394 = 0 &Rightarrow s^4-170s^2+3697 = 0 &Rightarrow s^2 = dfrac{170 pm sqrt{170^2 - 4cdot3697}}{2} &Rightarrow s^2 = dfrac{170 pm sqrt{28900 - 14788}}{2} &Rightarrow s^2 = dfrac{170 pm sqrt{14112}}{2} &Rightarrow s^2 = dfrac{170 pm sqrt{2^5cdot3^2cdot7^2}}{2} &Rightarrow s^2 = dfrac{170 pm 84sqrt{2}}{2} = 85 pm 42sqrt2 end{align*}$ Note that we know that we want the solution with $s^2 > 85$ since we know that $sin(alpha) > 0$ . Thus, $a+b=85+42=boxed{127}$ .

Solution 2

Rotate triangle $PAC$ 90 degrees counterclockwise about $C$ so that the image of $A$ rests on $B$ . Now let the image of $P$ be $P'$ . Note that $P'C=6$ , meaning triangle $PCP'$ is right isosceles, and $angle PP'C=45^circ$ . Then $PP'=6sqrt{2}$ . Now because $PB=7$ and $P'B=11$ , we observe that $angle P'PB=90^circ$ , by the Pythagorean Theorem on $P'PB$ . Now we have that $angle APC=angle BP'C=angle BP'P + angle PP'C$ . So we take the cosine of the second equality, using that fact that $angle PP'C=45^circ$ , to get $cos(BP'C)=frac{6sqrt{2}-7}{11sqrt{2}}$ . Finally, we use the fact that $cos(BP'C)=cos(CPA)$ and use the Law of Cosines on triangle $CPA$ to arrive at the value of $s^2$ .

Or notice that since $angle P'PB=90^circ$ and $angle PP'C=45^circ$ , we have $angle BPC=135^circ$ , and Law of Cosines on triangle $BPC$ gives the value of $s^2$ .

Solution 3 (coordinate bash)

Let point $P$ have coordinates $(x,y)$ and $C$ have coordinates $(0,0).$ Then, $A$ has $(s,0)$ and $B$ has $(0,s)$ .

By distance formula, we have $[x^2+y^2=36 tag{1}.]$ $[x^2+(y-s)^2=49 tag{2}.]$ $[(x-s)^2+y^2=121 tag{3}]$

Expanding $(2)$ and $(3)$ gives $[x^2+y^2-2ys+s^2=49,]$ and $[x^2+y^2-2sx+s^2=121,]$ respectively. Then, using equation $(1)$ , we have that $[-2ys+s^2=49-36=13,]$ and $[-2sx+s^2=121-36=85.]$

Then, solving for $x$ and $y$ gives $[x=frac{85-s^2}{-2s}=frac{s^2-85}{2s},]$ and $[y=frac{13-s^2}{-2s}=frac{s^2-13}{2s}.]$ Plugging these values of $x$ and $y$ into equation $(1)$ yields $[left(frac{s^2-85}{2s}right)^2+left(frac{s^2-13}{2s}right)^2=36.]$

We multiply both sides by $4s^2$ and expand, yielding the equation $[s^4-170s^2+7225+s^4-26s^2+169=144s^2.]$ Simplifying gives the equation $[2s^4-340s^2+7394=0.]$ Solving this quadratic gives $s^2=85pm 42sqrt{2}.$ Now, if this were the actual test, we stop here, noting that the question tells us $a$ and $b$ are positive, so $s^2$ must be $85+42sqrt{2}$ , and our solution is $127$ .

However, here is why $s^2$ cannot be $85-42sqrt{2}$ :

If $s^2=85-42sqrt{2}$ , using $1.4$ as an approximation for $sqrt{2}$ , $s^2approx 85-42cdot 1.4=85-58.8=26.2$ , and $s$ is slightly greater than $5$ . Also note that this implies that $ABapprox 7$ .

Note that at least one of $angle BPA$ , $angle APC$ and $angle BPC$ must be obtuse, since they sum to $360^{circ}$ . Then, note the well known fact that if $angle A$ is the largest angle in $Delta ABC$ , $BC$ must be the largest side. However, combined with the first fact, implies that either $AB$ is the largest side of $Delta ABP$ , $BC$ is the largest side of $Delta BPC$ , or $AC$ is the largest side of $Delta APC$ . By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so $s^2$ cannot equal $85-42sqrt{2}$ , and $s^2=85+42sqrt{2}$ .

The answer is $85+42=boxed{127}$

We start out by solving the equality first.
$begin{align*} sin^2x - sin x sin y + sin^2y &= frac34 sin x &= frac{sin y pm sqrt{sin^2 y - 4 ( sin^2y - frac34 ) }}{2} &= frac{sin y pm sqrt{3 - 3 sin^2 y }}{2} &= frac{sin y pm sqrt{3 cos^2 y }}{2} &= frac12 sin y pm frac{sqrt3}{2} cos y sin x &= sin (y pm frac{pi}{3}) end{align*}$ We end up with three lines that matter: $x = y + fracpi3$ , $x = y - fracpi3$ , and $x = pi - (y + fracpi3) = frac{2pi}{3} - y$ . We plot these lines below.
$[asy] size(5cm); D((0,0)--(3,0)--(3,3)--(0,3)--cycle); D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1)); D((-0.1,1)--(0.1,1)); D((-0.1,2)--(0.1,2)); D((2,0)--(3,1)--(1,3)--(0,2)); MP("frac{pi}{6}", (1,0), plain.S); MP("frac{pi}{3}", (2,0), plain.S); MP("frac{pi}{2}", (3,0), plain.S); MP("frac{pi}{6}", (0,1), plain.W); MP("frac{pi}{3}", (0,2), plain.W); MP("frac{pi}{2}", (0,3), plain.W); [/asy]$
Note that by testing the point $(pi/6,pi/6)$ , we can see that we want the area of the pentagon. We can calculate that by calculating the area of the square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.) $begin{align*} A &= left(frac{pi}{2}right)^2 - 2 cdot frac12 cdot left(frac{pi}{6}right)^2 - frac12 cdot left(frac{pi}{3}right)^2 &= pi^2 left ( frac14 - frac1{36} - frac1{18}right ) &= pi^2 left ( frac{9-1-2}{36} right ) = boxed{text{(C)} frac{pi^2}{6}} end{align*}$

We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.Note that if $d = gcd(999, a_2) neq 1$ , then $d|a_n$ for all $n geq 1$ , and $d$ does not divide $1$ , so if $gcd(999, a_2) neq 1$ , then $(a_n)$ does not complete. (Also, $a_{2006}$ cannot be 1 in this case since $d$ does not divide $1$ , so we do not care about these $a_2$ at all.)From now on, suppose $gcd(999, a_2) = 1$ .We will now show that $(a_n)$ completes at $i$ for some $i leq 2006$ . We will do this with 3 lemmas.Lemma: If $a_j neq a_{j + 1}$ , and neither value is $0$ , then $max(a_j, a_{j + 1}) > max(a_{j + 2}, a_{j + 3})$ .Proof: There are 2 cases to consider.If $a_j > a_{j + 1}$ , then $a_{j + 2} = a_j - a_{j + 1}$ , and $a_{j + 3} = |a_j - 2a_{j + 1}|$ . So $a_j > a_{j + 2}$ and $a_j > a_{j + 3}$ .

If $a_j < a_{j + 1}$ , then $a_{j + 2} = a_{j + 1} - a_j$ , and $a_{j + 3} = a_j$ . So $a_{j + 1} > a_{j + 2}$ and $a_{j + 1} > a_{j + 3}$ .

In both cases, $max(a_j, a_{j + 1}) > max(a_{j + 2}, a_{j + 3})$ , as desired.

Lemma: If $a_i = a_{i + 1}$ , then $a_i = 1$ . Moreover, if instead we have $a_i = 0$ for some $i > 2$ , then $a_{i - 1} = a_{i - 2} = 1$ .

Proof: By the way $(a_n)$ is constructed in the problem statement, having two equal consecutive terms $a_i = a_{i + 1}$ implies that $a_i$ divides every term in the sequence. So $a_i | 999$ and $a_i | a_2$ , so $a_i | gcd(999, a_2) = 1$ , so $a_i = 1$ . For the proof of the second result, note that if $a_i = 0$ , then $a_{i - 1} = a_{i - 2}$ , so by the first result we just proved, $a_{i - 2} = a_{i - 1} = 1$ .

Lemma: $(a_n)$ completes at $i$ for some $i leq 2000$ .

Proof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma and the fact that neither $999$ nor $a_2$ are $0$ , none of the pairs $(a_1, a_2), ..., (a_{1999}, a_{2000})$ can have a $0$ or be equal to $(1, 1)$ . So the first lemma implies $[max(a_1, a_2) > max(a_3, a_4) > cdots > max(a_{1999}, a_{2000}) > 0,]$ so $999 = max(a_1, a_2) geq 1000$ , a contradiction. Hence $(a_n)$ completes at $i$ for some $i leq 2000$ .

Now we're ready to find exactly which values of $a_2$ we want to count.

Let's keep in mind that $2006 equiv 2 pmod 3$ and that $a_1 = 999$ is odd. We have two cases to consider.

Case 1: If $a_2$ is odd, then $a_3$ is even, so $a_4$ is odd, so $a_5$ is odd, so $a_6$ is even, and this pattern must repeat every three terms because of the recursive definition of $(a_n)$ , so the terms of $(a_n)$ reduced modulo 2 are $[1, 1, 0, 1, 1, 0, ...,]$ so $a_{2006}$ is odd and hence $1$ (since if $(a_n)$ completes at $i$ , then $a_k$ must be $0$ or $1$ for all $k geq i$ ).

Case 2: If $a_2$ is even, then $a_3$ is odd, so $a_4$ is odd, so $a_5$ is even, so $a_6$ is odd, and this pattern must repeat every three terms, so the terms of $(a_n)$ reduced modulo 2 are $[1, 0, 1, 1, 0, 1, ...,]$ so $a_{2006}$ is even, and hence $0$ .

We have found that $a_{2006} = 1$ is true precisely when $gcd(999, a_2) = 1$ and $a_2$ is odd. This tells us what we need to count.

There are $phi(999) = 648$ numbers less than $999$ and relatively prime to it ( $phi$ is the Euler totient function). We want to count how many of these are even. Note that $[t mapsto 999 - t]$ is a 1-1 correspondence between the odd and even numbers less than and relatively prime to $999$ . So our final answer is $648/2 = 324$ , or $boxed{text{B}}$ .