AMC12 2002 B

We wish to find $\frac{9+99+\cdots +999999999}{9}$, or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$. This doesn't have the digit 0, so the answer is $\boxed{\mathrm{(A)}\ 0}$

By the distributive property, \[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}\].

Factoring, we get $n^2 - 3n + 2 = (n-2)(n-1)$. Either $n-1$ or $n-2$ is odd, and the other is even. Their product must yield an even number. The only prime that is even is $2$, which is when $n$ is $3$ or $0$. Since $0$ is not a positive number, the answer is $\boxed{\mathrm{(B)}\ \text{one}}$.

Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$, $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$ From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$. The only answer choice that is not true is $\boxed{\mathrm{(E)}\ n>84}$.

The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into $5- 2 = 3$ triangles) is $3 \cdot 180 = 540^{\circ}$. If we let $v = x - 2d, w = x - d, y = x + d, z = x+2d$, it follows that \[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}\] Note that since $x$ is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence. You can always assume the values are the same so $\frac{540}{5}=108$ Let $v$, $w$, $x$, $y$, $z$ be $v$, $v + d$, $v+2d$, $v+3d$, $v+4d$, respectively. Then we have \[v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}\] Dividing the equation by $5$, we have \[v + 2d = x = 108^{\circ} \mathrm {(D)}\]

Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$, it follows by comparing coefficients that $-a - b = a$ and that $ab = b$. Since $b$ is nonzero, $a = 1$, and $-1 - b = 1 \Longrightarrow b = -2$. Thus $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$.

Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. Since the mean is $a$, the sum of the integers is $3a$. So $8$ times the sum is just $24a$. With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$. $24=4\times6$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}$.

If there are five Mondays, there are only three possibilities for their dates: $(1,8,15,22,29)$, $(2,9,16,23,30)$, and $(3,10,17,24,31)$. In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August. In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August. In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August. The only day of the week that is guaranteed to appear five times is therefore $\boxed{\textrm{(D)}\ \text{Thursday}}$.

We can let $a=1$, $b=2$, $c=3$, and $d=4$. $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$ As $a, b, d$ is a geometric sequence, let $b=ka$ and $d=k^2a$ for some $k>0$. Now, $a, b, c, d$ is an arithmetic sequence. Its difference is $b-a=(k-1)a$. Thus $d=a + 3(k-1)a = (3k-2)a$. Comparing the two expressions for $d$ we get $k^2=3k-2$. The positive solution is $k=2$, and $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$. Letting $n$ be the common difference of the arithmetic progression, we have $b = a + n$, $c = a + 2n$, $d = a + 3n$. We are given that $b / a$ = $d / b$, or \[\frac{a + n}{a} = \frac{a + 3n}{a + n}.\] Cross-multiplying, we get \[a^2 + 2an + n^2 = a^2 + 3an\] \[n^2 = an\] \[n = a\] So $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.

Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\{-3, -2, -1, 0, 1, 2, 3\}$. It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\boxed{\mathrm{(A) } 13}$ integers.

Since $A-B$ and $A+B$ must have the same parity, and since there is only one even prime number, it follows that $A-B$ and $A+B$ are both odd. Thus one of $A, B$ is odd and the other even. Since $A+B > A > A-B > 2$, it follows that $A$ (as a prime greater than $2$) is odd. Thus $B = 2$, and $A-2, A, A+2$ are consecutive odd primes. At least one of $A-2, A, A+2$ is divisible by $3$, from which it follows that $A-2 = 3$ and $A = 5$. The sum of these numbers is thus $17$, a prime, so the answer is $\boxed{\mathrm{(E)}\ \text{prime}}$.

Let $x^2 = \frac{n}{20-n}$, with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$. Since $\text{gcd}(x^2, x^2 + 1) = 1$, it follows that $x^2 + 1$ divides $20$. Listing the factors of $20$, we find that $x = 0, 1, 2 , 3$ are the only $\boxed{\mathrm{(D)}\ 4}$ solutions (respectively yielding $n = 0, 10, 16, 18$).

Let $a, a+1, \ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + \frac{17(18)}{2} = 9(2a+17)$, is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$, and the sum is $225 \Rightarrow \mathrm{(B)}$. Notice that all five choices given are perfect squares. Let $a$ be the smallest number, we have \[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\] Subtract $153$ from each of the choices and then check its divisibility by $18$, we have $225$ as the smallest possible sum. $\mathrm {(B)}$

For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$. We can construct such a situation as below, so the answer is $\boxed{\mathrm{(D)}\ 12}$. Error creating thumbnail: Unable to save thumbnail to destination

Let $N = \overline{abcd} = 1000a + \overline{bcd}$, such that $\frac{N}{9} = \overline{bcd}$. Then $1000a + \overline{bcd} = 9\overline{bcd} \Longrightarrow 125a = \overline{bcd}$. Since $100 \le \overline{bcd} < 1000$, from $a = 1, \ldots, 7$ we have $7$ three-digit solutions, and the answer is $\mathrm{(D)}$.

On both dice, only the faces with the numbers $3,6$ are divisible by $3$. Let $P(a) = \frac{2}{8} = \frac{1}{4}$ be the probability that Juan rolls a $3$ or a $6$, and $P(b) = \frac{2}{6} = \frac 13$ that Amal does. By the Principle of Inclusion-Exclusion, \[P(a \cup b) = P(a) + P(b) - P(a \cap b) = \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2} \Rightarrow \mathrm{(C)}\] Alternatively, the probability that Juan rolls a multiple of $3$ is $\frac{1}{4}$, and the probability that Juan does not roll a multiple of $3$ but Amal does is $\left(1 - \frac{1}{4}\right) \cdot \frac{1}{3} = \frac{1}{4}$. Thus the total probability is $\frac 14 + \frac 14 = \frac 12$. The probability that neither Juan nor Amal rolls a multiple of $3$ is $\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}$; using complementary counting, the probability that at least one does is $1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}$. The product will be a multiple of 3 if and only if at least one of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is $2/8 = 1/4$. The probability that Juan does not roll 3 or 6, but Amal does is $(3/4) (1/3) = 1/4$. Thus, the probability that the product of the rolls is a multiple of 3 is \[\frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}.\]

We say Andy's lawn has an area of $x$. Beth's lawn thus has an area of $\frac{x}{2}$, and Carlos's lawn has an area of $\frac{x}{3}$. We say Andy's lawn mower cuts at a speed of $y$. Carlos's cuts at a speed of $\frac{y}{3}$, and Beth's cuts at a speed $\frac{2y}{3}$. Each person's lawn is cut at a time of $\frac{\text{area}}{\text{rate}}$, so Andy's is cut in $\frac{x}{y}$ time, as is Carlos's. Beth's is cut in $\frac{3}{4}\times\frac{x}{y}$, so the first one to finish is $\boxed{\mathrm{(B)}\ \text{Beth}}$.

Assume that the point $P$ is randomly chosen within the rectangle with vertices $(0,0)$, $(3,0)$, $(3,1)$, $(0,1)$. In this case, the region for $P$ to be closer to the origin than to point $(3,1)$ occupies exactly $\frac{1}{2}$ of the area of the rectangle, or $1.5$ square units. If $P$ is chosen within the square with vertices $(2,0)$, $(3,0)$, $(3,1)$, $(2,1)$ which has area $1$ square unit, it is for sure closer to $(3,1)$. Now if $P$ can only be chosen within the rectangle with vertices $(0,0)$, $(2,0)$, $(2,1)$, $(0,1)$, then the square region is removed and the area for $P$ to be closer to $(3,1)$ is then decreased by $1$ square unit, left with only $0.5$ square unit. Thus the probability that $P$ is closer to $(3.1)$ is $\frac{0.5}{2}=\frac{1}{4}$ and that of $P$ is closer to the origin is $1-\frac{1}{4}=\frac{3}{4}$. $\mathrm{(C)}$ First, join points $(0,0)$ and $(3,1)$. This line $l_1$ has equation $y = \frac{1}{3}x$.

Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242$. Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$. Taking their product, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720} \Rightarrow \mathrm{(D)}$.

Error creating thumbnail: Unable to save thumbnail to destination Let $OM = x$, $ON = y$. By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*} Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$. By the Pythagorean Theorem again, we have \[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}.\] Alternatively, we could note that since we found $x^2 + y^2 = 169$, segment $MN=13$. Right triangles $\triangle MON$ and $\triangle XOY$ are similar by Leg-Leg with a ratio of $\frac{1}{2}$, so $XY=2(MN)=\boxed{\mathrm{(B)}\ 26}$.

Since $2002 = 11 \cdot 13 \cdot 14$, it follows that \begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array} \right. \end{eqnarray*} Thus $\sum_{n=1}^{2001} a_n = 11 \cdot 10 + 13 \cdot 12 + 14 \cdot 13 = 448 \Rightarrow \mathrm{(A)}$. $\begin{array}{lr} 11, & \text{if\ }n=13 \cdot 14 \cdot k, \quad k = 1,2,\cdots 10;\\ 13, & \text{if\ }n=14 \cdot 11 \cdot k, \quad k = 1,2,\cdots 12;\\ 14, & \text{if\ }n=11 \cdot 13 \cdot k, \quad k = 1,2,\cdots 13;\\ \end{array}$.

By the change of base formula, $a_n = \frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n$. Thus \begin{align*}b- c &= \left(\frac{1}{\log 2002}\right)(\log 2 + \log 3 + \log 4 + \log 5 - \log 10 - \log 11 - \log 12 - \log 13 - \log 14)\\ &= \left(\frac{1}{\log 2002}\right)\left(\log \frac{2 \cdot 3 \cdot 4 \cdot 5}{10 \cdot 11 \cdot 12 \cdot 13 \cdot 14}\right)\\ &= \left(\frac{1}{\log 2002}\right) \log 2002^{-1} = -\left(\frac{\log 2002}{\log 2002}\right) = -1 \Rightarrow \mathrm{(B)}\end{align*}

Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ extended past $B$. Let $AD = x$ and $BD = y$. Using the Pythagorean Theorem, we obtain the equations \begin{align*} x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*} Subtracting $(1)$ equation from $(2)$ and $(3)$, we get \begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*} Then, subtracting $2 \times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$, so $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$ Error creating thumbnail: Unable to save thumbnail to destination

We have \[[ABCD] = 2002 \le \frac 12 (AC \cdot BD)\] (This is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$), with equality only if $\overline{AC} \perp \overline{BD}$. By the triangle inequality, \begin{align*}AC &\le PA + PC = 52\\ BD &\le PB + PD = 77\end{align*} with equality if $P$ lies on $\overline{AC}$ and $\overline{BD}$ respectively. Thus \[2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002\] Since we have the equality case, $\overline{AC} \perp \overline{BD}$ at point $P$, as shown below. By the Pythagorean Theorem, \begin{align*} AB = \sqrt{PA^2 + PB^2} & = \sqrt{24^2 + 32^2} = 40\\ BC = \sqrt{PB^2 + PC^2} & = \sqrt{32^2 + 28^2} = 4\sqrt{113}\\ CD = \sqrt{PC^2 + PD^2} & = \sqrt{28^2 + 45^2} = 53\\ DA = \sqrt{PD^2 + PA^2} & = \sqrt{45^2 + 24^2} = 51 \end{align*} The perimeter of $ABCD$ is $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$.

The first condition gives us that \[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\] which is a circle centered at $(-3,-3)$ with radius $4$. The second condition gives us that \[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\] Thus either \[x - y \ge 0,\quad x+y+6 \le 0\] or \[x - y \le 0,\quad x+y+6 \ge 0\] Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$, as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{\textbf{(E) }25}$.