AMC12 2000 A

First, we need to recognize that a number is going to be largest only if, of the $3$ factors, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$. It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$. Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors. Hence, we have $667 + 3 + 1 = \boxed{\text{(E) 671.}}$

We can use an elementary exponents rule to solve our problem. We know that $a^b\cdot a^c = a^{b+c}$. Hence, $2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}$. Solution edited by armang32324 and integralarefun

We can begin by labeling the number of initial jellybeans $x$. If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$ After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$, so $x = \boxed{(B) 50}$ Solution By: armang32324

Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$: $1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$ The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{\mathrm{C}}$.

When $x < 2,$ $x-2$ is negative so $|x - 2| = 2-x = p$ and $x = 2-p$. Thus $x-p = (2-p)-p = 2-2p$. $\boxed{\mathbf{(C)}}$

Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$. Thus, we can eliminate E. So, the answer must be $\boxed{\textbf{(C) }119}$.

If $\log_{b} 729 = n$, then $b^n = 729$. Since $729 = 3^6$, $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{\mathrm{E}}$.

We can attempt $0^2+1^2=1$ and $1^2+2^2=5$, so the pattern here looks like the number of squares in the $n$-th figure is $n^2+(n+1)^2$. When we plug in 100 for $n$, we get $100^2+101^2=10000+10201=20201$, or option $\textbf{(C)}$.

The first number is divisible by $1$. The sum of the first two numbers is even. The sum of the first three numbers is divisible by $3.$ The sum of the first four numbers is divisible by $4.$ The sum of the first five numbers is $400.$ Since $400$ is divisible by $4,$ the last score must also be divisible by $4.$ Therefore, the last score is either $76$ or $80.$ Case 1: $76$ is the last number entered. Since $400\equiv 76\equiv 1\pmod{3}$, the fourth number must be divisible by $3,$ but none of the scores are divisible by $3.$ Case 2: $80$ is the last number entered. Since $80\equiv 2\pmod{3}$, the fourth number must be $2\pmod{3}$. The only number which satisfies this is $71$. The next number must be $91$ since the sum of the first two numbers is even. So the only arrangement of the scores $76, 82, 91, 71, 80$ or $82, 76, 91, 71, 80$ $\Rightarrow \text{(C)}$

Step 1: Reflect in the $xy$-plane. Replace $z$ with its additive inverse: $(1,2,-3)$ Step 2: Rotate around $x$-axis 180 degrees. Replace $y$ and $z$ with their respective additive inverses. $(1, -2, 3)$ Step 3: Translate $5$ units in positive-$y$ direction. Replace $y$ with $y+5$. $(1,3,3) \Rightarrow \text {(E) }$

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$. Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$

It is not hard to see that \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+A+M+C+1\] Since $A+M+C=12$, we can rewrite this as \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] So we wish to maximize \[(A+1)(M+1)(C+1)-13\] Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$, we set $A=M=C=4$ Which gives us \[(4+1)(4+1)(4+1)-13=112\] so the answer is $\boxed{\textbf{(E) }112}.$

Let $c$ be the total amount of coffee, $m$ of milk, and $p$ the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so \[\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m\] Regrouping, we get $2c(6-p)=3m(p-4)$. Since both $c,m$ are positive, it follows that $6-p$ and $p-4$ are also positive, which is only possible when $p = 5\ \mathrm{(C)}$.

- The mean is $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$. - Arranged in increasing order, the list is $2,2,2,4,5,10$, so the median is either $2,4$ or $x$ depending upon the value of $x$. - The mode is $2$, since it appears three times. We apply casework upon the median: - If the median is $2$ ($x \le 2$), then the arithmetic progression must be constant. - If the median is $4$ ($x \ge 4$), because the mode is $2$, the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$. - If the median is $x$ ($2 \le x \le 4$), we must have the arithmetic progression $2, x, \frac{25+x}{7}$. Thus, we find that $2x=2+\frac{25+x}{7}$ so $x=3$. The answer is $3 + 17 = \boxed{20\ \mathrm{(E)}}$.

Let $y = \frac{x}{3}$; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$. Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$, and $z = -\frac{1}{3}, \frac{2}{9}$. These sum up to $\boxed{\textbf{(B) }-\frac19}$.

Index the rows with $i = 1, 2, 3, ..., 13$ Index the columns with $j = 1, 2, 3, ..., 17$ For the first row number the cells $1, 2, 3, ..., 17$ For the second, $18, 19, ..., 34$ and so on So the number in row = $i$ and column = $j$ is $f(i, j) = 17(i-1) + j = 17i + j - 17$ Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i, j) = i + 13j - 13$ So we need to solve $f(i, j) = g(i, j)$ $17i + j - 17 = i + 13j - 13$ $16i = 4 + 12j$ $4i = 1 + 3j$ $i = (1 + 3j)/4$ We get $(i, j) = (1, 1), f(i, j) = g(i, j) = 1$ $(i, j) = (4, 5), f(i, j) = g(i, j) = 56$ $(i, j) = (7, 9), f(i, j) = g(i, j) = 111$ $(i, j) = (10, 13), f(i, j) = g(i, j) = 166$ $(i, j) = (13, 17), f(i, j) = g(i, j) = 221$ $\boxed{D}$ $555$

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$, $AB = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, \[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] We multiply both sides by $\cos \theta$ to simplify the trigonometric functions, \[AC=OC \sin \theta\] Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$. Therefore, the answer is $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$.

There are either \[65 + 200 = 265\] or \[66 + 200 = 266\] days between the first two dates depending upon whether or not year $N+1$ is a leap year (since the February 29th of the leap year would come between the 300th day of year $N$ and 200th day of year $N + 1$). Since $7$ divides into $266$ but not $265$, for both days to be a Tuesday, year $N$ must be a leap year. Hence, year $N-1$ is not a leap year, and so since there are \[265 + 300 = 565\] days between the date in years $N,\text{ }N-1$, this leaves a remainder of $5$ upon division by $7$. Since we are subtracting days, we count 5 days before Tuesday, which gives us $\boxed{\mathbf{(A)} \ \text{Thursday}.}$

The answer is exactly $3$, choice $\mathrm{(C)}$. We can find the area of triangle $ADE$ by using the simple formula $\frac{bh}{2}$. Dropping an altitude from $A$, we see that it has length $12$ (we can split the large triangle into a $9-12-15$ and a $5-12-13$ triangle). Then we can apply the Angle Bisector Theorem on triangle $ABC$ to solve for $BE$. Solving $\frac{13}{BE}=\frac{15}{14-BE}$, we get that $BE=\frac{13}{2}$. $D$ is the midpoint of $BC$ so $BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\frac{1}{2}$ units long. Applying the formula $\frac{bh}{2}$, we get $\frac{12*\frac{1}{2}}{2}=3$.

We multiply all given expressions to get: \[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\] Adding all the given expressions gives that \[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\] We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$. Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$. \[\]

WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.

Note that there are 3 maxima/minima. Hence we know that the rest of the graph is greater than 10. We approximate each of the above expressions: 1. According to the graph, $P(-1) > 4$ 2. The product of the roots is $d$ by Vieta’s formulas. Also, $d = P(0) > 5$ according to the graph. 3. The product of the real roots is $>5$, and the total product is $<6$ (from above), so the product of the non-real roots is $< \frac{6}{5}$. 4. The sum of the coefficients is $P(1) > 2.5$ 5. The sum of the real roots is $> 5$. Clearly $\mathrm{(C)}$ is the smallest.

The product of the numbers has to be a power of $10$ in order to have an integer base ten logarithm. Thus all of the numbers must be in the form $2^m5^n$. Listing out such numbers from $1$ to $46$, we find $1,2,4,5,8,10,16,20,25,32,40$ are the only such numbers. Immediately it should be noticed that there are a larger number of powers of $2$ than of $5$. Since a number in the form of $10^k$ must have the same number of $2$s and $5$s in its factorization, we require larger powers of $5$ than we do of $2$. To see this, for each number subtract the power of $5$ from the power of $2$. This yields $0,1,2,-1,3,0,4,1,-2,5,2$, and indeed the only non-positive terms are $0,0,-1,-2$. Since there are only two zeros, the largest number that Professor Gamble could have picked would be $2$. Thus Gamble picks numbers which fit $-2 + -1 + 0 + 0 + 1 + 2$, with the first four having already been determined to be $\{25,5,1,10\}$. The choices for the $1$ include $\{2,20\}$ and the choices for the $2$ include $\{4,40\}$. Together these give four possible tickets, which makes Professor Gamble’s probability $1/4\ \mathrm{(B)}$.

First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$. $\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}$. Next, connect the center of the circle to side $AB$, and call this length $r$, and call the foot $M$. Since $ABC$ is equilateral, it follows that $MB=\frac{18}{\pi}$, and $OA$ (where $O$ is the center of the circle) is $\frac{36}{\pi}-r$. By the Pythagorean Theorem, you get $r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}$. Finally, we see that the circumference is $2{\pi}\cdot \frac{27}{2\pi}=\boxed{(D)27}$.

Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red. There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$.