1985 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $2x+1=8,$ then $4x+1=$

$15$

$16$

$17$

$18$

$19$

Concepts:linear equation substitution

Difficulty rating: 840

Small Hint:

First solve the given equation for $x$

Big Hint:

Substitute $x=\frac{7}{2}$ into $4x+1$

Solution:

The given equation gives $2x=7,$ so $x=\frac{7}{2}.$ Therefore $4x+1=4(\frac{7}{2})+1=15.$

Thus the correct answer is A.

2.

In an arcade game, the “monster” is the shaded sector of a circle of radius $1$ cm, as shown in the figure. The missing piece (the mouth) has central angle $60^\circ.$ What is the perimeter of the monster in cm?

$\pi+2$

$2\pi$

$\frac53\pi$

$\frac56\pi+2$

$\frac53\pi+2$

Concepts:arc circle circumference perimeter

Difficulty rating: 1000

Small Hint:

The remaining circular arc has central angle $300^\circ$

Big Hint:

Add the length of that arc to the two exposed radii

Solution:

The curved part is $\frac{300^\circ}{360^\circ}=\frac{5}{6}$ of a unit circle, so its length is $(\frac{5}{6})(2\pi)=\frac{5\pi}{3}.$ The two straight sides of the mouth are radii of total length $2.$ Hence the perimeter is $\frac{5\pi}{3}+2.$

Thus the correct answer is E.

3.

In right $\triangle ABC$ with legs $5$ and $12,$ arcs of circles are drawn, one with center $A$ and radius $12,$ the other with center $B$ and radius $5.$ They intersect the hypotenuse in $M$ and $N.$ Then $MN$ has length

$2$

$\frac{13}{5}$

$3$

$4$

$\frac{24}{5}$

Concepts:circle distance formula Pythagorean Theorem

Difficulty rating: 1330

Small Hint:

First find the hypotenuse of the $5$ - $12$ -right triangle

Big Hint:

Measure the positions of both arc intersections from $A$

Solution:

The hypotenuse has length $13.$ Since $AM=12,$ point $M$ is $12$ units from $A.$ Also $BN=5,$ so $AN=AB-BN=13-5=8.$ Therefore $MN=AM-AN$ $=12-8=4.$

Thus the correct answer is D.

4.

A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is

$\$306$

$\$333$

$\$342$

$\$348$

$\$360$

Concepts:divisibility money ratio and proportion

Difficulty rating: 1270

Small Hint:

Let the number of pennies be $p$

Big Hint:

Express the total value in cents as a multiple of $p$

Solution:

If there are $p$ pennies, there are $2p$ dimes and $6p$ quarters. Their total value is $p+10(2p)+25(6p)=171p$ cents. Of the listed dollar amounts, $34200=171\cdot200$ cents is possible.

Thus the correct answer is C.

5.

Which terms must be removed from the sum $\frac12+\frac14+\frac16+\frac18+\frac1{10}+\frac1{12}$ if the sum of the remaining terms is to equal $1?$

$\frac14$ and $\frac18$

$\frac14$ and $\frac1{12}$

$\frac18$ and $\frac1{12}$

$\frac16$ and $\frac1{10}$

$\frac18$ and $\frac1{10}$

Concepts:fraction whole number operations

Difficulty rating: 1470

Small Hint:

Compute how much larger the full sum is than $1$

Big Hint:

Use a common denominator of $120$ to compare the candidate pairs

Solution:

The full sum is $\begin{aligned} \frac{\substack{60+30+20\\{}+15+12+10}}{120} &=\frac{147}{120}. \end{aligned}$ The removed terms must therefore total $\frac{27}{120}=\frac{9}{40}.$ Since $\frac{1}{8}+\frac{1}{10}$ $=\frac{5}{40}+\frac{4}{40}$ $=\frac{9}{40},$ those are the required terms.

Thus the correct answer is E.

6.

One student in a class of boys and girls is to be chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is $\frac23$ of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is

$\frac13$

$\frac25$

$\frac12$

$\frac35$

$\frac23$

Concepts:basic probability ratio and proportion

Difficulty rating: 1100

Small Hint:

Equal likelihood makes the selection probabilities proportional to the class counts

Big Hint:

A boys-to-girls ratio of $2:3$ uses five equal parts in all

Solution:

Because every student is equally likely to be selected, the number of boys is $\frac{2}{3}$ of the number of girls. Thus the boys-to-girls ratio is $2:3,$ and boys make up $\frac{2}{2+3}=\frac{2}{5}$ of the class.

Thus the correct answer is B.

7.

In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from right to left. Thus, $a\times b-c$ in such languages means the same as $a(b-c)$ in ordinary algebraic notation. If $a\div b-c+d$ is evaluated in such a language, the result in ordinary algebraic notation would be

$\frac ab-c+d$

$\frac ab-c-d$

$\frac{d+c-b}{a}$

$\frac{a}{b-c+d}$

$\frac{a}{b-c-d}$

Concepts:algebraic manipulation order of operations

Difficulty rating: 1360

Small Hint:

Begin with the rightmost operation, $c+d$

Big Hint:

Next subtract that result from $b,$ and only then divide $a$

Solution:

Grouping from the right first gives $c+d.$ Subtracting this result from $b$ gives $b-(c+d)=b-c-d.$ Finally, dividing $a$ by that quantity gives $\frac{a}{b-c-d}.$

Thus the correct answer is E.

8.

Let $a,$ $a',$ $b,$ $b'$ be real numbers with $a$ and $a'$ nonzero. The solution to $ax+b=0$ is less than the solution to $a'x+b'=0$ if and only if

$a'b\lt ab'$

$ab'\lt a'b$

$ab\lt a'b'$

$\frac ba\lt\frac{b'}{a'}$

$\frac{b'}{a'}\lt\frac ba$

Concepts:inequality linear equation

Difficulty rating: 1360

Small Hint:

Write each equation’s solution without cross-multiplying

Big Hint:

Multiplying an inequality by $-1$ reverses its direction

Solution:

The two solutions are $-\frac{b}{a}$ and $-\frac{b'}{a'}.$ Thus the given comparison is $-\frac ba\lt-\frac{b'}{a'}.$ Multiplying both sides by $-1$ reverses the inequality and gives $\frac{b}{a}\gt \frac{b'}{a'},$ equivalently $\frac{b'}{a'}\lt \frac{b}{a}.$ No multiplication by the unknown-sign quantity $aa'$ is valid.

Thus the correct answer is E.

9.

The odd positive integers, $1,$ $3,$ $5,$ $7,$ $\ldots,$ are arranged in five columns continuing with the pattern shown. Counting from the left, the column in which $1985$ appears is the $\begin{array}{ccccc} &1&3&5&7\\ 15&13&11&9&\\ &17&19&21&23\\ 31&29&27&25&\\ &33&35&37&39\\ 47&45&43&41&\\ &49&51&53&55 \end{array}$

first

second

third

fourth

fifth

Concepts:arithmetic sequence modular arithmetic pattern recognition

Difficulty rating: 1310

Small Hint:

Look at the entries that begin each left-to-right row

Big Hint:

Relate $1985$ to a nearby multiple of $16$

Solution:

The last number in each right-to-left row is $16k-1,$ and the next row begins with $16k+1$ in the second column. Since $1985=16\cdot124+1,$ it begins such a row and lies in the second column.

Thus the correct answer is B.

10.

An arbitrary circle can intersect the graph of $y=\sin x$ in

at most $2$ points

at most $4$ points

at most $6$ points

at most $8$ points

more than $16$ points

Concepts:circle counting intersections trigonometry

Difficulty rating: 2240

Small Hint:

Consider a circle tangent to the $x$ -axis at the origin

Big Hint:

A very large radius keeps the lower arc close to the axis across many sine waves

Solution:

Take a circle tangent to the $x$ -axis at the origin with its center high above the axis. As its radius grows, its lower arc stays positive but arbitrarily close to the $x$ -axis over an arbitrarily long interval. On each positive arch of $y=\sin x,$ the sine curve is $0$ at the endpoints and rises well above this circular arc in between, producing two crossings. Choosing a sufficiently large radius therefore produces more than $16$ intersections.

Thus the correct answer is E.

11.

How many distinguishable rearrangements of the letters in CONTEST have both the vowels first? (For instance, OETCNST is one such arrangement, but OTETSNC is not.)

$60$

$120$

$240$

$720$

$2520$

Concepts:multiplication principle multiset permutations permutations

Difficulty rating: 1480

Small Hint:

Order the two vowels in the first two positions

Big Hint:

Permute the remaining five letters while accounting for the repeated $T$

Solution:

The vowels $O,E$ can be ordered first in $2!$ ways. The remaining letters $C,N,T,S,T$ can be arranged in $\frac{5!}{2!}=60$ distinguishable ways because the two $T$ ’s agree. The total is $2\cdot60=120.$

Thus the correct answer is B.

12.

Let $p,$ $q$ and $r$ be distinct prime numbers, where $1$ is not considered a prime. Which of the following is the smallest positive perfect cube having $n=pq^2r^4$ as a divisor?

$p^8q^8r^8$

$(pq^2r^2)^3$

$(p^2q^2r^2)^3$

$(pqr^2)^3$

$4p^3q^3r^3$

Concepts:divisibility perfect power prime factorization

Difficulty rating: 1660

Small Hint:

Every prime exponent in a perfect cube is a multiple of $3$

Big Hint:

Raise the exponents $1,$ $2,$ and $4$ to the least larger multiples of $3$

Solution:

A perfect cube has prime exponents divisible by $3.$ The least multiples of $3$ that are at least $1,$ $2,$ and $4$ are $3,$ $3,$ and $6,$ respectively. Thus the smallest possible cube is $p^3q^3r^6=(pqr^2)^3.$ Therefore the correct answer is D.

13.

Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is

$4$

$4.5$

$5$

$5.5$

$6$

Concepts:coordinate geometry lattice point shoelace formula

Difficulty rating: 1420

Small Hint:

Assign integer coordinates to the four corner pegs

Big Hint:

Use the shoelace formula on $(1,3),$ $(4,1),$ $(3,0),$ and $(0,1)$

Solution:

With the lower-left peg as $(0,0),$ the quadrilateral’s vertices in order are $(1,3),$ $(4,1),$ $(3,0),$ and $(0,1).$ The shoelace formula gives $\begin{aligned} &\frac12\left|(1+0+3+0)\right.\\ &\quad\left.{}-(12+3+0+1)\right|=6. \end{aligned}$ Therefore the correct answer is E.

14.

Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?

$4$

$5$

$6$

$7$

$8$

Concepts:angle sum extremal argument inequality

Difficulty rating: 1960

Small Hint:

Bound the three obtuse angles by $180^\circ$ and every other angle by $90^\circ$

Big Hint:

Compare that upper bound with the interior-angle sum $(n-2)180^\circ$

Solution:

For an $n$ -gon, the three obtuse angles have total less than $3\cdot180^\circ,$ while the other $n-3$ angles have total at most $(n-3)90^\circ.$ Hence $\begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ, \end{aligned}$ which simplifies to $n\lt7.$ Six sides are attainable, for example with interior angles $150^\circ,$ $150^\circ,$ $150^\circ,$ $90^\circ,$ $90^\circ,$ and $90^\circ.$ Thus the maximum is $6.$

Therefore the correct answer is C.

15.

If $a$ and $b$ are positive numbers such that $a^b=b^a$ and $b=9a,$ then the value of $a$ is

$9$

$\frac19$

$\sqrt[9]{9}$

$\sqrt[3]{9}$

$\sqrt[4]{3}$

Concepts:exponent radical substitution

Difficulty rating: 1960

Small Hint:

Substitute $b=9a$ into $a^b=b^a$

Big Hint:

Because $a\gt0,$ take the $a$ -th root and simplify

Solution:

Substitution gives $a^{9a}=(9a)^a.$ Taking the positive $a$ -th root yields $a^9=9a,$ so $a^8=9.$ Therefore $a=9^{\frac{1}{8}}=3^{\frac{1}{4}}=\sqrt[4]{3}.$

Thus the correct answer is E.

16.

If $A=20^\circ$ and $B=25^\circ,$ then the value of $(1+\tan A)(1+\tan B)$ is

$\sqrt3$

$2$

$1+\sqrt2$

$2(\tan A+\tan B)$

none of these

Concepts:trigonometric identity trigonometry

Difficulty rating: 1920

Small Hint:

Use $A+B=45^\circ$ in the tangent addition formula

Big Hint:

Express $\tan A+\tan B$ in terms of $\tan A\tan B$

Solution:

Because $\tan(A+B)=\tan45^\circ=1,$ $\frac{\tan A+\tan B}{1-\tan A\tan B}=1,$ so $\tan A+\tan B$ $=1-\tan A\tan B.$ Therefore $\begin{aligned} &(1+\tan A)(1+\tan B)\\ &\quad=1+(1-\tan A\tan B)\\ &\qquad+\tan A\tan B\\ &\quad=2. \end{aligned}$ Thus the correct answer is B.

17.

Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB.$ The area of $ABCD,$ rounded to one decimal place, is

$4.1$

$4.2$

$4.3$

$4.4$

$4.5$

Concepts:area diagonal rectangle vector

Difficulty rating: 2240

Small Hint:

Let the rectangle’s side lengths be $w,h;$ then its diagonal has length $3$

Big Hint:

Project the vertical and horizontal sides onto the diagonal to obtain $\frac{h^2}{3}=1$ and $\frac{w^2}{3}=2$

Solution:

Let $D=(0,0),$ $B=(w,h),$ $A=(0,h)$ and $C=(w,0).$ Since $DB=3,$ we have $w^2+h^2=9.$ The projection of $DA$ onto $DB$ has length $\frac{h^2}{3},$ and the first marked segment has length $1,$ so $h^2=3.$ Similarly, the projection of $DC$ reaches the second division point, so $\frac{w^2}{3}=2$ and $w^2=6.$ Hence the area is $wh=\sqrt{18}=3\sqrt2\approx4.2426,$ which rounds to $4.2.$

Therefore the correct answer is B.

18.

Six bags of marbles contain $18,$ $19,$ $21,$ $23,$ $25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?

$18$

$19$

$21$

$23$

$25$

Concepts:divisibility modular arithmetic partitions and compositions

Difficulty rating: 1790

Small Hint:

Add all six bag sizes, then remove the chipped bag

Big Hint:

Jane’s and George’s totals are in the ratio $2:1,$ so their combined total is divisible by $3$

Solution:

The six bags total $140.$ If the chipped bag contains $c$ marbles, then the other bags total three times George’s amount, so $140-c$ is divisible by $3.$ Of the choices, only $c=23$ has this property. It is attainable: George can take $18+21=39,$ while Jane takes $19+25+34=78.$

Thus the correct answer is D.

19.

Consider the graphs of $y=Ax^2$ and $y^2+3=x^2+4y,$ where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?

exactly $4$

exactly $2$

at least $1,$ but the number varies for different positive values of $A$

$0$ for at least one positive value of $A$

none of these

Concepts:algebraic manipulation counting intersections quadratic

Difficulty rating: 2350

Small Hint:

Use $x^2=\frac{y}{A}$ to eliminate $x^2$

Big Hint:

Show that the resulting quadratic has two distinct positive $y$ -roots

Solution:

Substituting $x^2=\frac{y}{A}$ into the second equation gives $Ay^2-(4A+1)y+3A=0.$ Its discriminant is $4A^2+8A+1\gt0.$ Its roots have positive product $3$ and positive sum $\frac{4A+1}{A},$ so both roots are positive and distinct. For each root $y,$ the equation $x^2=\frac{y}{A}$ gives two distinct values of $x.$ Thus there are exactly four intersection points.

Therefore the correct answer is A.

20.

A wooden cube with edge length $n$ units (where $n$ is an integer $\gt2$ ) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $n?$

$5$

$6$

$7$

$8$

none of these

Concepts:algebraic manipulation basic counting cube geometry

Difficulty rating: 1660

Small Hint:

Count the unpainted cubes in the interior cube of side length $n-2$

Big Hint:

On each face, exactly $(n-2)^2$ cubes avoid the edges

Solution:

There are $(n-2)^3$ completely unpainted cubes. Each of the six faces contributes $(n-2)^2$ cubes with exactly one painted face, so equality gives $6(n-2)^2=(n-2)^3.$ Since $n\gt2,$ division by $(n-2)^2$ gives $n-2=6,$ hence $n=8.$

Therefore the correct answer is D.

21.

How many integers $x$ satisfy the equation $(x^2-x-1)^{x+2}=1?$

$2$

$3$

$4$

$5$

none of these

Concepts:casework exponent quadratic

Difficulty rating: 2070

Small Hint:

An integer power can equal $1$ when the base is $1,$ when the base is $-1$ with even exponent, or when a nonzero base has exponent $0$

Big Hint:

Solve those three cases separately and check every candidate

Solution:

If the base is $1,$ then $x^2-x-1=1,$ giving $x=-1$ and $x=2.$ If the base is $-1,$ then $x(x-1)=0;$ only $x=0$ makes the exponent $x+2$ even. If the exponent is $0,$ then $x=-2,$ and its base is $5,$ so it also works. Thus the four solutions are $-2,$ $-1,$ $0,$ and $2.$

Therefore the correct answer is C.

22.

In a circle with center $O,$ $AD$ is a diameter, $ABC$ is a chord, $BO=5$ and $\angle ABO=\overset{\frown}{CD}=60^\circ.$ Then the length of $BC$ is

$3$

$3+\sqrt3$

$5-\frac{\sqrt3}{2}$

$5$

none of the above

Concepts:circle inscribed angle special right triangle

Difficulty rating: 2130

Small Hint:

The $60^\circ$ arc $CD$ makes $\angle CAD=30^\circ$

Big Hint:

Use the resulting $30^\circ$ - $60^\circ$ - $90^\circ$ triangles to compare $AB$ and $AC$

Solution:

Arc $CD$ gives $\angle CAD=30^\circ.$ Since $A,B,C$ are collinear and $A,O,D$ are collinear, $\angle BAO=30^\circ.$ With $\angle ABO=60^\circ,$ triangle $ABO$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle. Because $BO=5,$ we get $AB=10$ and $AO=5\sqrt3.$

Also $\angle ACD=90^\circ$ because $AD$ is a diameter. In the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle $ACD,$ the hypotenuse is $AD=10\sqrt3,$ so $AC=15.$ Therefore $BC=AC-AB=5.$

Thus the correct answer is D.

23.

If $\begin{aligned} x&=\frac{-1+i\sqrt3}{2},\\ y&=\frac{-1-i\sqrt3}{2}, \end{aligned}$ where $i^2=-1,$ then which of the following is not correct?

$x^5+y^5=-1$

$x^7+y^7=-1$

$x^9+y^9=-1$

$x^{11}+y^{11}=-1$

$x^{13}+y^{13}=-1$

Concepts:complex number modular arithmetic roots of unity

Difficulty rating: 2130

Small Hint:

Recognize $x$ and $y$ as the two nonreal cube roots of unity

Big Hint:

Their powers depend only on the exponent modulo $3$

Solution:

The numbers are the two nonreal cube roots of unity, so $x^n+y^n=2\cos\left(\frac{2\pi n}{3}\right).$ This equals $-1$ when $n$ is not divisible by $3,$ but equals $2$ when $n$ is divisible by $3.$ Of the listed exponents, only $9$ is divisible by $3,$ so its displayed equation is the one that is not correct.

Therefore the correct answer is C.

24.

A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}(d+1)-\log_{10}d.$ The probability that the digit $2$ is chosen is exactly $\frac{1}{2}$ the probability that the digit chosen is in the set

$\{2,3\}$

$\{3,4\}$

$\{4,5,6,7,8\}$

$\{5,6,7,8,9\}$

$\{4,5,6,7,8,9\}$

Concepts:basic probability logarithm telescoping

Difficulty rating: 1800

Small Hint:

Twice the probability of digit $2$ is $2\log_{10}(\frac{3}{2})$

Big Hint:

Sum the probabilities over a consecutive set by telescoping its logarithms

Solution:

The probability of $2$ is $\log_{10}(\frac{3}{2}),$ so twice that probability is $\log_{10}(\frac{9}{4}).$ For the digits $4$ through $8,$ the sum telescopes: $\begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94. \end{aligned}$ Therefore that set has exactly twice the probability of digit $2.$

Thus the correct answer is C.

25.

The volume of a certain rectangular solid is $8\text{ cm}^3,$ its total surface area is $32\text{ cm}^2,$ and its three dimensions are in geometric progression. The sum of the lengths in cm of all the edges of this solid is

$28$

$32$

$36$

$40$

$44$

Concepts:geometric sequence rectangular prism surface area volume

Difficulty rating: 1920

Small Hint:

Write the three dimensions as $\frac{t}{r},t,tr$

Big Hint:

Use the volume to find $t,$ then use the surface area to find $r+\frac{1}{r}$

Solution:

Write the dimensions as $\frac{2}{r},2,2r,$ since their product is $8.$ Half the surface area is the sum of the three pairwise products, so $\frac4r+4+4r=16,$ giving $r+\frac{1}{r}=3.$ The sum of the dimensions is $2(r+1+\frac{1}{r})=8.$ Since each dimension occurs on four edges, the sum of all edge lengths is $4\cdot8=32.$

Thus the correct answer is B.

26.

Find the least positive integer $n$ for which $\frac{n-13}{5n+6}$ is a non-zero reducible fraction.

$45$

$68$

$155$

$226$

none of these

Concepts:greatest common divisor modular arithmetic

Difficulty rating: 2130

Small Hint:

Any common divisor of $n-13$ and $5n+6$ also divides a suitable linear combination

Big Hint:

Compute $(5n+6)-5(n-13)$

Solution:

Euclid’s algorithm gives $\begin{aligned} &\gcd(n-13,5n+6)\\ &\quad=\gcd(n-13,71). \end{aligned}$ Since $71$ is prime, the nonzero fraction is reducible exactly when $n-13$ is a nonzero multiple of $71.$ The least positive possibility is $n=13+71=84,$ which is not among the four numerical choices.

Thus the correct answer is E.

27.

Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\ldots,$ defined by $\begin{aligned} x_1&=\sqrt[3]{3},\\ x_2&=(\sqrt[3]{3})^{\sqrt[3]{3}}. \end{aligned}$ and in general $x_n=(x_{n-1})^{\sqrt[3]{3}}$ for $n\gt1.$ What is the smallest value of $n$ for which $x_n$ is an integer?

$2$

$3$

$4$

$9$

$27$

Concepts:exponent inequality radical recursion

Difficulty rating: 2380

Small Hint:

Let $c=\sqrt[3]{3}$ and track the exponent of $3$

Big Hint:

Show that $x_n=3^{\frac{c^{\,n-1}}{3}}$ and examine the first four terms

Solution:

Let $c=\sqrt[3]{3}.$ Repeated application of the recurrence gives $x_n=3^{\frac{c^{\,n-1}}{3}},$ so $x_4=3^{\frac{c^3}{3}}=3.$ It remains to rule out the first three terms. The sequence is strictly increasing. Also $c\lt\frac{3}{2},$ so $x_2=c^c \lt\left(\frac32\right)^{\frac{3}{2}} \lt2.$ Finally $x_3=3^{\frac{1}{c}}.$ Since $2^c\lt2^{\frac{3}{2}}\lt3,$ we have $x_3\gt2,$ while $\frac{1}{c}\lt1$ gives $x_3\lt3.$ Thus $x_1,x_2,x_3$ are not integers, and the first integral term is $x_4.$

Therefore the correct answer is C.

28.

In $\triangle ABC,$ we have $\angle C=3\angle A,$ $a=27$ and $c=48.$ What is $b?$

$33$

$35$

$37$

$39$

not uniquely determined

Concepts:law of sines trigonometric identity

Difficulty rating: 2460

Small Hint:

Use the sine law to write $\frac{48}{27}=\frac{\sin3A}{\sin A}$

Big Hint:

After finding $\cos A,$ use $B=180^\circ-4A$ and $\sin4A=4\sin A\cos A\cos2A$

Solution:

By the sine law and the triple-angle identity, $\frac{48}{27} =\frac{\sin3A}{\sin A} =3-4\sin^2A.$ Hence $\sin^2A=\frac{11}{36}.$ Since $A\lt60^\circ,$ $\cos A=\frac{5}{6},$ and therefore $\cos2A=\frac{7}{18}.$ Also $B=180^\circ-4A,$ so another application of the sine law gives $\begin{aligned} b &=27\frac{\sin B}{\sin A}\\ &=27\frac{\sin4A}{\sin A}\\ &=27\cdot4\cos A\cos2A\\ &=35. \end{aligned}$ Thus the correct answer is B.

29.

In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab?$

$15880$

$17856$

$17865$

$17874$

$19851$

Concepts:algebraic manipulation digits pattern recognition place value

Difficulty rating: 2530

Small Hint:

Write a string of $k$ identical digits using $R=\frac{10^k-1}{9}$

Big Hint:

After removing the terminal zero, express the product as a string of $2k$ fours minus a string of $k$ eights

Solution:

Let $k=1985$ and $R=\frac{10^k-1}{9}.$ Then $a=8R,$ $b=5R,$ and $9ab$ ends in $0.$ Removing that zero does not change the digit sum and leaves $\begin{aligned} N=\frac{9ab}{10} &=36R^2\\ &=\frac49(10^{2k}-1)\\ &\quad{}-\frac89(10^k-1). \end{aligned}$ Thus $N$ is a string of $2k$ fours minus a string of $k$ eights. The subtraction produces $k-1$ fours, then $3,$ then $k-1$ fives, then $6.$ Its digit sum is $\begin{aligned} &4(k-1)+3+5(k-1)+6\\ &\quad=9k=17865. \end{aligned}$ Therefore the correct answer is C.

30.

Let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x.$ Then the number of real solutions to $4x^2-40\lfloor x\rfloor+51=0$ is

$0$

$1$

$2$

$3$

$4$

Concepts:casework floor and ceiling functions inequality quadratic

Difficulty rating: 2380

Small Hint:

Set $n=\lfloor x\rfloor$ and solve the equation for $x^2$

Big Hint:

Enforce $n\le x\lt n+1$ to determine which integer values of $n$ work

Solution:

Let $n=\lfloor x\rfloor.$ A solution must be positive and satisfy $\begin{aligned} x&=\sqrt{10n-\frac{51}{4}},\\ n&\le x\lt n+1. \end{aligned}$ The lower inequality is equivalent to $4n^2-40n+51\le0,$ so $n=2,3,\ldots,8.$ The upper inequality is equivalent to $4n^2-32n+55\gt0.$ Among $2,\ldots,8,$ this holds exactly for $n=2,6,7,8.$ Each interval contributes one root, so there are four real solutions.

Therefore the correct answer is E.