AMC12 2014 A

We have \[10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}\] Making the denominators equal gives \[\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{8}{10}\right)^{-1}\] \[\implies 10\cdot\left(\frac{4}{5}\right)^{-1}\] \[\implies 10\cdot\frac{5}{4}\] \[\implies \frac{50}{4}\] Finally, simplifying gives \[\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\]

Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\frac{x}{2}$. \begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\ x &=& 3.50\\ \end{eqnarray*} Plug in for 8 adult tickets and 6 child tickets. \begin{eqnarray*} 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\ &=&\boxed{\textbf{(B)}\ \ 38.50}\\ \end{eqnarray*}

Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house. Case 1: $\text{Y}$ is the $3^\text{rd}$ house. The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$ Case 2: $\text{Y}$ is the last house. There are two possible arrangements: $\text{B}-\text{O}-\text{R}-\text{Y}$ $\text{O}-\text{B}-\text{R}-\text{Y}$ The answer is $1+2=\boxed{\textbf{(B) } 3}$

We need to multiply $b$ by $\frac{d}{a}$ for the new cows and $\frac{e}{c}$ for the new time, so the answer is $b\cdot \frac{d}{a}\cdot \frac{e}{c}=\frac{bde}{ac}$, or $\boxed{\textbf{(A) } \frac{bde}{ac}}$.

WLOG, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean is $87$ and the median is $90$. Thus, the solution is \[90-87=3\implies\boxed{\textbf{(C)} \ 3}\]

Let the two digits be $a$ and $b$. Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$, or $2a = 7b$. This yields $a = 7$ and $b = 2$ because $a, b < 10$. Then, $72 + 27 = \boxed{\textbf{(D) }99}.$

The terms are $\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$, which are equivalent to $3^{\frac{3}{6}}$, $3^{\frac{2}{6}}$, and $3^{\frac{1}{6}}$. So the next term will be $3^{\frac{0}{6}}=1$, so the answer is $\boxed{\textbf{(A)}}$.

Let the listed price be $x$. Since all the answer choices are above $\$100$, we can assume $x > 100$. Thus the discounts after the coupons are used will be as follows: Coupon 1: $x\times10\%=.1x$ Coupon 2: $20$ Coupon 3: $18\%\times(x-100)=.18x-18$ For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$. The only choice that satisfies such conditions is $\boxed{\textbf{(C)}\ \$219.95}$

Let $a=1$. Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$. Our next set starting with $3$ is $\{3,4,5,6,7\}$. Our average is $25\div 5=5$. Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{\textbf{(B)}\ a+4}$.

Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or $\boxed{\dfrac{\sqrt3}3\textbf{ (B)}}$.

Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives. Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early. Simplifying gives $7d+525=10d$, or $d=175$. Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{\textbf{(C) } 210}$ miles.

Let the radius of the larger and smaller circles be $x$ and $y$, respectively. Also, let their centers be $O_1$ and $O_2$, respectively. Then the ratio we need to find is \[\dfrac{\pi x^2}{\pi y^2} = \dfrac{x^2}{y^2}\] Draw the radii from the centers of the circles to $A$ and $B$. We can easily conclude that the $30^{\circ}$ belongs to the larger circle, and the $60$ degree arc belongs to the smaller circle. Therefore, $m\angle AO_1B = 30^{\circ}$ and $m\angle AO_2B = 60^{\circ}$. Note that $\Delta AO_2B$ is equilateral, so when chord AB is drawn, it has length $y$. Now, applying the Law of Cosines on $\Delta AO_1B$: \[y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2\] \[\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}\]

We can discern three cases. Case 1: Each room houses one guest. In this case, we have $5$ guests to choose for the first room, $4$ for the second, ..., for a total of $5!=120$ assignments. Case 2: Three rooms house one guest; one houses two. We have $\binom{5}{3}$ ways to choose the three rooms with $1$ guest, and $\binom{2}{1}$ to choose the remaining one with $2$. There are $5\cdot4\cdot3$ ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of $\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200$ ways. Case 3: Two rooms house two guests; one houses one. We have $\binom{5}{2}$ to choose the two rooms with two people, and $\binom{3}{1}$ to choose one remaining room for one person. Then there are $5$ choices for the lonely person, and $\binom{4}{2}$ for the two in the first two-person room. The last two will stay in the other two-room, so there are $\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900$ ways. In total, there are $120+1200+900=2220$ assignments, or $\boxed{\textbf{(B)}}$.

We have $b-a=c-b$, so $a=2b-c$. Since $a,c,b$ is geometric, $c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0$. Since $a<b<c$, we can't have $b=c$ and thus $c=-2b$. Then our arithmetic progression is $4b,b,-2b$. Since $4b < b < -2b$, $b < 0$. The smallest possible value of $c=-2b$ is $(-2)(-1)=2$, or $\boxed{\textbf{(C)}}$.

For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$) palindromes. So the $a$s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum. For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$) palindromes. So the $b$s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum. Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum. It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $\boxed{\textbf{(B)}\; 18}$.

We can list the first few numbers in the form $8 \cdot (8....8)$ (Hard problem to do without the multiplication, but you can see the pattern early on) $8 \cdot 8 = 64$ $8 \cdot 88 = 704$ $8 \cdot 888 = 7104$ $8 \cdot 8888 = 71104$ $8 \cdot 88888 = 711104$ By now it's clear that the numbers will be in the form $7$, $k-2$ $1$'s, and $04$. We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$. Solving, we get $k = 991$, meaning the answer is $\boxed{\textbf{(D) } 991}$ Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$, so we just do $1000-9=\boxed{\textbf{(D) } 991}$.

Let one of the corners be $(0, 0, 0)$. We can orient the box such that the center of the small sphere closest to the corner is $(1,1,1)$, and the center of the large sphere is $(2, 2, h/2)$. Since the two spheres are tangent, the distance between their centers is $1+2 = 3$, so $\sqrt{(2-1)^2+(2-1)^2+(h/2-1)^2} = 3$. Solving, $h=2 + 2\sqrt{7}=\boxed{\textbf{(A)}}$

For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that: 1. $\log_b a$ is defined if and only if $a>0.$ 2. For $0<b<1,$ we conclude that: $\log_b a<c$ if and only if $a>b^c.$ $\log_b a>c$ if and only if $0<a<b^c.$ For $b>1,$ we conclude that: $\log_b a<c$ if and only if $0<a<b^c.$ $\log_b a>c$ if and only if $a>b^c.$ Therefore, we have \begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*} The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{\textbf{(C) }271}.$

Factor the quadratic into \[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\] where $-n$ is our integer solution. Then, \[k = \frac{12}{n} + 5n,\] which takes rational values between $-200$ and $200$ when $|n| \leq 39$, excluding $n = 0$. This leads to an answer of $2 \cdot 39 = \boxed{\textbf{(E) } 78}$.

Let $C_1$ be the reflection of $C$ across $\overline{AB}$, and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$. Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$. (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As $A$ lies on both $AB$ and $AC$, we have $C_2A=C_1A=CA=6$. Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$. Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}.\]

Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$. Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$. In order for this to be less than or equal to $1$, we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$. Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$, and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$. We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]

Between any two consecutive powers of $5$ there are either $2$ or $3$ powers of $2$ (because $2^2<5^1<2^3$). Consider the intervals $(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})$. We want the number of intervals with $3$ powers of $2$. From the given that $2^{2013}<5^{867}<2^{2014}$, we know that these $867$ intervals together have $2013$ powers of $2$. Let $x$ of them have $2$ powers of $2$ and $y$ of them have $3$ powers of $2$. Thus we have the system \[x+y=867\]\[2x+3y=2013\] from which we get $y=279$, so the answer is $\boxed{\textbf{(B)}}$.

the fraction $\dfrac{1}{99}$ can be written as \[\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}\]. similarly the fraction $\dfrac{1}{99^2}$ can be written as $\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}$ which is equivalent to \[\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}\] and we can see that for each $n+m=k$ there are $k-1$ $(n,m)$ combinations so the above sum is equivalent to: \[\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}\] we note that the sequence starts repeating at $k = 102$ yet consider \[\sum^{101}_{k=99}\dfrac{k-1}{10^{2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}} \]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})\]\[=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)\] so the decimal will go from 1 to 99 skipping the number 98 and we can easily compute the sum of the digits from 0 to 99 to be \[45\cdot10\cdot2=900\] subtracting the sum of the digits of 98 which is 17 we get \[900-17=883\textbf{(B) }\qquad\]

1. Draw the graph of $f_0(x)$ by dividing the domain into three parts. 2. Apply the recursive rule a few times to find the pattern. Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same. 3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$. $f_{100}(x)$ reaches $0$ at $x = -300$, then zigzags between $0$ and $-1$, hitting $0$ at every even $x$, before leaving $0$ at $x = 300$. This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$. This is a $601$-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{\textbf{(C) }301}$.

The parabola is symmetric through $y=- \frac{4}{3}x$, and the common distance is $5$, so the directrix is the line through $(1,7)$ and $(-7,1)$, which is the line \[3x-4y = -25.\] Using the point-line distance formula, the parabola is the locus \[x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2}\] which rearranges to $(4x+3y)^2 = 25(6x-8y+25)$. Let $m = 4x+3y \in \mathbb Z$, $\left\lvert m \right\rvert \le 1000$. Put $m = 25k$ to obtain \[25k^2 = 6x-8y+25\]\[25k = 4x+3y.\] and accordingly we find by solving the system that $x = \frac{1}{2} (3k^2-3) + 4k$ and $y = -2k^2+3k+2$. One can show that the values of $k$ that make $(x,y)$ an integer pair are precisely odd integers $k$.* For $\left\lvert 25k \right\rvert \le 1000$ this is $k= -39,-37,-35,\dots,39$, so $40$ values work and the answer is $\boxed{\textbf{(B)}}$.