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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4,001 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | 1 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\text... | Intersect at the origin and select a point on each line to define vectors $\mathbf{v}_{i}=(x_{i},y_{i})$ .
Note that $\theta=45^{\circ}$ gives equal magnitudes of the vector products \[\mathbf{v}_1\cdot\mathbf{v}_2 = v_{1}v_{2}\cos\theta \quad\mathrm{and}\quad |\mathbf{v}_1\times\mathbf{v}_2| = v_{1}v_{2}\sin\theta .\]... | 32 |
4,002 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | 2 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\text... | Place on coordinate plane.
Lines are $y=mx, y=6mx.$ The intersection point at the origin.
Goes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \implies ... | 32 |
4,003 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | 3 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\text... | Let one of the lines have equation $y=ax$ . Let $\theta$ be the angle that line makes with the x-axis, so $\tan(\theta)=a$ . The other line will have a slope of $\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}$ . Since the slope of one line is $6$ times the... | 32 |
4,004 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | 5 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\text... | If you have this formula memorized then it will be very easy to do: If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$ , then $\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}$
Now let the smaller slope be $m$ , thus the other slope is $6m$ . Using our formula above: \[\tan(45^\circ)=\frac{6m-m}{1+6m^2} \impli... | 32 |
4,005 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | 1 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging ... | 4 |
4,006 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | 2 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are ... | 4 |
4,007 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | 3 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power... | 4 |
4,008 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | 4 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $... | 4 |
4,009 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | 5 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This g... | 4 |
4,010 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 1 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^... | 100 |
4,011 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 2 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$ . Draw triangle $OCD$ , and median $OX$ . Because $OC = OD$ $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$ $OE = 5\sqrt2 - 2\sqrt5$ , and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\s... | 100 |
4,012 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 3 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | Let $O$ be the center of the circle. Notice how $OC = OD = r$ , where $r$ is the radius of the circle. By applying the law of cosines on triangle $OCE$ \[r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}.\]
Similarly, by applying the law of cosines on triangle $ODE$ \[r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2... | 100 |
4,013 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 4 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250); pair O, A, B, C, D, E, D1; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); D1 = (D.x,-D.y); draw(p); dot... | 100 |
4,014 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 5 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | Basically, by PoP, you have that \[CE \times DE = (10\sqrt{2}-2\sqrt{5})(2\sqrt{5}) = 20\sqrt{10} - 20.\] Therefore, as $CE^2 + DE^2 = (CD)^2 - 2(CE \times DE),$ basically, once you find $CD^2,$ the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to fin... | 100 |
4,015 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 6 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250); pair O, A, B, C, D, E, P, Q; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); P = (5*sqrt(2), -4.5); Q=(5*sqrt(2)-2*sqrt(5)-0.5, 0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectio... | 100 |
4,016 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | 7 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(... | Perhaps not reliable in general, but very useful as a last resort.
The choice of the radius $5\sqrt2$ is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition $BE=2\sqrt5$ , therefore it will likely be present in the answer as ... | 100 |
4,017 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_15 | 2 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know... | If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply ... | 13 |
4,018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 1 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.
Now, suppose that we have this deck again, with only one black card. Each time we pic... | 15 |
4,019 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 3 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | We know that we need to find the probability of adding 2 red and 2 blue balls in some order.
There are 6 ways to do this, since there are $\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.
We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted b... | 15 |
4,020 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 4 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red b... | 15 |
4,021 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 5 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\t... | 15 |
4,022 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 6 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | Here $X$ stands for R or B, and $Y$ for the remaining color.
After 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \[\frac{2}{3} \... | 15 |
4,023 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | 7 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the ur... | We can use dynamic programming to solve this problem.
We let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.
Notice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$... | 15 |
4,024 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17 | 1 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \q... | Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$ . We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$ . That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$ $\frac{-1+i\sqrt{3}}{2}$ $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$
In ord... | 2 |
4,025 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17 | 2 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \q... | Let $x_1=r$ , then \[x_2=\frac{-1+i\sqrt{3}}{2} r,\] \[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,\] \[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,\] which means $x_4$ is the same as $x_1$
Now we have 3 different roots of the polynomial, $x_1$ $x_2$ , and $x_3$ . Next, ... | 2 |
4,026 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | 1 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = ... | 6 |
4,027 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | 3 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separat... | 6 |
4,028 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | 4 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lf... | 6 |
4,029 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | 5 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le 70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which ... | 6 |
4,030 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | 6 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not squ... | 6 |
4,031 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_23 | 1 | How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that
\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\] then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?
$\textbf{(A)}\ 1 \qquad\textbf{... | For $n=2$ , we see that if $z_{1}+z_{2}=0$ , then $z_{1}=-z_{2}$ , so they are evenly spaced along the unit circle.
For $n=3$ , WLOG, we can set $z_{1}=1$ . Notice that now Re $(z_{2}+z_{3})=-1$ and Im $\{z_{2}\}$ $-$ Im $\{z_{3}\}$ . This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{... | 2 |
4,032 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 1 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmet... | 112 |
4,033 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 2 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization nee... | 112 |
4,034 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 3 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)... | 112 |
4,035 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 4 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2... | 112 |
4,036 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 6 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to... | 112 |
4,037 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 7 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}... | 112 |
4,038 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 8 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\c... | 112 |
4,039 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | 9 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are co... | Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero... | 112 |
4,040 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_1 | 1 | The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$
$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$ | The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{78}$ | 78 |
4,041 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | 1 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Since $a=1.5b$ , that means $b=\frac{a}{1.5}$ . We multiply by $3$ to get a $3b$ term, yielding $3b=2a$ , and $2a$ is $\boxed{200}$ of $a$ | 200 |
4,042 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | 2 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Without loss of generality, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\frac{3b}{a}=\frac{300}{150}=2$ , so $3b$ is $200\%$ of $a$ . Hence the answer is $\boxed{200}$ | 200 |
4,043 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | 3 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | As before, $a = 1.5b$ . Multiply by 2 to obtain $2a = 3b$ . Since $2 = 200\%$ , the answer is $\boxed{200}$ | 200 |
4,044 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | 4 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Without loss of generality, let $b=2$ . Then, we have $a=3$ and $3b=6$ . This gives $\frac{3b}{a}=\frac{6}{3}=2$ , so $3b$ is $200\%$ of $a$ , so the answer is $\boxed{200}$ | 200 |
4,045 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_3 | 1 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) }... | We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$... | 76 |
4,046 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_4 | 1 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\b... | 90 |
4,047 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_4 | 2 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$ | 90 |
4,048 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 1 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ .
Now we find the i... | 6 |
4,049 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 2 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\boxed{6}$ | 6 |
4,050 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 3 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\boxed{6}$ | 6 |
4,051 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 4 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-... | 6 |
4,052 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{6}$ | 6 |
4,053 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 6 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8... | 6 |
4,054 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 ... | 6 |
4,055 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 8 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \... | 6 |
4,056 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 9 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\sqrt2$ and equal length $2\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\frac{b\sqrt{4l^2-b^2}}{4}$ . ... | 6 |
4,057 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 10 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8... | 6 |
4,058 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | 12 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$ . Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqr... | 6 |
4,059 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_8 | 1 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures: [asy] unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0... | 19 |
4,060 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | 1 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$
Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]
By summing the geometric series and simplifying, w... | 16 |
4,061 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | 2 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Let $a = 0.2323\dots_k$ . Therefore, $k^2a=23.2323\dots_k$
From this, we see that $k^2a-a=23_k$ , so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$
Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is... | 16 |
4,062 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | 3 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$
Therefore now, we can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$ . Notice that $2k+3=2(k+1)+1=2(k-1)+5$ . As $17$ is a prime number, we therefore must have that one of $k-1$ and $k+1$ is divisible by $17$ ... | 16 |
4,063 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | 4 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$ . Now we want our base, $k$ , to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$ , the reason being that we wish to convert the number from base $10$ to base $k$ . Given the first equation, we know that $k$ must e... | 16 |
4,064 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | 5 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \im... | 16 |
4,065 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | 2 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | After obtaining $k + \frac{4}{k} = 6$ , notice that the required answer is $\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{20}$ , as before. | 20 |
4,066 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | 3 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | From the given data, $\log_2(x) = \frac{1}{\log_{16}(y)}$ , or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$
We know that $xy=64$ , so $x= \frac{64}{y}$
Thus $\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}$ , so $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$ , so $6(\log_2(y))-(\log_2(y))^2=4$
Solving for $\log_2(y)$ , we ... | 20 |
4,067 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | 1 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$ , but without los... | 432 |
4,068 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | 2 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | We note that the primes can be colored any of the $3$ colors since they don't have any proper divisors other than $1$ , which is not in the list. Furthermore, $6$ is the only number in the list that has $2$ distinct prime factors (namely, $2$ and $3$ ), so we do casework on $6$
Case 1 $2$ and $3$ are the same color
In ... | 432 |
4,069 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | 3 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | $2,4,8$ require different colors each, so there are $6$ ways to color these.
$5$ and $7$ can be any color, so there are $3\times 3$ ways to color these.
$6$ can have $2$ colors once $2$ is colored, and thus $3$ also has $2$ colors following $6$ , which leaves another $2$ for $9$
All together: $6\times 3 \times 3 \times... | 432 |
4,070 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_14 | 1 | For a certain complex number $c$ , the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\] has exactly 4 distinct roots. What is $|c|$
$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$ | The polynomial can be factored further broken down into
$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$
by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not n... | 10 |
4,071 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | 1 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \te... | Let $p, q$ , and $r$ be the roots of the polynomial. Then,
$p^3 - 5p^2 + 8p - 13 = 0$
$q^3 - 5q^2 + 8q - 13 = 0$
$r^3 - 5r^2 + 8r - 13 = 0$
Adding these three equations, we get
$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$
$s_3 - 5s_2 + 8s_1 = 39$
$39$ can be written as $13s_0$ , giving
$s_3 = 5s_2 -... | 10 |
4,072 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | 2 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \te... | Let $p, q$ , and $r$ be the roots of the polynomial. By Vieta's Formulae, we have
$p+q+r = 5$
$pq+qr+rp = 8$
$pqr=13$
We know $s_k = p^k + q^k + r^k$ . Consider $(p+q+r)(s_k) =5s_k$
$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$
Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^... | 10 |
4,073 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | 3 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \te... | Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$ . Then:
$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$
If we multiply both sides of the equation by $r^k$ , where $k$ is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can tr... | 10 |
4,074 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | 1 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac... | 9 |
4,075 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | 2 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | $\angle ACB$ is obtuse since its cosine is negative, so we let the foot of the altitude from $C$ to $AB$ be $H$ . Let $AH=11x$ $AC=16x$ $BH=7y$ , and $BC=8y$ . By the Pythagorean Theorem, $CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}$ and $CH=\sqrt{64y^2-49y^2}=y\sqrt{15}$ . Thus, $y=3x$ . The sides of the triangle are then $16... | 9 |
4,076 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | 3 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | Using the law of cosines, we get the following equations:
\[c^2=a^2+b^2+\frac{ab}{2}\] \[b^2=a^2+c^2-\frac{7ac}{4}\] \[a^2=b^2+c^2-\frac{11bc}{8}\]
Substituting $a^2+c^2-\frac{7ac}{4}$ for $b^2$ in $a^2=b^2+c^2-\frac{11bc}{8}$ and simplifying, we get the following: \[14a+11b=16c\]
Note that since $a, b, c$ are integers... | 9 |
4,077 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | 1 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 7... | Note that $z = \mathrm{cis }(45^{\circ})$
Also note that $z^{k} = z^{k + 8}$ for all positive integers $k$ because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo $8$
$1^2, 5^2,$ and $9^2$ are all $1 \pmod{8}$
$2^2, 6^2,$ and $10^2$ are all $4 \pmod{8}$
$3^2, 7^2,$ and $11^2$ are... | 36 |
4,078 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | 2 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 7... | It is well known that if $|z|=1$ then $\bar{z}=\frac{1}{z}$ . Therefore, we have that the desired expression is equal to \[\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)\] We know that $z=e^{\frac{i\pi}{4}}$ so $\bar{z}=e^{\frac{i7\pi}{4}}$ . Then, by De Moivre's Theore... | 36 |
4,079 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | 3 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 7... | We first calculate that $z^4 = -1$ . After a bit of calculation for the other even powers of $z$ , we realize that they cancel out add up to zero. Now we can simplify the expression to $\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)$ . Then, ... | 36 |
4,080 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | 4 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 7... | $z=\mathrm{cis }(\pi/4)$
Perfect squares mod 8: $1,4,1,0,1,4,1,0,1,4,1,0$
$1/z=\overline{z}=\mathrm{cis }(7\pi/4)$
$6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}$ | 36 |
4,081 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_22 | 1 | Circles $\omega$ and $\gamma$ , both centered at $O$ , have radii $20$ and $17$ , respectively. Equilateral triangle $ABC$ , whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$ , has vertex $A$ on $\omega$ , and the line containing side $\overline{BC}$ is tangent to $\gamma$ . Segments $\ove... | [asy] size(20cm); draw(circle((0,0), 20)); label("$\omega$", (0,0), 4.05*20*dir(149)*20/21); draw(circle((0,0), 17)); label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21); dot((0,0)); label("$O$", (0,0), E); pair aa = (-20, 0); dot(aa); label("$A$", aa, W); draw((-20,0)--(0,0)); real a = (-20 + (80/sqrt(13) - 34/sqrt(3))*(... | 130 |
4,082 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | 1 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsu... | By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$ . By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$ . Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$ . Thus, for each positive integer $m \geq 3$... | 11 |
4,083 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | 2 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsu... | Using the recursive definition, $a_4 = (4 \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$ . Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$ . Ther... | 11 |
4,084 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | 3 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsu... | We are given that \[a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] \[a_n=(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Since we are asked to find $\log_7(a_{2019})$ , we directly apply \[\log_7(a_n)=\log_7(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Using the property that $\log_ab^c=c\log_ab$ \[\log_7(a_... | 11 |
4,085 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | 1 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | The main insight is that
\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ . Thus,
\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$ , or in oth... | 34 |
4,086 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | 2 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | We can use the P-Adic Valuation (more info could be found here: Mathematicial notation ) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we... | 34 |
4,087 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | 3 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So... | 34 |
4,088 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | 1 | Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_... | For all nonnegative integers $n$ , let $\angle C_nA_nB_n=x_n$ $\angle A_nB_nC_n=y_n$ , and $\angle B_nC_nA_n=z_n$
Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$ ; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$ . By a similar argument, $\angle A_0A_1C_1=\angle... | 15 |
4,089 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | 2 | Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_... | We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of $\triangle A_nB_nC_n$ and $60^{\circ}$ (we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometr... | 15 |
4,090 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_2 | 1 | Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?
$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$ | Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{27}$ | 27 |
4,091 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | 1 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | \[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$... | 10 |
4,092 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | 2 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$ | 10 |
4,093 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | 3 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$ | 10 |
4,094 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | 4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=... | 10 |
4,095 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | 5 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we s... | 10 |
4,096 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | 1 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } ... | If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$ | 21 |
4,097 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | 2 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } ... | We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value o... | 21 |
4,098 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | 3 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } ... | We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$ | 21 |
4,099 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_6 | 1 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infini... | Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$ ,... | 0 |
4,100 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_6 | 2 | In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infini... | Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $C... | 0 |
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