id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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4,301 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | 1 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$ . So
$z^6=e^{\frac{ni\pi}{2}}$
This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{12}$ | 12 |
4,302 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | 2 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | $z = \sqrt[24]{1} = 1^{\frac{1}{24}}$
By Euler's identity $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$ , where $k$ is an integer.
Using De Moivre's Theorem , we have $z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}$ , where $0 \leq k<24$ that produce $24$ unique results.
Using De M... | 12 |
4,303 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | 3 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$ ), or notice that the question is simply referring to the 24th roots of unity, of which we know there mus... | 12 |
4,304 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_17 | 4 | There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | Let $a\in\mathbb{R}$ and $a = z^6.$ We have \[a^4 = 1 \implies a = 1,-1.\] $z^6 = \pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\boxed{12}.$ \[\] -svyn | 12 |
4,305 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | 1 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | Note that $n\equiv S(n)\bmod 9$ , so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$ . So, since $S(n)=1274\equiv 5\bmod 9$ , we have that $S(n+1)\equiv 6\bmod 9$ . The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{1239}$ | 239 |
4,306 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | 2 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | One possible value of $S(n)$ would be $1275$ , but this is not any of the choices. Therefore, we know that $n$ ends in $9$ , and after adding $1$ , the last digit $9$ carries over, turning the last digit into $0$ . If the next digit is also a $9$ , this process repeats until we get to a non- $9$ digit. By the end, the ... | 239 |
4,307 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_18 | 3 | Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$ | Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$ , we get $5$ . Adding one to that, we get $6$ . So, if we assess each option to see which one attains $6$ , we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9... | 239 |
4,308 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_20 | 1 | How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$ | By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$ . If $x\neq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=\pm \root{2016}\of{2017}$ . Hence, we have $3$ possible values $x$ , namely \[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-... | 597 |
4,309 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_21 | 1 | A set $S$ is constructed as follows. To begin, $S = \{0,10\}$ . Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$ , all of whose coefficients $a_i$ are elements of $S$ , then $x$ is put into $S$ . When no more elements ca... | At first, $S=\{0,10\}$
\[\begin{tabular}{r c l c l} \(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\ \(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\ \(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x... | 9 |
4,310 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_22 | 1 | A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ $(-2, 2)$ $(-2, -2)$ $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous m... | We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $\frac{4}{8}$ chance of moving to an inside edge and a $\frac{4}{8}$ chance of moving to ... | 39 |
4,311 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 1 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formul... | 7 |
4,312 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 2 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | Since all of the roots of $g(x)$ are distinct and are roots of $f(x)$ , and the degree of $f$ is one more than the degree of $g$ , we have that
\[f(x) = C(x-k)g(x)\]
for some number $k$ . By comparing $x^4$ coefficients, we see that $C=1$ . Thus,
\[x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)\]
Expanding and equating coeff... | 7 |
4,313 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 3 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | $f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x-r)\]
where $r\in\mathbb{C}$ is the fourth root of $f(x)$ . Substituting $g(x)$... | 7 |
4,314 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 4 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | A faster ending to Solution 1 is as follows.
We shall solve for only $a$ and $r$ . Since $10-r=100$ $r=-90$ , and since $a-r=1$ $a=-89$ . Then, \begin{align*} f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &=\boxed{7007} | 7 |
4,315 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 5 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored fo... | 7 |
4,316 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 6 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\ r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formul... | 7 |
4,317 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 7 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | Let the roots of $g(x)$ be $r_1$ $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ $r_2$ $r_3$ , and $r_4$ . From Vieta's, we have: \begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*} The fourth root is $a-1$ . Since $r_1$ $r_2$ , and $r_3$ are common roots, we have: \begin{align*} f(x)=g(x)(... | 7 |
4,318 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 8 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we g... | 7 |
4,319 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_23 | 9 | For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}... | Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after... | 7 |
4,320 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | 1 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ ... | Using the given ratios, note that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$
By AA Similarity, $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}$ and $\triangle ACX \sim \triangle EFX$ with a ratio of $\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}$ , so $\frac{XF}{CX}... | 17 |
4,321 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | 2 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ ... | We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point.
Let $Z$ be the intersection of $AC$ and $BD$ . First, from $ABCD$ being a cyclic quadrilateral, we have that $\triangle BCZ \sim \triangle AZD$ $\triangle BZA \sim \triangle CDZ$ . Therefore, $\frac{2}{... | 17 |
4,322 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | 3 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ ... | Denote $P$ to be the intersection between line $AE$ and circle $O$ . Note that $\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE$ , making $\angle GFE + \angle GDE = 180$ . Thus, $PEFG$ is a cyclic quadrilateral. Using Power of a Point on $X$ gives $XP \cdot XE = XG \cdot XF$
Since $\triangle ADX \sim \triangle ... | 17 |
4,323 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems/Problem_24 | 4 | Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ .
Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ ... | $\because$ $AC \parallel EF$ $\quad \therefore$ $\triangle ACX \sim \triangle EFX$ $\quad \frac{XF}{XC} = \frac{XE}{XA}$
By Power of a Point, $XG \cdot XC = XD \cdot XB$
By multiplying the $2$ equations we get $XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB$
$\because$ $YE \parallel AD$ $\quad \therefore$ $\triangle E... | 17 |
4,324 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_1 | 1 | Kymbrea's comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn's collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn's... | Kymbrea has $30$ comic books initially and every month, she adds two. This can be represented as $30 + 2x$ where x is the number of months elapsed. LaShawn's collection, similarly, is $10 + 6x$ . To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation $2(2x + 30) = 6x... | 25 |
4,325 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | 1 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\implies x=y$ .
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{2}$ | 2 |
4,326 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | 2 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\frac{x+3y}{3x-y}=\boxed{2}$ | 2 |
4,327 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | 3 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Let $y=ax$ . The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{2}$ - mathleticguyyy | 2 |
4,328 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_3 | 4 | Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\boxed{2}$ | 2 |
4,329 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_5 | 1 | The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$ , first quartile $Q_1 = 33$ , and third quartile $Q_3 = 43$ . An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ( $Q_1$ ) or more than $1.5$ times the interquartile range above th... | The interquartile range is defined as $Q3 - Q1$ , which is $43 - 33 = 10$ $1.5$ times this value is $15$ , so all values more than $15$ below $Q1$ $33 - 15 = 18$ is an outlier. The only one that fits this is $6$ . All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is... | 1 |
4,330 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_9 | 1 | A circle has center $(-10, -4)$ and has radius $13$ . Another circle has center $(3, 9)$ and radius $\sqrt{65}$ . The line passing through the two points of intersection of the two circles has equation $x+y=c$ . What is $c$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\... | The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$ . Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$ , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=(x-3)... | 3 |
4,331 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_9 | 2 | A circle has center $(-10, -4)$ and has radius $13$ . Another circle has center $(3, 9)$ and radius $\sqrt{65}$ . The line passing through the two points of intersection of the two circles has equation $x+y=c$ . What is $c$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\... | Note the specificity of the radii, $13$ and $\sqrt{65}$ , and that specificity is often deliberately added to simplify the solution to a problem.
One may recognize $13$ as the hypotenuse of the $\text{5-12-13}$ right triangle and $\sqrt{65}$ as the hypotenuse of the right triangle with legs $1$ and $8$ . We can suppose... | 3 |
4,332 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | 1 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a ... | 524 |
4,333 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | 2 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude ... | 524 |
4,334 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | 3 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in ... | 524 |
4,335 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | 4 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\bino... | 524 |
4,336 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_11 | 5 | Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?... | We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number o... | 524 |
4,337 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | 1 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A,... | Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields $6$ possible arrangements for... | 12 |
4,338 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_13 | 2 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(100); pair A,... | Burnside's Lemma can be applied to the problem. Number the disks 1-6 from top to bottom and left to right. First, we count the number of group actions. There are $6$ in total: $3$ rotations, $0$ °, $120$ °, and $240$ °. Additionally, there are $3$ different reflections, which are over lines passing through the middle o... | 12 |
4,339 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 1 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$ . Therefore, $BB' = 3$ and $BC' = 4$ . Also, $\angle B'BC' = 120^{\circ}$ , so by the Law of Cosines, $B'C' = \sqrt{37}$ . Therefore, the ... | 37 |
4,340 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 2 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$ . Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$ . Note that $\triangle A'DA$ is a 30-60-90 with right angle at $D$ . Since $AA'=6$ $AD=3$ and $A'D=3\sqrt{3}$ . So we know that $DB'=11$ . Note t... | 37 |
4,341 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 3 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Let $AB=BC=CA=x$ . We start by noting that we can just write $AB'$ as just $AB+BB'=4AB$ .
Similarly $BC'=4BC$ , and $CA'=4CA$ . We can evaluate the area of triangle $ABC$ by simply using Heron's formula, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\left(\frac{3x}{2}-x\right)}^3}=\frac{x^2\sqrt{3}}{4}$ .
Next in order to evaluate $... | 37 |
4,342 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 4 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Looking at the answer choices, we see that all but ${\textbf{(E)}}$ has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick $\boxed{37}$ | 37 |
4,343 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 5 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | We use barycentric coordinates wrt $\triangle ABC$ , to which we can easily obtain that $A'=(4,0,-3)$ $B'=(-3,4,0)$ , and $C'=(0,-3,4)$ . Now, since the coordinates are homogenized ( $-3+4=1$ ), we can directly apply the area formula to obtain that \[[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0... | 37 |
4,344 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 6 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | First, comparing bases yields that $[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12$ . By congruent triangles, \[[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],\] so $[A'B'C']:[ABC]=\boxed{37}$ | 37 |
4,345 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 8 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Drawing the diagram, we see that the large triangle, $A'B'C'$ , is composed of three congruent triangles with the triangle $ABC$ at the center. Let each of the sides of triangle $ABC$ be $x$ . Therefore, using the equilateral triangle area formula, the $[ABC] = \frac{x^2\sqrt{3}}{4}$ . We also know now that the sides o... | 37 |
4,346 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 9 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Solution by HydroQuantum
Let $AB=BC=CA=x$
Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$ \[y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(\cos 120) =\] \[(3x)^2+(4x)^2-2(3x)(4x)(\cos 120)=9x^2+16x^2-24x(\cos 120)=25x^2+12x^2=37x^2.\] Since both $\triangle ABC$ and $\triangle A'B'C'$ are both equilateral triangles, they must be... | 37 |
4,347 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 10 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Note that the height and base of $\triangle A'CC'$ are respectively 4 times and 3 times that of $\triangle ABC$ . Therefore the area of $\triangle A'CC'$ is 12 times that of $\triangle ABC$
By symmetry, $\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'$ . Adding the areas of these three triangles and $\trian... | 37 |
4,348 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 11 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | First we note that $A'B'C'\sim ABC$ due to symmetry. WLOG, let $B = (0, 0)$ and $AB = 1$ Therefore, $C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)$ . Using the condition that $CC' = 3$ , we get $C' = (4, 0)$ and $B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)$ . It is easy to check that $B'C' = \sq... | 37 |
4,349 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_15 | 12 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of ... | Note that angle $C'BB'$ is $120$ °, as it is supplementary to the equilateral triangle. Then, using area $= \frac{1}{2}ab\sin\theta$ and letting side $AB = 1$ for ease, we get: $4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}$ as the area of $C'BB'$ . Then, the area of $ABC$ is $\frac{\sqrt{3}}{4}$ , so the ratio is $\frac{3... | 37 |
4,350 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | 1 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We will consider this number $\bmod\ 5$ and $\bmod\ 9$ . By looking at the last digit, it is obvious that the number is $\equiv 4\bmod\ 5$ . To calculate the number $\bmod\ 9$ , note that
\[123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,\]
so it is equivalent to
... | 9 |
4,351 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | 2 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We know that this number is divisible by $9$ because the sum of the digits is $270$ , which is divisible by $9$ . If we subtracted $9$ from the integer we would get $1234 \cdots 4335$ , which is also divisible by $5$ and by $45$ . Thus the remainder is $9$ , or $\boxed{9}$ | 9 |
4,352 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_19 | 4 | Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We notice that $10^{k}\equiv 10 \pmod {45}$
Hence $N = 44+43\cdot10^{2}+42\cdot10^{4}+\cdots+10^{78} \equiv 44+10\cdot(1+2+3+\cdots+43)\equiv 9 \pmod {45}.$
Choose $\boxed{9}$ | 9 |
4,353 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 1 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | First, remove all the 90s, since they make no impact. So, we have numbers from $1$ to $10$ . Then, $5$ is the 7th number. Let the sum of the first $6$ numbers be $k$ . Then, $k\equiv 0 \mod 6$ and $k\equiv 3 \mod 7$ . We easily solve this as $k \equiv 24 \mod 42$ . Clearly, the sum of the first $6$ numbers must be $570... | 100 |
4,354 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 2 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sum of $90$ and an integer between $1$ and $10$ , we rewrite the problem into receiving scores between $1$ and $10$ . Later, we can add $90$ to her score to obtain the real answer.
From this point of view, the problem states that I... | 100 |
4,355 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 3 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | Let $n$ be Isabella's average after $6$ tests. $6n+95 \equiv 0 \pmod{7}$ , so $n \equiv 4 \pmod{7}$ . The only integer between $90$ and $100$ that satisfies this condition is $95$ . Let $m$ be Isabella's average after $5$ tests, and let $a$ be her sixth test score. $5m+a= 6\cdot95 \equiv 0 \pmod{5}$ , so $a$ is a multi... | 100 |
4,356 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 4 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | Let $T$ be the total sum of Isabella's first five test scores, and let $S$ be her score on the sixth test. It follows that $T\equiv 0\pmod {5}$ $T+S\equiv 0\pmod {6}$ , and $T+S+95\equiv 0\pmod {7}$ , since at each step, her average score was an integer. Using the last equivalence, $T+S\equiv -95\equiv 3\pmod{7}$ , so ... | 100 |
4,357 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 5 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | We know that Isabella's score on the 7th test is 95, and we work backwards. Let $a_n$ denote the average of the first $n$ scores. We are given that all $a_n$ are integers. If the $a_6$ is different from the $a_7$ , then the last score must be a multiple of 7 away from $a_6$ . However, $95-7=88$ and $95+7=102$ clearly c... | 100 |
4,358 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_21 | 6 | Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\... | Say the sum of the first 6 scores is $n$ . Then, we know that $n + 95 \equiv 0 \pmod{7}$ . Thus, $n \equiv 3 \pmod{7}$ . Additionally, since the average of these 6 scores was an integer, we know that $n \equiv 0 \pmod{6}$ . We find that the smallest $n$ to satisfy both of these inequalities is 24 (series below). \[3, 1... | 100 |
4,359 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_12B_Problems/Problem_25 | 1 | A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the numb... | Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is
\[T{n-5\choose 4}.\]
Thus, the expected value of the number of full teams in a random set of $9$ players is
\[\frac{T{n-5\choose 4}}{{n\ch... | 557 |
4,360 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | 1 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We can use subtraction of fractions to get \[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{100}.\] | 100 |
4,361 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | 2 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{100}$ | 100 |
4,362 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_1 | 3 | What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$ | We are given the equation $\frac{11!-10!}{9!}$
This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$ , which equals $10 \cdot 10$
Therefore, the answer is $10^2$ $\boxed{100}$ | 100 |
4,363 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_2 | 1 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$ | 3 |
4,364 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_2 | 2 | For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$ | 3 |
4,365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_4 | 1 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Since $x$ is the mean, \begin{align*} x&=\frac{60+100+x+40+50+200+90}{7}\\ &=\frac{540+x}{7}. \end{align*}
Therefore, $7x=540+x$ , so $x=\boxed{90}.$ | 90 |
4,366 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_4 | 2 | The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$ | Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the greater one. $x=\boxed{90}.$ ~bjc | 90 |
4,367 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_6 | 1 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | We are trying to find the value of $N$ such that \[1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.\] Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{9}.$ | 9 |
4,368 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_6 | 2 | A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .
Notice that $1 + 2 + 3 \cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \cdots + 20 = 155$ and $21 + \cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 6... | 9 |
4,369 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_9 | 1 | The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What i... | Let $s$ be the side length of the small squares.
The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$
Solving for $s$ , we get $s = \frac{4 - \sqrt{2}}{7}$ , so our answer is $4 + 7 \Rightarrow \boxed{11}$ | 11 |
4,370 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_9 | 2 | The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What i... | The diagonal of the small square can be written in two ways: $s \sqrt(2)$ and $2*(1-2s).$ Equating and simplifying gives $s = \frac{4 - \sqrt{2}}{7}$ . Hence our answer is $4 + 7 \Rightarrow \boxed{11}.$ | 11 |
4,371 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | 1 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in ... | 2 |
4,372 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | 2 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$ | 2 |
4,373 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_10 | 3 | Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved tw... | Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$ | 2 |
4,374 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_11 | 1 | Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\tex... | Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.
Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.
From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$
Adding these... | 64 |
4,375 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | 1 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the re... | 12 |
4,376 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | 2 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{... | 12 |
4,377 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_13 | 3 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3... | 12 |
4,378 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | 1 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ ... | Setting the first two equations equal to each other, $\log_3 x = \log_x 3$
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$
Now, by setting the first and third equations equal to each other, w... | 5 |
4,379 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_16 | 2 | The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ ... | Note that $\log_b a =\log_c a / \log_c b$
Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$
$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$
$\log_\frac{1}{b} a = -\log_a b$
Therefore, the system of equations can be simplified to:
$y = t$
$y = -t$
$y = \frac{1}{t}$
$y = -\frac{1}{t}$
where $t = \lo... | 5 |
4,380 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_18 | 1 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ . This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ . This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors... | 325 |
4,381 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_18 | 2 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | $110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a pr... | 325 |
4,382 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | 1 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relati... | For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37$
For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$
For $5$ heads, we either start off with $4$ heads, which gives us $_4\text{C}_1=4$ ways to arrange the ot... | 151 |
4,383 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | 2 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relati... | Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases:
(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ possibilities work.
(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The t... | 151 |
4,384 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_19 | 3 | Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relati... | Notice every $2$ flips, there is a $\dfrac{1}{4}$ chance to go $2$ steps left $(L),$ $\dfrac{1}{2}$ chance to stay put $(P),$ and $\dfrac{1}{4}$ chance to move $2$ steps right $(R).$ We have $4$ choices for how to get to $4:$ \[RR-1\text{ arrangement}-\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\] \[PRR-2\text{ arrangem... | 151 |
4,385 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | 1 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | We see that $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . So... | 109 |
4,386 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | 2 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | If the given conditions hold for all nonzero numbers $a, b,$ and $c$
Let $a=b=c.$ From the first two givens, this implies that
\[a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.\]
From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$
Let $c=b.$ Substituting this into the fi... | 109 |
4,387 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | 5 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x = 100$
$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$
$2016 \diamondsuit 2016 = 1$ $2016 \diamondsuit (2016 \diamondsuit 1) = 1$ , so $2016 \diamondsuit 1 = 2016$
$2016 \diamondsuit 1 = (2016 \diamondsu... | 109 |
4,388 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_20 | 6 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$ . Hence, $2016 \diamondsuit 6 = 336$ . Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$ . Therefore, the answer is $\boxed{109}$ | 109 |
4,389 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 1 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram dr... | 500 |
4,390 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 2 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(14... | 500 |
4,391 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 3 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
4,392 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 4 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
4,393 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 5 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -... | 500 |
4,394 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 7 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100... | 500 |
4,395 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 8 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM... | 500 |
4,396 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 10 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresp... | 500 |
4,397 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 11 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin... | 500 |
4,398 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_21 | 12 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
4,399 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | 1 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,... | 15 |
4,400 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_22 | 2 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\tex... | 15 |
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