AMC12 2010 B

The total number of minutes in her $9$-hour work day is $9 \times 60 = 540.$ The total amount of time spend in meetings in minutes is $45 + 45 \times 2 = 135.$ The answer is then $\frac{135}{540}$ $= \boxed{25\%}$ or $\boxed{(C)}$

We find the area of the big rectangle to be $8 \times 5 = 40$, and the area of the smaller rectangle to be $(8 - 2) \times (5 - 2) = 18$. The answer is then $40 - 18 = 22$ $(A)$.

We find the greatest common factor of $48$ and $64$ to be $16$. The number of factors of $16$ is $5$ which is the answer $(E)$.

$31 \equiv 3 \pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\boxed{B}$.

We simply plug in the numbers \[1 - 2 - 3 - 4 + e = 1 - (2 - (3 - (4 + e)))\] \[-8 + e = -2 - e\] \[2e = 6\] \[e = 3 \;\;(D)\]

The minimum possible value would be $70 - 50 = 20\%$. The maximum possible value would be $30 + 50 = 80\%$. The difference is $80 - 20 = \boxed{\textbf{(D) }60}$.

Let $x$ be the time it is not raining, and $y$ be the time it is raining, in hours. We have the system: $30x+20y=16$ and $x+y=\frac{2}{3}$ Solving gives $x=\frac{4}{15}$ and $y=\frac{2}{5}$ We want $y$ in minutes, $\frac{2}{5} \cdot 60=24 \Rightarrow C$

There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$, because there wouldn't be a $64$th place if $x$ was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $\boxed{\mathbf{(B)}\ 23}$.

We know that $n^2 = k^3$ and $n^3 = m^2$. Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$. Let $a = n^6$. Now we know $a$ must be a perfect $36$-th power because $lcm(9,4) = 36$, which means that $n$ must be a perfect $6$-th power. The smallest number whose sixth power is a multiple of $20$ is $10$, because the only prime factors of $20$ are $2$ and $5$, and $10 = 2 \times 5$. Therefore our is equal to number $10^6 = 1000000$, with $7$ digits $\Rightarrow \boxed {E}$.

We first sum the first $99$ numbers: $\frac{99(100)}{2}=99\cdot50$. Then, we know that the sum of the series is $99\cdot50+x$. There are $100$ terms, so we can divide this sum by $100$ and set it equal to $100x$: \[\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x\] Using difference of squares: \[x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}\] Thus, the answer is $\boxed{\textbf{(B) }\frac{50}{101}}$.

View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$. But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$. Recombining this yields $1001a_3 + 110a_2$. 1001 is divisible by 7, which means that as long as $a_2 = 0$, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ($9 \cdot 10$) possibilities for palindromes. However, if $a_2 = 7$, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

\[\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40\] \[\frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40\] \[\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40\] \[5\log_2x = 40\] \[\log_2x = 8\] \[x = 256 \;\; (D)\]

We note that $-1$ $\le$ $\sin x$ $\le$ $1$ and $-1$ $\le$ $\cos x$ $\le$ $1$. Therefore, there is no other way to satisfy this equation other than making both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is clear that $\triangle ABC$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle with $BC=2$ $\Longrightarrow$ $(C)$.

We want to try make $a+b$, $b+c$, $c+d$, and $d+e$ as close as possible so that $M$, the maximum of these, is smallest. Notice that $2010=670+670+670$. In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$. We see that in both cases, the value of $M$ is $671$, so the answer is $671 \Rightarrow \boxed{B}$.

We have either $i^{x}=(1+i)^{y}\neq z$, $i^{x}=z\neq(1+i)^{y}$, or $(1+i)^{y}=z\neq i^x$. $i^{x}=(1+i)^{y}$ only occurs when it is $1$. $(1+i)^{y}=1$ has only one solution, namely, $y=0$. $i^{x}=1$ has five solutions between 0 and 19, $x=0, x=4, x=8, x=12$, and $x=16$. $z\neq 1$ has nineteen integer solutions between zero and nineteen. So for $i^{x}=(1+i)^{y}\neq z$, we have $5\cdot 1\cdot 19=95$ ordered triples. For $i^{x}=z\neq(1+i)^{y}$, again this only occurs at $1$. $(1+i)^{y}\neq 1$ has nineteen solutions, $i^{x}=1$ has five solutions, and $z=1$ has one solution, so again we have $5\cdot 1\cdot 19=95$ ordered triples. For $(1+i)^{y}=z\neq i^x$, this occurs at $1$ and $16$. $(1+i)^{y}=1$ and $z=1$ both have one solution while $i^{x}\neq 1$ has fifteen solutions. $(1+i)^{y}=16$ and $z=16$ both have one solution, namely, $y=8$ and $z=16$, while $i^{x}\neq 16$ has twenty solutions ($i^x$ only cycles as $1, i, -1, -i$). So we have $15\cdot 1\cdot 1+20\cdot 1\cdot 1=35$ ordered triples. In total we have ${95+95+35=\boxed{\text{(D) }225}}$ ordered triples

We group this into groups of $3$, because $3|2010$. This means that every residue class mod 3 has an equal probability. If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens. Otherwise, we have $3|bc+b+1$, which means that $b(c+1) \equiv 2\pmod{3}$. So either \[b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}\] or \[b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3\] which will lead to the property being true. There is a $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$. The grand total is \[\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}\]

Answer (D): Let $a_{ij}$ denote the entry in row $i$ and column $j$. he given conditions imply that $a_{11}=1$, $a_{33}=9$, and $a_{22}=4,5,$ or $6$. If $a_{22}=4$, then $\{a_{12},a_{21}\}=\{2,3\}$, and the sets $\{a_{31},a_{32}\}$ and $\{a_{13},a_{23}\}$ are complementary subsets of $\{5,6,7,8\}$. There are $\binom{4}{2}=6$ ways to choose $\{a_{31},a_{32}\}$ and $\{a_{13},a_{23}\}$, and only one way to order the entries. There are $2$ ways to order $\{a_{12},a_{21}\}$, so $12$ arrays with $a_{22}=4$ meet the given conditions. Similarly, the conditions are met by $12$ arrays with $a_{22}=6$. If $a_{22}=5$, then $\{a_{12},a_{13},a_{23}\}$ and $\{a_{21},a_{31},a_{32}\}$ are complementary subsets of $\{2,3,4,6,7,8\}$ subject to the conditions $a_{12}<5$, $a_{21}<5$, $a_{32}>5$, and $a_{23}>5$. Thus $\{a_{12},a_{13},a_{23}\}\ne\{2,3,4\}$ or $\{6,7,8\}$, so its elements can be chosen in $\binom{6}{3}-2=18$ ways. Both the remaining entries and the ordering of all entries are then determined, so $18$ arrays with $a_{22}=5$ meet the given conditions. Altogether, the conditions are met by $12+12+18=42$ arrays.

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The starts on the origin. It is relatively easy to show that exactly one element in the set \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] has magnitude less than or equal to $1$. (Can you show how?) Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.

Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer. Suppose not, and say $r = m/n$ where $m>n>1$, and $\gcd(m,n)=1$. Then $n, n^2, n^3$ must all divide $a$ so $a=n^3k$ for some integer $k$. Then $S_R = n^3k + n^2mk + nm^2k + m^3k < 100$ and we see that even if $k=1$ and $n=2$, we get $m < 4$, which means that the only option for $r$ is $r=3/2$. A quick check shows that even this doesn't work. Thus $r$ must be an integer. Let $a, a+d, a+2d, a+3d$ be the quarterly scores for the Wildcats. Let $S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d$. Let $S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)$. Then $S_R<100$ implies that $r<5$, so $r\in \{2, 3, 4\}$. The Raiders win by one point, so\[a(1+r)(1+r^2) = 4a+6d+1.\] - If $r=4$ we get $85a = 4a+6d+1$ which means $3(27a-2d) = 1$, which is not possible with the given conditions. - If $r=3$ we get $40a = 4a+6d+1$ which means $6(6a-d) = 1$, which is also not possible with the given conditions. - If $r=2$ we get $15a = 4a+6d+1$ which means $11a-6d = 1$. Reducing modulo 6 we get $a \equiv 5\pmod{6}$. Since $15a<100$ we get $a<7$. Thus $a=5$. It then follows that $d=9$. Then the quarterly scores for the Raiders are $5, 10, 20, 40$, and those for the Wildcats are $5, 14, 23, 32$. Also $S_R = 75 = S_W + 1$. The total number of points scored by the two teams in the first half is $5+10+5+14=\boxed{\textbf{(E)}\ 34}$. Note if you don't realize while taking the test that $r$ might not be an integer: since an answer is achieved through casework on the integer value of $r$ and since there is only one right answer, the proof of $r$ being an integer can be skipped on the test (it takes up time).

By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2x$. The common ratio of the sequence is $\frac{\cos x}{\sin x}$, so we can write \[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\sin^2x}=1\] \[a_5=\frac{\cos x}{\sin x}\] \[a_6=\frac{\cos^2x}{\sin^2x}\] \[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\] \[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\] Since $\cos^3x=\sin^2x=1-\cos^2x$, we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$, which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$.

We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$. Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$. Then, noting that $P(2)-a=-a-a=-2a$ etc., and plugging in values of $2,4,6,8,$ we get \[P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a\] \[P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a\] \[P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a\] \[P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a\] $-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$. Solving, we receive $315$, so our answer is $\boxed{\textbf{(B)}\ 315}$. To complete the solution, we can let $a = 315$, and then try to find $Q(x)$. We know from the above calculation that $Q(2)=42, Q(4)=-70, Q(6)=42$, and $Q(8)=-6$. Then we can let $Q(x) = T(x)(x-2)(x-6)+42$, getting $T(4)=28, T(8)=-4$. Let $T(x)=L(x)(x-8)-4$, then $L(4)=-8$. Therefore, it is possible to choose $T(x) = -8(x-8)-4 = -8x + 60$, so the goal is accomplished. As a reference, the polynomial we get is \[P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315\] \[= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325\]

Let $AB = a$, $BC = b$, $CD = c$, and $AD = d$. We see that by the Law of Cosines on $\triangle ABD$ and $\triangle CBD$, we have: $BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$. $BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$. We are given that $ad = bc$ and $ABCD$ is a cyclic quadrilateral. As a property of cyclic quadrilaterals, opposite angles are supplementary so $\angle BAD = 180 - \angle BCD$, therefore $\cos{\angle BAD} = -\cos{\angle BCD}$. So, $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$. Adding, we get $2BD^2 = a^2 + b^2 + c^2 + d^2$. We now look at the equation $ad = bc$. Suppose that $a = 14$. Then, we must have either $b$ or $c$ equal $7$. Suppose that $b = 7$. We let $d = 6$ and $c = 12$. $2BD^2 = 196 + 49 + 36 + 144 = 425$, so our answer is $\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}$.

$P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$. Notice that $P(x)$ has roots $a\pm \sqrt {b}$, so that the roots of $P(Q(x))$ are the roots of $Q(x) = a + \sqrt {b}, a - \sqrt {b}$. For each individual equation, the sum of the roots will be $2c$ (symmetry or Vieta's). Thus, we have $4c = - 23 - 21 - 17 - 15$, or $c = - 19$. Doing something similar for $Q(P(x))$ gives us $a = - 54$. We now have $P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d$. Since $Q$ is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther" from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$. Thus, we have $Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}$, or $16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}$. Adding these gives us $20 - 2d = - 108$, or $d = 64$. Plugging this into $16 - d = - 54 + \sqrt {b}$, we get $b = 36$. The minimum value of $P(x)$ is $- b$, and the minimum value of $Q(x)$ is $- d$. Thus, our answer is $- (b + d) = - 100$, or answer $\boxed{\textbf{(A)}}$.

Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where \[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1\] We shall say that $f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}$. $f(x)$ has three vertical asymptotes at $x=\{-1,0,1\}$. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at $y=0$. The function intersects $1$ at some point from $x=-1$ to $x=0$, and at some point from $x=0$ to $x=1$, and at some point to the right of $x=1$. The intervals where the function is greater than $1$ are between the points where the function equals $1$ and the vertical asymptotes. If $p$, $q$, and $r$ are values of x where $f(x)=1$, then the sum of the lengths of the intervals is $(p-(-1))+(q-0)+(r-1)=p+q+r$. \[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1\] \[\implies x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)\] \[\implies x^3-3x^2-x+1=0\] And now our job is simply to find the sum of the roots of $x^3-3x^2-x+1$. Using Vieta's formulas, we find this to be $3$ $\Rightarrow\boxed{C}$. NOTE': For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose $C$.

Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$, there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is incorporated into the giant product. All numbers $n=67 \cdot x$, given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$, prime $p_x$ must be less than $67$, contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$, the possible values of x are $1,2,...,70,72,74,...78$, since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$.