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2001AIME I真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月11日下午5:45

2001AIME I真题及答案解析

答案解析请参考文末

Problem 1

Find the sum of all positive two-digit integers that are divisible by each of their digits.

Problem 2

A finite set $mathcal{S}$ of distinct real numbers has the following properties: the mean of $mathcal{S}cup{1}$ is $13$ less than the mean of $mathcal{S}$ , and the mean of $mathcal{S}cup{2001}$ is $27$ more than the mean of $mathcal{S}$ . Find the mean of $mathcal{S}$ .

Problem 3

Find the sum of the roots, real and non-real, of the equation $x^{2001}+left(frac 12-xright)^{2001}=0$ , given that there are no multiple roots.

Problem 4

In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+bsqrt{c}$ , where $a$ , $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$ .

Problem 5

An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the length of each side is $sqrt{frac mn}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 6

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 7

Triangle $ABC$ has $AB=21$ , $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $overline{AB}$ and $overline{AC}$ , respectively, such that $overline{DE}$ is parallel to $overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 8

Call a positive integer $N$ a $textit{7-10 double}$ if the digits of the base-7 representation of $N$ form a base-10 number that is twice $N$ . For example, $51$ is a 7-10 double because its base-7 representation is $102$ . What is the largest 7-10 double?

Problem 9

In triangle $ABC$ , $AB=13$ , $BC=15$ and $CA=17$ . Point $D$ is on $overline{AB}$ , $E$ is on $overline{BC}$ , and $F$ is on $overline{CA}$ . Let $AD=pcdot AB$ , $BE=qcdot BC$ , and $CF=rcdot CA$ , where $p$ , $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Problem 10

Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0le xle2,$ $0le yle3,$ and $0le zle4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Problem 11

In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$

Problem 12

A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Problem 13

In a certain circle, the chord of a $d$ -degree arc is 22 centimeters long, and the chord of a $2d$ -degree arc is 20 centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$

Problem 14

A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?

Problem 15

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

2001AIME I详细解析

Let our number be , . Then we have two conditions: and , or divides into and divides into . Thus or (note that if , then would not be a digit).
- For $b = a$ , we have $n = 11a$ for nine possibilities, giving us a sum of $11 cdot frac {9(10)}{2} = 495$ .
- For $b = 2a$ , we have $n = 12a$ for four possibilities (the higher ones give $b > 9$ ), giving us a sum of $12 cdot frac {4(5)}{2} = 120$ .
- For $b = 5a$ , we have $n = 15a$ for one possibility (again, higher ones give $b > 9$ ), giving us a sum of $15$ .
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = boxed{630}$ .
Let $x$ be the mean of $mathcal{S}$ . Let $a$ be the number of elements in $mathcal{S}$ . Then, the given tells us that $frac{ax+1}{a+1}=x-13$ and $frac{ax+2001}{a+1}=x+27$ . Subtracting, we have $begin{align*}frac{ax+2001}{a+1}-40=frac{ax+1}{a+1} Longrightarrow frac{2000}{a+1}=40 Longrightarrow a=49end{align*}$ We plug that into our very first formula, and get: $begin{align*}frac{49x+1}{50}&=x-13 \ 49x+1&=50x-650 \ x&=boxed{651}.end{align*}$
From Vieta's formulas, in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + cdots + a_0 = 0$ , then the sum of the roots is $frac{-a_{n-1}}{a_n}$ .From the Binomial Theorem, the first term of $left(frac 12-xright)^{2001}$ is $-x^{2001}$ , but $x^{2001}+-x^{2001}=0$ , so the term with the largest degree is $x^{2000}$ . So we need the coefficient of that term, as well as the coefficient of $x^{1999}$ . $begin{align*}binom{2001}{1} cdot (-x)^{2000} cdot left(frac{1}{2}right)^1&=frac{2001x^{2000}}{2}\ binom{2001}{2} cdot (-x)^{1999} cdot left(frac{1}{2}right)^2 &=frac{-x^{1999}*2001*2000}{8}=-2001 cdot 250x^{1999} end{align*}$ Applying Vieta's formulas, we find that the sum of the roots is $-frac{-2001 cdot 250}{frac{2001}{2}}=250 cdot 2=boxed{500}$ .We find that the given equation has a $2000^{text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $frac{1}{2}$ to get a sum of $boxed{500}$ .
After chasing angles, $angle ATC=75^{circ}$ and $angle TCA=75^{circ}$ , meaning $triangle TAC$ is an isosceles triangle and $AC=24$ .Using law of sines on $triangle ABC$ , we can create the following equation: $frac{24}{sin(angle ABC)}$ $=$ $frac{BC}{sin(angle BAC)}$ $angle ABC=45^{circ}$ and $angle BAC=60^{circ}$ , so $BC = 12sqrt{6}$ .We can then use the Law of Sines area formula $frac{1}{2} cdot BC cdot AC cdot sin(angle BCA)$ to find the area of the triangle.
$sin(75)$ can be found through the sin addition formula.

$sin(75)$ $=$ $frac{sqrt{6} + sqrt{2}}{4}$

Therefore, the area of the triangle is $frac{sqrt{6} + sqrt{2}}{4}$ $cdot$ $24$ $cdot$ $12sqrt{6}$ $cdot$ $frac{1}{2}$

$72sqrt{3} + 216$

$72 + 3 + 216 =$ $boxed{291}$
Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$ Note that the slope of $overline{AC}$ is $tan 60^circ = sqrt {3}.$ Hence, the equation of the line containing $overline{AC}$ is $[y = xsqrt {3} + 1.]$ This will intersect the ellipse when $begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(xsqrt {3} + 1)^{2} \ & = & x^{2} + 4(3x^{2} + 2xsqrt {3} + 1) implies x(13x+8sqrt 3)=0implies x = frac { - 8sqrt {3}}{13}. end{eqnarray*}$ We ignore the $x=0$ solution because it is not in quadrant 3.Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $left(frac {8sqrt {3}}{13},y_{0}right)$ and $left(frac { - 8sqrt {3}}{13},y_{0}right),$ respectively, for some value of $y_{0}.$ It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$ . Our answer is $[BC = 2*frac {8sqrt {3}}{13}=sqrt {4left(frac {8sqrt {3}}{13}right)^{2}} = sqrt {frac {768}{169}}implies m + n = boxed{937}.]$
Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$ . In the diagram below, the lowest $y$ -coordinate at each of $a$ , $b$ , $c$ , and $d$ corresponds to the value of the roll.The red path corresponds to the sequence of rolls $2, 3, 5, 5$ . This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths.The solution to this problem is therefore $dfrac{binom{9}{4}}{6^4} = {dfrac{7}{72}}$ . So the answer is $boxed{79}$ .
$[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,NW)); // D((A.x,0)--A,linetype("4 4")+linewidth(0.7)); D((I.x,0)--I,linetype("4 4")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); D(B--I--C); MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); [/asy]$ Let $I$ be the incenter of $triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $angle ABC$ and $angle ACB$ respectively. Then, $angle BID = angle CBI = angle DBI,$ so $triangle BDI$ is isosceles, and similarly $triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $triangle ADE$ and $triangle ABC$ is $frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = frac{43}{63} times 20 = frac{860}{63}$ . Thus, $m + n = boxed{923}$ .
We let $N_7 = overline{a_na_{n-1}cdots a_0}_7$ ; we are given that $[2(a_na_{n-1}cdots a_0)_7 = (a_na_{n-1}cdots a_0)_{10}]$ m (This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)Expanding, we find that $[2 cdot 7^n a_n + 2 cdot 7^{n-1} a_{n-1} + cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + cdots + a_0]$ or re-arranging,
$[a_0 + 4a_1 = 2a_2 + 314a_3 + cdots + (10^n - 2 cdot 7^n)a_n]$

Since the $a_i$ s are base- $7$ digits, it follows that $a_i < 7$ , and the LHS is less than or equal to $30$ . Hence our number can have at most $3$ digits in base- $7$ . Letting $a_2 = 6$ , we find that $630_7 = boxed{315}_{10}$ is our largest 7-10 double.
$[asy] /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle); D(D(MP("D",D))--D(MP("E",E,NE))--D(MP("F",F,NW))--cycle); [/asy]$ We let $[ldots]$ denote area; then the desired value is $frac mn = frac{[DEF]}{[ABC]} = frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$ Using the formula for the area of a triangle $frac{1}{2}absin C$ , we find that $frac{[ADF]}{[ABC]} = frac{frac 12 cdot p cdot AB cdot (1-r) cdot AC cdot sin angle CAB}{frac 12 cdot AB cdot AC cdot sin angle CAB} = p(1-r)$ and similarly that $frac{[BDE]}{[ABC]} = q(1-p)$ and $frac{[CEF]}{[ABC]} = r(1-q)$ . Thus, we wish to find $begin{align*}frac{[DEF]}{[ABC]} &= 1 - frac{[ADF]}{[ABC]} - frac{[BDE]}{[ABC]} - frac{[CEF]}{[ABC]} \ &= 1 - p(1-r) - q(1-p) - r(1-q)\ &= (pq + qr + rp) - (p + q + r) + 1 end{align*}$ We know that $p + q + r = frac 23$ , and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) Longleftrightarrow pq + qr + rp = frac{left(frac 23right)^2 - frac 25}{2} = frac{1}{45}$ . Substituting, the answer is $frac 1{45} - frac 23 + 1 = frac{16}{45}$ , and $m+n = boxed{061}$ .
The distance between the , , and coordinates must be even so that the midpoint can have integer coordinates. Therefore,
- For $x$ , we have the possibilities $(0,0)$ , $(1,1)$ , $(2,2)$ , $(0,2)$ , and $(2,0)$ , $5$ possibilities.
- For $y$ , we have the possibilities $(0,0)$ , $(1,1)$ , $(2,2)$ , $(3,3)$ , $(0,2)$ , $(2,0)$ , $(1,3)$ , and $(3,1)$ , $8$ possibilities.
- For $z$ , we have the possibilities $(0,0)$ , $(1,1)$ , $(2,2)$ , $(3,3)$ , $(4,4)$ , $(0,2)$ , $(0,4)$ , $(2,0)$ , $(4,0)$ , $(2,4)$ , $(4,2)$ , $(1,3)$ , and $(3,1)$ , $13$ possibilities.
However, we have $3cdot 4cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $frac {5cdot 8cdot 13 - 60}{60cdot 59} = frac {23}{177}Longrightarrow m+n = boxed{200}$ .
Let each point $P_i$ be in column $c_i$ . The numberings for $P_i$ can now be defined as follows. $begin{align*}x_i &= (i - 1)N + c_i\ y_i &= (c_i - 1)5 + i end{align*}$ We can now convert the five given equalities. $begin{align}x_1&=y_2 & Longrightarrow & & c_1 &= 5 c_2-3\ x_2&=y_1 & Longrightarrow & & N+c_2 &= 5 c_1-4\ x_3&=y_4 & Longrightarrow & & 2 N+c_3 &= 5 c_4-1\ x_4&=y_5 & Longrightarrow & & 3 N+c_4 &= 5 c_5\ x_5&=y_3 & Longrightarrow & & 4 N+c_5 &= 5 c_3-2 end{align}$ Equations $(1)$ and $(2)$ combine to form $[N = 24c_2 - 19]$ Similarly equations $(3)$ , $(4)$ , and $(5)$ combine to form $[117N +51 = 124c_3]$ Take this equation modulo 31 $[24N+20equiv 0 pmod{31}]$ And substitute for N $[24 cdot 24 c_2 - 24 cdot 19 +20equiv 0 pmod{31}]$ $[18 c_2 equiv 2 pmod{31}]$ Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 cdot 7 - 19 = 149$ The column values can also easily be found by substitution $begin{align*}c_1&=32\ c_2&=7\ c_3&=141\ c_4&=88\ c_5&=107 end{align*}$ As these are all positive and less than $N$ , $boxed{149}$ is the solution.
$[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label("$I$",I,W); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W*-1); label("$D$",D,W*-1); [/asy]$ The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$ The signed distance between a plane and a point $I$ can be calculated as $frac{(I-G) cdot P}{|P|}$ , where G is any point on the plane, and P is a vector perpendicular to ABC.A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)times(B-C)=langle 8, 12, 24 rangle$ Thus $frac{(I-C) cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$
$begin{align*}frac{(I-C) cdot P}{|P|}&=-r\ frac{(langle r, r, r rangle-langle 0, 0, 2 rangle) cdot P}{|P|}&=-r\ frac{langle r, r, r-2 rangle cdot langle 8, 12, 24 rangle}{langle 8, 12, 24 rangle}&=-r\ frac{44r -48}{28}&=-r\ 44r-48&=-28r\ 72r&=48\ r&=frac{2}{3} end{align*}$

Finally $2+3=boxed{005}$
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ , $AC$ , and $BD$ be the chords of the $d$ -degree arcs, and let $CD$ be the chord of the $3d$ -degree arc. Also let $x$ be equal to the chord length of the $3d$ -degree arc. Hence, the length of the chords, $AD$ and $BC$ , of the $2d$ -degree arcs can be represented as $x + 20$ , as given in the problem.Using Ptolemy's theorem, $[AB(CD) + AC(BD) = AD(BC)]$ $[22x + 22(22) = (x + 20)^2]$ $[22x + 484 = x^2 + 40x + 400]$ $[0 = x^2 + 18x - 84]$ We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. $[x = frac{-18 + sqrt{18^2 + 4(84)}}{2}]$ $[x = frac{-18 + sqrt{660}}{2}]$ $x$ simplifies to $frac{-18 + 2sqrt{165}}{2},$ which equals $-9 + sqrt{165}.$ Thus, the answer is $9 + 165 = boxed{174}$ .
Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$ -digit strings of $0$ 's and $1$ 's such that there are no two consecutive $1$ 's and no three consecutive $0$ 's.The last two digits of any $n$ -digit string can't be $11$ , so the only possibilities are $00$ , $01$ , and $10$ .Let $a_n$ be the number of $n$ -digit strings ending in $00$ , $b_n$ be the number of $n$ -digit strings ending in $01$ , and $c_n$ be the number of $n$ -digit strings ending in $10$ .If an $n$ -digit string ends in $00$ , then the previous digit must be a $1$ , and the last two digits of the $n-1$ digits substring will be $10$ . So $[a_{n} = c_{n-1}.]$ If an $n$ -digit string ends in $01$ , then the previous digit can be either a $0$ or a $1$ , and the last two digits of the $n-1$ digits substring can be either $00$ or $10$ . So $[b_{n} = a_{n-1} + c_{n-1}.]$
If an $n$ -digit string ends in $10$ , then the previous digit must be a $0$ , and the last two digits of the $n-1$ digits substring will be $01$ . So $[c_{n} = b_{n-1}.]$

Clearly, $a_2=b_2=c_2=1$ . Using the recursive equations and initial values: $[begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} multicolumn{19}{c}{}\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\hline end{array}]$

As a result $a_{19}+b_{19}+c_{19}=boxed{351}$ .
Choose one face of the octahedron randomly and label it with . There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.Clearly, the labels for the A-faces must come from the set , since these faces are all adjacent to . There are thus ways to assign the labels for the A-faces.The labels for the B-faces and C-face are the two remaining numbers from the above set, plus and . The number on the C-face must not be consecutive to any of the numbers on the B-faces.From here it is easiest to brute force the possibilities for the numbers on the B and C faces:
- 2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here.
- 2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here.
- 2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here.
- 2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility.
- 2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible.
- 2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible.
There is a total of $10$ possibilities. There are $3!=6$ permutations (more like "rotations") of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$ . There are $7!=5040$ ways to randomly fill in the rest of the octahedron. So the probability is $frac {60}{5040} = frac {1}{84}$ . The answer is $boxed{085}$ .