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4,701
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_22
| 3
|
Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$
|
For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: \[(0,0), (60,0), (60,60), (0,60).\]
Now, let a point in the square have coordinates $(x, y).$
In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles with area $60^2/100 = 36.$ For it to not be $60$ -ray partitional, we cannot make $60$ triangles with area $60^2/60 = 60.$
When we draw such a triangle, it's base will always be on one of the sides of the square. If it is on the bottom side, then the height must be $y$ . So, the base of each triangle must be $\frac{72}{y}.$ So, there will be $\frac{60}{\frac{72}{y}} = \frac{60y}{72}$ total triangles on that side.
If the triangle is on the right side of the square, then the height has to be $60 - x.$ So, the base will be $\frac{72}{60-x}$ and there will be $\frac{60 (60-x)}{72}$ triangles.
If the triangle is on the top, the height will be $60-y$ , the base will be $\frac{72}{60-y}$ and there will be $\frac{60 (60-y)}{72}$ triangles along this side.
The triangles on the left will have height $x$ , base $\frac{72}{x}$ and $\frac{60x}{72}$ such triangles must exist.
Simplifying, we get $\frac{5y}{6}, \frac{5 (60-x)}{6}, \frac{5(60-y)}{6}, \frac{5x}{6}$ triangles on the respective sides of the square.
All of these numbers must be integers. Let $x = 60a$ and $y = 60b$ where $0 < a, b < 1.$
Then, the amounts become: \[50a, 50b, 50 - 50a, 50 - 50b.\]
As long as $50a$ and $50b$ are integers, $50 - 50a$ and $50 - 50b$ will also be integers.
For this to happen, $a$ and $b$ must be of the form $\frac{x}{50}$ where $1 \le x \le 49.$ So, we have a total of $49^2 = 2401$ points that are $100$ -ray partitional.
Now, we must calculate the number of $100$ - ray partitional points that are also $60$ -ray partitional.
Using a method similar to before, if a point is $60$ -ray partitional, then we must be able to make $30a, 30b, 30 - 30a, 30 -30b$ triangles on the different sides.
So, $30a$ and $30b$ must be integers. This means $a$ and $b$ have to be of the form $\frac{y}{30}$ where $0<y<30.$
If a point is both $100$ -ray partitional and $60$ -ray partitional, then it can be written as \[\frac{x}{50} = \frac{y}{30}\] . Note that whenever $x$ is divisible by $5$ , a $y$ will always exist to satisfy the above equation.
So, $x$ has to be in the range $(0, 50)$ and must be divisible by $5$ . There are $9$ possibilities, namely $5, 10, 15, 20, 25, 30, 35, 40, 45.$
The $x$ -coordinate of the point and the $y$ -coordinate of the point can each use any of these $9$ possibilities, giving $9^2 = 81$ numbers that are both $100$ -ray partitional and $60$ -ray partitional.
Overall, we are left with $2401 - 81 = \boxed{2320}$ solutions.
| 320
|
4,702
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_25
| 1
|
Triangle $ABC$ has $\angle BAC = 60^{\circ}$ $\angle CBA \leq 90^{\circ}$ $BC=1$ , and $AC \geq AB$ . Let $H$ $I$ , and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$ , respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$
$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$
|
Let $\angle CAB=A$ $\angle ABC=B$ $\angle BCA=C$ for convenience.
It's well-known that $\angle BOC=2A$ $\angle BIC=90+\frac{A}{2}$ , and $\angle BHC=180-A$ (verifiable by angle chasing). Then, as $A=60$ , it follows that $\angle BOC=\angle BIC=\angle BHC=120$ and consequently pentagon $BCOIH$ is cyclic. Observe that $BC=1$ is fixed, hence the circumcircle of cyclic pentagon $BCOIH$ is also fixed. Similarly, as $OB=OC$ (both are radii), it follows that $O$ and also $[BCO]$ is fixed. Since $[BCOIH]=[BCO]+[BOIH]$ is maximal, it suffices to maximize $[BOIH]$
Verify that $\angle IBC=\frac{B}{2}$ $\angle HBC=90-C$ by angle chasing; it follows that $\angle IBH=\angle HBC-\angle IBC=90-C-\frac{B}{2}=\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}$ since $A+B+C=180\implies\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90$ by Triangle Angle Sum. Similarly, $\angle OBC=(180-120)/2=30$ (isosceles base angles are equal), hence \[\angle IBO=\angle IBC-\angle OBC=\frac{B}{2}-30=60-\frac{A}{2}-\frac{C}{2}=30-\frac{C}{2}\] Since $\angle IBH=\angle IBO$ $IH=IO$ by Inscribed Angles.
There are two ways to proceed.
Letting $O'$ and $R$ be the circumcenter and circumradius, respectively, of cyclic pentagon $BCOIH$ , the most straightforward is to write $[BOIH]=[OO'I]+[IO'H]+[HO'B]-[BO'O]$ , whence \[[BOIH]=\frac{1}{2}R^2(\sin(60-C)+\sin(60-C)+\sin(2C-60)-\sin(60))\] and, using the fact that $R$ is fixed, maximize $2\sin(60-C)+\sin(2C-60)$ with Jensen's Inequality.
A more elegant way is shown below.
Lemma: $[BOIH]$ is maximized only if $HB=HI$
Proof by contradiction: Suppose $[BOIH]$ is maximized when $HB\neq HI$ . Let $H'$ be the midpoint of minor arc $BI$ be and $I'$ the midpoint of minor arc $H'O$ . Then $[BOIH']=[IBO]+[IBH']>[IBO]+[IBH]=[BOIH]$ since the altitude from $H'$ to $BI$ is greater than that from $H$ to $BI$ ; similarly $[BH'I'O]>[BOIH']>[BOIH]$ . Taking $H'$ $I'$ to be the new orthocenter, incenter, respectively, this contradicts the maximality of $[BOIH]$ , so our claim follows. $\blacksquare$
With our lemma( $HB=HI$ ) and $IH=IO$ from above, along with the fact that inscribed angles that intersect the same length chords are equal, \[\angle ABC=2\angle IBC=2(\angle OBC+\angle OBI)=2(30+\frac{1}{3}\angle OCB)=80\implies\boxed{80}\]
| 80
|
4,703
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_2
| 1
|
Josanna's test scores to date are $90, 80, 70, 60,$ and $85.$ Her goal is to raise her test average at least $3$ points with her next test. What is the minimum test score she would need to accomplish this goal?
$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 85 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 95$
|
Take the average of her current test scores, which is \[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\]
This means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is
\[\frac{90+80+70+60+85+x}{6} = 80\]
\[385+x = 480\]
\[x = \boxed{95}\]
| 95
|
4,704
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_3
| 1
|
LeRoy and Bernardo went on a week-long trip together and agreed to share the costs equally. Over the week, each of them paid for various joint expenses such as gasoline and car rental. At the end of the trip it turned out that LeRoy had paid A dollars and Bernardo had paid B dollars, where $A < B.$ How many dollars must LeRoy give to Bernardo so that they share the costs equally?
$\textbf{(A)}\ \frac{A+B}{2} \qquad \textbf{(B)}\ \frac{A-B}{2} \qquad \textbf{(C)}\ \frac{B-A}{2} \qquad \textbf{(D)}\ B-A \qquad \textbf{(E)}\ A+B$
|
The total amount of money that was spent during the trip was \[A + B\] So each person should pay \[\frac{A+B}{2}\] if they were to share the costs equally. Because LeRoy has already paid $A$ dollars of his part, he still has to pay \[\frac{A+B}{2} - A =\] \[= \boxed{2}\]
| 2
|
4,705
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_4
| 1
|
In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161.$ What is the correct value of the product of $a$ and $b$
$\textbf{(A)}\ 116 \qquad \textbf{(B)}\ 161 \qquad \textbf{(C)}\ 204 \qquad \textbf{(D)}\ 214 \qquad \textbf{(E)}\ 224$
|
Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields
\[a*b = 32*7 =\]
\[= \boxed{224}\]
| 224
|
4,706
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_5
| 1
|
Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$ . What is the sum of the digits of $N$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$
|
$N$ must be divisible by every positive integer less than $7$ , or $1, 2, 3, 4, 5,$ and $6$ . Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$ , so each number divisible by these is a multiple of $60$ . The smallest multiple of $60$ is $60$ , so the second smallest multiple of $60$ is $2\times60=120$ . Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3}$
| 3
|
4,707
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6
| 1
|
Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$
|
In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°) $=\boxed{36}.$
| 36
|
4,708
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6
| 2
|
Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$
|
Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is $\frac{3x-2x}2 = \frac {x} 2$ by the arc length formula. Also note that $3x + 2x = 360$ , which simplifies to $x= 72.$ Hence the angle formed by the tangents is equal to $\boxed{36}$
| 36
|
4,709
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_6
| 3
|
Two tangents to a circle are drawn from a point $A$ . The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$ . What is the degree measure of $\angle{BAC}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 60$
|
Let the center of the circle be $O$ . The radii extending from $O$ to the points of tangency $B$ and $C$ are both perpendicular to lines $AB$ and $AC$ . Thus $\angle ABO = \angle ACO = 90^{\circ}$ , and quadrilateral $ABOC$ is cyclic, since the opposite angles add to $180^{\circ}$
Since the ratio of the arc lengths is $2 : 3$ , the shorter arc (the arc cut off by $\angle BOC$ ) is $144^{\circ}$
$\angle BOC$ and $\angle BAC$ must add up to $180^{\circ}$ since quadrilateral $ABOC$ is cyclic, so $\angle BAC = 180 - 144 = \boxed{36}$
| 36
|
4,710
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_8
| 1
|
Keiko walks once around a track at the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad \textbf{(B)}\ \frac{2\pi}{3} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ \frac{4\pi}{3} \qquad \textbf{(E)}\ \frac{5\pi}{3}$
|
To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance. We are given the difference in time, so all we need to find is the difference between the distances.
The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves.
The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle.
The formula for the circumference of a circle is $C = 2 * \pi * r$ where $r$ is the radius of the circle.
Let's define the circumference of the inside circle as $C_1$ and the circumference of the outside circle as $C_2$
If the radius of the inside circle ( $r_1$ ) is $n$ , then given the thickness of the track is 6 meters, the radius of the outside circle ( $r_2$ ) is $n + 6$
Using this, the difference in the circumferences is:
$C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$
$12\pi$ is the difference between the inside and outside lengths of the track. Divided by the time differential, we get:
$12\pi \div 36 = \boxed{3}$
| 3
|
4,711
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_10
| 1
|
Rectangle $ABCD$ has $AB=6$ and $BC=3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD$ . What is the degree measure of $\angle AMD$
$\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75$
|
Since $AB \parallel CD$ $\angle AMD = \angle CDM$ , so $\angle AMD = \angle CMD = \angle CDM$ , so $\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$ . Therefore, $\angle BMC = 30^\circ$ . Therefore $\angle AMD=\boxed{75}$
| 75
|
4,712
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_11
| 1
|
A frog located at $(x,y)$ , with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$ . What is the smallest possible number of jumps the frog makes?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
|
Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle).
Because either $1$ $5$ , or $7$ is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, the frog cannot reach $(1,0)$ in two moves.
However, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$
Thus, the answer is $= \boxed{3}$
| 3
|
4,713
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_13
| 1
|
Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$ . What is the sum of the possible values of $w$
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$
|
Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$ .
This means $a+b+c=9$ $a,b,c$ must be in the set ${1,3,4,5,6}$ .
The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$ $a+b$ , and $b+c$ then must be the remaining two numbers which are $4$ and $6$ .
The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$
\begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*}
\begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*}
The sum of the two w's is $15+16=31$ $\boxed{31}$
| 31
|
4,714
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_15
| 1
|
How many positive two-digit integers are factors of $2^{24}-1$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$
~ pi_is_3.14
|
Repeating difference of squares
$2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$
$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$
$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$
The sum of cubes formula gives us:
$2^{12}+1=(2^4+1)(2^8-2^4+1)$
$2^{12}+1 = 17\cdot241$
A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$ , since multiplying by $241$ will make any factor too large.
Multiplying $17$ by $3$ or $5$ will give a two-digit factor; $17$ itself will also work. The next smallest factor, $7$ , gives a three-digit number. Thus, there are $3$ factors that are multiples of $17$
Multiplying $13$ by $3$ $5$ , or $7$ will also give a two-digit factor, as well as $13$ itself. Higher numbers will not work, giving $4$ additional factors.
Multiply $7$ by $3$ $5$ , or $3^2$ for a two-digit factor. There are no more factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors.
Multiply $5$ by $3$ or $3^2$ for a two-digit factor. All higher factors have been counted already, so there are $2$ more factors.
Thus, the total number of factors is $3+4+3+2=\boxed{12}$
| 12
|
4,715
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17
| 1
|
Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$
|
$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$
$h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$
Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$
For $n=1$ $h_{1}(x)=10x - 1$
Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n:
\begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*}
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.
$h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$ , which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089}$
| 89
|
4,716
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_17
| 2
|
Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$
$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$
|
As before, $h_1(x)=10x-1$ . Compute $h_1(x)$ $h_2(x)$ , and $h_3(x)$ to yield 9, 89, and 889. Notice how this trend will repeat this trend (multiply by 10, subtract 1, repeat). As such, $h_{2011}$ is just 2010 8's followed by a nine. $2010(8)+9=\boxed{16089}$
| 89
|
4,717
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_23
| 1
|
A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A = (-3, 2)$ and $B = (3, -2)$ . Consider all possible paths of the bug from $A$ to $B$ of length at most $20$ . How many points with integer coordinates lie on at least one of these paths?
$\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255$
|
We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20$ , then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$ and $|x + 3| + |y - 2|$ is the length of the shortest path from $(x,y)$ to $(-3, 2)$
If $-3\le x \le 3$ , then $-7\le y \le 7$ satisfy the property. there are $15 \times 7 = 105$ lattice points here.
else let $3< x \le 8$ (and for $-8 \le x < -3$ because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from $(-3, 2)$ to $(x, y)$ added to the shortest distance from $(x, y)$ to $(3, -2)$ is $|x - 3| + |y + 2| + |x + 3| + |y - 2|$ . Since the minimum value for the difference between the y-coordinates is at $y = 0$ , we get $2x + 4 = 20$ or $-2x + 4 = 20$ . Thus, the upper and lower bounds for $x$ are $8$ and $-8$ , respectively.
Now we test each value for x satisfying $3< x \le 8$ and double the result because of symmetry.
For $x = 4$ , the possibles values of y are such that $|2y| \le 12$ for a total of $13$ lattice points,
for $x = 5$ , the possibles values of y are such that $|2y| \le 10$ for a total of $11$ lattice points,
for $x = 6$ , the possibles values of y are such that $|2y| \le 8$ for a total of $9$ lattice points,
for $x = 7$ , the possibles values of y are such that $|2y| \le 6$ for a total of $7$ lattice points,
for $x = 8$ , the possibles values of y are such that $|2y| \le 4$ for a total of $5$ lattice points,
Hence, there are a total of $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{195}$ lattice points.
| 195
|
4,718
|
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_23
| 2
|
A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$ -axis or $y$ -axis. Let $A = (-3, 2)$ and $B = (3, -2)$ . Consider all possible paths of the bug from $A$ to $B$ of length at most $20$ . How many points with integer coordinates lie on at least one of these paths?
$\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255$
|
(Anyone mind making a diagram for this)
Notice that the bug is basically moving from A to B (length 10) but going on a detour in the middle.
Specifically the detour would be of length 5 to some point and then going back by retracing its path (also length 5). Just to simplify things we can observe that the coordinates don't matter here and all we need to remember is that A and B are the diagonally opposite vertices of a 4 by 6 rectangle. The bug can start the "detour" from any point on or inside the rectangle. Notice that the bug can go 5 steps in the y direction, 5 steps in the x direction, or anything in between, so the points covered by possible detours from any point would look like a rhombus or square rotated 45 degrees (with centre at a point on or inside the rectangle). Drawing this out we would get an octagon.
Finding the final answer is then easy, for this solution I will slice the octagon into 4 rectangles (2 of which are squares) and 4 isosceles triangles. There are $(4+1)\cdot (6+1) = 35$ points on or inside the original triangle, $5\cdot 7$ points covered by the rectangles above and below the original one and $5\cdot 5$ points for the squares to the right and left of the original triangle. Lastly each of the four isosceles triangles cover $4+3+2+1 = 10$ points. (Notice that although the length of the detour is 5, the points on the edge of the triangles were already counted).
Adding these up, we get $3\cdot 35 + 2\cdot 25 + 4\cdot 10 = 195 => \boxed{195}$
| 195
|
4,719
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_1
| 1
|
What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$
$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$
|
$20-2010+201+2010-201+20=20+20=\boxed{40}$
| 40
|
4,720
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_2
| 1
|
A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?
$\textbf{(A)}\ 585 \qquad \textbf{(B)}\ 594 \qquad \textbf{(C)}\ 672 \qquad \textbf{(D)}\ 679 \qquad \textbf{(E)}\ 694$
|
It is easy to see that the ferry boat takes $6$ trips total. The total number of people taken to the island is
\begin{align*}&100+99+98+97+96+95\\ &=6(100)-(1+2+3+4+5)\\ &=600 - 15\\ &=\boxed{585}
| 585
|
4,721
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_3
| 1
|
Rectangle $ABCD$ , pictured below, shares $50\%$ of its area with square $EFGH$ . Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$ . What is $\frac{AB}{AD}$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$
|
From the problem statement, we know that \[\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}\]
If we let $a = EF$ and $b = AD$ , we see \[[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}\] . Hence, $\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{10}$
| 10
|
4,722
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_5
| 1
|
Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next $n$ shots are bullseyes she will be guaranteed victory. What is the minimum value for $n$
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 46$
|
Let $k$ be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.
\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ 6n &> 250\end{align*}
The lowest integer value that satisfies the inequality is $\boxed{42}$
| 42
|
4,723
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_6
| 1
|
$\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$
|
$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$
| 24
|
4,724
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_8
| 1
|
Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$
|
Let $\angle BAE = \angle ACD = x$
\begin{align*}\angle BCD &= \angle AEC = 60^\circ\\ \angle EAC + \angle FCA + \angle ECF + \angle AEC &= \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ \angle EAC &= 60^\circ - x\\ \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}
Since $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$ , triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90}$
| 90
|
4,725
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9
| 1
|
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
|
Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$
| 7
|
4,726
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9
| 2
|
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
|
We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.
Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$
| 7
|
4,727
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_9
| 3
|
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
|
We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7}$
| 7
|
4,728
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_10
| 1
|
The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$
|
$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$
\begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*}
The common difference is $4$ . The first term is $5$ and the $2010^\text{th}$ term is
\[5+4(2009) = \boxed{8041}\]
| 41
|
4,729
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_10
| 2
|
The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$
|
Since all the answer choices are around $2010 \cdot 4 = 8040$ , the common difference must be $4$ . The first term is therefore $9 - 4 = 5$ , so the $2010^\text{th}$ term is $5 + 4 \cdot 2009 = \boxed{8041}$
| 41
|
4,730
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_12
| 1
|
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
|
Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
Hence we must have one toad and $\boxed{3}$ frogs.
| 3
|
4,731
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_12
| 2
|
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
|
Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$ , or $n=0, 1$ ). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.
Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{3}$
| 3
|
4,732
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 1
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
The image below shows the two curves for $k=4$ . The blue curve is $x^2+y^2=k^2$ , which is clearly a circle with radius $k$ , and the red curve is a part of the curve $xy=k$
[asy] import graph; size(200); real f(real x) {return 4/x;}; real g1(real x) {return sqrt(4*4-x*x);}; real g2(real x) {return -sqrt(4*4-x*x);}; draw(graph(f,-20./3,-0.6),red); draw(graph(f,0.6,20./3),red); draw(graph(g1,-4,4),blue); draw(graph(g2,-4,4),blue); axes("$x$","$y$"); [/asy]
In the special case $k=0$ the blue curve is just the point $(0,0)$ , and as $0\cdot 0=0$ , this point is on the red curve as well, hence they intersect.
The case $k<0$ is symmetric to $k>0$ : the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$
Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$ . Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$
At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution.
For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$ . To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$ . By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$ , and that equality holds whenever $x^2 = (k/x)^2$ , i.e., $x=\pm\sqrt k$
Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$ . Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$
Clearly the only such integer is $k=1$ , hence the two curves are only disjoint for $k=1$ and $k=-1$ .
This is a total of $\boxed{2}$ values.
| 2
|
4,733
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 2
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.
Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates $(x, k/x)$ derived from the equation. Then, all coordinates that satisfy $\sqrt{ x^2+ (k/x)^2 } \leq k$ intersect the circle.
Squaring, we find $x^2+(k/x)^2 \leq k^2.$ After multiplying through by $x^2$ and rearranging, we find $x^4-x^2k^2+k^2 \leq 0$ .
We see this is a quadratic in $x^2$ and consider taking the determinant, which tells us that solutions are real when, after factoring: $k^2(k^2-4) \geq 0$ We plot this inequality on the number line to find it is satisfied for all values except: $(-1, 0, 1)$ .
We then eliminate 0 because it is extraneous as both $xy=0$ and $x^2+y^2=0$ are points which coincide.
Therefore, there are a total of $\boxed{2}$ values.
| 2
|
4,734
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 3
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
Since $xy=k$ , multiply the equation by 2 on both sides to get $2xy=2k$ . Now we can add the two equations to get $(x+y)^2=k^2+2k$ , for which the only value of $k$ that does not satisfy the equation is $-1$ , as that makes the RHS negative. Similarly, if we subtract the two equations, we obtain $(x-y)^2=k^2-2k$ , for which the only value of $k$ that does not satisfy the equation is $1$ , for the same reason above.
Thus, the only values are $k = 1, -1$ , giving us a total of $\boxed{2}$ values.
| 2
|
4,735
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 4
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
Multiply $k=xy$ by and substitute it into $k^2=x^2+y^2$ . Then, $k=\frac{x^2+y^2}{xy}$ . Recognize it? It's also $k=\frac{x}{y}+\frac{y}{x}$ . The minimum of this function (more accurately the minimum absolute value of the function) is k=2, -2 (when x=y or x=-y). As long as k>2 or k<-2, the function is valid. As such, $k\neq1,-1 \implies \boxed{2}$ . Elegant, huh?
| 2
|
4,736
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 5
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
Assume that $k\ge 0$ since if $k$ works then $-k$ also works. Let $x = ka$ and $y = kb$ . Then the given equations become $ab = \frac{1}{k}$ and $a^2 + b^2 = 1$ , which we don't want to intersect. The points that are the closest on this graph are $(1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{k})(1/\sqrt{k})$ . We don't want the hyperbola to intersect or go inside the circle, so we require $1/\sqrt{k} > 1/\sqrt{2}$ , so $k< 2\implies k = 0, 1$ . Also, by symmetry $k = -1$ also works.Obviously $k = 0$ doesn't work, so we can discard that, leaving $\boxed{2}$ integral solutions for $k$
| 2
|
4,737
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 6
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
To know when the system does not intersect, we first need to find when they do intersect. To do so, we substitute $xy=k$ into $x^{2}+y^{2}=k^{2}$ giving:
$x^2+y^2=(xy)^2$
By Simon's Favorite Factoring Trick, we get $1=(1-x)(1+x)(1-y)(1+y)$
Notice that other than for $x$ and/or $y$ equal to $-1$ and/or $1$ because for any $x$ or $y$ we can get some coordinates that work even if they're fractions which through $xy=k$ it can become an integer as $\dfrac{1}{2}\times2=1$
This means we have to exclude $xy=k=1$ or $-1$ . This is $\boxed{2}$ values that do not work.
| 2
|
4,738
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_13
| 7
|
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
|
The first equation is describing a circle with radius $k$ centered at $(0,0)$ , so we can rewrite it in terms of a new variable $\theta$
\[(k\cos(\theta))^2+(k\sin(\theta))^2=k^2.\]
A quick check shows that this identity holds since $\sin^2+\cos^2=1$ . Now, with an alternate expression for $x$ and $y$ , we can plug this into the second equation to get:
\[(k\sin(\theta))(k\cos(\theta))=k\rightarrow k^2\sin(\theta)\cos(\theta)=k.\]
Since the two equations does intersect when $k=0$ at $(x,y)=(0,0)$ , we do not need to consider the $k=0$ case, so we can safely divide both sides by $k^2$ to get:
\[\sin(\theta)\cos(\theta)=\frac{1}{k}.\]
We can see that the intersections of these two equations will remain unchanged, only now that the solutions will be in terms of $k$ and $\theta$ . Because the range of both $\sin$ and $\cos$ are limited to $[0,1]$ , the range of $\sin(\theta)\cos(\theta)$ will not exceed $1$ . However, the only way that the product could equal 1 is if both $\sin(\theta)$ and $\cos(\theta)$ to equal 1 at the same time, so since the maxima of the functions don't match up, this is an impossible case. Therefore, this equation will only have valid solutions when $k>1$ , not at $k=1$ . A quick sanity check with $\theta=\frac{\pi}{4}$ shows that $\frac{1}{k}=\frac{1}{2}$ and all values less than $\frac{1}{2}$ is indeed contained in the range.
For all $k<0$ , symmetry about the $y$ -axis will give an identical argument with the equation $\sin(\theta)\cos(\theta)=-\frac{1}{k}$ , with $k<-1$ producing valid solutions for our system. Then, the only two values of $k$ such that the system of equations won't intersect are $1$ and $-1$ , a total of $\boxed{2}$ values.
| 2
|
4,739
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_14
| 1
|
Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$
|
By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$
| 33
|
4,740
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_17
| 1
|
Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$
|
It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore
$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$
Based on the initial conditions,
\[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\]
Simplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{6}$
| 6
|
4,741
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18
| 1
|
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
|
Each path must go through either the second or the fourth quadrant.
Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ $(-3,3)$ , and $(-2,2)$
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$ $(3,-3)$ , and $(2,-2)$ .
Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$
| 698
|
4,742
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_18
| 2
|
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
|
As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$ . Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).
For the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$
For each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:
right, right, up, right,..., etc.
Also notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to
$\binom{8}{0}$
Now we consider the number of ways to get from $(4, -4)$ to $(4, 4)$ . by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$
Moving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ "right"s and $1$ "up". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ "up" into $8$ spots. This is
$\binom{8}{1} = 8$
Similarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$
Finally, for the third case (ant passes through $(2, -2)$ ) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ "ups" in a sequence of $8$ movements, or
$\binom{8}{2} = 28$
Again, we square $28$ $28^2 = 784$ . Adding up all of these cases we get
$1+64+784 = 849$
paths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have
$849*2=1698 \Rightarrow \boxed{1698}$
| 698
|
4,743
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_20
| 1
|
Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1<a_2 \le b_2$ and $a_n b_n = 2010$ for some $n$ . What is the largest possible value of $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$
|
Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$ , we can write the terms of each sequence as
\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ &\left(b_n\right) \Rightarrow \{1, y+1, 2y+1, 3y+1, ...\}\end{align*}
where $x$ and $y$ $x\leq y$ ) are the common differences of each, respectively.
Since
\begin{align*}a_n &= (n-1)x+1\\ b_n &= (n-1)y+1\end{align*}
it is easy to see that
$a_n \equiv b_n \equiv 1 \mod{(n-1)}$
Hence, we have to find the largest $n$ such that $\frac{a_n-1}{n-1}$ and $\frac{b_n-1}{n-1}$ are both integers; equivalently, we want to maximize $\gcd(a_n-1, b_n-1)$
The prime factorization of $2010$ is $2\cdot{3}\cdot{5}\cdot{67}$ . We list out all the possible pairs that have a product of $2010$ , noting that these are the possible values of $(a_n, b_n)$ and we need $a_n \leq b_n$
\[(2,1005), (3, 670), (5,402), (6,335), (10,201),(15,134),(30,67)\]
and soon find that the largest $n-1$ value is $7$ for the pair $(15, 134)$ , and so the largest $n$ value is $\boxed{8}$
| 8
|
4,744
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_20
| 2
|
Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1<a_2 \le b_2$ and $a_n b_n = 2010$ for some $n$ . What is the largest possible value of $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$
|
Since
$a_n*b_n = 2010,$
and
$a_n \le b_n$ ,
blue+yellow=green
it follows that
$a_n \le \sqrt{2010} \Rightarrow a_n \le 44$
But $a_n$ and $b_n$ are also integers, so $a_n$ must be a factor of $2010$ smaller than $44$ . Notice that $2010 = 2*3*5*67$ . Therefore $a_n = 2, 3, 5, 6, 112, 15,$ or $30$ and $b_n = 1005, 670, 402, 335, 201, 134,$ or $67$ ; respectively.
Notice that the term $a_m$ is equivalent to the first term $a_1 = 1$ plus $(m-1)$ times the common difference for that particular arithmetic sequence. Let the common difference of $(a_n)$ be $k$ and the common difference of $(b_n)$ be $i$ (not $\sqrt{-1}$ ). Then
$a_n$ (the $n$ th term, not the sequence itself) $=1 + k(n-1)$
and
$b_n = 1 + i(n-1)$
Subtracting one from all the possible values listed above for $a_n$ and $b_n$ , we get
$k(n-1) = 1, 2, 4, 5, 9, 14, 29$
and
$i(n-1) = 1004, 669, 401, 334, 200, 133, 66$
In order to maximize $n$ , we must maximize $n-1$ . Therefore $k$ and $i$ are coprime and $n-1$ is the GCF of any corresponding pair. Inspecting all of the pairs, we see that the GCF is always $1$ except for the pair $(14, 133),$ which has a GCF of $7$ . Therefore the maximum value of $n$ is $8 \Rightarrow \boxed{8}$
| 8
|
4,745
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21
| 1
|
The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
|
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$
Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$
Suppose we let $p$ $q$ , and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$ $q$ , and $r$
\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}
In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$
\[(x^3-ux^2+vx-w)^2\] \[= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2\]
[Quick note: Since we don't know $a$ $b$ , and $c$ , we really don't even need the last 3 terms of the expansion.]
\begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}
All that's left is to find the largest root of $x^3-5x^2+2x+8$
\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{4}
| 4
|
4,746
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21
| 2
|
The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
|
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$ .
We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2.
Let the function be $(x-p)^2(x-q)^2(x-r)^2$
Applying Vieta's formulas, we get $2p+2q+2r = 10$ or $p+q+r = 5$ .
Applying it again, we get, after simplification, $p^2+q^2+r^2+4pq+4pr+4qr = 29$
Notice that squaring the first equation yields $p^2+q^2+r^2+2pq+2qr+2pr= 25$ , which is similar to the second equation.
Subtracting this from the second equation, we get $2pq+2pr+2qr = 4$ . Now that we have the $pq+pr+qr$ term, we can manpulate the equations to
yield the sum of squares. $2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4$ or $2p^2+2q^2+2r^2+2pq+2qr+2pr = 46$ . We finally reach $(p+q)^2+(q+r)^2+(p+r)^2 = 46$
Since the answer choices are integers, we can guess and check squares to get $\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}$ in some order. We can check that this works by adding then and seeing $2p+2q+2r = 10$ . We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get $\boxed{4}$
| 4
|
4,747
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_21
| 3
|
The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
|
First, $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$ has exactly $3$ roots. Therefore, $y = (kx^3+lx^2+mx+n)^2 = 0$
So, $k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0$
By matching the coefficients of the first $4$ terms, we have $k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4$
Solving the equations above, we have $2$ sets of solutions; first set of which is $k = 1, l = -5, m = 2, n = 8$ . Second set of which is $k = -1, l = 5, m = -2, n = -8$ . After squaring both sets, they are the same i.e. $x^3-5x^2+2x+8 = 0$
This is equal to $(x-4)(x-2)(x+1) = 0$ . The largest root is $\boxed{4}$
| 4
|
4,748
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_22
| 1
|
What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$
|
If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$ , where $m$ is any integer from $1$ to $119$ , inclusive: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$
The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$ , since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$
The sum of the slopes at $x = \frac{1}{m}$ is
\begin{align*}&\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i\\ &=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i\\ &=-m^2-m+7140\end{align*}
Now we want to minimize $-m^2-m+7140$ . The zeros occur at $-85$ and $84$ , which means the slope is $0$ where $m = 84, 85$
We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49}$
| 49
|
4,749
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_22
| 2
|
What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$
|
By the triangle inequality, $|x-1|+|2x-1|+|3x-1|+\cdots + |119x-1| \geq |(x-1)+(2x-1)+\cdots+(119x)-1|.$ However, we may change signs of some of these terms to cancel out the $x$ 's.
Since the minimum exists, we want all the $x$ s to cancel out. Thus, we want to find some $n$ such that \[1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119\] \[\frac{n(n+1)}{2}=\frac{119\cdot120}{2}-\frac{n(n+1)}{2}\] \[n^2+n-7140=0\] \[n=84\]
Then, $x=\frac{1}{n}= \frac{1}{84}$ . The answer(expression's value) is then $84*1+(119-85+1)*(-1)$ , which becomes $84-35=\boxed{49}$
| 49
|
4,750
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24
| 1
|
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
|
The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$
First, let us find the number of zeros of the inside of the logarithm.
\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*}
After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval $\left(0, \frac{1}{8}\right)$ is obviously positive. This means the next interval $\left(\frac{1}{8}, \frac{1}{7}\right)$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$ . Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$
And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12}$
| 12
|
4,751
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24
| 2
|
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
|
Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)$ . It is easy to see that for $y_i$ to be positive for $1\leq i\leq8$ $\lfloor i x_i \rfloor$ must be even for $1 \leq i\leq8$ . Since an even number of positives times an even number of negatives equals a positive, there can be $2, 4, 6,$ or $8$ positive values of $y_i$ for $1 \leq i\leq 8$ for a given value of $x$ . (since $y_1$ is always positive on the range $[0, 1]$ )
Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from $0$ to $1$ . (I recommend increments of at most $0.1$ .) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for $y_1, y_2, ..., y_8$ , respectively. Draw a large bar (label it with $+$ so you know it's positive) for all values of $x_i$ such that $\lfloor i x_i \rfloor$ is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at $0$ , and slowly work your way to $1$ , marking all disjoint intervals in which your straightedge touches $2, 4, 6,$ or $8$ boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. )
If done correctly, you should obtain $\boxed{12}$ as your answer.
| 12
|
4,752
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_24
| 3
|
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
|
You should be able to somewhat visualize what the $\sin$ function looks like (if you can't then you should look it up and try to memorize it). To summarize, the graph of $y= \sin{x}$ is positive from the interval $(0, \pi)$ and negative from the interval $(\pi, 2\pi)$ (notice how the intervals use parentheses instead of brackets, as brackets would denote inclusive bounds and that it incorrect).
All of the various $\sin$ functions that we are multiplying are in the form of $k\pi x$ , which means the intervals have now changed to integer values.
Also, the period of the function $y =\sin{k \pi x}$ is equivalent to $\frac{2}{k}$ ; For example the period of $y= \sin{4\pi x}$ is $\frac{1}{2}$
Let a "wave" be a full period of a $\sin$ curve. Then, a "half-wave" is half of a full period of a $\sin$ curve. Over the interval $[0, 1],$ $y= \sin{\pi x}$ will display $1$ "half-wave", $y= \sin{2\pi x}$ will display $2$ "half-waves", and $y= \sin{k\pi x}$ will display $k$ "half-waves". Also notice that the graph of $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
The problem asks for the number of disjoint intervals, not the values of the intervals themselves. This means we only want the positive values of
$\prod_{k=1}^8 \sin{k\pi x}$
(the product of all of the various $\sin$ functions)
We don't care about the specific values themselves, we just need to know what the sign (+/-) of the product is. Notice that if an even number of the functions are negative, the product will be positive; conversely, if the number of negative functions is odd, the product will be negative. Therefore we only need to count the number of intervals during which an even number of functions are negative/positive. By definition, the sign of a function changes after it crosses one of its roots (because trig functions don't have multiplicities). Notice that the roots of $y = \sin{k\pi x}$ are in the form
$\frac{1}{k}, \frac{2}{k}, \frac{3}{k},...$
(this is in agreement with the fact that $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
Therefore, across the interval $[0, 1]$ , the sign will change every time the function hits a root of one of the $\sin$ functions. We must be careful, however, as some roots (such as $\frac{1}{2}$ and $\frac{2}{4}$ ) are duplicates.
Notice that the duplicates always come in pairs. For example, $\frac{1}{2}$ is also $\frac{2}{4}, \frac{3}{6},$ and $\frac{4}{8}$ ; there are four equivalent roots. Across the interval $[0, 1],$ there are no duplicate roots that come in triplets, or in fives. The roots are always single, doubled up, or quadrupled up. So every time the function reaches a single root, it switches signs, and every time it reaches a duplicate root, the sign remains the same (this still counts as 2 distinct intervals, though, as the zeroes "separate" them). Listing out the roots (not including $x=0$ and $x=1$ ):
$1$ : n/a
$2: \frac{1}{2}$
$3: \frac{1}{3}, \frac{2}{3}$
$4: \frac{1}{4}, \frac{2}{4}, \frac{3}{4}$
$5: \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$
$6: \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$
$7: \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$
$8: \frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}$
Then in order (showing duplicates as well):
$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$
Then creating a sequence of signs based on the roots:
$+,-,+,-,+,+,-,-,+,-,+,+,-,+,-,-,+,+,-,+,-,+$
We see that there are $12$ positive intervals $\Rightarrow \boxed{12}$
| 12
|
4,753
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25
| 1
|
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$
|
It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic.
Proof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\overline{BD}$ . When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\angle DAB$ or $\angle BCD$ will be degenerate and measure $0^\circ$ , so opposite angles will sum to less than $180^\circ$ . When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^\circ$ , so opposite angles will sum to more than $180^\circ$ . Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^\circ$
Denote $a$ $b$ $c$ , and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\ge b \ge c \ge d$
Since $a+b+c+d = 32$ , the Triangle Inequality implies that $a \le 15$
We will now split into $5$ cases.
Case $1$ $a = b = c = d$ $4$ side lengths are equal)
Clearly there is only $1$ way to select the side lengths $(8,8,8,8)$ , and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed.
Case $2$ $a = b = c > d$ or $a > b = c = d$ $3$ side lengths are equal)
If $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals.
Case $3$ $a = b > c = d$ $2$ pairs of side lengths are equal)
$a$ and $b$ can be any integer from $9$ to $15$ , and likewise $c$ and $d$ can be any integer from $1$ to $7$ . However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\cdot{2} = 14$
Case $4$ $a = b > c > d$ or $a > b = c > d$ or $a > b > c = d$ $2$ side lengths are equal)
If the $2$ equal side lengths are each $1$ , then the other $2$ sides must each be $15$ , which we have already counted in an earlier case. If the equal side lengths are each $2$ , there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1 = 45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\cdot{45} = 135$
Case $5$ $a > b > c > d$ (no side lengths are equal)
Using the same counting principles starting from $a = 15$ and eventually reaching $a = 9$ , we find that the total number of possible side lengths is $69$ . There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\frac{4!}{4} = 6$ . The total number of quadrilaterals is $6\cdot{69} = 414$
And so, the total number of quadrilaterals that can be made is $414 + 135 + 14 + 4 + 1 = \boxed{568}$
| 568
|
4,754
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25
| 2
|
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$
|
As with solution $1$ we would like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let $a \ge b \ge c\ge d$ be the sides of the quadrilateral.
There are $\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three. This occurs when $a \ge 16$ . For $a=16$ $b+c+d=16$ . There are $\binom{15}{2}$ ways to partition $16$ . Since $a$ could be any of the four sides, we have counted $4\binom{15}{2}$ degenerate quadrilaterals. Similarly, there are $4\binom{14}{2}$ $4\binom{13}{2} \cdots 4\binom{2}{2}$ for other values of $a$ . Thus, there are $\binom{31}{3} - 4\left(\binom{15}{2}+\binom{14}{2}+\cdots+\binom{2}{2}\right) = \binom{31}{3} - 4\binom{16}{3} = 2255$ non-degenerate partitions of $32$ by the hockey stick theorem. We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is $1$ square case, and $7$ rectangle cases, there are $2255-1-2\cdot7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\frac{2240}{4} = \boxed{568}$ total quadrilaterals.
| 568
|
4,755
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12A_Problems/Problem_25
| 3
|
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$
|
As with solution $1$ we find that there are $2255$ ways to form a quadrilateral if we don't account for rotations. We now apply Burnside's_Lemma . There are four types of actions in the group acting on the set of quadrilaterals. We will consider each individually:
Identity: maps a quadrilateral with sides $a,b,c,d$ in that order to $a,b,c,d$ . Obviously all members of the set of quadrilaterals are fixed points, for a total of $2255$ .
Rotation by one: maps a quadrilateral from $a,b,c,d$ to $b,c,d,a$ . For this to have a fixed point we need $a=b,b=c,c=d,d=a$ , so the only quadrilateral that is a fixed point is the square with side length $8$ , for a total of $1$ .
Rotation by two: maps a quadrilateral from $a,b,c,d$ to $c,d,a,b$ . For this to be a fixed point we need $a=c$ and $b=d$ . Thus the quadrilateral is of the form $x,y,x,y$ —a rectangle. We can count that there are $15$ rectangle cases, namely $(a, b, c, d) = \{ (1, 15, 1, 15), (2, 14, 2, 14), \cdots, (15, 1, 15, 1) \}$ .
Roration by three: maps a quadrilateral from $a,b,c,d$ to $d,a,b,c$ . Similarly to the rotation by one case, there is one fixed point here.
Summing up, we get that the total number of groups is $2255+1+15+1=2272$ . Since there are $4$ members of the group our final answer is $\frac{2272}{4}=\boxed{568}$
| 568
|
4,756
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_1
| 1
|
Makarla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$
|
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 $\frac{135}{540}$ $= \boxed{25}$
| 25
|
4,757
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_4
| 1
|
A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$
|
$31 \equiv 3 \pmod {7}$ 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 $\boxed{3}$
| 3
|
4,758
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_6
| 1
|
At the beginning of the school year, $50\%$ of all students in Mr. Well's class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$
|
Clearly, the minimum possible value would be $70 - 50 = 20\%$ . The maximum possible value would be $30 + 50 = 80\%$ . The difference is $80 - 20 = \boxed{60}$
| 60
|
4,759
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8
| 1
|
Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$
|
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 $23 \Rightarrow \boxed{23}$
| 23
|
4,760
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_8
| 2
|
Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$
|
Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$
Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$
| 23
|
4,761
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_9
| 1
|
Let $n$ be the smallest positive integer such that $n$ is divisible by $20$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
|
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 $\Rightarrow \boxed{7}$
| 7
|
4,762
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_12
| 1
|
For what value of $x$ does
\[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024$
|
Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$ , we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$ . Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$ , so $x = 2^8 = 256$ , or $\boxed{256}$
| 256
|
4,763
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14
| 1
|
Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$
|
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 $671 \Rightarrow \boxed{671}$
| 671
|
4,764
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14
| 2
|
Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$
|
Since $a + b \le M$ $d + e \le M$ , and $c < b + c \le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \ge 671$
For the values $(a,b,c,d,e) = (669,1,670,1,669)$ $M = 671$ , so the smallest possible value of $M$ is $\boxed{671}$ . The answer is (B).
| 671
|
4,765
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_14
| 3
|
Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$
|
Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.)
Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
1. (high, low, high, low, high)
or
2. (low, high, low, high, low)
In the extreme case that each "low" = $0$ $2010$ will be divided into either $3$ or $2$ numbers for cases 1 and 2, respectively. Obviously dividing by $3$ will yield a lower number, so we consider case 1.
Dividing $2010$ by $3$ yields $670$ , or
$670, 0, 670, 0, 670$
However, all five numbers must be positive, and the closest we can get to this is
$668, 3, 668, 3, 668$
The lowest possible sum of two adjacent numbers then becomes $671$ , or $\boxed{671}$
| 671
|
4,766
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_15
| 1
|
For how many ordered triples $(x,y,z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\}$ , where $i=\sqrt{-1}$
$\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235$
|
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 ${95+95+35=\boxed{225}$ ordered triples
| 225
|
4,767
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_17
| 1
|
The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$
|
Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.
\[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\]
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$
\[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\]
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$
\[12+18+12=\boxed{42}\]
| 42
|
4,768
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_17
| 2
|
The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$
|
This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$
| 42
|
4,769
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_19
| 1
|
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$
|
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 \[a(1+r)(1+r^2) = 4a+6d+1.\]
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 $5+10+5+14=\boxed{34}$
| 34
|
4,770
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_19
| 2
|
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$
|
Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$ . Now we can narrow our search for the values of $a,d$ , and $r$ . Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:
\[a(1+r+r^{2}+r^{3})=4a+6d+1\]
Now we can start trying out some values of $r$ . We try $r=2$ , which gives
\[15a=4a+6d+1\]
\[11a=6d+1\]
We need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\equiv 1 \pmod{6}$ . We see that this is $55$ , and therefore $a=5$ and $d=9$
So the Raiders' first two scores were $5$ and $10$ and the Wildcats' first two scores were $5$ and $14$
\[5+10+5+14=34 \longrightarrow \boxed{34}\]
| 34
|
4,771
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20
| 1
|
A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
|
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 $8 \Rightarrow \boxed{8}$
| 8
|
4,772
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20
| 2
|
A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
|
Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$ ; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$ . Then, working backwards we have $a_3=\frac{1}{r}$ $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$ . Now notice that since $a_1=\sin(x)$ and $a_2=\cos(x)$ , we need $a_1^2+a_2^2=1$ , so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$ . Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$ , which the left side is equal to $1+\cos(x)$ ; we see as well that the right hand side is equal to $a_8$ given $a_4=1$ , so the answer is $\boxed{8}$ . - mathleticguyyy
| 8
|
4,773
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_21
| 1
|
Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
|
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, 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 $\boxed{315}$
| 315
|
4,774
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_21
| 2
|
Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
|
The evenly-spaced data suggests using discrete derivatives to tackle this problem. First, note that any polynomial of degree $n$
can also be written as
Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ . This latter form is convenient for calculating discrete derivatives of $P(x)$
The discrete derivative of a function $f(x)$ is the related function $\Delta f(x)$ defined as
With this definition, it's easy to see that for any positive integer $k$ we have
This in turn allows us to use successive discrete derivatives evaluated at $x=1$ to calculate all of the coefficients $b_i$ using
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
Thus we can read down the column for $x=1$ to find that $k! b_k = (-2)^k a$ for $k = 0, 1, \ldots, 7$ . Interestingly, even if we choose $P(x)$ to have degree greater than $7$ , the $8$ coefficients of lowest order always satisfy these conditions. Moreover, it's straightforward to show that the $P(x)$ of degree $7$ with $b_k$ satisfying these conditions will fit the data given in the problem statement. Thus, to find minimal necessary and sufficient conditions on the value of $a$ , we need only consider these $8$ equations. As a result, $P(x)$ with integer coefficients fitting the given data exists iff $k!$ divides $2^k a$ for $k = 0, 1, \ldots, 7$ . In other words, it's necessary and sufficient that
The last condition holds if $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ . Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{315}$
| 315
|
4,775
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23
| 1
|
Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$
|
$P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$ . 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 $\boxed{100}$
| 100
|
4,776
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23
| 3
|
Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$
|
Let $P(x) = x^2 - (a+b)x + ab$ and $Q(x) = x^2 - (c+d)x + cd$ . Notice that the roots of $P(x)$ are $a,b$ and the roots of $Q(x)$ are $c,d.$ Then we get:
\begin{align*} P(Q(x)) &= a, b \\ x^2 - (c+d)x + cd &= a, b \end{align*} The two possible equations are then $x^2 - (c+d)x + cd-a=0$ and $x^2 - (c+d)x + cd-b=0$ . The solutions are $-23, -21, -17, -15$ . From Vieta's we know that the total sum $2(c+d) = -76 \implies c+d = -38$ so the roots are paired $-23, -15$ and $-21, -17$ . Let $cd - a = 23*15$ and $cd - b = 21*17$
We can similarly get that $ab - c = 59*49$ and $ab - d = 57*51$ , and $a+b = -108$ . Add the first two equations to get \[2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.\] This means $Q(x) = x^2 + 38x + 297$
Once more, we can similarly obtain \[ab = \frac{59*49 + 57*51 - 38}{2} = 2880.\] Therefore $P(x) = x^2 + 108x + 2880$
Now we can find the minimums to be \[19^2 - 19*38 + 297 = -64\] and \[54^2 - 54*108 + 2880 = -36.\] Summing, the answer is $\boxed{100}.$
| 100
|
4,777
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_23
| 4
|
Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$
|
Let $P(x) = (x+a)(x+b)$ $Q(x) = (x+c)(x+d)$
$P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)$
$Q(P(x)) = (x^2 + ax + bx + ab + c)(x^2 + ax + bx + ab + d)$
Notice how the coefficient for $x$ has to be the same for the two quadratics that are multiplied to create $P(Q(x))$ , and $Q(P(x))$
$P(Q(x)) = (x+ 23)(x+ 21)(x+ 17)(x+ 15) = (x^2 + 38x + 345)(x^2 + 38x + 357)$
$Q(P(x)) = (x+ 59)(x+ 57)(x+ 51)(x+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2907)$
$c + d = 38$ $cd + a = 345$ $cd + b = 357$ $a + b = 108$ $ab + c = 2891$ $ab + d = 2907$
$cd = \frac{345 + 357 - 108}{2} = 297$ $297 = 3^3 \cdot 11$ $c = 27$ $d = 11$
$a = 345 - 27*11 = 48$ $b = 357 - 27*11 = 60$
$P(x) = (x+48)(x+60) = x^2 + 108x + 2880 = (x+54)^2 - 36$
$Q(x) = (x+27)(x+11) = x^2 + 38x + 297 = (x+19)^2 -64$
$-36 -64 = \boxed{100}$
| 100
|
4,778
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_25
| 1
|
For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$
|
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 $70 + 1 + 6 = \boxed{77}$
| 77
|
4,779
|
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_25
| 2
|
For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$
|
After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$ , divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$ $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$ $73$ , and $79$ are prime numbers greater than $67$ and less than or equal to $80$ , subtract $3$ from $80$ to get the answer $80-3=\boxed{77}$
| 77
|
4,780
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_4
| 1
|
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55$
|
As all five options are divisible by $5$ , we may not use any pennies. (This is because a penny is the only coin that is not divisible by $5$ , and if we used between $1$ and $4$ pennies, the sum would not be divisible by $5$ .)
Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is $4\cdot 5 = 20$ . Therefore the option that is not reachable is $\boxed{15}$ $\Rightarrow$ $(A)$
| 15
|
4,781
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_5
| 1
|
One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216$
|
Let the original cube have edge length $a$ . Then its volume is $a^3$ .
The new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ .
The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{125}$
| 125
|
4,782
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_6
| 1
|
Suppose that $P = 2^m$ and $Q = 3^n$ . Which of the following is equal to $12^{mn}$ for every pair of integers $(m,n)$
$\textbf{(A)}\ P^2Q \qquad \textbf{(B)}\ P^nQ^m \qquad \textbf{(C)}\ P^nQ^{2m} \qquad \textbf{(D)}\ P^{2m}Q^n \qquad \textbf{(E)}\ P^{2n}Q^m$
|
We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{2}$
| 2
|
4,783
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_7
| 1
|
The first three terms of an arithmetic sequence are $2x - 3$ $5x - 11$ , and $3x + 1$ respectively. The $n$ th term of the sequence is $2009$ . What is $n$
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 1506 \qquad \textbf{(E)}\ 8037$
|
As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$ . This solves to $x=4$ . The first three terms then are $5$ $9$ , and $13$ . In general, the $n$ th term is $1+4n$ . Solving $1+4n=2009$ , we get $n=\boxed{502}$
| 502
|
4,784
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_8
| 1
|
Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$
|
Let the side length of the smaller square be $1$ , and let the smaller side of the rectangles be $y$ . Since the larger square's area is four times larger than the smaller square's, the larger square's side length is $2$ $2$ is equivalent to $2y+1$ , giving $y=1/2$ . Then, the longer side of the rectangles is $3/2$ $\frac{\frac{3}{2}}{\frac{1}{2}}=\boxed{3}$
| 3
|
4,785
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_9
| 1
|
Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$
|
As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$
To compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$
| 2
|
4,786
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_9
| 2
|
Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$
|
Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$
We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$
Comparing coefficients we have $a=3$ $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \boxed{2}$
| 2
|
4,787
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_10
| 1
|
In quadrilateral $ABCD$ $AB = 5$ $BC = 17$ $CD = 5$ $DA = 9$ , and $BD$ is an integer. What is $BD$
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$
|
By the triangle inequality we have $BD < DA + AB = 9 + 5 = 14$ , and also $BD + CD > BC$ , hence $BD > BC - CD = 17 - 5 = 12$
We get that $12 < BD < 14$ , and as we know that $BD$ is an integer, we must have $BD=\boxed{13}$
| 13
|
4,788
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11
| 1
|
The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$
|
Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$ th triangular number is $\frac{n(n+1)}{2}$ . Putting this together for $F_{20}$ this gives:
$\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$
| 761
|
4,789
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11
| 2
|
The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$
|
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle; draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); [/asy]
In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$
| 761
|
4,790
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11
| 3
|
The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$
|
When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761}
| 761
|
4,791
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_11
| 4
|
The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$
|
The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$ . The first finite differences go $4, 8, 12, 16, \dots$ . The second finite differences go $4, 4, 4, \dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we already know $d_1 = 1$ $d_2 = 3$ , and $d_3 = 13$ . Thus, \[a + b + c = d_1 = 1\] \[4a + 2b + c = d_2 = 3\] \[9a + 3b + c = d_3 = 13\] Solving this system for $a$ $b$ , and $c$ gives $a = 2$ $b = -2$ $c = 1$ . Finally, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{761}$
| 761
|
4,792
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_12
| 2
|
How many positive integers less than $1000$ are $6$ times the sum of their digits?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$
|
The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Since the number is $6$ times the sum of its digits, it must be divisible by $6$ , therefore also by $3$ , therefore the sum of its digits must be divisible by $3$ . With this in mind we can conclude that the number must be divisible by $18$ , not just by $6$ . Since the number is divisible by $18$ , it is also divisible by $9$ , therefore the sum of its digits is divisible by $9$ , therefore the number is divisible by $54$ , which leaves us with $54$ $108$ and $162$ . Only $54$ is $6$ times its digits, hence the answer is $\boxed{1}$
| 1
|
4,793
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15
| 1
|
For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$
|
We know that $i^x$ cycles every $4$ powers so we group the sum in $4$ s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\]
We can postulate that every group of $4$ is equal to $2-2i$ .
For 24 groups we thus, get $48-48i$ as our sum.
We know the solution must lie near
The next term is the $24*4+1=97$ th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$ ) and our sum is now $48+49i$ so $n=97\Rightarrow\boxed{97}$ is our answer
| 97
|
4,794
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15
| 2
|
For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$
|
Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary.
Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$ , and
the imaginary part is $i + 3i^3 + 5i^5 + \cdots$
Let's take a look at the real part first. We have $i^2=-1$ , hence the real part simplifies to $-2+4-6+8-10+\cdots$ .
If there were an odd number of terms, we could pair them as follows: $-2 + (4-6) + (8-10) + \cdots$ , hence the result would be negative. As we need the real part to be $48$ , we must have an even number of terms. If we have an even number of terms, we can pair them as $(-2+4) + (-6+8) + \cdots$ . Each parenthesis is equal to $2$ , thus there are $24$ of them, and the last value used is $96$ . This happens for $n=96$ and $n=97$ . As $n=96$ is not present as an option, we may conclude that the answer is $\boxed{97}$
| 97
|
4,795
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15
| 3
|
For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$
|
Some may know the equation:
\[\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\]
(For those curious, this comes from differentiating the equation for finite geometric sums)
Using this equation, we have
\[48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}\] \[=\frac{1-(n+1)i^n+ni^{n+1}}{-2}\] \[=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}\]
Since the imaginary and the real part must be positive, we know that $i^{n+1}=-1$ or $i^{n+1}=-i$ . By the same line of reason, $i^{n}=1,i$ . This only works for $n\equiv 1 \mod 4$ . Therefore, we have:
\[\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i\]
Solving either the real or imaginary part gives $\boxed{97}$
| 97
|
4,796
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_16
| 1
|
A circle with center $C$ is tangent to the positive $x$ and $y$ -axes and externally tangent to the circle centered at $(3,0)$ with radius $1$ . What is the sum of all possible radii of the circle with center $C$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
|
Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$ . For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$ , the distance between $C$ and $(3,0)$ must be exactly $r+1$
By the Pythagorean theorem the distance between $(r,r)$ and $(3,0)$ is $\sqrt{ (r-3)^2 + r^2 }$ , hence we get the equation $(r-3)^2 + r^2 = (r+1)^2$
Simplifying, we obtain $r^2 - 8r + 8 = 0$ . By Vieta's formulas the sum of the two roots of this equation is $\boxed{8}$
| 8
|
4,797
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17
| 1
|
Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$
|
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$
Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$
| 1
|
4,798
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_17
| 2
|
Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$
|
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
Which can be further rewritten as $r_1-r_1^2 = r_2-r_2^2$ .
Rearranging the equation we get $r_1-r_2 = r_1^2-r_2^2$ .
Expressing this as a difference of squares we get $r_1-r_2 = (r_1-r_2)(r_1+r_2)$
Dividing by like terms we finally get $r_1+r_2 = \boxed{1}$ as desired.
| 1
|
4,799
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18
| 1
|
For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$
|
The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$
This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ $N(4) = \boxed{7}$
| 7
|
4,800
|
https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_18
| 2
|
For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$
|
Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\boxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ $N(k)$ must be less than or equal to $7$
| 7
|
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