1990 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $\dfrac{\frac{x}{4}}{2}=\dfrac4{\frac{x}{2}},$ then $x=$

$\pm\frac12$

$\pm1$

$\pm2$

$\pm4$

$\pm8$

Concepts:complex fraction quadratic equation

Difficulty rating: 1090

Small Hint:

Simplify both complex fractions first

Big Hint:

After clearing denominators, solve the resulting equation in $x^2$

Solution:

The equation simplifies to $\frac{x}{8}=\frac8x.$ Since $x\ne0,$ multiplying by $8x$ gives $x^2=64,$ so $x=\pm8.$

Thus the correct answer is E.

2.

$\left(\dfrac14\right)^{-\frac{1}{4}}=$

$-16$

$-\sqrt2$

$-\frac1{16}$

$\frac1{256}$

$\sqrt2$

Concepts:fractional exponent negative exponent

Difficulty rating: 1000

Small Hint:

A negative exponent takes the reciprocal

Big Hint:

Rewrite $4$ as a power of $2$

Solution:

Taking the reciprocal gives $4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{1}{2}}=\sqrt2.$

Thus the correct answer is E.

3.

The consecutive angles of a trapezoid form an arithmetic sequence. If the smallest angle is $75^\circ,$ then the largest angle is

$95^\circ$

$100^\circ$

$105^\circ$

$110^\circ$

$115^\circ$

Concepts:quadrilateral arithmetic sequence angle sum

Difficulty rating: 1260

Small Hint:

Write the four angles as $75,75+d,75+2d,75+3d$

Big Hint:

Use the angle sum of a quadrilateral

Solution:

The four angles sum to $360^\circ,$ so $\begin{aligned} &75+(75+d)\\ &\quad +(75+2d)\\ &\quad +(75+3d)=360. \end{aligned}$ Thus $d=10,$ and the largest angle is $75+3(10)=105^\circ.$

Thus the correct answer is C.

4.

Let $ABCD$ be a parallelogram with $\angle ABC=120^\circ,$ $AB=16,$ and $BC=10.$ Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4.$ If $\overline{BE}$ intersects $\overline{AD}$ at $F,$ then $FD$ is closest to

$1$

$2$

$3$

$4$

$5$

Concepts:parallelogram similar triangles segment ratio

Difficulty rating: 1280

Small Hint:

Use that $\overline{AB}\parallel\overline{ED}$

Big Hint:

Triangles $ABF$ and $DEF$ are similar

Solution:

Because $AB\parallel ED,$ triangles $ABF$ and $DEF$ are similar. Hence $\frac{AF}{FD}=\frac{AB}{DE}=\frac{16}{4}=4.$ Since $AF+FD=AD=BC=10,$ we get $5FD=10,$ so $FD=2.$

Thus the correct answer is B.

5.

Which of these numbers is largest?

$\sqrt{\sqrt[3]{5\cdot6}}$

$\sqrt{6\sqrt[3]5}$

$\sqrt{5\sqrt[3]6}$

$\sqrt[3]{5\sqrt6}$

$\sqrt[3]{6\sqrt5}$

Concepts:radical exponent comparing expressions

Difficulty rating: 1360

Small Hint:

Every choice is positive, so raise each one to the sixth power

Big Hint:

Express each sixth power as $5^a6^b$

Solution:

Raising the five positive choices to the sixth power preserves their order and gives, respectively, $\begin{aligned} 30,\qquad &6^3\cdot5=1080,\\ 5^3\cdot6&=750,\\ 5^2\cdot6&=150,\qquad 6^2\cdot5=180. \end{aligned}$ The largest is the second.

Thus the correct answer is B.

6.

Points $A$ and $B$ are $5$ units apart. How many lines in a given plane containing $A$ and $B$ are $2$ units from $A$ and $3$ units from $B?$

$0$

$1$

$2$

$3$

more than $3$

Concepts:distance from point to line tangent line circle

Difficulty rating: 1500

Small Hint:

Replace the distance conditions by tangencies to two circles

Big Hint:

The circles have radii $2$ and $3$ and are externally tangent

Solution:

A qualifying line is a common tangent to the circle centered at $A$ with radius $2$ and the circle centered at $B$ with radius $3.$ Their center distance equals the sum of their radii, so they are externally tangent. They have two external common tangents and one common tangent at their point of contact, for a total of $3.$

Thus the correct answer is D.

7.

A triangle with integral sides has perimeter $8.$ The area of the triangle is

$2\sqrt2$

$\frac{16}{9}\sqrt3$

$2\sqrt3$

$4$

$4\sqrt2$

Concepts:triangle inequality Heron’s formula integer sides

Difficulty rating: 1410

Small Hint:

List the unordered triples of positive integers summing to $8$

Big Hint:

Only one triple satisfies the strict triangle inequality

Solution:

The only unordered positive integral side lengths summing to $8$ and satisfying the triangle inequality are $2,3,3.$ Their semiperimeter is $4,$ so Heron’s formula gives $\begin{aligned} K&=\sqrt{4(4-2)(4-3)(4-3)}\\ &=2\sqrt2. \end{aligned}$

Thus the correct answer is A.

8.

The number of real solutions of the equation $|x-2|+|x-3|=1$ is

$0$

$1$

$2$

$3$

more than $3$

Concepts:absolute value interval

Difficulty rating: 1150

Small Hint:

Interpret the two absolute values as distances on a number line

Big Hint:

Consider every $x$ between $2$ and $3$

Solution:

For every $x\in[2,3],$ $\begin{aligned} &|x-2|+|x-3|\\ &\quad=(x-2)+(3-x)\\ &=1. \end{aligned}$ Thus the entire interval $[2,3]$ consists of solutions, so there are more than $3.$

Thus the correct answer is E.

9.

Each edge of a cube is colored either red or black. Every face of the cube has at least one black edge. The smallest possible number of black edges is

$2$

$3$

$4$

$5$

$6$

Concepts:cube covering argument construction

Difficulty rating: 1450

Small Hint:

Each edge belongs to exactly two faces

Big Hint:

For the matching upper bound, choose three mutually nonadjacent edges

Solution:

Each black edge can cover only its two incident faces, so covering all $6$ faces requires at least $3$ black edges. Choose edges incident to the face-pairs top/front, bottom/left, and back/right. These three edges cover all six faces, so the minimum is $3.$

Thus the correct answer is B.

10.

An $11\times11\times11$ wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point?

$328$

$329$

$330$

$331$

$332$

Concepts:cube inclusion-exclusion counting

Difficulty rating: 1510

Small Hint:

At most three faces of a cube are visible from one point

Big Hint:

Use inclusion-exclusion on three mutually adjacent $11\times11$ faces

Solution:

From a suitable point one can see three mutually adjacent faces. They contain $\begin{aligned} &3(11^2)-3(11)+1\\ &=363-33+1\\ &=331 \end{aligned}$ distinct unit cubes: subtract the three shared edges and restore the corner cube.

Thus the correct answer is D.

11.

How many positive integers less than $50$ have an odd number of positive integer divisors?

$3$

$5$

$7$

$9$

$11$

Concepts:divisors perfect square

Difficulty rating: 1140

Small Hint:

Divisors normally pair as $d$ and $\frac{n}{d}$

Big Hint:

An unpaired divisor occurs exactly for a perfect square

Solution:

A positive integer has an odd number of divisors exactly when it is a perfect square. The squares below $50$ are $1,4,9,16,25,36,49,$ so there are $7.$

Thus the correct answer is C.

12.

Let $f$ be the function defined by $f(x)=ax^2-\sqrt2$ for some positive $a.$ If $f(f(\sqrt2))=-\sqrt2,$ then $a=$

$\frac{2-\sqrt2}{2}$

$\frac12$

$2-\sqrt2$

$\frac{\sqrt2}{2}$

$\frac{2+\sqrt2}{2}$

Concepts:function composition radical equation

Difficulty rating: 1340

Small Hint:

For positive $a,$ when can $f(y)$ equal $-\sqrt2?$

Big Hint:

Set the inner value $f(\sqrt2)$ equal to zero

Solution:

Because $a\gt0,$ the equation $f(y)=ay^2-\sqrt2=-\sqrt2$ forces $y=0.$ Hence $f(\sqrt2)=2a-\sqrt2=0,$ so $a=\frac{\sqrt2}{2}.$

Thus the correct answer is D.

13.

If the following instructions are carried out by a computer, which value of $X$ will be printed because of instruction $5?$

$1.$ START $X$ AT $3$ AND $S$ AT $0.$

$2.$ INCREASE THE VALUE OF $X$ BY $2.$

$3.$ INCREASE THE VALUE OF $S$ BY THE VALUE OF $X.$

$4.$ IF $S$ IS AT LEAST $10000,$ THEN GO TO INSTRUCTION $5;$ OTHERWISE, GO TO INSTRUCTION $2$ AND PROCEED FROM THERE.

$5.$ PRINT THE VALUE OF $X.$

$6.$ STOP.

$19$

$21$

$23$

$199$

$201$

Concepts:algorithm arithmetic sequence summation

Difficulty rating: 1630

Small Hint:

After $k$ passes through instructions $2$ and $3,$ find $X$ and $S$

Big Hint:

The values added to $S$ are consecutive odd numbers beginning with $5$

Solution:

After $k$ passes, $X=2k+3,$ and $\begin{aligned} S&=5+7+\cdots+(2k+3)\\ &=k(k+4). \end{aligned}$ Now $98(102)=9996\lt10000,$ while $99(103)=10197\ge10000.$ Thus the loop stops at $k=99,$ when $X=2(99)+3=201.$

Thus the correct answer is E.

14.

An acute isosceles triangle, $ABC,$ is inscribed in a circle. Through $B$ and $C,$ tangents to the circle are drawn, meeting at point $D.$ If $\angle ABC=\angle ACB=2\angle D$ and $x$ is the radian measure of $\angle A,$ then $x=$

$\frac37\pi$

$\frac49\pi$

$\frac5{11}\pi$

$\frac6{13}\pi$

$\frac7{15}\pi$

Concepts:circle tangent inscribed angle angle chasing

Difficulty rating: 1910

Small Hint:

Express each base angle in terms of $x$

Big Hint:

The angle between the tangents is $\pi-2x$

Solution:

The two base angles are each $\frac{\pi-x}{2}.$ The minor arc $BC$ has central angle $2x,$ so the angle between the tangents is $\pi-2x.$ The given relation yields $\frac{\pi-x}{2}=2(\pi-2x).$ Therefore $7x=3\pi,$ or $x=\frac{3\pi}{7}.$

Thus the correct answer is A.

15.

Four whole numbers, when added three at a time, give the sums $180,$ $197,$ $208,$ and $222.$ What is the largest of the four numbers?

$77$

$83$

$89$

$95$

cannot be determined from the given information

Concepts:system of equations sums

Difficulty rating: 1360

Small Hint:

Add the four given triple-sums

Big Hint:

Each original number appears in exactly three of those sums

Solution:

If $T$ is the sum of the four numbers, adding the four triple-sums gives $\begin{aligned} 3T&=180+197+208+222\\ &=807, \end{aligned}$ so $T=269.$ The omitted numbers are $269-180,$ $269-197,$ $269-208,$ and $269-222,$ whose largest is $89.$

Thus the correct answer is C.

16.

At one of George Washington’s parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If $13$ married couples attended, how many handshakes were there among these $26$ people?

$78$

$185$

$234$

$312$

$325$

Concepts:combinations complementary counting

Difficulty rating: 1440

Small Hint:

Begin with all unordered pairs of the $26$ guests

Big Hint:

Remove spouse pairs and pairs consisting of two women

Solution:

There are $\binom{26}{2}=325$ possible pairs. Exclude the $13$ married pairs and the $\binom{13}{2}=78$ pairs of women. The number of handshakes is $325-13-78=234.$

Thus the correct answer is C.

17.

How many of the numbers $100,$ $101,$ $\ldots,$ $999$ have three different digits in increasing order or in decreasing order?

$120$

$168$

$204$

$216$

$240$

Concepts:combinations digit counting

Difficulty rating: 1700

Small Hint:

Choosing three distinct digits fixes their increasing or decreasing order

Big Hint:

Treat the digit $0$ carefully in the increasing case

Solution:

An increasing three-digit number cannot use $0,$ so there are $\binom93=84$ increasing numbers. Any three digits chosen from $0$ through $9$ form a valid decreasing three-digit number, because the largest digit comes first; this gives $\binom{10}{3}=120.$ The total is $84+120=204.$

Thus the correct answer is C.

18.

First $a$ is chosen at random from the set $\{1,2,3,\ldots,99,100\},$ and then $b$ is chosen at random from the same set. The probability that the integer $3^a+7^b$ has units digit $8$ is

$\frac1{16}$

$\frac18$

$\frac3{16}$

$\frac15$

$\frac14$

Concepts:modular arithmetic probability exponent cycle

Difficulty rating: 2060

Small Hint:

The units digits of both powers repeat with period $4$

Big Hint:

List the residue pairs $(a\bmod4,b\bmod4)$ that produce a units digit of $8$

Solution:

The units digits of $3^a$ for $a\equiv1,2,3,0\pmod4$ are $3,9,7,1,$ while those of $7^b$ are $7,9,3,1.$ A sum ending in $8$ occurs for the residue pairs $\begin{aligned} (a,b)\equiv{}&(2,2),(3,0),\\ &(0,1)\pmod4. \end{aligned}$ Each residue occurs $25$ times among $1,\ldots,100,$ so the probability is $\frac{3}{16}.$

Thus the correct answer is C.

19.

For how many integers $N$ between $1$ and $1990$ is the improper fraction $\frac{N^2+7}{N+4}$ not in lowest terms?

$0$

$86$

$90$

$104$

$105$

Concepts:greatest common divisor modular arithmetic counting integers in a range

Difficulty rating: 1900

Small Hint:

Reduce $N^2+7$ modulo $N+4$

Big Hint:

The only possible common prime factor is a divisor of $23$

Solution:

Modulo $N+4,$ we have $N\equiv-4,$ so $N^2+7\equiv16+7=23.$ Thus the fraction is reducible exactly when $N+4$ is divisible by $23,$ or $N\equiv19\pmod{23}.$ The values are $19,42,\ldots,1974,$ a total of $86.$

Thus the correct answer is B.

20.

In the figure, $ABCD$ is a quadrilateral with right angles at $A$ and $C.$ Points $E$ and $F$ are on $\overline{AC},$ and $\overline{DE}$ and $\overline{BF}$ are perpendicular to $\overline{AC}.$ If $AE=3,$ $DE=5,$ and $CE=7,$ then $BF=$

$3.6$

$4$

$4.2$

$4.5$

$5$

Concepts:coordinate geometry perpendicular lines system of equations

Difficulty rating: 1910

Small Hint:

Place $\overline{AC}$ on the $x$ -axis with $A=(0,0)$

Big Hint:

Use dot products for the right angles at $A$ and $C$

Solution:

Set $A=(0,0),$ $E=(3,0),$ $C=(10,0),$ $D=(3,-5),$ and $B=(f,h),$ where $h=BF.$ Since $AB\perp AD,$ $(f,h)\cdot(3,-5)=0,$ so $3f=5h.$ Since $BC\perp DC,$ $(f-10,h)\cdot(-7,-5)=0,$ so $7f+5h=70.$ Substitution gives $f=7$ and $h=\frac{21}{5}=4.2.$

Thus the correct answer is C.

21.

Consider a pyramid $P\text{-}ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,$ $B,$ $C,$ and $D.$ If $AB=1$ and $\angle APB=2\theta,$ then the volume of the pyramid is

$\frac{\sin\theta}{6}$

$\frac{\cot\theta}{6}$

$\frac1{6\sin\theta}$

$\frac{1-\sin2\theta}{6}$

$\frac{\sqrt{\cos2\theta}}{6\sin\theta}$

Concepts:pyramid trigonometry circumradius volume

Difficulty rating: 2380

Small Hint:

In isosceles triangle $APB,$ express $PA$ using the chord $AB$

Big Hint:

Relate $PA$ to the pyramid height and the center-to-vertex distance of the square

Solution:

Let $R=PA=PB.$ In isosceles triangle $APB,$ $1=AB=2R\sin\theta,$ so $R=\frac{1}{2\sin\theta}.$ If $h$ is the pyramid height, the horizontal distance from the square’s center to $A$ is $\frac{1}{\sqrt2},$ hence $\begin{aligned} h^2 &=R^2-\frac12\\ &=\frac{\cos2\theta}{4\sin^2\theta}. \end{aligned}$ Thus $h=\frac{\sqrt{\cos2\theta}}{2\sin\theta},$ and the volume is $\frac13(1)h=\frac{\sqrt{\cos2\theta}}{6\sin\theta}.$

Thus the correct answer is E.

22.

If the six solutions of $x^6=-64$ are written in the form $a+bi,$ where $a$ and $b$ are real, then the product of those solutions with $a\gt0$ is

$-2$

$0$

$2i$

$4$

$16$

Concepts:complex roots polar form

Difficulty rating: 2230

Small Hint:

Write $-64$ in polar form and list its six sixth roots

Big Hint:

The roots with positive real part form a conjugate pair

Solution:

The roots have modulus $2$ and arguments $\frac\pi6+\frac{k\pi}{3},\qquad k=0,1,\ldots,5.$ The roots with positive real part have arguments $\frac{\pi}{6}$ and $\frac{11\pi}{6}.$ They are conjugates of modulus $2,$ so their product is $2^2=4.$

Thus the correct answer is D.

23.

If $x,$ $y\gt0,$ $\log_yx+\log_xy=\frac{10}{3},$ and $xy=144,$ then $\frac{x+y}{2}=$

$12\sqrt2$

$13\sqrt3$

$24$

$30$

$36$

Concepts:logarithm quadratic equation symmetric expression

Difficulty rating: 2130

Small Hint:

Let $t=\log_yx,$ so that $\log_xy=\frac{1}{t}$

Big Hint:

The resulting values of $t$ show that one of $x,y$ is the cube of the other

Solution:

Let $t=\log_yx.$ Then $t+\frac1t=\frac{10}{3},$ so $3t^2-10t+3=0,$ giving $t=3$ or $\frac{1}{3}.$ Thus one of $x,y$ is the cube of the other. Let the smaller be $u.$ Then $u^4=xy=144,$ so $u=2\sqrt3$ and $u^3=24\sqrt3.$ Therefore $\frac{x+y}{2}=\frac{2\sqrt3+24\sqrt3}{2}=13\sqrt3.$

Thus the correct answer is B.

24.

All students at Adams High School and at Baker High School take a certain exam. The average scores for boys, for girls, and for boys and girls combined, at Adams HS and Baker HS are shown in the table, as is the average for boys at the two schools combined. What is the average score for the girls at the two schools combined?

	Adams	Baker	Adams & Baker
Boys:	$71$	$81$	$79$
Girls:	$76$	$90$	?
Boys & Girls:	$74$	$84$

$81$

$82$

$83$

$84$

$85$

Concepts:weighted average ratio system of equations

Difficulty rating: 1720

Small Hint:

Use each school’s combined average to find its ratio of girls to boys

Big Hint:

Use the combined boys’ average to relate the numbers of boys at the two schools

Solution:

Let Adams have $x$ boys and $y$ girls. From its average, $71x+76y=74(x+y),$ so $y=\frac{3x}{2}.$ If Baker has $u$ boys and $v$ girls, its average gives $u=2v.$ The combined boys’ average gives $71x+81u=79(x+u),$ so $u=4x$ and $v=2x.$ Hence the combined girls’ average is $\frac{76(\frac{3x}{2})+90(2x)}{\frac{3x}{2}+2x}=84.$

Thus the correct answer is D.

25.

Nine congruent spheres are packed inside a unit cube in such a way that one of them has its center at the center of the cube and each of the others is tangent to the center sphere and to three faces of the cube. What is the radius of each sphere?

$1-\frac{\sqrt3}{2}$

$\frac{2\sqrt3-3}{2}$

$\frac{\sqrt2}{6}$

$\frac14$

$\frac{\sqrt3(2-\sqrt2)}4$

Concepts:three-dimensional geometry distance formula tangent spheres

Difficulty rating: 2260

Small Hint:

Place one corner sphere’s center at $(r,r,r)$

Big Hint:

Its distance from $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ equals $2r$

Solution:

If the radius is $r,$ a corner sphere has center $(r,r,r)$ and the central sphere has center $(\frac{1}{2},\frac{1}{2},\frac{1}{2}).$ Tangency gives $\sqrt3\left(\frac12-r\right)=2r.$ Solving and rationalizing, $r=\frac{\sqrt3}{2(2+\sqrt3)} =\frac{2\sqrt3-3}{2}.$

Thus the correct answer is B.

26.

Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to him in the circle. Then each person computes and announces the average of the numbers of his two neighbors. The figure shows the average announced by each person (not the original number the person picked). The number picked by the person who announced the average $6$ was

$1$

$5$

$6$

$10$

not uniquely determined from the given information

Concepts:recurrence linear system cyclic arrangement

Difficulty rating: 2500

Small Hint:

If $x_i$ is a picked number and $a_i$ the displayed average, then $x_{i-1}+x_{i+1}=2a_i$

Big Hint:

Start with two unknown adjacent picked numbers and propagate around the circle

Solution:

Index the displayed averages $a_0,a_1,\ldots,a_9$ clockwise as $1,2,\ldots,10,$ and let $x_i$ be the corresponding picked numbers. Write $x_0=t,\ x_1=u.$ From $x_{i-1}+x_{i+1}=2a_i,$ successive values are $\begin{aligned} x_2&=4-t, &x_3&=6-u,\\ x_4&=4+t, &x_5&=4+u,\\ x_6&=8-t, &x_7&=10-u,\\ x_8&=8+t, &x_9&=8+u. \end{aligned}$ The two closing equations give $t=6$ and $u=-3.$ Hence $x_5=4+u=1.$

Thus the correct answer is A.

27.

Which of these triples could not be the lengths of the three altitudes of a triangle?

$1,$ $\sqrt3,$ $2$

$3,$ $4,$ $5$

$5,$ $12,$ $13$

$7,$ $8,$ $\sqrt{113}$

$8,$ $15,$ $17$

Concepts:triangle inequality altitude reciprocal

Difficulty rating: 2380

Small Hint:

For fixed area $K,$ a side corresponding to altitude $h$ equals $\frac{2K}{h}$

Big Hint:

Test the triangle inequality on the reciprocals of each triple

Solution:

If the altitudes are $h_1,h_2,h_3,$ then the corresponding sides are proportional to $\frac{1}{h_1},\frac{1}{h_2},\frac{1}{h_3}.$ For $5,12,13,$ $\frac15\gt\frac1{12}+\frac1{13},$ so the reciprocals fail the triangle inequality. Direct checking shows that the reciprocals of each other listed triple satisfy all strict triangle inequalities.

Thus the correct answer is C.

28.

A quadrilateral that has consecutive sides of lengths $70,$ $90,$ $130,$ and $110$ is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of lengths $x$ and $y.$ Find $|x-y|.$

$12$

$13$

$14$

$15$

$16$

Concepts:tangential quadrilateral cyclic quadrilateral tangent lengths

Difficulty rating: 2760

Small Hint:

Assign one tangent length to each vertex, so adjacent pairs sum to the four side lengths

Big Hint:

For supplementary opposite angles, the products of the tangent lengths at opposite vertices are equal

Solution:

Let the tangent lengths from the four consecutive vertices be $u,v,w,z.$ Then $\begin{aligned} u+v&=70, &v+w&=90,\\ w+z&=130, &z+u&=110. \end{aligned}$ Thus $v=70-u,$ $w=20+u,$ and $z=110-u.$ If the inradius is $r,$ a vertex with angle $A$ has tangent length $r\cot(\frac{A}{2}).$ Opposite angles are supplementary, so $uw=vz.$ Therefore $u(20+u)=(70-u)(110-u),$ giving $u=38.5.$ Hence the two segments of the $130$ -side are $w=58.5$ and $z=71.5,$ whose difference is $13.$

Thus the correct answer is B.

29.

A subset of the integers $1,$ $2,$ $\ldots,$ $100$ has the property that none of its members is $3$ times another. What is the largest number of members such a subset can have?

$50$

$66$

$67$

$76$

$78$

Concepts:extremal combinatorics powers independent set

Difficulty rating: 2340

Small Hint:

Group integers into chains $m,3m,9m,\ldots$ where $m$ is not a multiple of $3$

Big Hint:

Within each chain, alternating entries give a largest allowed selection

Solution:

Partition the integers into chains $m,3m,9m,\ldots,$ with $m$ not a multiple of $3.$ In each chain, no two adjacent terms may both be selected, so a maximum selection takes alternating terms starting with $m.$ Equivalently, select the integers whose exponent of $3$ is even. There are $\begin{aligned} &\left(100-\left\lfloor\frac{100}{3}\right\rfloor\right)\\ &\quad+\left(\left\lfloor\frac{100}{9}\right\rfloor -\left\lfloor\frac{100}{27}\right\rfloor\right)\\ &\quad+\left\lfloor\frac{100}{81}\right\rfloor\\ &=67+8+1=76. \end{aligned}$ The chain argument also proves no larger selection is possible.

Thus the correct answer is D.

30.

If $R_n=\frac12(a^n+b^n),$ where $a=3+2\sqrt2,$ $b=3-2\sqrt2,$ and $n=0,$ $1,$ $2,$ $\ldots,$ then $R_{12345}$ is an integer. Its units digit is

$1$

$3$

$5$

$7$

$9$

Concepts:recurrence modular arithmetic conjugates

Difficulty rating: 2260

Small Hint:

Use $a+b=6$ and $ab=1$ to obtain a recurrence for $R_n$

Big Hint:

Compute the recurrence modulo $10$ and look for a short period

Solution:

Because $a,b$ are roots of $t^2-6t+1=0,$ $R_n=6R_{n-1}-R_{n-2}.$ Starting with $R_0=1,\ R_1=3,$ the units digits are $1,3,7,9,7,3,1,\ldots,$ with period $6.$ Since $12345\equiv3\pmod6,$ the units digit is the same as that of $R_3,$ namely $9.$

Thus the correct answer is E.