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1988AIME 真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月12日上午11:45

1988AIME 真题及答案解析

答案解析请参考文末

Problem 1

One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons. The sample shown below has ${1, 2, 3, 6, 9}$ as its combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?

Problem 2

For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ . For $n ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ . Find $f_{1988}(11)$ .

Problem 3

Find $(log_2 x)^2$ if $log_2 (log_8 x) = log_8 (log_2 x)$ .

Problem 4

Suppose that $|x_i| < 1$ for $i = 1, 2, dots, n$ . Suppose further that

$|x_1| + |x_2| + dots + |x_n| = 19 + |x_1 + x_2 + dots + x_n|$ .What is the smallest possible value of $n$ ?

Problem 5

Let $frac{m}{n}$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ . Find $m + n$ .

Problem 6

It is possible to place positive integers into the vacant twenty-one squares of the 5 times 5 square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).

Problem 7

In triangle $ABC$ , $tan angle CAB = 22/7$ , and the altitude from $A$ divides $BC$ into segments of length $3$ and $17$ . What is the area of triangle $ABC$ ?

Problem 8

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: $begin{align*} f(x,x) &=x, \ f(x,y) &=f(y,x), quad text{and} \ (x + y) f(x,y) &= yf(x,x + y). end{align*}$ Calculate $f(14,52)$ .

Problem 9

Find the smallest positive integer whose cube ends in $888$ .

Problem 10

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

Problem 11

Let $w_1, w_2, dots, w_n$ be complex numbers. A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, dots, z_n$ such that

$sum_{k = 1}^n (z_k - w_k) = 0.$ For the numbers $w_1 = 32 + 170i$ , $w_2 = -7 + 64i$ , $w_3 = -9 +200i$ , $w_4 = 1 + 27i$ , and $w_5 = -14 + 43i$ , there is a unique mean line with y-intercept $3$ . Find the slope of this mean line.

Problem 12

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ , $b$ , $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$ .

Problem 13

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ .

Problem 14

Let $C$ be the graph of $xy = 1$ , and denote by $C^*$ the reflection of $C$ in the line $y = 2x$ . Let the equation of $C^*$ be written in the form $[12x^2 + bxy + cy^2 + d = 0.]$ Find the product $bc$ .

Problem 15

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order 1, 2, 3, 4, 5, 6, 7, 8, 9.

While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

1988AIME 详细解析

Currently there are ${10 choose 5}$ possible combinations. With any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $sum^{9}_{k=1}{10 choose k}$ . Now we can use the identity $sum^{n}_{k=0}{n choose k}=2^{n}$ . So the number of additional combinations is just $2^{10}-{10choose 0}-{10choose 10}-{10 choose 5}=1024-1-1-252=boxed{770}$ .
We see that $f_{1}(11)=4$ $f_2(11) = f_1(4)=16$ $f_3(11) = f_1(16)=49$ $f_4(11) = f_1(49)=169$ $f_5(11) = f_1(169)=256$ $f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = boxed{169}$ .
Raise both as exponents with base 8: $begin{align*} 8^{log_2 (log_8 x)} &= 8^{log_8 (log_2 x)}\ 2^{3 log_2(log_8x)} &= log_2x\ (log_8x)^3 &= log_2x\ left(frac{log_2x}{log_28}right)^3 &= log_2x\ (log_2x)^2 &= (log_28)^3 = boxed{27}\ end{align*}$
A quick explanation of the steps: On the 1st step, we use the property of logarithms that $a^{log_a x} = x$ . On the 2nd step, we use the fact that $k log_a x = log_a x^k$ . On the 3rd step, we use the change of base formula, which states $log_a b = frac{log_k b}{log_k a}$ for arbitrary $k$ .
Since $|x_i| < 1$ then $[|x_1| + |x_2| + dots + |x_n| = 19 + |x_1 + x_2 + dots + x_n| < n.]$ So $n ge 20$ . We now just need to find an example where $n = 20$ : suppose $x_{2k-1} = frac{19}{20}$ and $x_{2k} = -frac{19}{20}$ ; then on the left hand side we have $left|frac{19}{20}right| + left|-frac{19}{20}right| + dots + left|-frac{19}{20}right| = 20left(frac{19}{20}right) = 19$ . On the right hand side, we have $19 + left|frac{19}{20} - frac{19}{20} + dots - frac{19}{20}right| = 19 + 0 = 19$ , and so the equation can hold for $n = boxed{020}$ .
$10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $frac{m}{n} = frac{144}{10000} = frac{9}{625}$ , and $m + n = boxed{634}$ .
Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ and $b$ , and then calculate $(*)$ . $begin{tabular}[b]{|c|c|c|c|c|}hline 4a & & & * & \ hline 3a & 74 & & & \ hline 2a & & & & 186 \ hline a & & 103 & & \ hline 0 & b & 2b & 3b & 4b \ hline end{tabular}$ We can compute the squares at the intersections of two existing numbers in terms of $a$ and $b$ ; two such equations will give us the values of $a$ and $b$ . On the fourth row from the bottom, the common difference is $74 - 3a$ , so the square at $(2,3)$ has a value of $148 - 3a$ . On the third column from the left, the common difference is $103 - 2b$ , so that square also has a value of $2b + 3(103 - 2b) = 309 - 4b$ . Equating, we get $148 - 3a = 309 - 4b Longrightarrow 4b - 3a = 161$ .Now we compute the square $(2,2)$ . By rows, this value is simply the average of $2a$ and $186$ , so it is equal to $frac{2a + 186}{2} = a + 93$ . By columns, the common difference is $103 - 2b$ , so our value is $206 - 2b$ . Equating, $a + 93 = 206 - 2b Longrightarrow a + 2b = 113$ .Solving $begin{align*}4b - 3a &= 161\ a + 2b &= 113 end{align*}$
gives $a = 13$ , $b = 50$ . Now it is simple to calculate $(4,3)$ . One way to do it is to see that $(2,2)$ has $206 - 2b = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = frac{(3,0) + (3,4)}{2} Longrightarrow (3,4) = * = boxed{142}$ .

First, let $a =$ the number to be placed in the first column, fourth row. Let $b =$ the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of $a$ and $b$ :

$begin{tabular}[b]{|c|c|c|c|c|}hline 4a & & & & \ hline 3a & & & & \ hline 2a & & & & \ hline a & & & & \ hline 0 & b & 2b & 3b & 4b \ hline end{tabular}$

Next, let $a + b + c =$ the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of $a$ , $b$ , and $c$ :

$begin{tabular}[b]{|c|c|c|c|c|}hline 4a & 4a + b + 4c & & & \ hline 3a & 3a + b + 3c & & & \ hline 2a & 2a + b + 2c & & & \ hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \ hline 0 & b & 2b & 3b & 4b \ hline end{tabular}$

We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:

$begin{tabular}[b]{|c|c|c|c|c|}hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \ hline 3a & 3a + b + 3c & 3a + 2b + 6c & 3a + 3b + 9c & 3a + 4b + 12c \ hline 2a & 2a + b + 2c & 2a + 2b + 4c & 2a + 3b + 6c & 2a + 4b + 8c \ hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \ hline 0 & b & 2b & 3b & 4b \ hline end{tabular}$

We now have a system of equations.

$3a + b + 3c = 74$
$2a + 4b + 8c = 186$ $a + 2b + 2c = 103$

Solving, we find that $(a,b,c) = (13,50, - 5)$ . The number in the square marked by the asterisk is $4a + 3b + 12c = boxed{142}$
Let $D$ be the intersection of the altitude with $overline{BC}$ , and $h$ be the length of the altitude. Without loss of generality, let $BD = 17$ and $CD = 3$ . Then $tan angle DAB = frac{17}{h}$ and $tan angle CAD = frac{3}{h}$ . Using the tangent sum formula, $begin{align*} tan CAB &= tan (DAB + CAD)\ frac{22}{7} &= frac{tan DAB + tan CAD}{1 - tan DAB cdot tan CAD} \ &=frac{frac{17}{h} + frac{3}{h}}{1 - left(frac{17}{h}right)left(frac{3}{h}right)} \ frac{22}{7} &= frac{20h}{h^2 - 51}\ 0 &= 22h^2 - 140h - 22 cdot 51\ 0 &= (11h + 51)(h - 11) end{align*}$ The positive value of $h$ is $11$ , so the area is $frac{1}{2}(17 + 3)cdot 11 = boxed{110}$ .
Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.Note that $[f(14,52) = f(14,14 + 38) = frac {52}{38}f(14,38)]$ Repeating the process several times, $begin{eqnarray*}f(14,52) & = & f(14,14 + 38) = frac {52}{38}f(14,38) \ & = & frac {52}{38}times frac {38}{24}f(14,14 + 24) = frac {52}{24}f(14,24) \ & = & frac {52}{10}f(10,14) \ & = & frac {52}{10}times frac {14}{4}f(10,4) = frac {91}{5}f(4,10) \ & = & frac {91}{3}f(4,6) \ & = & 91f(2,4) \ & = & 91times 2 f(2,2) = 364. end{eqnarray*}$ Notice that $f(x,y) = text{lcm}(x,y)$ satisfies all three properties:Clearly, $text{lcm}(x,x) = x$ and $text{lcm}(x,y) = text{lcm}(y,x)$ .
Using the identities $text{gcd}(x,y) cdot text{lcm}(x,y) = xy$ and $text{gcd}(x,x+y) = text{gcd}(x,y)$ , we have:

$y cdot text{lcm}(x,x+y)$ $= dfrac{y cdot x(x+y)}{text{gcd}(x,x+y)}$ $= dfrac{(x+y) cdot xy}{text{gcd}(x,y)}$ $= (x+y) cdot text{lcm}(x,y)$ .

Hence, $f(x,y) = text{lcm}(x,y)$ is a solution to the functional equation.

Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of $f(14,52)$ .

Therefore, $f(14,52) = text{lcm}(14,52) = text{lcm}(2 cdot 7,2^2 cdot 13) = 2^2 cdot 7 cdot 13 = boxed{364}$ .
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of ; using the binomial theorem gives us . Since we are looking for the tens digit, we get . This is true if the tens digit is either or . Casework:
- $4$ : Then our cube must be in the form of $(100k + 42)^3 equiv 3(100k)(42)^2 + 42^3 equiv 200k + 88 pmod{1000}$ . Hence the lowest possible value for the hundreds digit is $4$ , and so $442$ is a valid solution.
- $9$ : Then our cube is $(100k + 92)^3 equiv 3(100k)(92)^2 + 92^3 equiv 200k + 688 pmod{1000}$ . The lowest possible value for the hundreds digit is $1$ , and we get $192$ . Hence, since $192 < 442$ , the answer is $fbox{192}$
$n^3 equiv 888 pmod{1000} implies n^3 equiv 0 pmod 8$ and $n^3 equiv 13 pmod{125}$ . $n equiv 2 pmod 5$ due to the last digit of $n^3$ . Let $n = 5a + 2$ . By expanding, $125a^3 + 150a^2 + 60a + 8 equiv 13 pmod{125} implies 5a^2 + 12a equiv 1 pmod{25}$ .

By looking at the last digit again, we see $a equiv 3 pmod5$ , so we let $a = 5a_1 + 3$ where $a_1 in mathbb{Z^+}$ . Plugging this in to $5a^2 + 12a equiv 1 pmod{25}$ gives $10a_1 + 6 equiv 1 pmod{25}$ . Obviously, $a_1 equiv 2 pmod 5$ , so we let $a_1 = 5a_2 + 2$ where $a_2$ can be any non-negative integer.

Therefore, $n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67$ . $n$ must also be a multiple of $8$ , so $125a_2 + 67 equiv 5a_2 + 3 pmod 8 implies a_2 = 1,9,17 ldots$ . Therefore, the minimum of $n$ is $125 + 67 = boxed{192}$ .

Let $x^3 = 1000a + 888$ . We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$ , where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$ . Taking mod $5$ , $25$ , and $125$ , we get $y^3 equiv 1pmod 5$ , $y^3 equiv 11pmod{25}$ , and $y^3 equiv 111pmod{125}$ .

We can work our way up, and find that $yequiv 1pmod 5$ , $yequiv 21pmod{25}$ , and finally $yequiv 96pmod{125}$ . This gives us our smallest value, $y = 96$ , so $x = boxed{192}$ , as desired. - Spacesam
The number of segments joining the vertices of the polyhedron is ${48choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face.Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 cdot 4 = 8 cdot 6 = 6 cdot 8 = 48$ .Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges $E$ is $frac{3}{2}V = 72$ .Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$ , and each octagon $20$ . The number of diagonals is thus $2 cdot 12 + 9 cdot 8 + 20 cdot 6 = 216$ .Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = 840$ .We first find the number of vertices on the polyhedron: There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $text{vertices} = frac{12 cdot 4 + 8 cdot 6 + 6 cdot 8}{3}=48$ .
We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints).

Counting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram. You can draw a diagram to count this better: Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex. The total number of segments joining vertices that aren't on the same face is $48cdot 35cdot frac 12 = 24 cdot 35 = boxed{840}$
$sum_{k=1}^5 z_k - sum_{k=1}^5 w_k = 0$ $sum_{k=1}^5 z_k = 3 + 504i$ Each $z_k = x_k + y_ki$ lies on the complex line $y = mx + 3$ , so we can rewrite this as $sum_{k=1}^5 z_k = sum_{k=1}^5 x_k + sum_{k=1}^n y_ki$ $3 + 504i = sum_{k=1}^5 x_k + i sum_{k=1}^5 (mx_k + 3)$ Matching the real parts and the imaginary parts, we get that $sum_{k=1}^5 x_k = 3$ and $sum_{k=1}^5 (mx_k + 3) = 504$ . Simplifying the second summation, we find that $msum_{k=1}^5 x_k = 504 - 3 cdot 5 = 489$ , and substituting, the answer is $m cdot 3 = 489 Longrightarrow m = 163$ .
Call the cevians AD, BE, and CF. Using area ratios ( $triangle PBC$ and $triangle ABC$ have the same base), we have: $frac {d}{a + d} = frac {[PBC]}{[ABC]}$ Similarily, $frac {d}{b + d} = frac {[PCA]}{[ABC]}$ and $frac {d}{c + d} = frac {[PAB]}{[ABC]}$ .Then, $frac {d}{a + d} + frac {d}{b + d} + frac {d}{c + d} = frac {[PBC]}{[ABC]} + frac {[PCA]}{[ABC]} + frac {[PAB]}{[ABC]} = frac {[ABC]}{[ABC]} = 1$ The identity $frac {d}{a + d} + frac {d}{b + d} + frac {d}{c + d} = 1$ is a form of Ceva's Theorem.Plugging in $d = 3$ , we get
$[frac{3}{a + 3} + frac{3}{b + 3} + frac{3}{c+3} = 1]$ $[3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)]$ $[3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27]$ $[9(a + b + c) + 54 = abc=boxed{441}]$
Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$ .Clearly, the constant term of $P(x)$ must be $- 1$ . Now, we have $(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + cdots + c_{15}x - 1)$ , where $c_{i}$ is some coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15} = 1$ .Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2 - x - 1)(cdots + c_{14}x^2 + x - 1) = text{something} + 0x^2 + text{something}$ . Therefore, $c_{14} = - 2$ . Undergoing a similar process, $c_{13} = 3$ , $c_{12} = - 5$ , $c_{11} = 8$ , and we see a nice pattern. The coefficients of $P(x)$ are just the Fibonacci sequence with alternating signs! Therefore, $a = c_1 = F_{16}$ , where $F_{16}$ denotes the 16th Fibonnaci number and $a = boxed{987}$ .
Given a point $P (x,y)$ on $C$ , we look to find a formula for $P' (x', y')$ on $C^*$ . Both points lie on a line that is perpendicular to $y=2x$ , so the slope of $overline{PP'}$ is $frac{-1}{2}$ . Thus $frac{y' - y}{x' - x} = frac{-1}{2} Longrightarrow x' + 2y' = x + 2y$ . Also, the midpoint of $overline{PP'}$ , $left(frac{x + x'}{2}, frac{y + y'}{2}right)$ , lies on the line $y = 2x$ . Therefore $frac{y + y'}{2} = x + x' Longrightarrow 2x' - y' = y - 2x$ .Solving these two equations, we find $x' = frac{-3x + 4y}{5}$ and $y' = frac{4x + 3y}{5}$ . Substituting these points into the equation of $C$ , we get $frac{(-3x+4y)(4x+3y)}{25}=1$ , which when expanded becomes $12x^2-7xy-12y^2+25=0$ .Thus, $bc=(-7)(-12)=boxed{084}$ .
Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$ , or the last element of $S$ can be removed. The question asks for the number of different orders in which the all of the remaining elements of $S$ can be removed, given that $8$ had been removed already.

Since $8$ had already been added to the pile, the numbers $1 ldots 7$ had already been added at some time to the pile; $9$ might or might not have been added yet. So currently $S$ is a subset of ${1, 2, ldots 7}$ , possibly with $9$ at the end. Given that $S$ has $k$ elements, there are $k+1$ intervals for $9$ to be inserted, or $9$ might have already been placed, giving $k+2$ different possibilities.

Thus, the answer is $sum_{k=0}^{7} {7 choose k}(k+2)$ $= 1 cdot 2 + 7 cdot 3 + 21 cdot 4 + 35 cdot 5 + 35 cdot 6 + 21 cdot 7 + 7 cdot 8 + 1 cdot 9$ $= boxed{704}$ .