AMC12 2003 A

The first $2003$ even counting numbers are $2,4,6,...,4006$. The first $2003$ odd counting numbers are $1,3,5,...,4005$. Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$. $(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$ $= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$ The answer is 2003

Since T-shirts cost $5$ dollars more than a pair of socks, T-shirts cost $5+4=9$ dollars. Since each member needs $2$ pairs of socks and $2$ T-shirts, the total cost for $1$ member is $2(4+9)=26$ dollars. Since $2366$ dollars was the cost for the club, and $26$ was the cost per member, the number of members in the League is $2366\div 26=\boxed{\mathrm{(B)}\ 91}$.

The volume of the original box is $15\cdot10\cdot8=1200.$ The volume of each cube that is removed is $3\cdot3\cdot3=27.$ Since there are $8$ corners on the box, $8$ cubes are removed. So the total volume removed is $8\cdot27=216$. Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}.$

Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km. Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours. Therefore her average speed in km/hr is $\frac{2}{\left(\frac{2}{3}\right)}=\boxed{\mathrm{(A)}\ 3}$.

$AMC10+AMC12=123422$ $AMC00+AMC00=123400$ $AMC+AMC=1234$ $2\cdot AMC=1234$ $AMC=\frac{1234}{2}=617$ Since $A$, $M$, and $C$ are digits, $A=6$, $M=1$, $C=7$. Therefore, $A+M+C = 6+1+7 = \boxed{\mathrm{(E)}\ 14}$.

We start by looking at the answers. Examining statement C, we notice: $x \heartsuit 0 = |x-0| = |x|$ $|x| eq x$ when $x<0$, but statement C says that it does for all $x$. Therefore the statement that is not true is $\boxed{\mathrm{(C)}\ x\heartsuit 0=x\ \text{for all}\ x}$

By the triangle inequality, no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$. Since all sides must be integers, the largest possible length of a side is $3$. Therefore, all such triangles must have all sides of length $1$, $2$, or $3$. Since $2+2+2=6<7$, at least one side must have a length of $3$. Thus, the remaining two sides have a combined length of $7-3=4$. So, the remaining sides must be either $3$ and $1$ or $2$ and $2$. Therefore, the number of triangles is $\boxed{\mathrm{(B)}\ 2}$.

For any positive integer $n$ which is not a perfect square, exactly half of its positive factors will be less than $\sqrt{n}$, since each such factor can be paired with one that is larger than $\sqrt{n}$. (By contrast, if $n$ is a perfect square, one of its factors will be exactly $\sqrt{n}$, which would therefore have to be paired with itself.) Since $60$ is indeed not a perfect square, it follows that half of its positive factors are less than $\sqrt{60} \approx 7.746$. This estimate clearly shows that there are not even any integers, let alone factors of $60$, between $7$ and $\sqrt{60}$. Accordingly, exactly half of the positive factors of $60$ are in fact less than $7$, so the answer is precisely $\boxed{\mathrm{(E)}\ \frac{1}{2}}$. Testing all positive integers less than $7$, we find that $1$, $2$, $3$, $4$, $5$, and $6$ all divide $60$. The prime factorization of $60$ is $2^2 \cdot 3 \cdot 5$, so using the standard formula for the number of divisors, the total number of divisors of $60$ is $3 \cdot 2 \cdot 2 = 12$. Therefore, the required probability is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$. Though this is not recommended for reasons of time, one can simply write out all the factors of $60$, eventually finding that \[60 = 1 \cdot 60 = 2 \cdot 30 = 3 \cdot 20 = 4 \cdot 15 = 5 \cdot 12 = 6 \cdot 10.\] Hence $60$ has $12$ factors, of which $6$ are less than $7$ (namely, $1$, $2$, $3$, $4$, $5$, and $6$), so the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.

If $(2,3)$ is in $S$, then its reflection in the line $y = x$, i.e. $(3,2)$, is also in $S$. Now by reflecting these points in both coordinate axes, we quickly deduce that every point of the form $(\pm 2, \pm 3)$ or $(\pm 3, \pm 2)$ must be in $S$. Moreover, by drawing out this set of $8$ points, we observe that it satisfies all of the required symmetry conditions, so no more points need to be added to $S$. Accordingly, the smallest possible number of points in $S$ is precisely $\boxed{\mathrm{(D)}\ 8}$.

Because the ratios are $3:2:1$, Al, Bert, and Carl believe that they need to take $1/2$, $1/3$, and $1/6$ of the pile when they each arrive, respectively. After each person comes, $1/2$, $2/3$, and $5/6$ of the pile's size (just before each came) remains. The pile starts at $1$, and at the end $\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6}\cdot 1 = \frac{5}{18}$ of the original pile goes unclaimed. (Note that because of the properties of multiplication, it does not matter what order the three come in.) Hence the answer is $\boxed{\mathrm{(D)}\ \dfrac{5}{18}}$.

Suppose that the common perimeter is $P$. Then the side lengths of the square and the triangle are $\frac{P}{4}$ and $\frac{P}{3}$, respectively. The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is therefore $\frac{P}{4} \cdot \sqrt{2} = \frac{P\sqrt{2}}{4}$, and so its radius is $\frac{\left(\frac{P\sqrt{2}}{4}\right)}{2} = \frac{P\sqrt{2}}{8}$. Hence the area of the circle is \[A = \pi\left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64} = \frac{\pi P^2}{32}.\] Now consider the circle circumscribed around the equilateral triangle. By symmetry, its center must be the same as that of the triangle, so its radius is simply the distance from the center of the triangle to a vertex. Recalling that the centroid of any triangle divides its medians in the ratio $2:1$, and that the medians of an equilateral triangle are the same as its altitudes, we deduce that the radius is $\frac{2}{3}$ of the total length of an altitude. Since the side length of this triangle is $\frac{P}{3}$, the length of an altitude is $\frac{P}{3}\sin\left(60^{\circ}\right) = \frac{P}{3} \cdot \frac{\sqrt{3}}{2} = \frac{P\sqrt{3}}{6}$, so finally the radius is \[\frac{2}{3} \cdot \frac{P\sqrt{3}}{6} = \frac{P\sqrt{3}}{9},\] and thus the area of this circle is \[B = \pi\left(\frac{P\sqrt{3}}{9}\right)^2 = \pi \cdot \frac{3P^2}{81} = \frac{\pi P^2}{27}.\] This gives \[\frac{A}{B}=\frac{\left(\frac{\pi P^2}{32}\right)}{\left(\frac{\pi P^2}{27}\right)} = \frac{\pi P^2}{32} \cdot \frac{27}{\pi P^2}= \boxed{\mathrm{(C) \ } \frac{27}{32}}.\]

Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$, respectively. $B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it. $R_1$ is the only other red card that evenly divides $B_5$, so $R_1$ must be the other card next to $B_5$. $B_4$ is the only blue card that $R_4$ evenly divides, so $R_4$ must be at one end of the stack and $B_4$ must be the card next to it. $R_2$ is the only other red card that evenly divides $B_4$, so $R_2$ must be the other card next to $B_4$. $R_2$ doesn't evenly divide $B_3$, so $B_3$ must be next to $R_1$, $B_6$ must be next to $R_2$, and $R_3$ must be in the middle. This yields the following arrangement from top to bottom: $\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}$ Therefore, the sum of the numbers on the middle three cards is $3+3+6=\boxed{\mathrm{(E)}\ 12}$.

Let the squares be labeled $A$, $B$, $C$, and $D$. When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$, and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$. So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing. Therefore, squares $1$, $2$, and $3$ will prevent the polygon from becoming a cube with one face missing. Squares $4$, $5$, $6$, $7$, $8$, and $9$ will allow the polygon to become a cube with one face missing when folded. Thus the answer is $\boxed{\mathrm{(E)}\ 6}$. Another way to think of it is that a cube missing one face has $5$ of its $6$ faces. Since the shape has $4$ faces already, we need another face. The only way to add another face is if the added square does not overlap any of the others. $1$,$2$, and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{\mathrm{(E)}\ 6}$ If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\boxed{\mathrm{(E)}\ 6}$. ~Sophia866

Since the area of square $\text{ABCD}$ is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of $\text{AKB}$, and thus $\text{DMC}$, is $2\sqrt{3}$. The diagonal of the square $\text{KNML}$ will then be $4+4\sqrt{3}$. From here there are 2 ways to proceed: First: Since the diagonal is $4+4\sqrt{3}$, the side length is $\frac{4+4\sqrt{3}}{\sqrt{2}}$, and the area is thus $\frac{16+48+32\sqrt{3}}{2}=\boxed{\textbf{(D) } 32+16\sqrt{3}}$. Second: Since a square is a rhombus, the area of the square is $\frac{d_1d_2}{2}$, where $d_1$ and $d_2$ are the diagonals of the rhombus. Since the diagonal is $4+4\sqrt{3}$, the area is $\frac{(4+4\sqrt{3})^2}{2}=\boxed{\textbf{(D) } 32+16\sqrt{3}}$. Because $ABCD$ has area $16$, its side length is simply $\sqrt{16}\implies 4$. Angle chasing, we find that the angle of $KBL=360-(90+2(60))=360-(210)=150$. We also know that $KB=4, BL=4$. Using Law of Cosines, we find that side $(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{3}$. However, the area of $KLMN$ is simply $(KL)^2$, hence the answer is $\boxed{\textbf{(D) } 32+16\sqrt{3}}$ First we show that $KLMN$ is a square. How? Show that it is a rhombus that has a right angle. $\angle{NAK}=\angle{KBL}=\angle{LCM}=\angle{DNM}=150^\circ$. All the sides of the equilateral triangles are equal, so the triangles are congruent. Notice that $\angle{KNL}=45^\circ$, etc, so $\angle{KNM}=90^\circ$. So we have a square. Instead of going to find $KN$ using Law of Cosines, we can inscribe the square in another bigger one and then subtract the four right triangles in the corners, so after all of this, we find that the answer is $\boxed{\textbf{(D) } 32+16\sqrt{3}}$

The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$. The area of the smaller semicircle is $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$. Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^\circ$. The area of the $60^\circ$ sector of the larger semicircle is $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$. The area of the triangle is $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$. So the shaded area is $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem.

After we pick point $P$, we realize that $ABC$ is symmetric for this purpose, and so the probability that $ACP$ is the greatest area, or $ABP$ or $BCP$, are all the same. Since they add to $1$, the probability that $ABP$ has the greatest area is $\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$ We will use geometric probability. Let us take point $P$, and draw the perpendiculars to $BC$, $CA$, and $AB$, and call the feet of these perpendiculars $D$, $E$, and $F$ respectively. The area of $\triangle ACP$ is simply $\frac{1}{2} * AC * PF$. Similarly we can find the area of triangles $BCP$ and $ABP$. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose $P, PD + PE + PF$ = the height of the triangle. Setting the area of triangle $ABP$ greater than $ACP$ and $BCP$, we want $PF$ to be the largest of $PF$, $PD$, and $PE$. We then realize that $PF = PD = PE$ when $P$ is the incenter of $\triangle ABC$. Let us call the incenter of the triangle $Q$. If we want $PF$ to be the largest of the three, by testing points we realize that $P$ must be in the interior of quadrilateral $QDCE$. So our probability (using geometric probability) is the area of $QDCE$ divided by the area of $ABC$. We will now show that the three quadrilaterals, $QDCE$, $QEAF$, and $QFBD$ are congruent. As the definition of point $Q$ yields, $QF$ = $QD$ = $QE$. Since $ABC$ is equilateral, $Q$ is also the circumcenter of $\triangle ABC$, so $QA = QB = QC$. By the Pythagorean Theorem, $BD = DC = CE = EA = AF = FB$. Also, angles $BDQ, BFQ, CEQ, CDQ, AFQ$, and $AEQ$ are all equal to $90^\circ$. Angles $DBF, FAE, ECD$ are all equal to $60$ degrees, so it is now clear that quadrilaterals $QDCE, QEAF, QFBD$ are all congruent. Summing up these areas gives us the area of $\triangle ABC$. $QDCE$ contributes to a third of that area so $\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$.

$APMD$ obviously forms a kite. Let the intersection of the diagonals be $E$. $AE+EM=AM=2\sqrt{5}$ Let $AE=x$. Then, $EM=2\sqrt{5}-x$. By Pythagorean Theorem, $DE^2=4^2-AE^2=2^2-EM^2$. Thus, $16-x^2=4-(2\sqrt{5}-x)^2$. Simplifying, $x=\frac{8}{\sqrt{5}}$. By Pythagoras again, $DE=\frac{4}{\sqrt{5}}$. Then, the area of $ADP$ is $DE\cdot AE=\frac{32}{5}$. Using $4$ instead as the base, we can drop a altitude from P. $\frac{32}{5}=\frac{bh}{2}$. $\frac{32}{5}=\frac{4h}{2}$. Thus, the horizontal distance is $\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}$

When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$. Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive. For each of the $9\cdot10\cdot10=900$ possible values of $q$, there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$. Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$, each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$. Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$.

If we take the parabola $ax^2 + bx + c$ and reflect it over the x - axis, we have the parabola $-ax^2 - bx - c$. Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then: \begin{align*} f(x) = a(x+5)^2 + b(x+5) + c = ax^2 + (10a+b)x + 25a + 5b + c \\ g(x) = -a(x-5)^2 - b(x-5) - c = -ax^2 + 10ax -bx - 25a + 5b - c \end{align*} Adding them up produces: \begin{align*} (f + g)(x) = ax^2 + (10a+b)x + 25a + 5b + c - ax^2 + 10ax -bx - 25a + 5b - c = 20ax + 10b \end{align*} This is a line with slope $20a$. Since $a$ cannot be $0$ (because $ax^2 + bx + c$ would be a line) we end up with $\boxed{\textbf{(D)} \text{ a non-horizontal line }}$

The answer is $\boxed{\textrm{(A)}}$. Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are $k$ B's in the first five letters, then there must be $5-k$ C's in the first five letters, so there must be $k$ C's and $5-k$ A's in the next five letters, and $k$ A's and $5-k$ B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with this restriction is $\binom{5}{k}^3$ (since there are $\binom{5}{k}$ ways to arrange $k$ B's and $5-k$ C's). Therefore the answer is $\sum_{k=0}^{5}\binom{5}{k}^{3}$.

Let the roots be $r_1=0, r_2, r_3, r_4, r_5$. According to Viète's formulae, we have $d=r_1r_2r_3r_4 + r_1r_2r_3r_5 + r_1r_2r_4r_5 + r_1r_3r_4r_5 + r_2r_3r_4r_5$. The first four terms contain $r_1=0$ and are therefore zero, thus $d=r_2r_3r_4r_5$. This is a product of four non-zero numbers, therefore $d$ must be non-zero $\Longrightarrow \mathrm{(D)}$. Clearly, since $(0,0)$ is an intercept, $e$ must be $0$. But if $d$ was $0$, $x^2$ would divide the polynomial, which means it would have a double root at $0$, which is impossible, since all five roots are distinct.

If $A$ and $B$ meet, their paths connect $(0,0)$ and $(5,7).$ There are $\binom{12}{5}=792$ such paths. Since the path is $12$ units long, they must meet after each travels $6$ units, so the probability is $\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}$. Note: The number of paths, $\binom{12}{5}$ comes from the fact that there must be 5 ups/downs and 7 lefts/rights in one path. WLOG, for Object A, the number of paths would be the amount of combinations of the sequence of letters with 5 "U"s 7 "R"s (i.e. UUUUURRRRRRR). This is $\frac{12!}{5!7!}$, which is equivalent to $\binom{12}{5}$.

We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$. We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$. To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$. To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{\mathrm{(B)}}$.

Using logarithmic rules, we see that \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\] Since $a$ and $b$ are both greater than $1$, using AM-GM gives that the term in parentheses must be at least $2$, so the largest possible values is $2-2=0 \Rightarrow \boxed{\textbf{B}}.$ Note that the maximum occurs when $a=b$.

The function $f(x) = \sqrt{x(ax+b)}$ has a codomain of all non-negative numbers, or $0 \le f(x)$. Since the domain and the range of $f$ are the same, it follows that the domain of $f$ also satisfies $0 \le x$. The function has two zeroes at $x = 0, \frac{-b}{a}$, which must be part of the domain. Since the domain and the range are the same set, it follows that $\frac{-b}{a}$ is in the codomain of $f$, or $0 \le \frac{-b}{a}$. This implies that one (but not both) of $a,b$ is non-positive. The problem states that there is at least one positive value of b that works, thus $a$ must be non-positive, $b$ is non-negative, and the domain of the function occurs when $x(ax+b) > 0$, or $0 \le x \le \frac{-b}{a}.$ Completing the square, $f(x) = \sqrt{a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}} \le \sqrt{\frac{-b^2}{4a}}$ by the Trivial Inequality (remember that $a \le 0$). Since $f$ is continuous and assumes this maximal value at $x = \frac{-b}{2a}$, it follows that the range of $f$ is $0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.$ As the domain and the range are the same, we have that $\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0$ (we can divide through by $b$ since it is given that $b$ is positive). Hence $a = 0, -4$, which both we can verify work, and the answer is $\mathbf{(C)}$.