ID stringlengths 8 10 | Year int64 1.98k 2.02k | Problem Number int64 1 15 | Question stringlengths 37 2.66k | Answer int64 0 997 | Part stringclasses 2
values | Solution listlengths 1 25 |
|---|---|---|---|---|---|---|
1990-I-13 | 1,990 | 13 | Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit? | 184 | null | [
"Lemma: For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9 (Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n. If there is exactly 1 n-digit power of 9, then such a number $m$ cann... |
1990-I-14 | 1,990 | 14 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$ . Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segm... | 594 | null | [
"Solution 1(Synthetic) [asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y)... |
1990-I-15 | 1,990 | 15 | Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*} | 20 | null | [
"Set $S = (x + y)$ and $P = xy$. Then the relationship \\[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})\\] can be exploited: \\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\\\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\\end{eqnarray*}... |
1991-I-1 | 1,991 | 1 | Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*} | 146 | null | [
"Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$, which factors to $(a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa... |
1991-I-2 | 1,991 | 2 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{}... | 840 | null | [
"[asy] real r = 0.35; size(220); pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8); pair A=(0,0),B=(4,0),C=(4,3),D=(0,3); D(A--B--C--D--cycle); pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0); pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r); D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4); MP(\... |
1991-I-4 | 1,991 | 4 | How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ ? | 159 | null | [
"The range of the sine function is $-1 \\le y \\le 1$. It is periodic (in this problem) with a period of $\\frac{2}{5}$. Thus, $-1 \\le \\frac{1}{5} \\log_2 x \\le 1$, and $-5 \\le \\log_2 x \\le 5$. The solutions for $x$ occur in the domain of $\\frac{1}{32} \\le x \\le 32$. When $x > 1$ the logarithm function ret... |
1991-I-5 | 1,991 | 5 | Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20_{}^{}!$ be the resulting product? | 128 | null | [
"If the fraction is in the form $\\frac{a}{b}$, then $a < b$ and $\\gcd(a,b) = 1$. There are 8 prime numbers less than 20 ($2, 3, 5, 7, 11, 13, 17, 19$), and each can only be a factor of one of $a$ or $b$. There are $2^8$ ways of selecting some combination of numbers for $a$; however, since $a<b$, only half of them... |
1991-I-8 | 1,991 | 8 | For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$ ? | 10 | null | [
"Let $x^2 + ax + 6a = (x - s)(x - r)$. Vieta's yields $s + r = - a, sr = 6a$. \\begin{eqnarray*}sr + 6s + 6r &=& 0\\\\ sr + 6s + 6r + 36 &=& 36\\\\ (s + 6)(r + 6) &=& 36 \\end{eqnarray*} Without loss of generality let $r \\le s$. The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( ... |
1991-I-9 | 1,991 | 9 | Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$ | 44 | null | [
"Use the two trigonometric Pythagorean identities $1 + \\tan^2 x = \\sec^2 x$ and $1 + \\cot^2 x = \\csc^2 x$. If we square the given $\\sec x = \\frac{22}{7} - \\tan x$, we find that \\begin{align*} \\sec^2 x &= \\left(\\frac{22}7\\right)^2 - 2\\left(\\frac{22}7\\right)\\tan x + \\tan^2 x \\\\ 1 &= \\left(\\frac{2... |
1991-I-10 | 1,991 | 10 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | 532 | null | [
"Solution 1 Let us make a chart of values in alphabetical order, where $P_a,\\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$, and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \\sum_{n=1}^{b} P_b$): \\[\\begin{array}{|r||r|r|r|} \\hline \\text{String}&P_... |
1991-I-11 | 1,991 | 11 | Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$ , no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelv... | 135 | null | [
"We wish to find the radius of one circle, so that we can find the total area. Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is ta... |
1991-I-12 | 1,991 | 12 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ , $Q^{}_{}$ , $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ , $BQ^{}_{}=20$ , $PR^{}_{}=30$ , and $QS^{... | 677 | null | [
"[asy]defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label(\"\\(A\\)\",A,NW);label(\"\\(B\\)\",B,NE);label(\"\\(C\\)\"... |
1991-I-13 | 1,991 | 13 | A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consisten... | 990 | null | [
"Solution 1 Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by \\[\\frac{r(r-1)}{(r+b)(r+b-1)}+\\frac{b(b-1)}{(r+b)(r+b-1)}=\\frac{r(r-1)+(t-r)(t-r-1)}{t(... |
1991-I-14 | 1,991 | 14 | A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by $\overline{AB}$ , has length 31. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ . | 384 | null | [
"[asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D... |
1992-I-1 | 1,992 | 1 | Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms. | 400 | null | [
"Solution 1 There are 8 fractions which fit the conditions between 0 and 1: $\\frac{1}{30},\\frac{7}{30},\\frac{11}{30},\\frac{13}{30},\\frac{17}{30},\\frac{19}{30},\\frac{23}{30},\\frac{29}{30}$ Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 ... |
1992-I-2 | 1,992 | 2 | A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there? | 502 | null | [
"Note that an ascending number is exactly determined by its digits: for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits. So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$... |
1992-I-3 | 1,992 | 3 | A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $0.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater ... | 164 | null | [
"Let $n$ be the number of matches won, so that $\\frac{n}{2n}=\\frac{1}{2}$, and $\\frac{n+3}{2n+4}>\\frac{503}{1000}$. Cross multiplying, $1000n+3000>1006n+2012$, so $n<\\frac{988}{6}=164 \\dfrac {4}{6}=164 \\dfrac{2}{3}$. Thus, the answer is $164.",
"Let $n$ be the number of matches she won before the weekend b... |
1992-I-4 | 1,992 | 4 | In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below. \[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} ... | 62 | null | [
"Consider what the ratio means. Since we know that they are consecutive terms, we can say \\[\\frac{\\dbinom{n}{k-1}}{3} = \\frac{\\dbinom{n}{k}}{4} = \\frac{\\dbinom{n}{k+1}}{5}.\\] Taking the first part, and using our expression for $n$ choose $k$, \\[\\frac{n!}{3(k-1)!(n-k+1)!} = \\frac{n!}{4k!(n-k)!}\\] \\[\\fr... |
1992-I-5 | 1,992 | 5 | Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$ , $0^{}_{}<r<1$ , that have a repeating decimal expansion in the form $0.abcabcabc\ldots=0.\overline{abc}$ , where the digits $a^{}_{}$ , $b^{}_{}$ , and $c^{}_{}$ are not necessarily distinct. To write the elements of $S^{}_{}$ as fractions in lowest terms, ho... | 660 | null | [
"We consider the method in which repeating decimals are normally converted to fractions with an example: $x=0.\\overline{176}$ $\\Rightarrow 1000x=176.\\overline{176}$ $\\Rightarrow 999x=1000x-x=176$ $\\Rightarrow x=\\frac{176}{999}$ Thus, let $x=0.\\overline{abc}$ $\\Rightarrow 1000x=abc.\\overline{abc}$ $\\Righta... |
1992-I-6 | 1,992 | 6 | For how many pairs of consecutive integers in $\{1000,1001,1002^{}_{},\ldots,2000\}$ is no carrying required when the two integers are added? | 156 | null | [
"For one such pair of consecutive integers, let the smaller integer be $\\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$ We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. \\[\\begin{array}{c|c|c|c|c|c|c} & & & ... |
1992-I-7 | 1,992 | 7 | Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron. | 320 | null | [
"Since the area $BCD=80=\\frac{1}{2}\\cdot10\\cdot16$, the perpendicular from $D$ to $BC$ has length $16$. The perpendicular from $D$ to $ABC$ is $16 \\cdot \\sin 30^\\circ=8$. Therefore, the volume is $\\frac{8\\cdot120}{3}=320.",
"The area of $ABC$ is 120 and $BC$=10, the slant height is 24. Height from $A$ to ... |
1992-I-8 | 1,992 | 8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{th}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ . | 819 | null | [
"Note that the $\\Delta$s are reminiscent of differentiation; from the condition $\\Delta(\\Delta{A}) = 1$, we are led to consider the differential equation \\[\\frac{d^2 A}{dn^2} = 1\\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \\[a_{n} = \\frac{1}{2}(n-19)(n-92)\\] as we ... |
1992-I-9 | 1,992 | 9 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ , $BC=50^{}_{}$ , $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime p... | 164 | null | [
"Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$ Then $XD=xy-70, XC=y(92-x)-50,$ thus \\[\\frac{xy-70}{y(92-x)-50} = \\frac{XD}{XC} = \\frac{ED}{EC}=\\frac{AP}{... |
1992-I-10 | 1,992 | 10 | Consider the region $A^{}_{}$ in the complex plane that consists of all points $z^{}_{}$ such that both $\frac{z^{}_{}}{40}$ and $\frac{40^{}_{}}{\overline{z}}$ have real and imaginary parts between $0^{}_{}$ and $1^{}_{}$ , inclusive. What is the integer that is nearest the area of $A^{}_{}$ ? | 572 | null | [
"Let $z=a+bi \\implies \\frac{z}{40}=\\frac{a}{40}+\\frac{b}{40}i$. Since $0\\leq \\frac{a}{40},\\frac{b}{40}\\leq 1$ we have the inequality \\[0\\leq a,b \\leq 40\\]which is a square of side length $40$. Also, $\\frac{40}{\\overline{z}}=\\frac{40}{a-bi}=\\frac{40a}{a^2+b^2}+\\frac{40b}{a^2+b^2}i$ so we have $0\\le... |
1992-I-11 | 1,992 | 11 | Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$ , the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$ , and the result... | 945 | null | [
"Let $l$ be a line that makes an angle of $\\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$. The angle between $l$ and $l_1$ is $\\theta - \\frac{\\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\\theta - \\frac{\\pi}{70}$... |
1992-I-12 | 1,992 | 12 | In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one of the remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For exampl... | 792 | null | [
"By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts. One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that g... |
1992-I-13 | 1,992 | 13 | Triangle $ABC^{}_{}$ has $AB=9^{}_{}$ and $BC: AC=40: 41^{}_{}$ . What's the largest area that this triangle can have? | 820 | null | [
"Solution 1 First, consider the triangle in a coordinate system with vertices at $(0,0)$, $(9,0)$, and $(a,b)$. Applying the distance formula, we see that $\\frac{ \\sqrt{a^2 + b^2} }{ \\sqrt{ (a-9)^2 + b^2 } } = \\frac{40}{41}$. We want to maximize $b$, the height, with $9$ being the base. Simplifying gives $-a^2 ... |
1992-I-14 | 1,992 | 14 | In triangle $ABC^{}_{}$ , $A'$ , $B'$ , and $C'$ are on the sides $BC$ , $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ , $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'... | 94 | null | [
"Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \\[\\frac{AO}{OA'}+1=\\frac{AA'}{OA'}=\\frac{[ABC]}{[BOC]}=\\frac{K_A+K_B+K_C}{K_A}.\\] Therefore, we have \\[\\frac{AO}{OA'}=\\frac{K_B+K_C}{K_A}\\] \\[\\frac{BO}{OB'}=\\frac{K_A+K_C}{K_B}\\] \\[\\frac{CO}{OC'}=\\f... |
1992-I-15 | 1,992 | 15 | Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? | 396 | null | [
"Let the number of zeros at the end of $m!$ be $f(m)$. We have $f(m) = \\left\\lfloor \\frac{m}{5} \\right\\rfloor + \\left\\lfloor \\frac{m}{25} \\right\\rfloor + \\left\\lfloor \\frac{m}{125} \\right\\rfloor + \\left\\lfloor \\frac{m}{625} \\right\\rfloor + \\left\\lfloor \\frac{m}{3125} \\right\\rfloor + \\cdots... |
1993-I-1 | 1,993 | 1 | How many even integers between 4000 and 7000 have four different digits? | 728 | null | [
"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 = 44... |
1993-I-2 | 1,993 | 2 | 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 $\frac{n^{2}}{2}$ miles on the $n^{\mbo... | 580 | null | [
"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 \\f... |
1993-I-4 | 1,993 | 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 $bc - ad = 93\,$ ? | 870 | null | [
"Solution 1 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 +... |
1993-I-5 | 1,993 | 5 | 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 $P_{20}(x)\,$ ? | 763 | null | [
"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... |
1993-I-6 | 1,993 | 6 | What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | 495 | null | [
"Solution 1 Denote the first of each of the series of consecutive integers as $a,\\ b,\\ c$. Therefore, $n = a + (a + 1) \\ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$. Simplifying, $9a = 10b + 9 = 11c + 19$. The relationship between $a,\\ b$ suggests that $b$ is divisible by $9$. Also, $10b -10 = 10(b-1) = 11c$,... |
1993-I-7 | 1,993 | 7 | Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimen... | 5 | null | [
"There is a total of $P(1000,6)$ possible ordered $6$-tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$ There are $C(1000,6)$ possible sets $\\{a_1,a_2,a_3,b_1,b_2,b_3\\}.$ We have five valid cases for the increasing order of these six elements: $aaabbb$ $aababb$ $aabbab$ $abaabb$ $ababab$ Note that the $a$'s are different from ... |
1993-I-8 | 1,993 | 8 | 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 $... | 365 | null | [
"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 $... |
1993-I-9 | 1,993 | 9 | Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction and label this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ ar... | 118 | null | [
"The label $1993$ will occur on the $\\frac12(1993)(1994) \\pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$. Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (199... |
1993-I-10 | 1,993 | 10 | Euler's formula states that for a convex polyhedron with $V\,$ vertices, $E\,$ edges, and $F\,$ faces, $V-E+F=2\,$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its $V\,$ vertices, $T\,$ triangular faces and $P^{}_{}$ pentagonal faces meet. What is the value... | 250 | null | [
"Solution 1 The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equila... |
1993-I-11 | 1,993 | 11 | Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the ... | 93 | null | [
"The probability that the $n$th flip in each game occurs and is a head is $\\frac{1}{2^n}$. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $\\frac{1}{2}+\\frac{1}{8}+\\frac{1}{32}+\\cdots = \\frac{\\frac{1}{2}}{1-\\frac{1}{4}}=\\fra... |
1993-I-12 | 1,993 | 12 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ , $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ , $P_3\,$ , $P_4, \dots$ are generated by rolling th... | 344 | null | [
"Solution 1 If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$, then $u=2r-p$ and $v=2s-q$. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \\[P_{n-1}=(x_{n-1},y_{n-1}) = (... |
1993-I-13 | 1,993 | 13 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 163 | null | [
"Solution 1 Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \\begin{align}y&=-\\frac{100}{t}x+200-\\frac... |
1993-I-14 | 1,993 | 14 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter ... | 448 | null | [
"Put the rectangle on the coordinate plane so its vertices are at $(\\pm4,\\pm3)$, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$. Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, an... |
1993-I-15 | 1,993 | 15 | Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ , $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relat... | 997 | null | [
"[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label(\"$A$\",A,SE); label(\"$B$\",B,SW); label(\"$C$\",C,N); label(\"$H$\",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label(\"$S$\",x[0],SW); draw(circle((4.29... |
1994-I-1 | 1,994 | 1 | The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000? | 63 | null | [
"One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$. Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \\equiv -1,\\ 1 \\equiv 2,\\ 1 \\pmod{3}$. Since 1994 is even, $n$ must be congruent to $1 \\pmod{3}$. It will be the $\\frac{1994}{2} = 997$th such term, so $n = 4 + (997... |
1994-I-2 | 1,994 | 2 | 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 $\overline{AB}\,$... | 312 | null | [
"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 $OE = AB - OQ = AB - 10$. The other leg, $AE$, is... |
1994-I-8 | 1,994 | 8 | The points $(0,0)\,$ , $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ . | 315 | null | [
"Solution 1 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{ali... |
1994-I-9 | 1,994 | 9 | 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... | 394 | null | [
"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: Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability $\\frac {1}{2k - 1}$, then we have $k - 1$ pairs left. So this contribu... |
1994-I-10 | 1,994 | 10 | 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 $m+n.\,$ | 450 | null | [
"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 =... |
1994-I-11 | 1,994 | 11 | 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 94 of the bric... | 465 | null | [
"Solution 1 We have the smallest stack, which has a height of $94 \\times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$'s, $6$'s, and $15$'s. Bec... |
1994-I-12 | 1,994 | 12 | 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 field. ... | 702 | null | [
"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 $52$, a... |
1994-I-14 | 1,994 | 14 | A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle... | 71 | null | [
"At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP(\"B\",(0,0),NW), C = MP(\"C\",D((1,0))), A = MP(\"A\",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * p... |
1994-I-15 | 1,994 | 15 | 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... | 597 | null | [
"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 circum... |
1995-I-1 | 1,995 | 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... | 255 | null | [
"The sum of the areas of the squares if they were not interconnected is a geometric sequence: $1^2 + \\left(\\frac{1}{2}\\right)^2 + \\left(\\frac{1}{4}\\right)^2 + \\left(\\frac{1}{8}\\right)^2 + \\left(\\frac{1}{16}\\right)^2$ Then subtract the areas of the intersections, which is $\\left(\\frac{1}{4}\\right)^2 +... |
1995-I-2 | 1,995 | 2 | Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2.$ | 25 | null | [
"Taking the $\\log_{1995}$ (logarithm) of both sides and then moving to one side yields the quadratic equation $2(\\log_{1995}x)^2 - 4(\\log_{1995}x) + 1 = 0$. Applying the quadratic formula yields that $\\log_{1995}x = 1 \\pm \\frac{\\sqrt{2}}{2}$. Thus, the product of the two roots (both of which are positive) is... |
1995-I-3 | 1,995 | 3 | Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n... | 67 | null | [
"It takes an even number of steps for the object to reach $(2,2)$, so the number of steps the object may have taken is either $4$ or $6$. If the object took $4$ steps, then it must have gone two steps N and two steps E, in some permutation. There are $\\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and... |
1995-I-4 | 1,995 | 4 | 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. | 224 | null | [
"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... |
1995-I-5 | 1,995 | 5 | For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$ | 51 | null | [
"Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$. Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$. Let $m'$ be the conjugate of $m$, and $n'$ be ... |
1995-I-6 | 1,995 | 6 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ ? | 589 | null | [
"We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\\frac{63\\times 39-1}{2} = 1228$ factors of $n^2$ that are less th... |
1995-I-8 | 1,995 | 8 | For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers? | 85 | null | [
"Since $y|x$, $y+1|x+1$, then $\\text{gcd}\\,(y,x)=y$ (the bars indicate divisibility) and $\\text{gcd}\\,(y+1,x+1)=y+1$. By the Euclidean algorithm, these can be rewritten respectively as $\\text{gcd}\\,(y,x-y)=y$ and $\\text{gcd}\\, (y+1,x-y)=y+1$, which implies that both $y,y+1 | x-y$. Also, as $\\text{gcd}\\,(y... |
1995-I-9 | 1,995 | 9 | 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 $a+b.$ AIME 1995 Problem 9.png | 616 | null | [
"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... |
1995-I-10 | 1,995 | 10 | What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer? | 215 | null | [
"The requested number $\\mod {42}$ must be a prime number. Also, every number that is a multiple of $42$ greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to $42$ and the numbers that are multiples of $42$ greater than them, unti... |
1995-I-11 | 1,995 | 11 | A right rectangular prism $P_{}$ (i.e., a rectangular parallelepiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P_{}$ cuts $P_{}$ into two prisms, one of which is similar to $P_{},$ and both of which have nonzero volume. Given that $b=1995,$ for how many order... | 40 | null | [
"Let $P'$ be the prism similar to $P$, and let the sides of $P'$ be of length $x,y,z$, such that $x \\le y \\le z$. Then \\[\\frac{x}{a} = \\frac{y}{b} = \\frac zc < 1.\\] Note that if the ratio of similarity was equal to $1$, we would have a prism with zero volume. As one face of $P'$ is a face of $P$, it follows ... |
1995-I-12 | 1,995 | 12 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$ | 5 | null | [
"Solution 1 (trigonometry) [asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A on... |
1995-I-13 | 1,995 | 13 | Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$ | 400 | null | [
"When $\\left(k - \\frac {1}{2}\\right)^4 \\leq n < \\left(k + \\frac {1}{2}\\right)^4$, $f(n) = k$. Thus there are $\\left \\lfloor \\left(k + \\frac {1}{2}\\right)^4 - \\left(k - \\frac {1}{2}\\right)^4 \\right\\rfloor$ values of $n$ for which $f(n) = k$. Expanding using the binomial theorem, \\begin{align*} \\le... |
1995-I-14 | 1,995 | 14 | 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 form $m\pi-... | 378 | null | [
"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 $\\overline{OF} \\perp \\overline{AB}$. [asy] size(200); pathpen = black + linewidth(0.7); pen d = dashed+linewidth(0... |
1995-I-15 | 1,995 | 15 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before one encounters a run of 2 tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ . | 37 | null | [
"Solution 1 Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the successful sequences are composed of blocks of $1$ to $4$ H's separated by T's and end with $5$ H's. Since the probability that the sequence starts with TH is $1/4$, the total probability is that $3/2$ of the probab... |
1996-I-2 | 1,996 | 2 | 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? | 340 | null | [
"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 $n < 1000$): $4\\leq n <8$ $16\\leq n<32$ $64\\leq n<128$ $256\\leq n<512$ There are $4$ for the first inequality, $16$ for the seco... |
1996-I-3 | 1,996 | 3 | Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms. | 44 | null | [
"Using Simon's Favorite Factoring Trick, we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$. Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \\... |
1996-I-4 | 1,996 | 4 | 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 a shadow, which does not include the area beneath the cube is 48 square centime... | 166 | null | [
"[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)); ... |
1996-I-5 | 1,996 | 5 | Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a$ , $b$ , and $c$ , and that the roots of $x^3+rx^2+sx+t=0$ are $a+b$ , $b+c$ , and $c+a$ . Find $t$ . | 23 | null | [
"By Vieta's formulas on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$, we have $a + b + c = s = -3$, $ab + bc + ca = 4$, and $abc = 11$. Then $t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$ This is just the definition for $-P(-3) = 23. Alternatively, we can expand the expression to ge... |
1996-I-6 | 1,996 | 6 | In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integer... | 49 | null | [
"Solution 1 We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games. No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games. Now we use PIE: The probability that one te... |
1996-I-7 | 1,996 | 7 | 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? | 300 | null | [
"There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. [asy] pathpen = black; pair O = (3.5,3.5); D(O); for... |
1996-I-8 | 1,996 | 8 | 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 $6^{20}$ ? | 799 | null | [
"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 sq... |
1996-I-9 | 1,996 | 9 | 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 encount... | 342 | null | [
"Solution 1 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 \\p... |
1996-I-10 | 1,996 | 10 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ . | 159 | null | [
"$\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-4... |
1996-I-11 | 1,996 | 11 | 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 $\theta$ . | 276 | null | [
"\\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{... |
1996-I-12 | 1,996 | 12 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum $|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|$ . The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 58 | null | [
"Solution 1 Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \\cdot 8!$, because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ place... |
1996-I-13 | 1,996 | 13 | In triangle $ABC$ , $AB=\sqrt{30}$ , $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio \[\dfrac{\text{Area}(\triangle ADB)}{\text{Area}(\triangle ABC)}\] can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$... | 65 | null | [
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP(\"A\",A)--MP(\"B\",B,SW)--MP(\"C\",C)--A--MP(\"D\",D)--B); D(MP(\"E\",E)); MP(\"\\sqrt{30}\",(A+B)/2,NW); MP(\"\\sqrt{6}\",(A+C)/2,SE); MP(\"\\frac{\\sqrt{15}}2\",(E+C)... |
1996-I-14 | 1,996 | 14 | A $150\times 324\times 375$ 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? | 768 | null | [
"Solution 1 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... |
1996-I-15 | 1,996 | 15 | 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 the greatest integer that does not exceed $1000r$ . | 777 | null | [
"Solution 1 (trigonometry) [asy]size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--MP(\"D\",D,N)--cycle); D(B--D); D(A--C); D(MP(\"O\",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5))... |
1997-I-1 | 1,997 | 1 | How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers? | 750 | null | [
"Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$, where $b$ is a positive integer. Numbers congruent to $2 \\pmod 4$ cannot be obtained because squares are $0, 1 \\pmod 4.$ Thus, the ans... |
1997-I-2 | 1,997 | 2 | The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares. The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 125 | null | [
"To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\\choose 2} = 36$. Similarly, there are ${9\\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles. For $s$, there are $8^2$ unit squares, $7^2$ of the $2\\times2$ squa... |
1997-I-3 | 1,997 | 3 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum ... | 126 | null | [
"Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \\Longrightarrow 9xy-1000x-y=0$. Using SFFT, this factorizes to $(9x-1)\\left(y-\\dfrac{1000}{9}\\right)=\\dfrac{1000}{9}$, and $(9x-1)(9y-1000)=1000$. Since $89 < 9x-1 < 890$, we can use trial and erro... |
1997-I-4 | 1,997 | 4 | Circles of radii 5, 5, 8, and $m/n$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 17 | null | [
"If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\\frac mn = r$, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$. Then we form two right triangles, of lengths $5, x, 5+r$ and $5, 8+r+x, 13$, wher ... |
1997-I-5 | 1,997 | 5 | The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits, any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of po... | 417 | null | [
"The nearest fractions to $\\frac 27$ with numerator $1$ are $\\frac 13, \\frac 14$; and with numerator $2$ are $\\frac 26, \\frac 28 = \\frac 13, \\frac 14$ anyway. For $\\frac 27$ to be the best approximation for $r$, the decimal must be closer to $\\frac 27 \\approx .28571$ than to $\\frac 13 \\approx .33333$ or... |
1997-I-6 | 1,997 | 6 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle. What is the largest value of $n$ for which $A_1$ , $A_n$ , and $B$ are consecutive vertices of a regular polygon? | 42 | null | [
"Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\\angle A_2A_1A_n = \\frac{(n-2)180}{n}$, $\\angle A_nA_1B = \\frac{(m-2)180}{m}$, and $\\angle A_2A_1B = 60^{\\circ}$. Since those three angles add up to $360^{\\circ}$, \\begin{eqnarray*} \\frac{(n-2)180... |
1997-I-7 | 1,997 | 7 | A car travels due east at $\frac 23$ miles per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ miles 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... | 198 | null | [
"We set up a coordinate system, with the starting point of the car at the origin. At time $t$, the car is at $\\left(\\frac 23t,0\\right)$ and the center of the storm is at $\\left(\\frac{t}{2}, 110 - \\frac{t}{2}\\right)$. Using the distance formula, \\begin{eqnarray*} \\sqrt{\\left(\\frac{2}{3}t - \\frac 12t\\rig... |
1997-I-8 | 1,997 | 8 | How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0? | 90 | null | [
"For more detailed explanations, see related problem (AIME I 2007, 10). The problem is asking us for all configurations of $4\\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns: The first two columns share no two numbers in the same row. There are ${4\\choose2} =... |
1997-I-9 | 1,997 | 9 | 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... | 233 | null | [
"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... |
1997-I-10 | 1,997 | 10 | 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... | 117 | null | [
"Case 1: All three attributes are the same. This is impossible since sets contain distinct cards. Case 2: Two of the three attributes are the same. There are ${3\\choose 2}$ ways to pick the two attributes in question. Then there are $3$ ways to pick the value of the first attribute, $3$ ways to pick the value of t... |
1997-I-11 | 1,997 | 11 | 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 $100x$ ? | 241 | null | [
"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} \\co... |
1997-I-12 | 1,997 | 12 | 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 $f$ . | 58 | null | [
"First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have $\\frac {a\\frac {ax + b}{cx + d} + b}{c\\frac {ax + b}{cx + d} + d} = x$, which reduces to \\[\\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\\frac {px + q}{rx + s} = x.\\] In order for thi... |
1997-I-14 | 1,997 | 14 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $m/n$ be the probability that $\sqrt{2+\sqrt{3}}\le |v+w|$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 582 | null | [
"Solution 1 $z^{1997}=1=1(\\cos 0 + i \\sin 0)$ Define $\\theta = 2\\pi/1997$. By De Moivre's Theorem the roots are given by $z=\\cos (k\\theta) +i\\sin(k\\theta), \\qquad k \\in \\{0,1,\\ldots,1996\\}$ Now, let $v$ be the root corresponding to $m\\theta=2m\\pi/1997$, and let $w$ be the root corresponding to $n\\th... |
1997-I-15 | 1,997 | 15 | 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 squa... | 554 | null | [
"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 ot... |
1998-I-1 | 1,998 | 1 | For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ and $8^8$ , and $k$ ? | 25 | null | [
"It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$. $6^6 = 2^6\\cdot3^6$ $8^8 = 2^{24}$ $12^{12} = 2^{24}\\cdot3^{12}$ The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$. Therefore $12^{12}... |
1998-I-2 | 1,998 | 2 | Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x \le 2y \le 60$ and $y \le 2x \le 60$ . | 480 | null | [
"Solution 1 Pick's theorem states that: $A = I + \\frac B2 - 1$ The conditions give us four inequalities: $x \\le 30$, $y\\le 30$, $x \\le 2y$, $y \\le 2x$. These create a quadrilateral, whose area is $\\frac 12$ of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of ... |
1998-I-3 | 1,998 | 3 | The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region? | 800 | null | [
"The equation given can be rewritten as: $40|x| = - y^2 - 2xy + 400$ We can split the equation into a piecewise equation by breaking up the absolute value: $40x = -y^2 - 2xy + 400\\quad\\quad x\\ge 0$ $40x = y^2 + 2xy - 400 \\quad \\quad x < 0$ Factoring the first one: (alternatively, it is also possible to complet... |
1998-I-4 | 1,998 | 4 | Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$ | 17 | null | [
"In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ eve... |
1998-I-5 | 1,998 | 5 | Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$ | 40 | null | [
"Though the problem may appear to be quite daunting, it is actually not that difficult. $\\frac {k(k-1)}2$ always evaluates to an integer (triangular number), and the cosine of $n\\pi$ where $n \\in \\mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$, and odd o... |
1998-I-6 | 1,998 | 6 | Let $ABCD$ be a parallelogram . Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$ | 308 | null | [
"Solution 1 There are several similar triangles. $\\triangle PAQ\\sim \\triangle PDC$, so we can write the proportion: $\\frac{AQ}{CD} = \\frac{PQ}{PC} = \\frac{735}{112 + 735 + RC} = \\frac{735}{847 + RC}$ Also, $\\triangle BRQ\\sim DRC$, so: $\\frac{QR}{RC} = \\frac{QB}{CD} = \\frac{112}{RC} = \\frac{CD - AQ}{CD}... |
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