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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6,701 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | 2 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the s... | First do the same process for assigning coordinates to the car. The car moves $\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $\left(\frac{2}{3}t, 0\right)$
Take the storm as circle. Given southeast movement, split the vector into component, getting position $\left(\frac{1}{2}t, 11... | 198 |
6,702 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | 3 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the s... | We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\left(\frac{1}{6}, \frac{1}{2}\right)$ . Labeling the car's starting position A, the storm center B, and the right triangle fo... | 198 |
6,703 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | 1 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . F... | Looking at the properties of the number, it is immediately guess-able that $a = \phi = \frac{1+\sqrt{5}}2$ (the golden ratio ) is the answer. The following is the way to derive that:
Since $\sqrt{2} < a < \sqrt{3}$ $0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1$ . Thus $\langle a^2 \rangle = a^{-1}$ , and it... | 233 |
6,704 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | 2 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . F... | Find $a$ as shown above. Note that, since $a$ is a root of the equation $a^3 - 2a - 1 = 0$ $a^3 = 2a + 1$ , and $a^{12} = (2a + 1)^4$ . Also note that, since $a$ is a root of $a^2 - a - 1 = 0$ $\frac{1}{a} = a - 1$ . The expression we wish to calculate then becomes $(2a + 1)^4 - 144(a - 1)$ . Plugging in $a = \frac{1 +... | 233 |
6,705 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | 4 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . F... | As Solution 1 stated, $a^3 - 2a - 1 = 0$ $a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)$ . So, $a^2 - a - 1 = 0$ $1 = a^2 - a$ $\frac1a = a-1$ $a^3 = 2a+1$ $a^2 = a+1$
$a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5$
$a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 14... | 233 |
6,706 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | 1 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of thre... | Adding the cases up, we get $27 + 54 + 36 = \boxed{117}$ | 117 |
6,707 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | 2 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of thre... | Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different t... | 117 |
6,708 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | 3 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of thre... | Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ possibilities. However if all three attributes are the same then the set must be c... | 117 |
6,709 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 1 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Note that $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}$ by the cofunction identities.(We could have also written it as $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n... | 241 |
6,710 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 2 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | \begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} \end{eqnarray*}
Using the identity $\sin a +... | 241 |
6,711 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 3 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | A slight variant of the above solution, note that
\begin{eqnarray*} \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\ \sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n \end{eqnarray*}
This is th... | 241 |
6,712 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 4 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)
$= - 1 + ... | 241 |
6,713 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 5 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written as $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ .
Consider the rhombus $OABC$ on the complex plane such that $O$ is the origin, $A$ represents $\te... | 241 |
6,714 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 6 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | We write $x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}$ since $\cos x = \sin (90^{\circ}-x).$ Now we by the sine angle sum we know that $\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).$ So the expression simplifies to $\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos ... | 241 |
6,715 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | 7 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | We can pair the terms of the summations as below.
\[\dfrac{(\cos{1} + \cos{44}) + (\cos{2} + \cos{43}) + (\cos{3} + \cos{42}) + \cdots + (\cos{22} + \cos{23})}{(\sin{1} + \sin{44}) + (\sin{2} + \sin{43}) + (\sin{3} + \sin{42}) + \cdots + (\sin{22} + \sin{23})}.\]
From here, we use the cosine and sine subtraction formul... | 241 |
6,716 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | 2 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Begin by finding the inverse function of $f(x)$ , which turns out to be $f^{-1}(x)=\frac{19d-b}{a-19c}$ . Since $f(f(x))=x$ $f(x)=f^{-1}(x)$ , so substituting 19 and 97 yields the system, $\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}$ , an... | 58 |
6,717 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | 3 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Because there are no other special numbers other than $19$ and $97$ , take the average to get $\boxed{58}$ . (Note I solved this problem the solution one way but noticed this and this probably generalizes to all $f(x)=x, f(y)=y$ questions like these) | 58 |
6,718 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | 5 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$ , we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$ . The only point that is undefined under an inversion is the center of the circ... | 58 |
6,719 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | 1 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | Define $\theta = 2\pi/1997$ . By De Moivre's Theorem the roots are given by
Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$ , and let $w$ be the root corresponding to $n\theta=2n\pi/ 1997$ . Then \begin{align*} |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\... | 582 |
6,720 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | 2 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | The solutions of the equation $z^{1997} = 1$ are the $1997$ th roots of unity and are equal to $\text{cis}(\theta_k)$ , where $\theta_k = \tfrac {2\pi k}{1997}$ for $k = 0,1,\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity $|v+w|$ is unchanged upon rotation... | 582 |
6,721 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | 3 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | We can solve a geometrical interpretation of this problem.
Without loss of generality, let $u = 1$ . We are now looking for a point exactly one unit away from $u$ such that the point is at least $\sqrt{2 + \sqrt{3}}$ units away from the origin. Note that the "boundary" condition is when the point will be exactly $\sqrt... | 582 |
6,722 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | 4 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | Since $z^{1997}=1$ , the roots will have magnitude $1$ . Thus, the roots can be written as $\cos(\theta)+i\sin(\theta)$ and $\cos(\omega)+i\sin(\omega)$ for some angles $\theta$ and $\omega$ . We rewrite the requirement as $\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|$ , which can now be ea... | 582 |
6,723 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_15 | 1 | The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square... | Consider points on the complex plane $A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the other tw... | 554 |
6,724 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_15 | 2 | The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square... | Since $\angle{BAD}=90$ and $\angle{EAF}=60$ , it follows that $\angle{DAF}+\angle{BAE}=90-60=30$ . Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$ . Let the image of $D$ be $D'$ . Since angles are preserved under rotation, $\angle{DAF}=\angle{D'AE}$ . It follows that $\angle{D'AE}+\ang... | 554 |
6,725 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | 1 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | Let's make a table.
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\]
\begin{eqnarray*} x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95 \end{eqnarray*}
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progre... | 200 |
6,726 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | 2 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | Use the table from above. Obviously $c = 114$ . Hence $a+e = 115$ . Similarly, $1+a = 96 + e \Rightarrow a = 95+e$
Substitute that into the first to get $2e = 20 \Rightarrow e=10$ , so $a=105$ , and so the value of $x$ is just $115+x = 210 + 105 \Rightarrow x = \boxed{200}$ | 200 |
6,727 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | 3 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\] The formula \[e=\frac{1+19}{2}\] can be used. Therefore, $e=10$ . Similarly, \[96=\frac{1+d}{2}\] So $d=191$
Now we have this table: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hli... | 200 |
6,728 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_2 | 1 | For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed x. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer? | For integers $k$ , we want $\lfloor \log_2 n\rfloor = 2k$ , or $2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}$ . Thus, $n$ must satisfy these inequalities (since $n < 1000$ ):
There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+... | 340 |
6,729 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_4 | 1 | A wooden cube , whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex , the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square cen... | [asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw... | 166 |
6,730 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7 | 2 | Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? | There are 4 cases:
1. The center square is occupied, in which there are $12$ cases.
2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases.
3. The center square isn't occupied and the two squares can rotate to each other ... | 300 |
6,731 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_8 | 1 | The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$ | The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}\cdot2^{19})$ , and by SFFT $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$ . Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ( $3... | 799 |
6,732 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | 1 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he en... | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$ , leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$ . Then he goes ahead and opens all lockers $2 \pmod {8}$ , leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$ . He the... | 342 |
6,733 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | 2 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he en... | We can also solve this with recursion. Let $L_n$ be the last locker he opens given that he started with $2^n$ lockers. Let there be $2^n$ lockers. After he first reaches the end of the hallway, there are $2^{n-1}$ lockers remaining. There is a correspondence between these unopened lockers and if he began with $2^{n-1}$... | 342 |
6,734 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | 3 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he en... | List all the numbers from $1$ through $1024$ , then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $\boxed{342}$ | 342 |
6,735 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | 1 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{... | 159 |
6,736 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | 2 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}$ which is the same as $\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141{^\circ}}$
So $19x = 141 +180n$ , for some integer $n$ .
Multiplying by $19$ g... | 159 |
6,737 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | 3 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | It seems reasonable to assume that $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}$ for some angle $\theta$ . This means \[\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}\] for some constant $... | 159 |
6,738 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | 1 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | \begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}
Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$
or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2}... | 276 |
6,739 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | 2 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | Divide through by $z^3$ . We get the equation $z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0$ . Let $x = z + \frac {1}{z}$ . Then $z^3 + \frac {1}{z^3} = x^3 - 3x$ . Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$ , with solutions $x = 1, \frac { - 1\pm\sqrt {5}}{2}$ . For $x = 1$ , ... | 276 |
6,740 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | 3 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | We recognize that $z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^7-1}{z-1}$ and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting $z^5$ and $z$ as shown below. $z^6+z^5+z^4+z^3+z^2+z+1-(z^5+z)=0\implies \frac{z^7-1}{z-1}-(z^5+z)=0$
Now, knowing that $z=1$ is not a roo... | 276 |
6,741 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | 4 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | Add 1 to both sides of the equation to get $x^6+x^4+x^3+x^2+1+1=1$ . We can rearrange to find that $(x^6+x^3+1)+(x^4+x^2+1)=1$ . Then, using sum of a geometric series, $\frac{x^9-1}{x^3-1}+\frac{x^6-1}{x^2-1}=1$
Combining the two terms of the LHS, we get that $\frac{x^{11}-x^9-x^2+1+x^9-x^6-x^3+1}{x^5-x^3-x^2+1}=1$ , s... | 276 |
6,742 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_14 | 1 | $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes | Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \in \mathbb{Z_{+}}$
We consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$ , where $m \in \mathbb{R}, 0 < m \le 1$
Consider... | 768 |
6,743 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_14 | 2 | $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes | Consider a point travelling across the internal diagonal, and let the internal diagonal have a length of $d$ . The point enters a new unit cube in the $x,y,z$ dimensions at multiples of $\frac{d}{150}, \frac{d}{324}, \frac{d}{375}$ respectively. We proceed by using PIE.
The point enters a new cube in the $x$ dimension ... | 768 |
6,744 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_15 | 1 | In parallelogram $ABCD$ , let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$ , and angle $ACB$ is $r$ times as large as angle $AOB$ . Find $\lfloor 1000r \rfloor$ | Let $\theta = \angle DBA$ . Then $\angle CAB = \angle DBC = 2\theta$ $\angle AOB = 180 - 3\theta$ , and $\angle ACB = 180 - 5\theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $\triangle ABO,\, \triangle BCO$
Dividing the two equalities yields
\[\frac{\sin 2\theta}{\sin \theta... | 777 |
6,745 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_1 | 1 | Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bi... | The sum of the areas of the squares if they were not interconnected is a geometric sequence
Then subtract the areas of the intersections, which is $\left(\frac{1}{4}\right)^2 + \ldots + \left(\frac{1}{32}\right)^2$
The majority of the terms cancel, leaving $1 + \frac{1}{4} - \frac{1}{1024}$ , which simplifies down to $... | 255 |
6,746 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_4 | 1 | Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$ . The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. | We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$ . Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency ). Then we note that $\overline{O_3A... | 224 |
6,747 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | 2 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$ . Then $n^2$ has $\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$
This simplifies to $\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$
The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$
Thus, our general formula for $n=p_1^{k_1}p_... | 589 |
6,748 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | 3 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | Consider divisors of $n^2: a,b$ such that $ab=n^2$ .
WLOG, let $b\ge{a}$ and $b=\frac{n}{a}$
Then, it is easy to see that $a$ will always be less than $b$ as we go down the divisor list of $n^2$ until we hit $n$
Therefore, the median divisor of $n^2$ is $n$
Then, there are $(63)(39)=2457$ divisors of $n^2$ . Exactly $\... | 589 |
6,749 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | 1 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | Let $x=\angle CAM$ , so $3x=\angle CDM$ . Then, $\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11$ . Expanding $\tan 3x$ using the angle sum identity gives \[\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.\] Thus, $\frac{3-\tan^2x}{1-3\tan^2x}=11$ . Solving, we get $\tan x= \frac 12$ . Hence, $CM=\frac{11}2$ and $AC... | 616 |
6,750 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | 2 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | In a similar fashion, we encode the angles as complex numbers, so if $BM=x$ , then $\angle BAD=\text{Arg}(11+xi)$ and $\angle BDM=\text{Arg}(1+xi)$ . So we need only find $x$ such that $\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)$ . This will happen when $\frac{363x-x^3}{1331-33x^2}=x$ , ... | 616 |
6,751 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | 3 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | Let $\angle BAD=\alpha$ , so $\angle BDM=3\alpha$ $\angle BDA=180-3\alpha$ , and thus $\angle ABD=2\alpha.$ We can then draw the angle bisector of $\angle ABD$ , and let it intersect $\overline{AM}$ at $E.$ Since $\angle BAE=\angle ABE$ $AE=BE.$ Let $AE=x$ . Then we see by the Pythagorean Theorem, $BM=\sqrt{BE^2-ME^2}=... | 616 |
6,752 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | 4 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | The triangle is symmetrical so we can split it in half ( $\triangle ABM$ and $\triangle ACM$ ).
Let $\angle BAM = y$ and $\angle BDM = 3y$ . By the Law of Sines on triangle $BAD$ $\frac{10}{\sin 2y} = \frac{BD}{\sin y}$ . Using $\sin 2y = 2\sin y\cos y$ we can get $BD = \frac{5}{\cos y}$ . We can use this information t... | 616 |
6,753 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | 5 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real... | Suppose $\angle BAM=\angle CAM =x$ , since $\angle BDC=3\angle BAC$ , we have $\angle BDM=\angle MDC = 3x$ . Therefore, $\angle DBC=\angle DCB = 90^\circ -3x$ and $\angle ABD=\angle DCA=2x$ . As a result, $\triangle KAC$ is isosceles, $KC=KA$
Let $H$ be a point on the extension of $CD$ through $D$ such that $\overline{... | 616 |
6,754 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_10 | 1 | What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer? | Let our answer be $n$ . Write $n = 42a + b$ , where $a, b$ are positive integers and $0 \leq b < 42$ . Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.
If $b$ is $0\mod{5}$ , then $b = 5$ because $5$ is the only prime divisible by $5$ . We get $n = 215$ as our largest possibility in this case.
If $b$ is $1... | 215 |
6,755 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_13 | 1 | Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$ | This is a pretty easy problem just to bash. Since the max number we can get is $7$ , we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$ . Then just do how many numbers there are times $\frac{1}{\lfloor n \rfloor}$ , which should be $5+17+37+65+101+145+30 = \boxed{400}$ | 400 |
6,756 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_14 | 1 | In a circle of radius $42$ , two chords of length $78$ intersect at a point whose distance from the center is $18$ . The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the for... | Let the center of the circle be $O$ , and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$ , such that $AE = CE < BE = DE$ . Let $F$ be the midpoint of $\overline{AB}$ . Then $\overline{OF} \perp \overline{AB}$
By the Pythagorean Theorem $OF = \sqrt{OB^2 - BF^2} = \sqrt{42^2 - 39^2} = 9\sqrt{3}$... | 378 |
6,757 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_2 | 1 | A circle with diameter $\overline{PQ}$ of length 10 is internally tangent at $P$ to a circle of radius 20. Square $ABCD$ is constructed with $A$ and $B$ on the larger circle, $\overline{CD}$ tangent at $Q$ to the smaller circle, and the smaller circle outside $ABCD$ . The length of $\overline{AB}$ can be written in the... | 1994 AIME Problem 2 - Solution.png
Call the center of the larger circle $O$ . Extend the diameter $\overline{PQ}$ to the other side of the square (at point $E$ ), and draw $\overline{AO}$ . We now have a right triangle , with hypotenuse of length $20$ . Since $OQ = OP - PQ = 20 - 10 = 10$ , we know that $OE = AB - OQ =... | 312 |
6,758 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_3 | 1 | The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$ | \begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ &= 4561 \end{align*}
So, the remainder is $\boxed{561}$ | 561 |
6,759 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_3 | 2 | The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$ | Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, \[T_{n-1} + T_n = n^2,\] where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n... | 561 |
6,760 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_4 | 1 | Find the positive integer $n\,$ for which \[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\] (For real $x\,$ $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$ | Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$ , then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$
Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$ . So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$
Let $k$ be the integer such that $2^k \le n<2^{k+... | 312 |
6,761 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_5 | 1 | Given a positive integer $n\,$ , let $p(n)\,$ be the product of the non-zero digits of $n\,$ . (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
What is the largest prime factor of $S\,$ | Suppose we write each number in the form of a three-digit number (so $5 \equiv 005$ ), and since our $p(n)$ ignores all of the zero-digits, replace all of the $0$ s with $1$ s. Now note that in the expansion of
we cover every permutation of every product of $3$ digits, including the case where that first $1$ represents... | 103 |
6,762 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_5 | 2 | Given a positive integer $n\,$ , let $p(n)\,$ be the product of the non-zero digits of $n\,$ . (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
What is the largest prime factor of $S\,$ | Note that $p(1)=p(11), p(2)=p(12), p(3)=p(13), \cdots p(19)=p(9)$ , and $p(37)=3p(7)$ . So $p(10)+p(11)+p(12)+\cdots +p(19)=46$ $p(10)+p(11)+\cdots +p(99)=46*45=2070$ . We add $p(1)+p(2)+p(3)+\cdots +p(10)=45$ to get 2115. When we add a digit we multiply the sum by that digit. Thus $2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115... | 103 |
6,763 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | 1 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.
[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(... | 660 |
6,764 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | 2 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two uni... | 660 |
6,765 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | 3 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of ... | 660 |
6,766 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_7 | 1 | For certain ordered pairs $(a,b)\,$ of real numbers , the system of equations
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there? | The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$ , centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$ $(\pm5,\pm5)$ , and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cd... | 72 |
6,767 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | 1 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
\[(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\... | 315 |
6,768 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | 2 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:
First, drop a perpendicular from $O$ to $AB$ . Call this midpoint of $AB M$ . Thus, $M=\left(\frac{a+b}{2}, 24\right)$ . The vector from $... | 315 |
6,769 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | 3 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Plot this equilateral triangle on the complex plane.
Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\left(\frac{a+b}{3}, 16i\right)$ . The new coordinates of the equilateral triangle a... | 315 |
6,770 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | 4 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$
$a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248$ . Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$ $b^2 = 2ab - 555 ==> a^2 = 2ab + 693$
We have $b^2 + 1248 = 2b\sqrt{b^2+1248} + 693$
Moving ... | 315 |
6,771 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_9 | 1 | A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two... | Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:
Therefore, we obtain the recursion $P_k = \frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $\frac {3^5}{11*9*7*5*3} = \frac {9}{385}$ ... | 394 |
6,772 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_9 | 2 | A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two... | Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here.
1. We pick the other A. There's a $\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this ... | 394 |
6,773 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | 1 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | Since $\triangle ABC \sim \triangle CBD$ , we have $\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$ , respectively, where x is an integer.
By the Pythagorean Theorem , we find that $AC^2 + BC^2 = AB^2 \L... | 450 |
6,774 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | 2 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | We will solve for $\cos B$ using $\triangle CBD$ , which gives us $\cos B = \frac{29^3}{BC}$ . By the Pythagorean Theorem on $\triangle CBD$ , we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$ . Trying out factors of $29^6$ , we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$... | 450 |
6,775 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | 3 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | Using similar right triangles, we identify that $CD = \sqrt{AD \cdot BD}$ . Let $AD$ be $29 \cdot k^2$ , to avoid too many radicals, getting $CD = k \cdot 29^2$ . Next we know that $AC = \sqrt{AB \cdot AD}$ and that $BC = \sqrt{AB \cdot BD}$ . Applying the logic with the established values of k, we get $AC = 29k \cdot ... | 450 |
6,776 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_11 | 2 | Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all ninety-four of... | Using bricks of dimensions $4''\times10''\times19''$ is comparable to using bricks of dimensions $0''\times6''\times15''$ which is comparable to using bricks of dimensions $0''\times2''\times5''$ . Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$
It follows ... | 465 |
6,777 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_12 | 1 | A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the fie... | Suppose there are $n$ squares in every column of the grid, so there are $\frac{52}{24}n = \frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$
Each vertical fence has length $24$ , and there are $\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and the... | 702 |
6,778 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_12 | 2 | A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the fie... | Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$ . (If you are confused, just draw the square out). This is clearly greater than $1994$ ,... | 702 |
6,779 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_13 | 1 | The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of | Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$
Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$
$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \tex... | 850 |
6,780 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_13 | 2 | The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of | Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]
Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]
Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.
We see that $\dfrac{1}{x}=13-\omega$ . Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \ome... | 850 |
6,781 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_15 | 1 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppo... | Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$ , and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$ . According to the problem statement, the circumcenter... | 597 |
6,782 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_1 | 1 | How many even integers between 4000 and 7000 have four different digits? | The thousands digit is $\in \{4,5,6\}$
Case $1$ : Thousands digit is even
$4, 6$ , two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$
Case $2$... | 728 |
6,783 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_1 | 2 | How many even integers between 4000 and 7000 have four different digits? | Firstly, we notice that the thousands digit could be $4$ $5$ or $6$ .
Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework.
Case $1$
Here we let thousands digit be $4$
4 _ _ _
We take care of restrictions first, and realize that there are 4 choi... | 728 |
6,784 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_2 | 1 | During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}... | On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\righ... | 580 |
6,785 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | 1 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,
Similarily, those who caught $12$ or fewer fish averaged $5$ fish pe... | 943 |
6,786 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | 2 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\text{(b)}$ , we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$ . From $\text{(c)}$ we have $\f... | 943 |
6,787 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 1 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$
Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ ... | 870 |
6,788 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 2 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $b = a + m$ and $c = a + m + n$ . From $a + d = b + c$ $d = b + c - a = a + 2m + n$
Substituting $b = a + m$ $c = a + m + n$ , and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$ \[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\] Hence, $(m,n) = (1,92)$ or $(3,28)$
For $(m,n) = (1,92)$ , we ... | 870 |
6,789 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 3 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \begin{align*} (a+d)^2 = (b+c)^2 &\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\ &\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\ &\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \end{align*} We can plug $93$ in for $bc - ad$ t... | 870 |
6,790 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Add the two conditions together to get $a+d+ad+93=b+c+bc$ . Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$ . This implies that for every quadruple $(a,b,c,d)$ , we can replace $a\longrightarrow a+1$ $b\longrightarrow b+1$ , etc. and this will still produce a valid quadruple. This means, that we can f... | 870 |
6,791 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 5 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $r = d-c$ . From the equation $a+d = b+c$ , we have \[r = d-c = b-a ,\] so $b = a+r$ and $c = d-r$ . We then have \[93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .\] Since $c > b$ $d-r > a+r$ , or $d-a-r > r$ . Since the prime factorization of 93 is $3 \cdot 31$ , we must either have $r=1$ and $d-a-r = 93$ , or $... | 870 |
6,792 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | 6 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Assume $d = x+m, a = x-m, c = x+n$ , and $b = x-n$ . This clearly satisfies the condition that $a+d = b+c$ since ( $2x = 2x$ ) . Now plug this into $bc-ad = 93$ . You get $(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93$
Since $m>n$ (as given by the condition that $a<b<c<d$ ), $m+n>m... | 870 |
6,793 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_5 | 1 | Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ | Notice that \begin{align*}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}
Using the formula for the sum of the first $n$ numbers, $1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210$ . Therefore, \[P_{20}(x) = P_... | 763 |
6,794 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_5 | 2 | Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ | Notice the transformation of $P_{n-1}(x)\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\cdots+20=210$ to them. (Indeed, this is very easy to check in general.)
Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/co... | 763 |
6,795 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | 1 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \ b, \ c$
$n = 9a + 36 = 10b + 45 = 11c + 55$
This can be rewritten as the following congruences:
$n \equiv 0 \pmod{9}$
$n \equiv 5 \pmod{10}$
$n \equiv 0 \pmod{11}$
Since 9 and 11 are relatively prime, n i... | 495 |
6,796 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | 2 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | Let $n$ be the desired integer. From the given information, we have \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$... | 495 |
6,797 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | 3 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$ . From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From... | 495 |
6,798 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | 4 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we tak... | 495 |
6,799 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 1 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | Call the two subsets $m$ and $n.$ For each of the elements in $S,$ we can assign it to either $m,n,$ or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ el... | 365 |
6,800 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 2 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | Given one of ${6 \choose k}$ subsets with $k$ elements, the other also has $2^k$ possibilities; this is because it must contain all of the "missing" $n - k$ elements and thus has a choice over the remaining $k.$ We want $\sum_{k = 0}^6 {6 \choose k}2^k = (2 + 1)^6 = 729$ by the Binomial Theorem. But the order of the se... | 365 |
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