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1984AIME 真題及答案解析

Category: AMC美國數學競賽, 國際競賽 Date: 2019年12月12日下午12:27

1984AIME 真題及答案解析

答案解析請參考文末

Problem 1

Find the value of? $a_2+a_4+a_6+a_8+ldots+a_{98}$ ?if? $a_1$ ,? $a_2$ ,? $a_3ldots$ ?is an arithmetic progression with common difference 1, and? $a_1+a_2+a_3+ldots+a_{98}=137$ .

Problem 2

The integer? $n$ ?is the smallest positive multiple of? $15$ ?such that every digit of? $n$ ?is either? $8$ ?or? $0$ . Compute? $frac{n}{15}$ .

Problem 3

A point? $P$ ?is chosen in the interior of? $triangle ABC$ ?such that when lines are drawn through? $P$ ?parallel to the sides of? $triangle ABC$ , the resulting smaller triangles? $t_{1}$ ,? $t_{2}$ , and? $t_{3}$ ?in the figure, have areas? $4$ ,? $9$ , and? $49$ , respectively. Find the area of? $triangle ABC$ .

$[asy] size(200); pathpen=black+linewidth(0.65);pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy]$

Problem 4

Let? $S$ ?be a list of positive integers - not necessarily distinct - in which the number? $68$ ?appears. The arithmetic mean of the numbers in? $S$ ?is? $56$ . However, if? $68$ ?is removed, the arithmetic mean of the numbers is? $55$ . What's the largest number that can appear in? $S$ ?

Problem 5

Determine the value of? $ab$ ?if? $log_8a+log_4b^2=5$ ?and? $log_8b+log_4a^2=7$ .

Problem 6

Three circles, each of radius? $3$ , are drawn with centers at? $(14, 92)$ ,? $(17, 76)$ , and? $(19, 84)$ . A line passing through? $(17,76)$ ?is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

Problem 7

The function f is defined on the set of integers and satisfies? $f(n)= begin{cases} n-3 & mbox{if }nge 1000 f(f(n+5)) & mbox{if }n<1000 end{cases}$

Find? $f(84)$ .

Problem 8

The equation? $z^6+z^3+1=0$ ?has complex roots with argument? $theta$ ?between? $90^circ$ ?and? $180^circ$ ?in the complex plane. Determine the degree measure of? $theta$ .

Problem 9

In tetrahedron? $ABCD$ , edge? $AB$ ?has length 3 cm. The area of face? $ABC$ ?is? $15mbox{cm}^2$ ?and the area of face? $ABD$ ?is? $12 mbox { cm}^2$ . These two faces meet each other at a? $30^circ$ ?angle. Find the volume of the tetrahedron in? $mbox{cm}^3$ .

Problem 10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over? $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over? $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of? $30$ ?multiple choice problems and that one's score,? $s$ , is computed by the formula? $s=30+4c-w$ , where? $c$ ?is the number of correct answers and? $w$ ?is the number of wrong answers. Students are not penalized for problems left unanswered.)

Problem 11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let? $frac m n$ ?in lowest terms be the probability that no two birch trees are next to one another. Find? $m+n$ .

Problem 12

A function? $f$ ?is defined for all real numbers and satisfies? $f(2+x)=f(2-x)$ ?and? $f(7+x)=f(7-x)$ ?for all? $x$ . If? $x=0$ ?is a root for? $f(x)=0$ , what is the least number of roots? $f(x)=0$ ?must have in the interval? $-1000leq x leq 1000$ ?

Problem 13

Find the value of? $10cot(cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21).$

Problem 14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Problem 15

Determine? $w^2+x^2+y^2+z^2$ ?if

$frac{x^2}{2^2-1}+frac{y^2}{2^2-3^2}+frac{z^2}{2^2-5^2}+frac{w^2}{2^2-7^2}=1$ $frac{x^2}{4^2-1}+frac{y^2}{4^2-3^2}+frac{z^2}{4^2-5^2}+frac{w^2}{4^2-7^2}=1$ $frac{x^2}{6^2-1}+frac{y^2}{6^2-3^2}+frac{z^2}{6^2-5^2}+frac{w^2}{6^2-7^2}=1$ $frac{x^2}{8^2-1}+frac{y^2}{8^2-3^2}+frac{z^2}{8^2-5^2}+frac{w^2}{8^2-7^2}=1$

One approach to this problem is to apply the formula for the sum of an?arithmetic series?in order to find the value of? $a_1$ , then use that to calculate? $a_2$ ?and sum another arithmetic series to get our answer.A somewhat quicker method is to do the following: for each? $n geq 1$ , we have? $a_{2n - 1} = a_{2n} - 1$ . We can substitute this into our given equation to get? $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + ldots + (a_{98} - 1) + a_{98} = 137$ . The left-hand side of this equation is simply? $2(a_2 + a_4 + ldots + a_{98}) - 49$ , so our desired value is? $frac{137 + 49}{2} = boxed{093}$ .
Any multiple of 15 is a multiple of 5 and a multiple of 3.Any multiple of 5 ends in 0 or 5; since? $n$ ?only contains the digits 0 and 8, the units?digit?of? $n$ ?must be 0.The sum of the digits of any multiple of 3 must be?divisible?by 3. If? $n$ ?has? $a$ ?digits equal to 8, the sum of the digits of? $n$ ?is? $8a$ . For this number to be divisible by 3,? $a$ ?must be divisible by 3. We also know that? $a>0$ ?since? $n$ ?is positive. Thus? $n$ ?must have at least three copies of the digit 8.The smallest number which meets these two requirements is 8880. Thus the answer is? $frac{8880}{15} = boxed{592}$ .
By the transversals that go through? $P$ , all four triangles are?similar?to each other by the? $AA$ ?postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity? $K = dfrac{absin C}{2}$ ?to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as? $2x, 3x, 7x$ . Thus, the corresponding side on the large triangle is? $12x$ , and the area of the triangle is? $12^2 = boxed{144}$ .
Suppose? $S$ ?has? $n$ ?members other than 68, and the sum of these members is? $s$ . Then we're given that? $frac{s + 68}{n + 1} = 56$ ?and? $frac{s}{n} = 55$ . Multiplying to clear?denominators, we have $s + 68 = 56n + 56$ ?and $s = 55n$ ?so $68 = n + 56$ , $n = 12$ ?and $s = 12cdot 55 = 660$ .Because the sum and number of the elements of? $S$ ?are fixed, if we want to maximize the largest number in? $S$ , we should take all but one member of? $S$ ?to be as small as possible. Since all members of? $S$ ?are positive integers, the smallest possible value of a member is 1. Thus the largest possible element is? $660 - 11 = boxed{649}$ .
Use the?change of base formula?to see that? $frac{log a}{log 8} + frac{2 log b}{log 4} = 5$ ; combine?denominators?to find that? $frac{log ab^3}{3log 2} = 5$ . Doing the same thing with the second equation yields that? $frac{log a^3b}{3log 2} = 7$ . This means that? $log ab^3 = 15log 2 Longrightarrow ab^3 = 2^{15}$ ?and that? $log a^3 b = 21log 2 Longrightarrow a^3 b = 2^{21}$ . If we multiply the two equations together, we get that? $a^4b^4 = 2^{36}$ , so taking the fourth root of that,? $ab = 2^9 = boxed{512}$ .We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become? $frac{ln a}{ln 8} + frac{2 ln b}{ln 4} = 5$ ?and? $frac{ln b}{ln 8} + frac{2 ln a}{ln 4} = 7$ . Adding the equations and factoring, we get? $(frac{1}{ln 8}+frac{2}{ln 4})(ln a+ ln b)=12$ . Rearranging we see that? $ln ab = frac{12}{frac{1}{ln 8}+frac{2}{ln 4}}$ . Again, we pull exponents out of our logarithms to get? $ln ab = frac{12}{frac{1}{3 ln 2} + frac{2}{2 ln 2}} = frac{12 ln 2}{frac{1}{3} + 1} = frac{12 ln 2}{frac{4}{3}} = 9 ln 2$ . This means that? $frac{ln ab}{ln 2} = 9$ . The left-hand side can be interpreted as a base-2 logarithm, giving us? $ab = 2^9 = boxed{512}$ .This solution is very similar to the above two, but it utilizes the well-known fact that? $log_{m^k}{n^k}= log_m{n}.$ ?Thus,? $log_8a+log_4b^2=5 Rightarrow log_{2^3}{(sqrt[3]{a})^3} + log_{2^2}{b^2} = 5 Rightarrow log_2{sqrt[3]{a}} + log_2{b} = 5 Rightarrow log_2{sqrt[3]{a}b} = 5.$ ?Similarly,? $log_8b+log_4a^2=7 Rightarrow log_2{sqrt[3]{b}a} = 7.$ ?Adding these two equations, we have? $log_2{a^{frac{4}{3}}b^{frac{4}{3}}} = 12 Rightarrow ab = 2^{12timesfrac{3}{4}} = 2^9 = boxed{512}$ .We can change everything to a common base, like so:? $log_8{a} + log_8{b^3} = 5,$ ? $log_8{b} + log_8{a^3} = 7.$ ?We set the value of? $log_8{a}$ ?to? $x$ , and the value of? $log_8{b}$ ?to? $y.$ ?Now we have a system of linear equations: $[x + 3y = 5,]$ $[y + 3x = 7.]$ We multiply the second equation by three and subtract the first equation from the second to get? $8x = 16,$ ?which gives? $x=2.$ ?We substitute our value into the first equation and find that? $y = 1.$ ?Now we know that? $log_8{a} = 2,$ ?and? $log_8{b} = 1,$ ?so we find that? $a = 64,$ ?and? $b=8.$ ?Multiplying together gives us? $ab = boxed{512}.$ Add the two equations to get? $log_8 {a}+ log_8 {b}+ log_{a^2}+log_{b^2}=12$ . This can be simplified with the log property? $log_n {x}+log_n {y}=log_n {xy}$ . Using this, we get? $log_8 {ab}+ log_4 {a^2b^2}=12$ . Now let? $log_8 {ab}=c$ ?and? $log_4 {a^2b^2}=k$ . Converting to exponents, we get? $8^c=ab$ ?and? $4^k=(ab)^2$ . Sub in the? $8^c$ ?to get? $k=3c$ . So now we have that? $k+c=12$ ?and? $k=3c$ ?which gives? $c=3$ ,? $k=9$ . This means? $log_4 {a^2b^2}=9$ ?so? $4^9=(ab)^2 implies ab=(2^2)^9 implies 2^9 implies boxed {512}$ Add the equations and use the facts that? $log{a} +log{b}=log{ab}$ ?and? $log{k^n} =nlog{k}$ ?to get $[log_8{ab} +2log_4{ab}=12]$ Now use the change of base identity with base as 2: $[dfrac{log_2{ab}}{log_2{8}}+dfrac{2log_2{ab}}{log_2{4}}=12]$ Which gives: $[frac{4}{3}log_2{ab}=12]$ Solving gives,? $boxed{ab=2^9=512}$ By properties of logarithms, we know that? $log_8 {a}+ log_4 {b ^ 2} = log_2 {a ^ {1/3}}+ log_2 {b} = 5$ .
Using the fact that? $log_a {b} + log_a {c} = log_a{b*c}$ , we get? $log_2 {a^{1/3} * b} = 5$ .

Similarly, we know that? $log_2 {a * b^{1/3}} = 7$ .

From these two equations, we get? $a^{1/3} * b = 2^5$ ?and? $a * b^{1/3} = 2^7$ .

Multiply the two equations to get? $a^{4/3} * b^{4/3} = 2^{12}$ . Solving, we get that? $a*b = 2^{12*3/4} = 2^9 =$ $boxed{512}$ .
The line passes through the center of the second circle; hence it is the circle's?diameter?and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle.Draw the?midpoint?of? $overline{AC}$ ?(the centers of the other two circles), and call it? $M$ . If we draw the feet of the?perpendiculars?from? $A,C$ ?to the line (call? $E,F$ ), we see that? $triangle AEMcong triangle CFM$ ?by?HA congruency; hence? $M$ ?lies on the line. The coordinates of? $M$ ?are? $left(frac{19+14}{2},frac{84+92}{2}right) = left(frac{33}{2},88right)$ .Thus, the slope of the line is? $frac{88 - 76}{frac{33}{2} - 17} = -24$ , and the answer is? $boxed{024}$ .Remark: Notice the fact that the radius is 3 is not used in this problem; in fact changing the radius does not affect the answer.
Define? $f^{h} = f(f(cdots f(f(x))cdots))$ , where the function? $f$ ?is performed? $h$ ?times. We find that? $f(84) = f(f(89)) = f^2(89) = f^3(94) = ldots f^{y}(1004)$ .? $1004 = 84 + 5(y - 1) Longrightarrow y = 185$ . So we now need to reduce? $f^{185}(1004)$ .Let’s write out a couple more iterations of this function: $begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000) &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)end{align*}$ So this function reiterates with a period of 2 for? $x$ . It might be tempting at first to assume that? $f(1004) = 1001$ ?is the answer; however, that is not true since the solution occurs slightly before that. Start at? $f^3(1004)$ : $[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=boxed{997}]$
We shall introduce another factor to make the equation easier to solve. If? $r$ ?is a root of? $z^6+z^3+1$ , then? $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial? $x^9-1$ ?has all of its roots with?absolute value? $1$ ?and argument of the form? $40m^circ$ ?for integer? $m$ ?(the ninth degree?roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation? $x^6 + x^3 + 1 = 0$ .This reduces? $theta$ ?to either? $120^{circ}$ ?or? $160^{circ}$ . But? $theta$ ?can't be? $120^{circ}$ ?because if? $r=cos 120^circ +isin 120^circ$ , then? $r^6+r^3+1=3$ . (When we multiplied by? $r^3 - 1$ ?at the beginning, we introduced some extraneous solutions, and the solution with? $120^circ$ ?was one of them.) This leaves? $boxed{theta=160}$ .The substitution? $y=z^3$ ?simplifies the equation to? $y^2+y+1 = 0$ . Applying the quadratic formula gives roots? $y=-frac{1}{2}pm frac{sqrt{3}i}{2}$ , which have arguments of? $120$ ?and? $240,$ ?respectively. This means? $arg(z) = frac{120 ; text{or} ;240}{3} + frac{360n}{3}$ , and the only one between 90 and 180 is? $boxed{theta=160}$ .
$[asy] /* modified version of olympiad modules */ import three; real markscalefactor = 0.03; path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 anglemark(triple A, triple B, triple C, real t=8 ... real[] s) { triple M,N,P[],Q[]; path3 mark; int n=s.length; M=t*markscalefactor*unit(A-B)+B; N=t*markscalefactor*unit(C-B)+B; for (int i=0; i<n; ++i) { P[i]=s[i]*markscalefactor*unit(A-B)+B; Q[i]=s[i]*markscalefactor*unit(C-B)+B; } mark=arc(B,M,N); for (int i=0; i<n; ++i) { if (i%2==0) { mark=mark--reverse(arc(B,P[i],Q[i])); } else { mark=mark--arc(B,P[i],Q[i]); } } if (n%2==0 && n!=0) mark=(mark--B--P[n-1]); else if (n!=0) mark=(mark--B--Q[n-1]); else mark=(mark--B--cycle); return mark; } size(200); import three; defaultpen(black+linewidth(0.7)); pen small = fontsize(10); triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0); currentprojection=perspective(16,-10,8); draw(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight); draw(surface(A--B--D--cycle),rgb(0.7,0.6,0.6),nolight); /* draw pyramid - other lines + angles */ draw(A--B--C--A--D--B--D--C); draw(D--Da--Db--cycle); draw(rightanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15)); /* labeling points */ label("$A$",A,SW);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$30^{circ}$",Db+(0,.35,0.08),(1.5,1.2),small); label("$3$",(A+B)/2,S); label("$15mathrm{cm}^2$",(Db+C)/2+(0,-0.5,-0.1),NE,small); label("$12mathrm{cm}^2$",(A+D)/2,NW,small); [/asy]$ Position face? $ABC$ ?on the bottom. Since? $[triangle ABD] = 12 = frac{1}{2} cdot AB cdot h_{ABD}$ , we find that? $h_{ABD} = 8$ . Because the problem does not specify, we may assume both? $ABC$ ?and? $ABD$ ?to be isosceles triangles. Thus, the height of? $ABD$ ?forms a? $30-60-90$ ?with the height of the tetrahedron. So,? $h = frac{1}{2} (8) = 4$ . The volume of the tetrahedron is thus? $frac{1}{3}Bh = frac{1}{3} cdot15 cdot 4 = boxed{020}$ .It is clear that? $DX=8$ ?and? $CX=10$ ?where? $X$ ?is the foot of the perpendicular from? $D$ ?and? $C$ ?to side? $AB$ . Thus? $[DXC]=frac{absin{c}}{2}=20=5*h rightarrow h = 4$ ?where h is the height of the tetrahedron from? $D$ . Hence, the volume of the tetrahedron is? $bh/3=15*4/3=boxed{020}$ ?~ ShreyJMake faces? $ABC$ ?and? $ABD$ ?right triangles. This makes everything a lot easier. Then do everything in solution 1.
Let Mary's score, number correct, and number wrong be? $s,c,w$ ?respectively. Then $s=30+4c-w=30+4(c-1)-(w-4)=30+4(c+1)-(w+4)$ .Therefore, Mary could not have left at least five blank; otherwise, 1 more correct and 4 more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have score above 80, or even 30.)It follows that? $c+wgeq 26$ ?and? $wleq 3$ , so? $cgeq 23$ ?and? $s=30+4c-wgeq 30+4(23)-3=119$ . So Mary scored at least 119. To see that no result other than 23 right/3 wrong produces 119, note that? $s=119Rightarrow 4c-w=89$ ?so? $wequiv 3pmod{4}$ . But if? $w=3$ , then? $c=23$ , which was the result given; otherwise? $wgeq 7$ ?and? $cgeq 24$ , but this implies at least 31 questions, a contradiction. This makes the minimum score? $boxed{119}$ .
First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.)The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making? ${8choose5} = 56$ ?different ways to arrange this.There are? ${12 choose 5} = 792$ ?total ways to arrange the twelve trees, so the probability is? $frac{56}{792} = frac{7}{99}$ .The answer is? $7 + 99 = boxed{106}$ .Let? $b$ ,? $n$ ?denote birch tree and not-birch tree, respectively. Notice that we only need? $4$ ? $n$ s to separate the? $5$ ? $b$ s. Specifically, $[b,n,b,n,b,n,b,n,b]$ Since we have? $7$ ? $n$ s, we are placing the extra? $3$ ? $n$ s into the? $6$ ?intervals beside the? $b$ s.Now doing simple casework.If all? $3$ ? $n$ s are in the same interval, there are? $6$ ?ways.
If? $2$ ?of the? $3$ ? $n$ s are in the same interval, there are? $6cdot5=30$ ?ways.

If the? $n$ s are in? $3$ ?different intervals, there are? ${6 choose 3} =20$ ?ways.

In total there are? $6+30+20=56$ ?ways.

There are? ${12choose5}=792$ ?ways to distribute the birch trees among all? $12$ ?trees.

Thus the probability equals? $frac{56}{792}=frac{7}{99}Longrightarrow m+n=7+99=boxed{106}$ .
If? $f(2+x)=f(2-x)$ , then substituting? $t=2+x$ ?gives? $f(t)=f(4-t)$ . Similarly,? $f(t)=f(14-t)$ . In particular, $[f(t)=f(14-t)=f(14-(4-t))=f(t+10)]$ Since? $0$ ?is a root, all multiples of? $10$ ?are roots, and anything congruent to? $4pmod{10}$ ?are also roots. To see that these may be the only integer roots, observe that the function $[f(x) = sin frac{pi x}{10}sin frac{pi (x-4)}{10}]$ satisfies the conditions and has no other roots.In the interval? $-1000leq xleq 1000$ , there are? $201$ ?multiples of? $10$ ?and? $200$ ?numbers that are congruent to? $4 pmod{10}$ , therefore the minimum number of roots is? $boxed{401}$ .
We know that? $tan(arctan(x)) = x$ ?so we can repeatedly apply the addition formula,? $tan(x+y) = frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$ . Let? $a = cot^{-1}(3)$ ,? $b=cot^{-1}(7)$ ,? $c=cot^{-1}(13)$ , and? $d=cot^{-1}(21)$ . We have $tan(a)=frac{1}{3},quadtan(b)=frac{1}{7},quadtan(c)=frac{1}{13},quadtan(d)=frac{1}{21}$ ,so $tan(a+b) = frac{frac{1}{3}+frac{1}{7}}{1-frac{1}{21}} = frac{1}{2}$ and $tan(c+d) = frac{frac{1}{13}+frac{1}{21}}{1-frac{1}{273}} = frac{1}{8}$ ,so $tan((a+b)+(c+d)) = frac{frac{1}{2}+frac{1}{8}}{1-frac{1}{16}} = frac{2}{3}$ .Thus our answer is? $10cdotfrac{3}{2}=boxed{015}$ .Apply the formula? $cot^{-1}x + cot^{-1} y = cot^{-1}left(frac {xy-1}{x+y}right)$ ?repeatedly. Using it twice on the inside, the desired sum becomes? $cot (cot^{-1}2+cot^{-1}8)$ . This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning.On the coordinate plane, let? $O=(0,0)$ ,? $A_1=(3,0)$ ,? $A_2=(3,1)$ ,? $B_1=(21,7)$ ,? $B_2=(20,10)$ ,? $C_1=(260,130)$ ,? $C_2=(250,150)$ ,? $D_1=(5250,3150)$ ,? $D_2=(5100,3400)$ , and? $H=(5100,0)$ . We see that? $cot^{-1}(angle A_2OA_1)=3$ ,? $cot^{-1}(angle B_2OB_1)=7$ ,? $cot^{-1}(angle C_2OC_1)=13$ , and? $cot^{-1}(angle D_2OD_1)=21$ . The sum of these four angles forms the angle of triangle? $OD_2H$ , which has a cotangent of? $frac{5100}{3400}=frac{3}{2}$ , which must mean that? $cot( cot^{-1}3+cot^{-1}7+cot^{-1}13+cot^{-1}21)=frac{3}{2}$ . So the answer is? $10cdotleft(frac{3}{2}right)=boxed{015}.$
Take an even positive integer? $x$ .? $x$ ?is either? $0 bmod{6}$ ,? $2 bmod{6}$ , or? $4 bmod{6}$ . Notice that the numbers? $9$ ,? $15$ ,? $21$ , ... , and in general? $9 + 6n$ ?for nonnegative? $n$ ?are odd composites. We now have 3 cases:If? $x ge 18$ ?and is? $0 bmod{6}$ ,? $x$ ?can be expressed as? $9 + (9+6n)$ ?for some nonnegative? $n$ . Note that? $9$ ?and? $9+6n$ ?are both odd composites.If? $xge 44$ ?and is? $2 bmod{6}$ ,? $x$ ?can be expressed as? $35 + (9+6n)$ ?for some nonnegative? $n$ . Note that? $35$ ?and? $9+6n$ ?are both odd composites.If? $xge 34$ ?and is? $4 bmod{6}$ ,? $x$ ?can be expressed as? $25 + (9+6n)$ ?for some nonnegative? $n$ . Note that? $25$ ?and? $9+6n$ ?are both odd composites.Clearly, if? $x ge 44$ , it can be expressed as a sum of 2 odd composites. However, if? $x = 42$ , it can also be expressed using case 1, and if? $x = 40$ , using case 3.? $38$ ?is the largest even integer that our cases do not cover. If we examine the possible ways of splitting? $38$ ?into two addends, we see that no pair of odd composites add to? $38$ . Therefore,? $boxed{038}$ ?is the largest possible number that is not expressible as the sum of two odd composite numbers.Let? $n$ ?be an integer that cannot be written as the sum of two odd composite numbers. If? $n>33$ , then? $n-9,n-15,n-21,n-25,n-27,$ ?and? $n-33$ ?must all be prime (or? $n-33=1$ , which yields? $n=34=9+25$ ?which does not work). Thus? $n-9,n-15,n-21,n-27,$ ?and? $n-33$ ?form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is? $5,11,17,23,$ ?and? $29$ , yielding a maximal answer of 38. Since? $38-25=13$ , which is prime, the answer is? $boxed{038}$ .
Rewrite the system of equations as? $frac{x^{2}}{t-1}+frac{y^{2}}{t-3^{2}}+frac{z^{2}}{t-5^{2}}+frac{w^{2}}{t-7^{2}}=1.$ ?This equation is satisfied when? $t = 4,16,36,64$ , as then the equation is equivalent to the given equations. After clearing fractions, for each of the values? $t=4,16,36,64$ , we have the?equation? $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ ? $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)$ . We can move the expression? $(t-1)(t-9)(t-25)(t-49)$ ?to the left hand side to obtain the difference of the polynomials:? $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ ? $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)$ ?and? $(t-1)(t-9)(t-25)(t-49)$ Since the polynomials are equal at? $t=4,16,36,64$ , we can express the difference of the two polynomials with a quartic polynomial that has roots at? $t=4,16,36,64$ , so
$x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ ? $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64)$

Note the leading coefficient of the RHS is? $-1$ ?because it must match the leading coefficient of the LHS, which is? $-1$ .

Now we can plug in? $t=1$ ?into the polynomial equation. Most terms drop, and we end up with

$[x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)]$

so that

$[x^2=frac{3cdot 15cdot 35cdot 63}{8cdot 24cdot 48}=frac{3^2cdot 5^2cdot 7^2}{2^{10}}]$

Similarly, we can plug in? $t=9,25,49$ ?and get

$begin{align*} y^2&=frac{5cdot 7cdot 27cdot 55}{8cdot 16cdot 40}=frac{3^3cdot 5cdot 7cdot 11}{2^{10}} z^2&=frac{21cdot 9cdot 11cdot 39}{24cdot 16cdot 24}=frac{3^2cdot 7cdot 11cdot 13}{2^{10}} w^2&=frac{45cdot 33cdot 13cdot 15}{48cdot 40cdot 24}=frac{3^2cdot 5cdot 11cdot 13}{2^{10}}end{align*}$

Now adding them up,

$begin{align*}z^2+w^2&=frac{3^2cdot 11cdot 13(7+5)}{2^{10}}=frac{3^3cdot 11cdot 13}{2^8} x^2+y^2&=frac{3^2cdot 5cdot 7(5cdot 7+3cdot 11)}{2^{10}}=frac{3^2cdot 5cdot 7cdot 17}{2^8}end{align*}$

with a sum of

$[frac{3^2(3cdot 11cdot 13+5cdot 7cdot 17)}{2^8}=3^2cdot 4=boxed{036}.]$

/*Lengthy proof that any two cubic polynomials in? $t$ ?which are equal at 4 values of? $t$ ?are themselves equivalent: Let the two polynomials be? $A(t)$ ?and? $B(t)$ ?and let them be equal at? $t=a,b,c,d$ . Thus we have? $A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0$ . Also the polynomial? $A(t) - B(t)$ ?is cubic, but it equals 0 at 4 values of? $t$ . Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into? $(t-a)(t-b)(t-c)(t-d)($ some nonzero polynomial $)$ ?which would have a degree greater than or equal to 4, contradicting the statement that? $A(t) - B(t)$ ?is cubic. Because? $A(t) - B(t) = 0, A(t)$ ?and? $B(t)$ ?are equivalent and must be equal for all? $t$ .

Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes? $x^2,y^2,z^2,$ ?and? $w^2$ ?separately before adding them to obtain the final answer is appealing because it gives the individual values of? $x^2,y^2,z^2,$ ?and? $w^2$ ?which can be plugged into the given equations to check.