AMC10 2005 A

(D) Since Mike tipped \$2, which was $10\% = \frac{1}{10}$ of his bill, his bill must have been $2 \cdot 10 = 20$ dollars. Similarly, Joe tipped \$2, which was $20\% = \frac{1}{5}$ of his bill, so his bill must have been $2 \cdot 5 = 10$ dollars. The difference between their bills is therefore \$10.

2. (C) First we have $(1 \star 2)=\dfrac{1+2}{1-2}=-3.$ Then $((1 \star 2)\star 3)=(-3 \star 3)=\dfrac{-3+3}{-3-3}=\dfrac{0}{-6}=0.$

Since $2x + 7 = 3$ we have $x = -2$. Hence $-2 = b(-2) -10$, so $2b = -8$, and $b = -4$.

Let $w$ be the width of the rectangle. Then the length is $2w$, and $x^2 = w^2 + (2w)^2 = 5w^2$. The area is consequently $w(2w) = 2w^2 = \frac{2}{5}x^2$.

(A) If Dave buys seven windows separately he will purchase six and receive one free, for a cost of \$600. If Doug buys eight windows separately, he will purchase seven and receive one free, for a total cost of \$700. The total cost to Dave and Doug purchasing separately will be \$1300. If they purchase fifteen windows together, they will need to purchase only 12 windows, for a cost of \$1200, and will receive 3 free. This will result in a savings of \$100.

The sum of the 50 numbers is $20 \cdot 30 + 30 \cdot 20 = 1200$. Their average is $1200/50 = 24$.

Because (rate)(time) = (distance), the distance Josh rode was $(4/5)(2) = 8/5$ of the distance that Mike rode. Let $m$ be the number of miles that Mike had ridden when they met. Then the number of miles between their houses is $13 = m + (8/5)m = (13/5)m$. Thus $m = 5$.

(C) The symmetry of the figure implies that $\triangle ABH$, $\triangle BCE$, $\triangle CDF$, and $\triangle DAG$ are congruent right triangles. So $$BH=CE=\sqrt{BC^2-BE^2}=\sqrt{50-1}=7,$$ and $EH=BH-BE=7-1=6$. Hence the square $EFGH$ has area $6^2=36$.

There are three X's and two O's, and the tiles are selected without replacement, so the probability is $\frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{10}$.

(A) The quadratic formula yields $$x=\frac{-(a+8)\pm\sqrt{(a+8)^2-4\cdot4\cdot9}}{2\cdot4}.$$ The equation has only one solution precisely when the value of the discriminant, $$(a+8)^2-144,$$ is 0. This implies that $a=-20$ or $a=4$, and the sum is $-16$. OR The equation has one solution if and only if the polynomial is the square of a binomial with linear term $\pm\sqrt{4x^2}=\pm2x$ and constant term $\pm\sqrt{9}=\pm3$. Because $(2x\pm3)^2$ has a linear term $\pm12x$, it follows that $a+8=\pm12$. Thus $a$ is either $-20$ or $4$, and the sum of those values is $-16$.

(B) The unit cubes have a total of $6n^3$ faces, of which $6n^2$ are red. Therefore $$ \frac{1}{4}=\frac{6n^2}{6n^3}=\frac{1}{n}, $$ so $n=4$.

(B) The trefoil is constructed of four equilateral triangles and four circular segments, as shown. These can be combined to form four $60^\circ$ circular sectors. Since the radius of the circle is $1$, the area of the trefoil is $$ \frac{4}{6}\left(\pi\cdot 1^2\right)=\frac{2}{3}\pi. $$

(E) The condition is equivalent to $130n>n^2>2^4=16,$ so $130n>n^2$ and $n^2>16.$ This implies that $130>n>4.$ So $n$ can be any of the 125 integers strictly between 130 and 4.

The first and last digits must be both odd or both even for their average to be an integer. There are $5 \cdot 5 = 25$ odd-odd combinations for the first and last digits. There are $4 \cdot 5 = 20$ even-even combinations that do not use zero as the first digit. Hence the total is 45.

(E) Written as a product of primes, we have $3!\cdot 5!\cdot 7!=2^8\cdot 3^4\cdot 5^2\cdot 7.$ A cube that is a factor has a prime factorization of the form $2^p\cdot 3^q\cdot 5^r\cdot 7^s$, where $p$, $q$, $r$, and $s$ are all multiples of $3$. There are $3$ possible values for $p$, which are $0$, $3$, and $6$. There are $2$ possible values for $q$, which are $0$ and $3$. The only value for $r$ and for $s$ is $0$. Hence there are $6=3\cdot 2\cdot 1\cdot 1$ distinct cubes that divide $3!\cdot 5!\cdot 7!$. They are $1=2^0 3^0 5^0 7^0,\quad 8=2^3 3^0 5^0 7^0,\quad 27=2^0 3^3 5^0 7^0,$ $64=2^6 3^0 5^0 7^0,\quad 216=2^3 3^3 5^0 7^0,\quad \text{and}\quad 1728=2^6 3^3 5^0 7^0.$

Let $10a+b$ be the two-digit number. When $a+b$ is subtracted the result is $9a$. The only two-digit multiple of 9 that ends in 6 is $36$, so $a=4$. The ten numbers between 40 and 49 have this property.

(D) Each number appears in two sums, so the sum of the sequence is $2(3+5+6+7+9)=60.$ The middle term of a five-term arithmetic sequence is the mean of its terms, so $60/5=12$ is the middle term. The figure shows an arrangement of the five numbers that meets the requirement.

(A) There are four possible outcomes, ABAA, ABABA, ABBAA, and BBAAA, but they are not equally likely. This is because, in general, the probability of any specific four-game series is $(1/2)^4 = 1/16$, whereas the probability of any specific five-game series is $(1/2)^5 = 1/32$. Thus the first listed outcome is twice as likely as each of the other three. Let $p$ be the probability of the occurrence $ABBAA$. Then the probability of $ABABA$ is also $p$, as is the probability of $BBAAA$, whereas the probability of $ABAA$ is $2p$. So $$2p + p + p + p = 1,\quad \text{and}\quad p = \frac{1}{5}.$$ The only outcome in which team B wins the first game is $BBAAA$, so the probability of this occurring is $1/5$.

(D) Consider the rotated middle square shown in the figure. It will drop until length \(DE\) is 1 inch. Thus \(FC = DF = FE = \frac{1}{2}\) and \(BC = \sqrt{2}\). Hence \(BF = \sqrt{2} - 1/2\). This is added to the 1 inch height of the supporting squares, so the overall height of point \(B\) above the line is \(1 + BF = \sqrt{2} + \frac{1}{2}\) inches.

(A) The octagon can be partitioned into five squares and four half squares, each with side length $\frac{\sqrt{2}}{2}$, so its area is $$\left(5+4\cdot\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)^2=\frac{7}{2}.$$

(B) Because $1+2+\cdots+n=\dfrac{n(n+1)}{2}$, $1+2+\cdots+n$ divides the positive integer $6n$ if and only if $\dfrac{6n}{n(n+1)/2}=\dfrac{12}{n+1}$ is an integer. There are five such positive values of $n$, namely, $1,2,3,5,$ and $11$.

(D) The sets $S$ and $T$ consist, respectively, of the positive multiples of $4$ that do not exceed $2005\cdot 4=8020$ and the positive multiples of $6$ that do not exceed $2005\cdot 6=12,030$. Thus $S\cap T$, the set of numbers that are common to $S$ and to $T$, consists of the positive multiples of $12$ that do not exceed $8020$. Let $[x]$ represent the largest integer that is less than or equal to $x$. Then the number of elements in the set $S\cap T$ is $$\left[\frac{8020}{12}\right]=\left[668+\frac{1}{3}\right]=668.$$

(C) Let $O$ be the center of the circle. Each of $\triangle DCE$ and $\triangle ABD$ has a diameter of the circle as a side. Thus the ratio of their areas is the ratio of the two altitudes to the diameters. These altitudes are $DC$ and the altitude from $C$ to $DO$ in $\triangle DCE$. Let $F$ be the foot of this second altitude. Since $\triangle CFO$ is similar to $\triangle DCO$, $$ \frac{CF}{DC}=\frac{CO}{DO}=\frac{AO-AC}{DO}=\frac{\frac{1}{2}AB-\frac{1}{3}AB}{\frac{1}{2}AB}=\frac{1}{3}, $$ which is the desired ratio.

(B) The conditions imply that both $n$ and $n+48$ are squares of primes. So for each successful value of $n$ we have primes $p$ and $q$ with $p^2=n+48$ and $q^2=n$, and $48=p^2-q^2=(p+q)(p-q).$ The pairs of factors of $48$ are $48$ and $1,\quad 24$ and $2,\quad 16$ and $3,\quad 12$ and $4,\quad \text{and}\ 8\ \text{and}\ 6.$ These give pairs $(p,q)$, respectively, of $\left(\frac{49}{2},\frac{47}{2}\right),\ (13,11),\ \left(\frac{19}{2},\frac{13}{2}\right),\ (8,4),\ \text{and}\ (7,1).$ Only $(p,q)=(13,11)$ gives prime values for $p$ and for $q$, with $n=11^2=121$ and $n+48=13^2=169.$

(D) We have $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(ABE)}=\dfrac{AD}{AB}=\dfrac{19}{25}$ and $\dfrac{\mathrm{Area}(ABE)}{\mathrm{Area}(ABC)}=\dfrac{AE}{AC}=\dfrac{14}{42}=\dfrac{1}{3},$ so $\dfrac{\mathrm{Area}(ABC)}{\mathrm{Area}(ADE)}=\dfrac{25}{19}\cdot\dfrac{3}{1}=\dfrac{75}{19},$ and $\dfrac{\mathrm{Area}(BCED)}{\mathrm{Area}(ADE)}=\dfrac{\mathrm{Area}(ABC)-\mathrm{Area}(ADE)}{\mathrm{Area}(ADE)}=\dfrac{75}{19}-1=\dfrac{56}{19}.$ Thus $\dfrac{\mathrm{Area}(ADE)}{\mathrm{Area}(BCED)}=\dfrac{19}{56}.$