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4,401
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22
| 3
|
How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$
|
As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S); [/asy]
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S); [/asy]
Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S); [/asy]
Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); [/asy]
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\]
| 15
|
4,402
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23
| 2
|
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
|
WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $\boxed{12}$
| 12
|
4,403
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23
| 3
|
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
|
This problem is going to require some geometric probability, so let's dive right in.
Take three integers $x$ $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true
\[x + y > z\] \[y + z > x\] \[x + z > y\]
Now, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.
Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.
\[y > -x + z\] \[y > x - z\] \[y < x + z\]
Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ $y \in [0, 1]$
I recommend drawing the following figure to get an idea of what is going on.
The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.
Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$
Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to
Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.
Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!
To find the area of the triangles, our plan of attack will be
Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$
The area of the triangles is
\[\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)\]
The integral from $0$ to $1$ is
\[\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}\]
The total area of all the possible unit squares is quite obviously
\[\int_0^11dx = 1\]
Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is
\[\frac{\frac{1}{2}}{1} = \boxed{12}\]
| 12
|
4,404
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_23
| 4
|
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
|
Consider a stick of length $1$ . Cutting the stick at two random points gives a triangle from the three new segments. These two random points must be on opposite sides of the halfway mark. Thus, after the first cut is made, there is $\boxed{12}$ probability that the second cut is on the opposite side.
| 12
|
4,405
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24
| 1
|
There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
|
The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*} So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum $a$
\[f(x)=\left(x-\frac{a}{3}\right)^3\] \[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\] Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that
\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\] \[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}\]
| 9
|
4,406
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24
| 2
|
There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
|
Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \[\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.\] Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields \[a^2 \ge 27 \quad \Rightarrow \quad a \ge 3\sqrt{3}.\] Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{9}$
| 9
|
4,407
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24
| 3
|
There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
|
We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{9}$
| 9
|
4,408
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25
| 1
|
Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$
$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$
|
Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\le (x+1)^2-x^2=2x+1$ , this first happens at $x\ge \lfloor 99/2\rfloor = 50$ . The perfect squares from here go: $2500, 2601, 2704, 2809\dots$ . Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$ . After the ones and tens make $100$ , the hundreds place will go up by $2$ , thus reaching our goal. Since $10^2=100$ , the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$ . The missing number is one less than the number of hundreds $(k=2)$ of $3600$ , or $35$
Now consider $f(4)$ . Instead of the difference between two squares needing to be $100$ , the difference must now be $10000$ . This first happens at $x\ge 5000$ . After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$ . This will take place at $\left(5000+100\right)^2=5100^2= 26010000$ . Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
In general, each new value of $f(k+2)$ will add two digits to the " $5$ " and one digit to the " $1$ ". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$ , the last for $k = 8$ will be $\left(50000000+10000\right)^2$ , and so on until $k=2016$ . Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$ $k=8$ is $50001^2$ , and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a " $25$ " with two more digits than the last number, a " $10$ " with one more digit, and a " $1$ ". The missing number is one less than that, so the "1" will be subtracted from $f(k)$ . In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$
Therefore:
\[f(2) =35 =25 +10\] \[f(4) =2600 =2500 +100\] \[f(6) =251000 =250000 +1000\] \[f(8) = 25010000 = 25000000 + 10000\]
And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is:
$2.52525252525\dots 2525\cdot 10^{2015}$ $1.11111\dots 110\cdot 10^{1008}$ , with $2016$ repetitions each of " $25$ " and " $1$ ".
There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits.
Since $2n = 2016$ $n = 1008$ , and $8n = 8064$ , or $\boxed{8064}$
| 64
|
4,409
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_25
| 2
|
Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$
$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$
|
We assume $n \geq 1$ for all claims.
We will let $g_k(x) = \left \lfloor \frac{x^2}{10^k}\right \rfloor$ . This is the result when the last k digits are truncated off $x^2$
Let $x_n$ = the smallest a, such that $g_{2n}(a) - g_{2n}(a-1) \geq 2$
We then have $\left \lfloor \frac{a^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(a-1)^2}{10^{2n}} \right \rfloor \geq 2$
Claim 1: $x_n > 5\cdot10^{2n-1}$
Proof of Claim 1:
Assume for the sake of contradiction, we have $x_n \leq 5 \cdot 10^{2n-1}$
Let $c = \frac{2x_n - 1}{10^{2n}}$ $d = \frac{(x_n-1)^2}{10^{2n}}$
Note that since $x_n \leq 5 \cdot 10^{2n-1}$ , we have $c < 1$
It is well known that $\lfloor i+j \rfloor \leq \lfloor i \rfloor + \lfloor j \rfloor + 1$ for any $i$ and $j$
Since $x_n$ satisfies the condition, we have:
$1 = \lfloor c \rfloor + 1 \geq \lfloor{c+d} \rfloor - \lfloor d \rfloor \geq 2$
This is a contradiction. $\blacksquare$
Claim 2: $x_n = 5 \cdot 10^{2n-1} + 10^n$
Proof of Claim 2:
We will show our choice of $x_n = 5 \cdot 10^{2n-1} + 10^n$ satisfies the criteria above.
$\left \lfloor \frac{{x_n}^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} + 10^{2n}}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} - 2 \cdot 10^n + 1}{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + 10^{n} + 1) - (25 \cdot 10^{2n-2} + 10^{n} - 1) = 2$
Now, we will show all values smaller than $5 \cdot 10^{2n-1} + 10^n$ don’t satisfy the criteria.
We will assume for the sake of contradiction, there exists an $x'_n$ which satisfies the criteria.
We will write $x'_n$ as $5 \cdot 10^{2n - 1} + k$ . By our hypothesis, we have $k < 10^n$ . We also have $k > 0$ by claim 1.
We have $\left \lfloor \frac{{x'}_n^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x'_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + k \cdot 10^{2n} + k^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + (k-1) \cdot 10^{2n} + (k-1)^2 }{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + k) - (25 \cdot 10^{2n-2} + k - 1) = 1 \geq 2.$ This gets a contradiction. $\blacksquare$
Claim 3: $f(2n) = 25 \cdot10^{2n - 2} + 10^{n}$
Proof of Claim 3: Because of claim 2, $x_n$ is $5 \cdot 10^{2n-1} + 10^n$ . Note that $g_n(x_n)$ and $g_n(x_n - 1)$ are the first numbers written that will differ by at least 2. Thus, $f(2n) = g_n(x_n - 1) + 1 = 25 \cdot 10^{2n-2} + 10^n$ $\blacksquare$
Thus, $f(2) + f(4) \dots f(2016) = \underbrace{(2525 \dots 25)}_{1008 \ 25s} + \underbrace{(111111 \dots 1111)}_{1008 \ 1s}$ . This addition has no regroups/carry overs, so we can just take the sums of the digits of each of the addends, and sum them together to get 8064. $\boxed{8064}$
| 64
|
4,410
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_1
| 1
|
What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$
|
By: Dragonfly
We find that $a^{-1}$ is the same as $2$ , since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be
\[\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}\]
We can then simplify the equation to get $\boxed{10}$
| 10
|
4,411
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3
| 1
|
Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$
|
By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{4032}$
| 32
|
4,412
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_3
| 2
|
Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$
|
Consider $x$ is negative.
We replace all instances of $x$ with $|x|$
$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$
$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$
$=$ $|x|$ $+|x|$
$=|2x|$
$=4032 \implies \boxed{4032}$
| 32
|
4,413
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4
| 1
|
The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$
|
By: dragonfly
We set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that
$4x=5y$
and
$2\times(90-x)=90-y$
since the larger the original angle, the smaller the complement.
We then find that $x=75$ and $y=60$ , and their sum is $\boxed{135}$
| 135
|
4,414
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_4
| 2
|
The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$
|
We can visualize the problem like so: [asy] path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N); draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N); draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S); draw((0,5)--(0,-15)); draw((90,5)--(90,-15)); [/asy] \[5x+1y = 90^\circ = 4x+2y\] Moving like terms to the same side gets $x = y$ , and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$ , so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{135}$ . ~ emerald_block
| 135
|
4,415
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_6
| 1
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All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$
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By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\boxed{8}$
| 8
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4,416
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_7
| 1
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Josh writes the numbers $1,2,3,\dots,99,100$ . He marks out $1$ , skips the next number $(2)$ , marks out $3$ , and continues skipping and marking out the next number to the end of the list. Then he goes back to the start of his list, marks out the first remaining number $(2)$ , skips the next number $(4)$ , marks out $6$ , skips $8$ , marks out $10$ , and so on to the end. Josh continues in this manner until only one number remains. What is that number?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 56 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 96$
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Following the pattern, you are crossing out...
Time 1: Every non-multiple of $2$
Time 2: Every non-multiple of $4$
Time 3: Every non-multiple of $8$
Following this pattern, you are left with every multiple of $64$ which is only $\boxed{64}$
| 64
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4,417
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_9
| 1
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Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$
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By Albert471
To start, use algebra to determine the number of posts on each side. You have (the long sides count for $2$ because there are twice as many) $6x = 20 + 4$ (each corner is double counted so you must add $4$ ) Making the shorter end have $4$ , and the longer end have $8$ $((8-1)*4)*((4-1)*4) = 28*12 = 336$ . Therefore, the answer is $\boxed{336}$
| 336
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4,418
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10
| 1
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A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
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Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$ . Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$ . Simplifying we get $(a-b)(a+b) = 8$ . Thus $a+b$ and $a-b$ have to be a factor of 8. The only way for them to be factors of $8$ and remain integers is if $a+b = 4$ and $a-b = 2$ . So the answer is $\boxed{4}$
| 4
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4,419
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_10
| 2
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A quadrilateral has vertices $P(a,b)$ $Q(b,a)$ $R(-a, -b)$ , and $S(-b, -a)$ , where $a$ and $b$ are integers with $a>b>0$ . The area of $PQRS$ is $16$ . What is $a+b$
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$
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Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is $2a^2 - 2b^2$ , so $a^2 - b^2 = 8$ . Since $a$ and $b$ are integers, $a = 3$ and $b = 1$ , so $a + b = \boxed{4}$
| 4
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4,420
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_11
| 1
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How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$
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Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
[asy] Label l; l.p=fontsize(8); xaxis(-1,8,Ticks(l, 1.0)); yaxis(-1,16,Ticks(l, 1.0)); real f(real x) { return x * pi; } D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { for(int y = 0; y < 16; ++y) { if(x * pi > y) { D((x,y)); } } } [/asy] (red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$ , and the limit for the $x$ -value is $5$ squares. First we count the $1*1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$ , and continuing on we have $9$ $6$ , and $3$ for $x$ -values for $1$ $2$ , and $3$ in the equation $y=\pi x$ . So there are $12+9+6+3 = 30$ squares with length $1$ in the figure.
For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi)$ , so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi)$ , so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure.
Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi)$ , so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$ . As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{50}$
| 50
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4,421
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14
| 1
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The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
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The sum of the geometric sequence is $\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \Rightarrow a = \frac{1}{r}.$ This means, \[S = \frac{a}{1 - r} = \frac{1/r}{1 - r} = \frac{1}{r(1 - r)}.\] In order to minimize $S,$ we maximize the denominator. By AM-GM, \[\frac{(r) + (1 - r)}{2} \ge \sqrt{r(1-r)} \Rightarrow \frac{1}{4} \ge r(1-r).\] Equality occurs at $r = 1-r \Rightarrow r = \frac{1}{2}.$ This gives the minimum value of $S$ as $\frac{\frac{1}{1/2}}{1 - \frac{1}{2}} = \boxed{4}.$
| 4
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4,422
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14
| 2
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The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
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A geometric sequence always looks like
\[a,ar,ar^2,ar^3,\dots\]
and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.
\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}
$\textbf{Solving in terms of \textit{a} then graphing}$
\[S=\frac{a^2}{a-1}\]
We seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\boxed{4}.$
| 4
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4,423
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_14
| 3
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The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
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\[\textbf{Completing the Square and Quadratics}\] Let $r$ be the common ratio. If the second term is $1$ , the first must be $\frac{1}{r}$ . By the infinite geometric series formula, the sum must be \[S=\frac{\frac{1}{r}}{1-r}\] This equals $\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$ , which is $\frac{1}{4}$ , completing the square. Thus, the minimum value of $S$ is $\boxed{4}$
| 4
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4,424
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15
| 1
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All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$
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First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{729}$
| 729
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4,425
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15
| 2
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All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$
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We'll proceed from the factoring process above.
By the AM-GM inequality,
\[\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}\]
Cubing both sides,
\[\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}\]
Let $a_1=(a+f)$ $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.
\[\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
$a+b+c+d+e+f$ is fixed at 27.
\[\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
\[\boxed{729}\]
| 729
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4,426
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_15
| 4
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All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$
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It is obvious to put $5$ $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the faces with $6$ and $5$
Therefore the answer is $7 \cdot 5 \cdot 6 + 7 \cdot 4 \cdot 6 + 7 \cdot 3 \cdot 5 + 7 \cdot 3 \cdot 4 + 2 \cdot 5 \cdot 6 + 2 \cdot 4 \cdot 6 + 2 \cdot 3 \cdot 5 + 2 \cdot 3 \cdot 4 = \boxed{729}$
| 729
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4,427
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16
| 1
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In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
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We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.
For the first case, we can cleverly choose the convenient form of our sequence to be \[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]
because then our sum will just be $(2n+1)a$ . We now have \[(2n+1)a = 345\] and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that \[2n+1 = 3, 5, 15, 23\] work, and no more, because $2n+1=1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$ , there must be negative integers in the sequence. Our solutions are $\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}$
For the even cases, we choose our sequence to be of the form: \[a-(n-1), \cdots, a, a+1, \cdots, a+n\] so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$ . In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$
We have found all 7 solutions and our answer is $\boxed{7}$
| 7
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4,428
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16
| 2
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In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
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The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is
$S=\dfrac{(a-b+1)(a+b)}{2}$
$345=\dfrac{(a-b+1)(a+b)}{2}$
$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$
Let $c=a-b+1$ and $d=a+b$
$2\cdot 3\cdot 5\cdot 23=c\cdot d$
If we factor $690$ into all of its factor groups $(\text{exg}~ (10,69) ~\text{or} ~(15,46))$ we will have several ordered pairs $(c,d)$ where $c<d$
The number of possible values for $c$ is half the number of factors of $690$ which is $\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8$
However, we have one extraneous case of $(1,690)$ because here, $a=b$ and we have the sum of one consecutive number which is not allowed by the question.
Thus the answer is $8-1=7$
$\boxed{7}$
| 7
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4,429
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16
| 3
|
In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
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The consecutive sums can be written as $345=kn+\sum_{i=1}^{k-1}{i}$ where $k$ is the number of terms in a sequence, and $n$ is the first term. Then, $\{k,n\}\in \mathbb{N}$ and $k\geq2$ . Evaluating the sum and rearranging yields $n=\frac{345}{k}-\frac{k-1}{2}$
The prime factorization of $345$ is $1\cdot3\cdot5\cdot23$ . Then, $3\cdot5$ $3\cdot23$ , and $5\cdot23$ are also divisors. As odd divisors of $345$ , note that they all produce integer solutions to $n$ as $k$ . Only $k!=1$ is not valid, as $k\geq2$ . Similarly, quickly notice that $k=2$ is a solution. Multiplying an odd divisor by $2$ always yields an integer solution (see below). As such, the even integer solutions are $k=2, 2\cdot3, 2\cdot5, 2\cdot15 ... 2\cdot345$
Note that the function is decreasing for increasing values of $k$ and that $\frac{345}{k}-\frac{k-1}{2}=4$ for $k=23$ . Thus, when $n$ is negative, $k$ is only slightly more than $k=23$ . Recall $\{k,n\}\in \mathbb{N}$ . Since the next highest solution, $k=2\cdot23=46$ is twice $k=23$ $k\leq23$ . Thus, the remaining solutions are when $k=2,6,10,3,5,15,23$ $\implies \boxed{7}$
| 7
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4,430
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16
| 5
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In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
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By the sum of an arithmetic sequence... this ultimately comes to $n+n+1+n+2....+n+p=345=(2n+p)(p+1)=690=23\cdot3\cdot5\cdot2$
Quick testing (would take you roughly a minute)
We see that the first 7 values of $p$ that work are
$p=1,2,4,5,9,14,22$
We see that each one of them works.
Hence, the answer is $\boxed{7}$
| 7
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4,431
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_16
| 6
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In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
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The sum of all integers from $x$ to $y$ inclusive is equal to
\[\frac{y(y+1)}{2}-\left(\frac{x(x+1)}{2}\right)+x.\]
In words, $x$ added to sum of the first $x$ positive inegers subtracted from the sum of the first positive $y$ integers.
Setting this equal to $345$ and multiplying by $2$ to clear fractions, we can see that \[y(y+1)-x(x+1)+2x\] \[=y^2-x^2+y+x\] \[=(y+x)(y-x+1)=690.\] Now we know that, $y+x$ and $y-x+1$ multiply together to $690$ , and they're both integers. Now, we can set $y+x$ and $y-x+1$ equal to factors of $690$ . Clearly, $x>\frac{1}{2},$ so $y+x$ will take the larger of the two factors.
We can factorize $690$ as $1\cdot2\cdot3\cdot5\cdot23$
We also know that when solving for $y$ , we can add the two systems in $y+x$ and $y-x+1$ to get a new equation in terms of $2y+1.$ In order for $y$ to be an integer, the sum of the two factors must be odd.
We also know that there is only one factor of $2$ , so either $y+x$ or $y-x+1$ will be even. The number of odd factors multiplied together can either be $1, 2$ or $3$ (0 doesn't work, since $(690, 1)$ doesn't). There are a total of $3$ odd numbers present in our new prime factorization, now excluding $1$
Therefore, the number of combinations that yield integral values for both $x$ and $y$ are \[{3\choose1}+{3\choose2}+{3\choose3}= \boxed{7}\]
| 7
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4,432
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_18
| 1
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What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$
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Consider the case when $x \geq 0$ $y \geq 0$ \[x^2+y^2=x+y\] \[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\] Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle: \[A = \frac{\pi}{4} +\frac{1}{2}\] [asy]draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));}[/asy] Because of symmetry, the area is the same in all four quadrants.
The answer is $\boxed{2}$
| 2
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4,433
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20
| 1
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A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$
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We use complementary counting. First, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.
There are $21$ ways to choose the team that beat the two other teams, and $\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{385})$ is our answer.
| 385
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4,434
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20
| 2
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A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$
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As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the “cycle-win” condition—e.g. here, A beats a team in X which beats a team in Y which beats A. The first and third part of the condition are already met by our wlog, so we just need to count of number of ways the second condition is true (a team in X beats a team in Y). These are the number of cycle-wins that include A, then multiply by 21 (for each team) and divide by 3 (since every set will be counted by each of the 3 teams that are a part of that set).
To do this, let X $=\{B, ..., L\}$ and Y $=\{M, ..., U\}$ . Since a total of $10*10=100$ losses total were suffered by teams in Y and $\binom{10}{2}=45^{*}$ losses were suffered by teams in Y from teams in Y, we have $100-45=55$ losses suffered by teams in Y from teams in X. Hence, for each of these $55$ losses, there is exactly one set of three teams that includes A that satisfies the problem conditions. Thus, the answer is $\frac{55\cdot 21}{3}=\boxed{385}$
| 385
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4,435
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https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_20
| 3
|
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$
|
Note that there are $21$ teams total and $\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\frac{1}{4}$ of all the arrangements satisfy the second case, and $\frac{1330}{4}=332.5,$ which is by far the closest to $\boxed{385}.$
| 385
|
4,436
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25
| 1
|
The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$
|
Let $b_i=19\text{log}_2a_i$ . Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\geq 2$ . The characteristic polynomial of this linear recurrence is $x^2-x-2=0$ , which has roots $2$ and $-1$
Therefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$ . Using the fact that $b_0=0, b_1=1,$ we can solve a pair of linear equations for $k_1, k_2$
$k_1+k_2=0$ $2k_1-k_2=1$
Thus $k_1=\frac{1}{3}$ $k_2=-\frac{1}{3}$ , and $b_n=\frac{2^n-(-1)^n}{3}$
Now, $a_1a_2\cdots a_k=2^{\frac{(b_1+b_2+\cdots+b_k)}{19}}$ , so we are looking for the least value of $k$ so that
$b_1+b_2+\cdots+b_k \equiv 0 \pmod{19}$
Note that we can multiply all $b_i$ by three for convenience, as the $b_i$ are always integers, and it does not affect divisibility by $19$
Now, for all even $k$ the sum (adjusted by a factor of three) is $2^1+2^2+\cdots+2^k=2^{k+1}-2$ . The smallest $k$ for which this is a multiple of $19$ is $k=18$ by Fermat's Little Theorem, as it is seen with further testing that $2$ is a primitive root $\pmod{19}$
Now, assume $k$ is odd. Then the sum (again adjusted by a factor of three) is $2^1+2^2+\cdots+2^k+1=2^{k+1}-1$ . The smallest $k$ for which this is a multiple of $19$ is $k=17$ , by the same reasons. Thus, the minimal value of $k$ is $\boxed{17}$
| 17
|
4,437
|
https://artofproblemsolving.com/wiki/index.php/2016_AMC_12B_Problems/Problem_25
| 3
|
The sequence $(a_n)$ is defined recursively by $a_0=1$ $a_1=\sqrt[19]{2}$ , and $a_n=a_{n-1}a_{n-2}^2$ for $n\geq 2$ . What is the smallest positive integer $k$ such that the product $a_1a_2\cdots a_k$ is an integer?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 21$
|
Like in Solution 2 , calculate the first few terms of the sequence, but also keep a running sum $c_n$ of the logarithms (not modulo $19$ here): \[0,1,2,5,10,21,42,\dots\] Notice that $c_n=2c_{n-1}+1$ for odd $n$ and $c_n=2c_{n-1}$ for even $n$ . Since $2$ is relatively prime to $19$ , we can ignore even $n$ and calculate odd $n$ using $c_1 = 1, c_{n} = 4c_{n-2}+1$ (modulo $19$ ): \[,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots\] $c_n$ is first a multiple of $19$ at $n = \boxed{17}$ . ~ emerald_block
| 17
|
4,438
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_2
| 1
|
Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?
$\textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72$
|
Letting $x$ be the third side, then by the triangle inequality, $20-15 < x < 20+15$ , or $5 < x < 35$ . Therefore the perimeter must be greater than 40 but less than 70. 72 is not in this range, so $\boxed{72}$ is our answer.
| 72
|
4,439
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3
| 1
|
Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$
|
If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ . When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ . So, Payton's score must be $1215-1120 = \boxed{95}$
| 95
|
4,440
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_3
| 2
|
Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$
|
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ more points, so Payton needs $80+15 = \boxed{95}$
| 95
|
4,441
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4
| 1
|
The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$
|
Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$
| 32
|
4,442
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_4
| 2
|
The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$
|
Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$
| 32
|
4,443
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_6
| 1
|
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$
|
This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$
| 4
|
4,444
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_7
| 1
|
Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$
|
Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$
| 21
|
4,445
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_10
| 1
|
Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$
|
Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\boxed{26}$
| 26
|
4,446
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_11
| 1
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On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$
|
Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\boxed{5}$ values of $k$
| 5
|
4,447
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15
| 1
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What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$
|
We can rewrite the fraction as $\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we divide by $2$ twenty-two times, so we will need $22$ more digits. Hence, the answer is $4 + 22 = \boxed{26}$
| 26
|
4,448
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15
| 2
|
What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$
|
Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\frac{5^{22} \cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \cdot 123456789) \div 10^{26}$ . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, $\boxed{26}$ is the minimum number of digits to the right of the decimal point needed.
| 26
|
4,449
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_15
| 3
|
What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$
|
The denominator is $10^4 \cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \boxed{26}$
| 26
|
4,450
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18
| 1
|
The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
|
The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.
The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$
$a^2-8a=k^2$
$a(a-8)=k^2$
$((a-4)+4)((a-4)-4)=k^2$
$(a-4)^2-4^2=k^2$
$(a-4)^2=k^2+4^2$
From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$
In the case $k=0$ , the equation simplifies to $|a-4|=4$ . From this equation, we have $a=0,8$ . For both $a=0$ and $a=8$ $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$ . Hence, $a=-1,9$ . Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$ , so these two values also satisfy the original constraints.
There are a total of four possible values for $a$ $-1,0,8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is $\boxed{16}$
| 16
|
4,451
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18
| 2
|
The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
|
By the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$ $a\in\mathbb{Z}$ , since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissas (mantissae?).
Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.
Positive $a$
$a\geq8$ and $a\leq0$ , since $1\geq a \geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\leq9$ . After $16$ and $36$ , the difference between the closest nonzero factor pairs of perfect squares exceeds $8$ . For $8\geq a \geq9$ $a=8,9$ . Checking both yields an integer.
Negative $a$
We can instead test with $\sqrt{-a(8-a)}$ . If $b=8-a$ , we have our original expression. Thus, for the same reasons, $b=8,9\implies 8,9=8-a$ $a=-1$ (0 does not affect the answer).
$-1+8+9=16\implies\boxed{16}$
| 16
|
4,452
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18
| 3
|
The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
|
Let $m$ and $n$ be the roots of $x^2-ax+2a$
By Vieta's Formulas, $n+m=a$ and $mn=2a$
Substituting gets us $n+m=\frac{mn}{2}$
$2n-mn+2m=0$
Using Simon's Favorite Factoring Trick:
$n(2-m)+2m=0$
$-n(2-m)-2m=0$
$-n(2-m)-2m+4=4$
$(2-n)(2-m)=4$
This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$ . Adding these up gets $\boxed{16}$
| 16
|
4,453
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18
| 4
|
The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
|
The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\] . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]
We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$ , we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$ . These are the only ones; if we want to make sure then we must hand check up to $b=8$ . Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find $\boxed{16}$ . as the answer.
| 16
|
4,454
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_18
| 5
|
The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$
|
First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.
We can complete the square and rearrange to get $(a-4)^2-b^2=16$ . Let's define $m=a-4$ , just to make things a little easier to write, so now we have $(m+b)(m-b)=16$ . We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$ . (Note that $m$ and $b$ must both be integers)
$(m+b)(m-b)=16$
$16*1=16$ $\Rightarrow$ Doesn't work
$8*2=16$ $\Rightarrow$ $m=5, b=3$
$4*4=16$ $\Rightarrow$ $m=4, b=0$
$2*8=16$ $\Rightarrow$ $m=5, b=-3$
$1*16=16$ $\Rightarrow$ Doesn't work
$-16*-1=16$ $\Rightarrow$ Doesn't work
$-8*-2=16$ $\Rightarrow$ $m=-5, b=-3$
$-4*-4=16$ $\Rightarrow$ $m=-4, b=0$
$-2*-8=16$ $\Rightarrow$ $m=-5, b=-3$
$-1*-16=16$ $\Rightarrow$ Doesn't work
We want the possible values of $m$ , which are $-5,-4,4,$ and $5$ . As $m+4=a$ $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\boxed{16}$
| 16
|
4,455
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19
| 1
|
For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$
|
Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{31}$
| 31
|
4,456
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_19
| 2
|
For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$
|
Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this down and testing, it appears that this holds for all $x$ . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers.
In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{31}$
| 31
|
4,457
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 1
|
Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
|
The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$
Thus $2a + b = 18$ , so $2a = 18 - b$
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ , so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$
We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$ , so $b^3 - 9b^2 + 64 = 0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{3}$
| 3
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4,458
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 2
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Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
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The area is $12$ , the semiperimeter is $9$ , and $a = 9 - \frac12b$ . Using Heron's formula, $\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12$ . Squaring both sides and simplifying, we have $-b^3+9b^2-64=0$ . Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$ , so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{3}$
| 3
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4,459
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 3
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Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
|
Triangle $T$ , being isosceles, has an area of $\frac{1}{2}(8)\sqrt{5^2-4^2}=12$ and a perimeter of $5+5+8=18$ .
Triangle $T'$ similarly has an area of $\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12$ and $2a+b=18$
Now we apply our computational fortitude.
\[\frac{1}{2}(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=12\] \[(b)\bigg(\sqrt{a^2-\frac{b^2}{4}}\bigg)=24\] \[(b)\sqrt{4a^2-b^2}=48\] \[b^2(4a^2-b^2)=48^2\] \[b^2(2a+b)(2a-b)=48^2\] Plug in $2a+b=18$ to obtain \[18b^2(2a-b)=48^2\] \[b^2(2a-b)=128\] Plug in $2a=18-b$ to obtain \[b^2(18-2b)=128\] \[2b^3-18b^2+128=0\] \[b^3-9b^2+64=0\] We know that $b=8$ is a valid solution by $T$ . Factoring out $b-8$ , we obtain \[(b-8)(b^2-b-8)=0 \Rightarrow b^2-b-8=0\] Utilizing the quadratic formula gives \[b=\frac{1\pm\sqrt{33}}{2}\] We clearly must pick the positive solution. Note that $5<\sqrt{33}<6$ , and so ${3<\frac{1+\sqrt{33}}{2}<\frac{7}{2}}$ , which clearly gives an answer of $\boxed{3}$ , as desired.
| 3
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4,460
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 4
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Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
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Triangle T has perimeter $5 + 5 + 8 = 18$ so $18 = 2a + b$
Using Heron's, we get $\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})\left(\frac{b}{2}\right)^2(\frac{2a-b}{2})}$
We know that $2a + b = 18$ from above so we plug that in, and we also know that then $2a - b = 18 - 2b$
$12 = \frac{3b}{2}\sqrt{9-b}$
$64 = 9b^2 - b^3$
We plug in 3 for $b$ in the LHS, and we get 54 which is too low. We plug in 4 for $b$ in the LHS, and we get 80 which is too high. We now know that $b$ is some number between 3 and 4.
If $b \geq 3.5$ , then we would round up to 4, but if $b < 3.5$ , then we would round down to 3. So let us plug in 3.5 for $b$
We get $67.375$ which is too high, so we know that $b < 3.5$
Thus the answer is $\boxed{3}$
| 3
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4,461
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 5
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Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
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For this new triangle, say its legs have length $d$ and the base length $2c$ . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that $c+d=9$ and $c*\sqrt{d^2-c^2}=12$ . It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!
Now, modify the square-root equation with $d=9-c$ ; you get $c^2*(81-18c)=144$ , so $-18c^3+81c^2=144$ . Divide by $-9$ to get $2c^3-9c^2+16=0$ . Obviously, $c=4$ is a root as established by triangle $T$ ! So, use synthetic division to obtain $2c^2-c-4=0$ , upon which $c=\frac{1+\sqrt{33}}{4}$ , which is closest to $\frac{3}{2}$ (as opposed to $2$ ). That's enough to confirm that the answer has to be $\boxed{3}$
| 3
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4,462
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_20
| 6
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Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ $5$ , and $8$ , while those of $T'$ have lengths $a$ $a$ , and $b$ . Which of the following numbers is closest to $b$
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
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Since triangles $T'$ and $T$ have the same area and the same perimeter, $2a+b=18$ and $9*(9-a)^2(9-b) = 9*4^2*1$ By trying each answer choice, it is clear that the answer is $\boxed{3}$
| 3
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4,463
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_22
| 2
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For each positive integer $n$ , let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$ , with no more than three $A$ s in a row and no more than three $B$ s in a row. What is the remainder when $S(2015)$ is divided by $12$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 10$
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We can start off by finding patterns in $S(n)$ . When we calculate a few values we realize either from performing the calculation or because the calculation was performed in the exact same way that $S(n) = 2^n - 2((n_4)- (n_5) \dots (n_n))$ . Rearranging the expression we realize that the terms aside from $2^{2015}$ are congruent to $0$ mod $12$ (Just put the equation in terms of 2^{2015} and the four combinations excluded and calculate the combinations mod $12$ ). Using patterns we can see that $2^{2015}$ is congruent to $8$ mod $12$ . Therefore $\boxed{8}$ is our answer.
| 8
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4,464
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12A_Problems/Problem_23
| 1
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Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$
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WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square $n$ , the distance from the point to the outside of the circle (the radius $0.5$ ), and any portion of the nearest side that lies within the circle as represented by the Pythagorean Theorem $\sqrt{\frac{1}{4}-n^2}$ . Thus, the total length the second point can exist in can be represented by $f(n)=4-(0.5+n+\sqrt{\frac{1}{4}-n^2})$ . Distributing, $f(n)=3.5-n-\sqrt{\frac{1}{4}-n^2}$
Then, we can find the average of this function through calculus (wow more calc?). This formula is as follows, \[\frac{1}{a_{f} -a_{i}} \int_{a_{i}}^{a_{f}} f(x)\;dx\]
For this case, the limits of integration are $0$ and $0.5$ $0\leq n\leq 0.5$ ). Then, we have,
\[2\int_{0}^{\frac{1}{2}} 3.5-n-\sqrt{\frac{1}{4}-n^2}\;dn\] \[2(\int_{0}^{\frac{1}{2}} 3.5\;dn -\int_{0}^{\frac{1}{2}}n\;dn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)\]
Even if you don't know how to integrate, as long as you know the idea of integration, you can figure these out. Graphing the first two, you can see that the first is a rectangle of length $0.5$ and width $3.5$ . The second is an isosceles right triangle of leg length $0.5$
\[\int_{0}^{\frac{1}{2}} 3.5\;dn=\frac{7}{4}, \int_{0}^{\frac{1}{2}}n\;dn=\frac{1}{8}\]
Recognize that the third integral is a semicircle of radius $0.5$ and centered at the origin. This is where $\pi$ comes in. From $0$ to $0.5$ , the integral is simply a quarter circle. $\frac{\frac{1}{2}^2\pi}{4}=\frac{\pi}{16}$ \[\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}\;dn)=\frac{\pi}{16}\]
If you want me to actually integrate these, look below. Do note that this is for those that have a limited knowledge of integration or those that have little time but are being very clever.
Overall, where second point could lie to satisfy the problem is a length of $2(\frac{7}{4}-\frac{\pi}{16}-\frac{1}{8})=\frac{26-\pi}{8}$ . By contrast, the total length where it could lie is the perimeter of the square $4$ . So the possible points that the second point could be make up $\frac{\frac{26-\pi}{8}}{4}=\frac{26-\pi}{32}$ of the square's perimeter. Obviously, $\gcd(32, 26, 1)=1$ $32+26+1=59\implies\boxed{59}$
| 59
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4,465
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_3
| 1
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Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$
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Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \boxed{8}$
| 8
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4,466
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_7
| 1
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A regular 15-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$
$\textbf{(A)}\; 24 \qquad\textbf{(B)}\; 27 \qquad\textbf{(C)}\; 32 \qquad\textbf{(D)}\; 39 \qquad\textbf{(E)}\; 54$
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From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence $R = 360^\circ / 15 = 24^\circ$ , the number of degrees between adjacent sides. Therefore the answer is $L + R = 15 + 24 = \boxed{39}$
| 39
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4,467
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8
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What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$
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$(625^{\log_5 2015})^\frac{1}{4} = ((5^4)^{\log_5 2015})^\frac{1}{4} = (5^{4 \cdot \log_5 2015})^\frac{1}{4} = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} = ((5^{\log_5 2015})^4)^\frac{1}{4} = (2015^4)^\frac{1}{4} = \boxed{2015}$
| 15
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4,468
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8
| 2
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What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$
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We can rewrite $\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$
| 15
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4,469
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8
| 3
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What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$
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$(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{2015}$
| 15
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4,470
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_8
| 4
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What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$
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We note that the year number is just $2015$ , so just guess $\boxed{2015}$
| 15
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4,471
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_10
| 1
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How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$
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Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$
When $a=2$ , similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives 3 triangles $(2,3,4), (2,4,5), (2,5,6)$ within our perimeter constraint.
When $a=3$ $c$ can be $b+1$ or $b+2$ , which gives triangles $(3,4,5), (3,4,6), (3,5,6)$ . Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.
All in all, this gives us $3+2 = \boxed{5}$ triangles.
| 5
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4,472
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_13
| 1
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Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$
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$\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$ , hence $\angle ACB = \angle ADB = 40^\circ$ . By considering $\triangle ABC$ , it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$ . Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{6}.$
| 6
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4,473
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18
| 1
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For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$
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This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$ . By evaluating $r(4)$ to be $4$ , we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$ , we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consists of composite numbers, which, by definition, are a product of at least two positive primes, the minimum value of $r(n)$ is $4$ , so $\textbf{(A)}$ is incorrect. That leaves us with $\boxed{3}$
| 3
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4,474
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_18
| 2
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For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$
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Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.
The lowest number we can get is $y = 2+2 = 4$ . For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can be obtained by adding 2 ten times and adding a 3; this corresponds to the argument $n = 2^{10} \times 3$ . Thus our answer is $\boxed{3}$
| 3
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4,475
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20
| 1
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For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$
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Simply take some time to draw a table of values of $f(i,j)$ for the first few values of $i$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & 0\\ \hline 4 & 3 & 1 & 3 & 1 & 3\\ \hline 5 & 1 & 1 & 1 & 1 & 1\\ \hline \end{array}\]
Now we claim that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$ . We will prove this by induction on $i$ and $j$ . The base cases for $i = 5$ , have already been proven.
For our inductive step, we must show that for all valid values of $j$ $f(i, j) = 1$ if for all valid values of $j$ $f(i - 1, j) = 1$
We prove this itself by induction on $j$ . For the base case, $j=0$ $f(i, 0) = f(i-1, 1) = 1$ . For the inductive step, we need $f(i, j) = 1$ if $f(i, j-1) = 1$ . Then, $f(i, j) = f(i-1, f(i, j-1)).$ $f(i, j-1) = 1$ by our inductive hypothesis from our inner induction and $f(i-1, 1) = 1$ from our outer inductive hypothesis. Thus, $f(i, j) = 1$ , completing the proof.
It is now clear that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$
Thus, $f(2015,2) = \boxed{1}$
| 1
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4,476
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_20
| 2
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For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$
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We are given that \[f(0,n) \equiv n+1\pmod{5} .\] Then, $f(1,n) = f(0, f(1, n-1)) \equiv f(1,n-1) + 1\pmod{5}$ . Thus $f(1,n) \equiv n+f(1,0)\pmod{5}$ . Since $f(1,0)=f(0,1)=2$ , we get \[f(1,n) \equiv n+2\pmod{5} .\] Then, $f(2,n) = f(1, f(2, n-1)) \equiv f(2,n-1) + 2\pmod{5}$ . Thus $f(2,n) \equiv 2n+f(2,0)\pmod{5}$ . Since $f(2,0)=f(1,1)=3$ , we get \[f(2,n) \equiv 2n+3\pmod{5} .\] Now $f(3,n) = f(2, f(3, n-1)) \equiv 2f(3,n-1) + 3\pmod{5}$ . Thus \begin{align*} f(3,n) &\equiv 2f(3,n-1) + 3 &\pmod{5} \\ 2f(3,n-1) &\equiv 2^2f(3,n-2) + 3\cdot 2 &\pmod{5} \\ \vdots \qquad &\quad\vdots \quad\qquad \vdots \qquad\qquad \vdots &\\ 2^{n-1}f(3,1) &\equiv 2^nf(3,0) + 3\cdot 2^{n-1} &\pmod{5} \end{align*} Adding them all up we get \[f(3,n) \equiv 3(2^n-1)\pmod{5} .\] This means that $f(3,0)=0$ $f(3,1)=3$ $f(3,2)=4$ $f(3,3)=1$ , and $f(3,4)=0$ . Thus $f(3,n)$ never takes the value 2.
Since $f(4,n)=f(3,f(4,n-1))$ , this implies that $f(4,n) \neq 2$ for any $n$ . By induction, $f(3,n) \neq f(3,2) = 4$ for any $n$ . It follows that $f(3,n) \neq f(3,4) = 0$ for any $n$ . Thus $f(4,n)$ only takes values in $\{1,3\}$ . In fact, it alternates between 1 and 3: $f(4,0)=f(3,1)=3$ , then $f(4,1)=f(3,f(4,0))=f(3,3)=1$ , then $f(4,2)=f(3,f(4,1))=f(3,1)=3$ , and so on.
Repeating the argument above, we see that $f(5,n) = f(4, f(5,n-1))$ can only take values in $\{1,3\}$ . However, $f(5,n-1)\neq 0$ for any $n$ implies that $f(5,n)\neq f(4,0)=3$ for any $n$ . Thus $f(5,n)=1$ for all $n$ . We can easily verify this: $f(5,0)=f(4,1)=1$ , then $f(5,1)=f(4,f(5,0))=f(4,1)=1$ , then $f(5,2)=f(4,f(5,1))=f(4,1)=1$ , and so on.
Then $f(6,0)=f(5,1)=1$ . Moreover, $f(6,n) = f(5,f(6,n-1)) = 1$ for all $n$ . Continuing in this manner we see that $f(m,n)=1$ for all $m\ge 5$
In particular, $f(2015,2) = \boxed{1}$
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4,477
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21
| 1
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Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$
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We can translate this wordy problem into this simple equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$
Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.
Case 2: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}$
Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$
Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$
| 13
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4,478
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21
| 2
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Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$
|
It can easily be seen that the problem can be expressed by the equation: \[\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19\]
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
\[\frac{s+a}{2} - \frac{s+b}{5} = 19\] Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get: \[5s+5a-2s-2b=190\] \[3s = 190+2b-5a\] \[s = 63 + \frac{1+2b-5a}{3}\]
Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$ . We solve using casework.
Case 1: $a = 0$
If $a = 0$ , we have $2b \equiv 2 \mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.
Case 2: $a = 1$
If $a = 1$ , we have $2b-5 \equiv 2 \mod 3$ , which can be rewritten as $2b \equiv 1 \mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$
Adding these together we get $64+66+63=193$ . The sum of the digits is $\boxed{13}$
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4,479
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21
| 3
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Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$
|
As before, we write the equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.\]
To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:
\[\frac{s}{2} - 19 = \frac{s}{5}.\]
This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$
Now, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18. If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21. When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$
Then, by inspection, $s=63, 64,$ or $66$ , so $\sum s = 193 \implies \boxed{13}.$
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4,480
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21
| 4
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Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$
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Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{13}.$
| 13
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4,481
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_21
| 5
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Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$
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We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \in \{1,2\}$ and $b \in \{1,2,3,4,5\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.
The reason for starting at 1 and ending at 5 instead of 0 through 4 is that the last step can be quite problematic to deal with, especially if it is possible to make it in one go, so we treat it as a different jump that can take all possible jump values. We know that D makes it in 19 fewer than C, so $c+1-(d+1) = 19 \implies c = 19+d \implies 5d + b = 2(19+d) + a \implies 3d = 38 + a - b$
Now that we have this nice equivalence, we can do the thing and it works.
If we take both sides mod 3 and rearrange, we get $b \equiv 2+a \pmod{3}$ This gives us the following satisfactory $(a,b)$ relational pairs: $\{(1,3),(2,1),(2,4)\}$ . We can now just find the corresponding $d$ value for each pair, sum it all up using $5d + b = n$ and sum the digits to reveal $\boxed{13}$ as the answer.
| 13
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4,482
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22
| 1
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Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$
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Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ways in which six people can sit down in a chair that is either the same chair or an adjacent chair in a circle.
Consider the similar problem of $n$ people sitting in a chair that is either the same chair or an adjacent chair in a row. Call the number of possibilities for this $F_n$ . Then if the leftmost person stays put, the problem is reduced to a row of $n-1$ chairs, and if the leftmost person shifts one seat to the right, the new person sitting in the leftmost seat must be the person originally second from the left, reducing the problem to a row of $n-2$ chairs. Thus, $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$ . Clearly $F_1 = 1$ and $F_2 = 2$ , so $F_3 = 3$ $F_4 = 5$ , and $F_5 = 8$
Now consider the six people in a circle and focus on one person. If that person stays put, the problem is reduced to a row of five chairs, for which there are $F_5 = 8$ possibilities. If that person moves one seat to the left, then the person who replaces him in his original seat will either be the person originally to the right of him, which will force everyone to simply shift over one seat to the left, or the person originally to the left of him, which reduces the problem to a row of four chairs, for which there are $F_4 = 5$ possibilities, giving $1 + 5 = 6$ possibilities in all. By symmetry, if that person moves one seat to the right, there are another $6$ possibilities, so we have a total of $8+6+6 = \boxed{20}$ possibilities.
| 20
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4,483
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22
| 2
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Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$
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Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):
\[\begin{tabular}{| c | c | c | c | c | c |} \hline - & - & A & A & A & - \\ \hline - & - & - & B & B & B \\ \hline C & - & - & - & C & C \\ \hline D & D & - & - & - & D \\ \hline E & E & E & - & - & - \\ \hline - & F & F & F & - & - \\ \hline \end{tabular}\]
The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:
$\begin{tabular}{c} CDEFAB \\ CEAFBD \\ CEFABD \\ CEFBAD \\ CFEABD \\ CFEBAD \\ DEAFBC \\ DEAFCB \\ DEFABC \\ DEFACB \\ DEFBAC \\ DFEABC \\ DFEACB \\ DFEBAC \\ EDAFBC \\ EDAFCB \\ EDFABC \\ EDFACB \\ EDFBAC \\ EFABCD \end{tabular}$
There are $\boxed{20}$ such permutations.
| 20
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4,484
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_22
| 3
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Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$
|
We can represent each rearrangement as a permutation of the six elements $\{1,2,3,4,5,6\}$ in cycle notation. Note that any such permutation cannot have a 1-cycle, so the only possible types of permutations are 2,2,2-cycles, 4,2-cycles, 3,3-cycles, and 6-cycles. We deal with each case separately.
For 2,2,2-cycles, suppose that one of the 2-cycles switches the people across from each other, i.e. $(14)$ $(25)$ , or $(36)$ . WLOG, we may assume it to be $(14)$ . Then we could either have both of the other 2-cycles be across from each other, giving the permutation $(14)(25)(36)$ or else neither of the other 2-cycles is across from each other, in which case the only possible permutation is $(14)(26)(35)$ . This can happen for $(25)$ and $(36)$ as well. So since the first permutation is not counted twice, we find a total of $1+3=4$ permutations that are 2,2,2-cycles where at least one of the 2-cycles switches people diametrically opposite from each other. Otherwise, since the elements in a 2-cycle cannot differ by 1, 3, or 5 mod 6, they must differ by 2 or 4 mod 6, i.e. they must be of the same parity. But since we have three odd and three even elements, this is impossible. Hence there are exactly 4 such permutations that are 2,2,2-cycles.
For 4,2-cycles, we assume for the moment that 1 is part of the 2-cycle. Then the 2-cycle can be $(13)$ $(15)$ , or $(14)$ . The first two are essentially the same by symmetry, and we must arrange the elements 2, 4, 5, 6 into a 4-cycle. However, 5 must have two neighbors that are not next to it, which is impossible, hence the first two cases yield no permutations. If the 2-cycle is $(14)$ , then we must arrange the elements 2, 3, 5, 6 into a 4-cycle. Then 2 must have the neighbors 5 and 6. We find that the 4-cycles $(2536)$ and $(2635)$ satisfy the desired properties, yielding the permutations $(14)(2536)$ and $(14)(2635)$ . This can be done for the 2-cycles $(25)$ and $(36)$ as well, so we find a total of 6 such permutations that are 4,2-cycles.
For 3,3-cycles, note that if 1 neighbor 4, then the third element in the cycle will neighbor one of 1 and 4, so this is impossible. Therefore, the 3-cycle containing 1 must consist of elements 1, 3, and 5. Therefore, we obtain the four 3,3-cycles $(135)(246)$ $(153)(246)$ $(135)(264)$ , and $(153)(264)$
For 6-cycles, note that the neighbors of 1 can be 3 and 4, 3 and 5, or 4 and 5. In the first case, we may assume that it looks like $(314\dots)$ -- the form $(413\dots)$ is also possible but equivalent to this case. Then we must place the elements 2, 5, and 6. Note that 5 and 6 cannot go together, so 2 must go in between them. Also, 5 cannot neighbor 4, so we are left with one possibility, namely $(314625)$ , which has an analogous possibility $(413526)$ . In the second case, we assume that it looks like $(315\dots)$ . The 2 must go next to the 5, and the 6 must go last (to neighbor the 3), so the only possibility here is $(315246)$ , with the analogous possibility $(513642)$ . In the final case, we may assume that it looks like $(415\dots)$ . Then the 2 and 3 cannot go together, so the 6 must go in between them. Therefore, the only possibility is $(415362)$ , with the analogous possibility $(514263)$ . We have covered all possibilities for 6-cycles, and we have found 6 of them.
Therefore, there are $4+6+4+6 = \boxed{20}$ such permutations.
| 20
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4,485
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23
| 1
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A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$
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We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b, ac \le bc$ , from the first equation we get $abc \le 6bc$ . Thus $a\le 6$ . From the second equation we see that $a > 2$ . Thus $a\in \{3, 4, 5, 6\}$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.
| 10
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4,486
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_23
| 2
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A rectangular box measures $a \times b \times c$ , where $a$ $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$
|
The surface area is $2(ab+bc+ca)$ , and the volume is $abc$ , so equating the two yields
\[2(ab+bc+ca)=abc.\]
Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]
First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\geqslant3$
Also note that $c \geq b \geq a > 0$ , hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$ .
Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$ , so $a \leq 6$
So we have $a=3, 4, 5$ or $6$
Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ . From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$ , we have $b \leq \frac{2}{k}$ . Thus $\frac{1}{k}<b \leq \frac{2}{k}$
When $a=3$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, 9, 10, 11, 12$ . We find the solutions $(a, b, c)=(3, 7, 42)$ $(3, 8, 24)$ $(3, 9, 18)$ $(3, 10, 15)$ $(3, 12, 12)$ , for a total of $5$ solutions.
When $a=4$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . We find the solutions $(a, b, c)=(4, 5, 20)$ $(4, 6, 12)$ $(4, 8, 8)$ , for a total of $3$ solutions.
When $a=5$ , we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$
When $a=6$ $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$
Thus, there are $5+3+1+1 = \boxed{10}$ solutions.
| 10
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4,487
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24
| 1
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Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
|
First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$ , with $O_1$ and $O_2$ lying on the same side of $PQ$ , and $O_1 O_2=39$ . Let $x=O_1 R$ $y=O_2 R$ , and suppose that the radius of circle $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$
Then the power of point $R$ with respect to $C_1$ is
\[(r+x)(r-x) = r^2 - x^2 = 24^2\]
and the power of point $R$ with respect to $C_2$ is
\[\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.\]
Also, note that $x-y=39$
Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$ . As $x-y=39$ , we find that $r^2=64(x+y) = 64(2y+39)$ . Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$ . This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$ , and as $y$ is a length, it is positive, hence $y=57$ , and $x=y+39=96$ . This is the only possibility if the two centers lie on the same side of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that $O_1 R + O_2 R = O_1 O_2 = 39$ . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{192}$
| 192
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4,488
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24
| 2
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Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
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Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that $A$ $B$ $C$ , and $D$ are all on the perpendicular bisector of $PQ$ . Let us say the distance from $A$ to the line $PQ$ is some $a$ . Therefore, the distance from $B$ to the line $PQ$ is $\sqrt{(\frac{8}{5}\sqrt{a^2 + 24^2})^2 - 24^2}$ , which comes out to be $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576}$ . Since $AB = 39$ , we have one of $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576} \pm a$ to be equal to $39$ . We can solve both equations to get that out of the four possible solutions, and only two are positive: $7$ and $57$ . Note that since no two circles can be congruent, we need the radius of one of $A$ or $C$ to be $7$ and the other to be $57$ . Plugging in to find the corresponding radii of $B$ and $D$ gives $32$ and $96$ , and adding everything up gives $\boxed{192}$
| 192
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4,489
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24
| 3
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Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
|
Let's start by drawing $PQ$ . Because all circles contain $P$ and $Q$ , all the centers lie on the perpendicular bisector of $PQ$ , and point $R$ is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of $PQ$ , and two are on the opposite side of $PQ$ . For the latter two circles--call them $A$ and $B$ -- $AR+BR=39$
Let's consider the next case, where $C$ and $D$ lie on the same side. Construct right triangles from the picture, and use the Pythagorean Theorem (divide by 3 to negate big numbers). You will get that the distance from $R$ to the closest circle center is $57$ . Therefore, the answer is $39+2\cdot57+39=\boxed{192}$
| 192
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4,490
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24
| 4
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Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
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Since the radical axis $PQ$ is perpendicular to the line connecting the center of the circles, we have that $A$ $B$ $C$ $D$ , and $R$ are collinear. WLOG, assume that $A$ and $B$ are on the same side of $R$ and let $AR=y$ and let $BP=x$ so that $AP=\frac{5}{8}x$
Then, using the Pythagorean Theorem on right triangles $PBR$ and $PAR$ \[(39+y)^2+24^2=x^2\qquad(1)\] \[y^2+24^2=\frac{25}{64}x^2\qquad(2)\] Subtracting the $(2)$ from $(1)$ gives \[x^2=64(2y+39)\qquad(3)\] Substituting $(3)$ into $(2)$ gives $y^2-50y-399=0=(y-57)(y+7).$ Taking the positive solution ( $y>0$ ), $y=AR=57$ and $BR=(57)+39=96$
Since none of the circles are congruent, $C$ and $D$ must be on the opposite side of $R$ so $CR+DR=CD=39$ . Hence, $AR+BR+CR+DR=57+96+39=\boxed{192}$
| 192
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4,491
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_24
| 5
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Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
|
Note that the four circles are coaxial, meaning $A,B,C,D,R$ are all collinear. Let $AR=x$ . By Pythagorean Theorem , the radius of circle $A$ squared would be $r_A^2 = x^2+24^2$ and the radius of circle $B$ squared would be $r_B^2 = (x+39)^2+24^2.$ Since $r_A^2 = \dfrac{25}{64}r_B^2$ \[x^2+24^2 = \dfrac{25}{64}((x+39)^2+24^2)\] Solving this will give $x^2-50x-399 = 0$ , or $x=57, -7$ $AR = 57$ . The same equation will apply to $CR$ $CR$ would be the other root: $CR=7$ (the negative $7$ signals that $C$ is the negative, or opposite, direction of $D$ , about $R$ ). Thus, $AR = 57$ $BR = 57+39 = 96$ $CR = 7$ $DR = 39-7 = 32$ , implying $AR+BR+CR+DR = \boxed{192}$
| 192
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4,492
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 1
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A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
Let $x=e^{i\pi/6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:
\[1+2x+3x^2+\cdots+(k+1)x^k\]
We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series
\begin{align*} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=x+2x^2+3x^3+\cdots+2015x^{2015} \\ (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\ S &= \frac{1-x^{2015}}{(1-x)^2}-\frac{2015x^{2015}}{1-x} \end{align*}
We want to find $|S|$ . First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$ . Therefore
\[S =\frac{1-\frac{1}{x}}{(1-x)^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.\]
Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$
Now we just have to find $|1-x|$ . This can just be computed directly:
\[1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i\]
\[|1-x|^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2\]
\[|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}\]
Therefore $|S|=2016\cdot\frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008 \sqrt{6}$
Thus the answer is $1008+2+1008+6=\boxed{2024}$
| 24
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4,493
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 2
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A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
Here is an alternate solution that does not use complex numbers:
The distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:
$x=1\cos{0}+2\cos{30}+3\cos{60}+\cdots+2014\cos{270}+2015\cos{300}$
A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than $180$ as follows:
$x=1\cos{0}+2\cos{30}+3\cos{60}+4\cos{90}+5\cos{120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdots-2015\cos{120}$
Grouping terms with like-cosines and factoring out the cosines:
$x=(1-7+13-\cdots+2005-2011)\cos{0}+\cdots+(6-12+18-\cdots-2004+2010)\cos{150}$
Each sum in the parentheses has $336$ terms (except the very last one, which has $335$ ). By pairing each term, we see there are $\frac{336}{2}$ pairs of $-6$ . Therefore, each sum equals $168\cdot-6=-1008$ except the very last sum, which has $167$ pairs of $-6$ plus an extra 2010 and equals $167\cdot-6+2010=1008$ . Plugging in these values:
$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$
We can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:
$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$
$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$
Finally, we use the Pythagorean to find the distance from $P_0$ . This distance is given by:
$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot1008^2\cdot(1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}=1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6=\boxed{2024}$
| 24
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4,494
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 3
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A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?
First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\overline{P_0P_1}$ and $\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\overline{P_3C_1}$ . We will also set the distance $\overline{P_0P_1} = n$ and thus $\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}$ and the length $\overline{C_1C_2} = \frac{n+1}{2}$ . We can also see that the triangle $\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\overline{C_0C_1} = \frac{n+2}{2}$ and $\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length
\[\overline{P_0C_1} = n + \frac{(n+1)\sqrt{3}}{2} + \frac{n+2}{2}\] \[\overline{C_1C_4} = \frac{(n+1)}{2} + \frac{(n+2)\sqrt{3}}{2} + (n+3) + \frac{(n+4)\sqrt{3}}{2} + \frac{n+5}{2}\] \[\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}\] \[\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}\] , and finally \[\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}\]
Here the point $C_4$ is defined as the intersection of lines $\overline{P_3P_4}$ and $\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\overline{P_6P_7}$ and $\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\overline{P_{9}P_{10}}$ and $\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\sqrt{3})(\sqrt{2})$ which gives us a final answer of $\boxed{2024}$
| 24
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4,495
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 4
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A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ$
We want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction. [asy] draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow); draw((1,0)--(3/2+sqrt(3)/2, 0), dashed); draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed); draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed); draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed); draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed); [/asy] Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$ ) is $-6\left\langle 1, 2+\sqrt{3}\right\rangle$
To compute the distance to $P_{2015}$ , we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$
The bee reaches $P_{2016}$ after $\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\left\langle 1, 2+\sqrt{3}\right\rangle = \left\langle -1008, -2006-1008\sqrt{3}\right\rangle$
The bee moves at an angle of $-30^\circ$ from $P_{2015}$ to $P_{2016}$ , so subtracting it means moving an angle of $150^\circ$ . Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\left\langle -1008\sqrt{3}, 1008\right\rangle$
Therefore the total displacement vector to $P_{2015}$ is $\left\langle -1008, -2006-1008\sqrt{3}\right\rangle + \left\langle -1008\sqrt{3}, 1008\right\rangle = \left\langle -1008-1008\sqrt{3}, -1008-1008\sqrt{3}\right\rangle$ . The displacement is thus $\sqrt{2}\left|-1008-1008\sqrt{3}\right| = 1008\sqrt{2}+1008\sqrt{6} \implies 1008+2+1008+6 = \boxed{2024}$
| 24
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4,496
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 5
|
A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\frac{i(n-1)\pi}{6}}$ . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.
Then we have \[P_n=e^{0}+2e^{\frac{i\pi}{6}}+\cdots+2015e^{\frac{2014i\pi}{6}},\] and this becomes \[(1-7+13-\cdots-2011)e^0+(2-8+\cdots-2012)e^{\frac{i\pi}{6}}+\cdots+(6-12+\cdots+2010)e^{\frac{5i\pi}{6}}.\]
Simplifying, we have \[-6\cdot168-6\cdot168e^{\frac{i\pi}{6}}-6\cdot168e^{\frac{2i\pi}{6}}-6\cdot168e^{\frac{3i\pi}{6}}-6\cdot168e^{\frac{4i\pi}{6}}-(2010-6\cdot167)e^{\frac{5i\pi}{6}},\] which eventually simplifies to \[-1008-(2+\sqrt3)1008i,\] and this is a $15-75-90$ triangle which has ratios of $1:(2+\sqrt3):(\sqrt2+\sqrt6)$ so the magnitude is $1008\sqrt2+1008\sqrt6$ and the answer is $1008+2+1008+6=\boxed{2024}$
| 24
|
4,497
|
https://artofproblemsolving.com/wiki/index.php/2015_AMC_12B_Problems/Problem_25
| 6
|
A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$
|
After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in the directions corresponding to the moves with lengths $n+7$ $n+8$ $n+9$ $n+10$ $n+11$ , and $n+12$ . This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\sqrt{3})$ , a move that repeats every 12 moves. Since $\frac{2016}{12} = 168$ , we have that $P_(2016) = (-1008, -2016-1008\sqrt{3})$ . It follows that $P_(2015) = (-1008-1008\sqrt{3}, -1008-1008\sqrt{3})$ so the distance to $P_0$ is $1008(\sqrt{2} + \sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \boxed{2024}$
| 24
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4,498
|
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4
| 4
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problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object
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Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\text{B}-\text{O}-\text{R}-\text{Y}$
$\text{O}-\text{B}-\text{R}-\text{Y}$
The answer is $1+2=\boxed{3}$
| 3
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4,499
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https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4
| 5
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problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object
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There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\cdot\frac{1}{2}\cdot\frac{1}{2}= \boxed{3}$
| 3
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4,500
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https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_4
| 6
|
problem_id
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2 Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object
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To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
Therefore, our answer is $\boxed{3}$
| 3
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