1980 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

The largest whole number such that seven times the number is less than $100$ is

$12$

$13$

$14$

$15$

$16$

Answer: C

Concepts:inequality whole number operations

Difficulty rating: 770

Small Hint:

Translate “seven times the number is less than $100$ ” into an inequality

Big Hint:

Locate $\frac{100}{7}$ between two consecutive whole numbers

Solution:

The number must be less than $\frac{100}{7}=14+\frac{2}{7}.$ The largest whole number below this value is $14.$

Therefore, the correct answer is C.

2.

The degree of $(x^2+1)^4(x^3+1)^3$ as a polynomial in $x$ is

$5$

$7$

$12$

$17$

$72$

Answer: D

Concepts:polynomial exponent

Difficulty rating: 1080

Small Hint:

Find the highest power contributed by each parenthesized factor

Big Hint:

Degrees add when nonzero polynomials are multiplied

Solution:

The first factor has degree $2\cdot4=8,$ and the second has degree $3\cdot3=9.$ Their product therefore has degree $8+9=17.$

Therefore, the correct answer is D.

3.

If the ratio of $2x-y$ to $x+y$ is $\frac23,$ what is the ratio of $x$ to $y?$

$\frac15$

$\frac45$

$1$

$\frac65$

$\frac54$

Answer: E

Concepts:ratio and proportion linear equation

Difficulty rating: 1100

Small Hint:

Write the stated ratio as an equation of two fractions

Big Hint:

Cross-multiply and collect the $x$ -terms and $y$ -terms separately

Solution:

The condition gives $\frac{2x-y}{x+y}=\frac23.$ Thus $6x-3y=2x+2y,$ so $4x=5y$ and $\frac{x}{y}=\frac{5}{4}.$

Therefore, the correct answer is E.

4.

In the adjoining figure, $CDE$ is an equilateral triangle and $ABCD$ and $DEFG$ are squares. The measure of $\angle GDA$ is

$90^\circ$

$105^\circ$

$120^\circ$

$135^\circ$

$150^\circ$

Answer: C

Concepts:square (geometry)equilateral triangle angle sum

Difficulty rating: 1310

Small Hint:

Identify the three known angles at $D$

Big Hint:

Use the full $360^\circ$ angle around point $D$

Solution:

At $D,$ the two square angles are $90^\circ$ each and the equilateral-triangle angle is $60^\circ.$ Hence $\begin{aligned} \angle GDA &=360^\circ-90^\circ\\ &\quad-60^\circ-90^\circ\\ &=120^\circ. \end{aligned}$

Therefore, the correct answer is C.

5.

If $AB$ and $CD$ are perpendicular diameters of circle $Q,$ $P$ in $AQ,$ and $\angle QPC=60^\circ,$ then the length of $PQ$ divided by the length of $AQ$ is

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{3}$

$\frac{\sqrt2}{2}$

$\frac12$

$\frac23$

Answer: B

Concepts:circle right triangle special right triangle

Difficulty rating: 1330

Small Hint:

Use the perpendicular diameters to identify a right angle in $\triangle PQC$

Big Hint:

Relate the short and long legs of the resulting $30^\circ$ - $60^\circ$ - $90^\circ$ triangle

Solution:

Triangle $PQC$ is right at $Q$ and has $\angle P=60^\circ,$ so it is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle. Since $QC=AQ,$ $\frac{PQ}{AQ}=\frac{PQ}{QC}=\frac1{\sqrt3} =\frac{\sqrt3}{3}.$

Therefore, the correct answer is B.

6.

A positive number $x$ satisfies the inequality $\sqrt x\lt2x$ if and only if

$x\gt\frac14$

$x\gt2$

$x\gt4$

$x\lt\frac14$

$x\lt4$

Answer: A

Concepts:inequality radical

Difficulty rating: 1180

Small Hint:

Use the fact that both sides are positive

Big Hint:

Divide by $\sqrt x$ before squaring or isolating $x$

Solution:

Because $x\gt0,$ division by $\sqrt x$ preserves the inequality: $1\lt2\sqrt x.$ Thus $\sqrt x\gt\frac{1}{2},$ which is equivalent to $x\gt\frac{1}{4}.$

Therefore, the correct answer is A.

7.

Sides $AB,$ $BC,$ $CD,$ and $DA$ of convex polygon $ABCD$ have lengths $3,$ $4,$ $12,$ and $13,$ respectively; and $\angle CBA$ is a right angle. The area of the quadrilateral is

$32$

$36$

$39$

$42$

$48$

Answer: B

Concepts:Pythagorean Theorem right triangle area decomposition

Difficulty rating: 1430

Small Hint:

Draw diagonal $AC$ and find its length from $\triangle ABC$

Big Hint:

Recognize a second right triangle using the side lengths $5,$ $12,$ and $13$

Solution:

Triangle $ABC$ is a $3$ - $4$ - $5$ right triangle, so $AC=5$ and its area is $6.$ Since $5^2+12^2=13^2,$ triangle $ACD$ is right at $C$ and has area $(\frac{1}{2})(5)(12)=30.$ The quadrilateral’s area is $6+30=36.$

Therefore, the correct answer is B.

8.

How many pairs $(a,b)$ of nonzero real numbers satisfy the equation $\frac1a+\frac1b=\frac1{a+b}?$

none

$1$

$2$

one pair for each $b\ne0$

two pairs for each $b\ne0$

Answer: A

Concepts:rational equation quadratic

Difficulty rating: 1590

Small Hint:

Clear the denominators, noting that $a,$ $b,$ and $a+b$ must be nonzero

Big Hint:

Treat the resulting homogeneous quadratic as an equation in the ratio $\frac{a}{b}$

Solution:

Clearing denominators gives $(a+b)^2=ab,$ or $a^2+ab+b^2=0.$ Since $b\ne0,$ let $t=\frac{a}{b}.$ Then $t^2+t+1=0,$ whose discriminant is $1-4=-3.$ Thus there are no real pairs.

Therefore, the correct answer is A.

9.

A man walks $x$ miles due west, turns $150^\circ$ to his left and walks $3$ miles in the new direction. If he finishes at a point $\sqrt3$ miles from his starting point, then $x$ is

$\sqrt3$

$2\sqrt3$

$\frac32$

$3$

not uniquely determined by the given information

Answer: E

Concepts:vector distance formula quadratic casework

Difficulty rating: 1590

Small Hint:

Resolve the second walk into horizontal and vertical components

Big Hint:

The distance equation is quadratic in $x$ ; check whether both positive roots are valid

Solution:

Take east as the positive horizontal direction. After the turn, the $3$ -mile displacement has components $(\frac{3\sqrt3}{2},-\frac{3}{2}).$ Therefore $\left(\frac{3\sqrt3}{2}-x\right)^2 +\left(\frac32\right)^2=3.$ Hence $\frac{3\sqrt3}{2}-x=\pm\frac{\sqrt3}{2},$ giving $x=\sqrt3$ or $x=2\sqrt3.$ The value is not unique.

Therefore, the correct answer is E.

10.

The number of teeth in three meshed circular gears $A,$ $B,$ $C$ are $x,$ $y,$ $z,$ respectively. (The teeth on all gears are the same size and regularly spaced as in the figure.) The angular speeds, in revolutions per minute, of $A,$ $B,$ $C$ are in the proportion

$x:y:z$

$z:y:x$

$y:z:x$

$yz:xz:xy$

$xz:yx:zy$

Answer: D

Concepts:ratio and proportion rate

Difficulty rating: 1380

Small Hint:

Meshed teeth have the same tangential speed at their points of contact

Big Hint:

Angular speed is inversely proportional to circumference and hence to the number of teeth

Solution:

The gears’ circumferences are proportional to $x,$ $y,$ and $z,$ while their tangential speeds have equal magnitude. Their angular speeds are therefore proportional to $\frac1x:\frac1y:\frac1z.$ Multiplying all three terms by $xyz$ gives $yz:xz:xy.$

Therefore, the correct answer is D.

11.

If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10,$ respectively, then the sum of the first $110$ terms is

$90$

$-90$

$110$

$-110$

$-100$

Answer: D

Concepts:arithmetic sequence summation system of equations

Difficulty rating: 1780

Small Hint:

Write $S_n=\frac n2(2a+(n-1)d)$

Big Hint:

Use the two known partial sums to find the expression $2a+109d$

Solution:

The two given sums yield $\begin{aligned} 2a+9d&=20,\\ 2a+99d&=\frac15. \end{aligned}$ Subtracting gives $90d=-\frac{99}{5},$ so $d=-\frac{11}{50}$ and $2a=\frac{1099}{50}.$ Therefore $\begin{aligned} S_{110}&=55(2a+109d)\\ &=55(-2)\\ &=-110. \end{aligned}$

Therefore, the correct answer is D.

12.

The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx,$ respectively. Suppose $L_1$ makes twice as large an angle with the horizontal (measured counterclockwise from the positive $x$ -axis) as does $L_2,$ and that $L_1$ has $4$ times the slope of $L_2.$ If $L_1$ is not horizontal, then $mn$ is

$\frac{\sqrt2}{2}$

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

$2$

$-2$

not uniquely determined by the given information

Answer: C

Concepts:slope trigonometric identity system of equations

Difficulty rating: 1980

Small Hint:

Let the angle of inclination of $L_2$ be $\theta$

Big Hint:

Substitute $m=4n$ into $\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}$

Solution:

Let $\tan\theta=n.$ Then $m=\tan(2\theta)$ and also $m=4n.$ Hence $4n=\frac{2n}{1-n^2}.$ The nonhorizontal condition gives $n\ne0,$ so $2(1-n^2)=1,$ or $n^2=\frac{1}{2}.$ Thus $mn=4n^2=2.$

Therefore, the correct answer is C.

13.

A bug (of negligible size) starts at the origin on the coordinate plane. First it moves $1$ unit right to $(1,0).$ Then it makes a $90^\circ$ turn counterclockwise and travels $\frac12$ a unit to $(1,\frac12).$ If it continues in this fashion, each time making a $90^\circ$ turn counterclockwise and traveling half as far as in the previous move, to which of the following points will it come closest?

$(\frac23,\frac23)$

$(\frac45,\frac25)$

$(\frac23,\frac45)$

$(\frac23,\frac13)$

$(\frac25,\frac45)$

Answer: B

Concepts:geometric sequence complex number summation

Difficulty rating: 1830

Small Hint:

Represent a $90^\circ$ counterclockwise turn by multiplication by $i$

Big Hint:

The displacement vectors form an infinite geometric series with ratio $\frac{i}{2}$

Solution:

As complex numbers, the successive displacement vectors are $1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots.$ Their sum is $\frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i.$ Thus the bug approaches $(\frac{4}{5},\frac{2}{5}).$

Therefore, the correct answer is B.

14.

If $c$ is a constant and the function $f$ defined by $\begin{gathered} f(x)=\frac{cx}{2x+3},\\ x\ne-\frac32, \end{gathered}$ satisfies $f(f(x))=x$ for all real numbers $x$ except $-\frac32,$ then $c$ is

$-3$

$-\frac32$

$\frac32$

$3$

not uniquely determined by the given information

Answer: A

Concepts:function rational equation algebraic manipulation

Difficulty rating: 1780

Small Hint:

Compute $f(f(x))$ as a single rational expression

Big Hint:

For the result to equal $x$ identically, compare the coefficient of $x$ in the denominator

Solution:

Direct composition gives $f(f(x)) =\frac{c^2x}{(2c+6)x+9}.$ For this to equal $x$ identically, the denominator must be the constant $c^2.$ Thus $2c+6=0$ and $c^2=9,$ both of which give $c=-3.$

Therefore, the correct answer is A.

15.

A store prices an item in dollars and cents so that when $4\%$ sales tax is added no rounding is necessary because the result is exactly $n$ dollars, where $n$ is a positive integer. The smallest value of $n$ is

$1$

$13$

$25$

$26$

$100$

Answer: B

Concepts:percentage divisibility money

Difficulty rating: 1860

Small Hint:

Let the untaxed price be an integer number $p$ of cents

Big Hint:

Reduce the equation $1.04p=100n$ to a divisibility condition on $n$

Solution:

Let the price be $p$ cents. Then $\frac{104}{100}p=100n,$ so $13p=1250n.$ Since $13$ and $1250$ are relatively prime, $13$ must divide $n.$ The smallest positive possibility is $n=13.$

Therefore, the correct answer is B.

16.

Four of the eight vertices of a cube are vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.

$\sqrt2$

$\sqrt3$

$\sqrt{\frac32}$

$\frac2{\sqrt3}$

$2$

Answer: B

Concepts:cube geometry surface area equilateral triangle

Difficulty rating: 1780

Small Hint:

If the cube edge is $s,$ each tetrahedron edge is a face diagonal

Big Hint:

Compare $6s^2$ with four equilateral triangles of side $s\sqrt2$

Solution:

Let the cube edge be $s.$ Each tetrahedron edge is a cube face diagonal, so it has length $s\sqrt2.$ The tetrahedron’s surface area is $4\left(\frac{\sqrt3}{4}(s\sqrt2)^2\right) =2\sqrt3\,s^2.$ The cube’s surface area is $6s^2,$ so the ratio is $\frac{6}{2\sqrt3}=\sqrt3.$

Therefore, the correct answer is B.

17.

Given that $i^2=-1,$ for how many integers $n$ is $(n+i)^4$ an integer?

none

$1$

$2$

$3$

$4$

Answer: D

Concepts:complex number binomial theorem zero product property

Difficulty rating: 1780

Small Hint:

Expand $(n+i)^4$ and isolate its imaginary part

Big Hint:

An integer has imaginary part zero, so solve the resulting cubic factorization

Solution:

Expansion gives $\begin{aligned} (n+i)^4 &=n^4-6n^2+1\\ &\quad+4n(n^2-1)i. \end{aligned}$ The imaginary part vanishes exactly when $n=0,$ $n=1,$ or $n=-1.$ For each of these three integers the real part is an integer, so there are $3$ values.

Therefore, the correct answer is D.

18.

If $b\gt1,$ $\sin x\gt0,$ $\cos x\gt0$ and $\log_b\sin x=a,$ then $\log_b\cos x$ equals

$2\log_b(1-b^{\frac{a}{2}})$

$\sqrt{1-a^2}$

$b^{a^2}$

$\frac12\log_b(1-b^{2a})$

none of these

Answer: D

Concepts:logarithm trigonometric identity radical

Difficulty rating: 1590

Small Hint:

Convert $\log_b\sin x=a$ into an exponential equation

Big Hint:

Use $\cos x=\sqrt{1-\sin^2x}$ and the positivity of $\cos x$

Solution:

We have $\sin x=b^a,$ so $\cos x=\sqrt{1-b^{2a}}.$ Therefore $\log_b\cos x =\frac12\log_b(1-b^{2a}).$

Therefore, the correct answer is D.

19.

Let $C_1,$ $C_2,$ and $C_3$ be three parallel chords of a circle on the same side of the center. The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3.$ The lengths of the chords are $20,$ $16,$ and $8.$ The radius of the circle is

$12$

$4\sqrt7$

$\frac{5\sqrt{65}}3$

$\frac{5\sqrt{22}}2$

not uniquely determined by the given information

Answer: D

Concepts:chord Pythagorean Theorem system of equations

Difficulty rating: 2100

Small Hint:

A perpendicular from the center bisects each chord

Big Hint:

Let the nearest chord be distance $u$ from the center and the common spacing be $v$

Solution:

Let $r$ be the radius, $u$ the distance to the $20$ -unit chord, and $v$ the common spacing. Then $\begin{aligned} r^2&=u^2+10^2,\\ r^2&=(u+v)^2+8^2,\\ r^2&=(u+2v)^2+4^2. \end{aligned}$ Consecutive subtraction gives $2uv+v^2=36$ and $2uv+3v^2=48,$ so $v^2=6$ and $u=\frac{15}{\sqrt6}.$ Hence $r^2=u^2+100=\frac{275}{2},$ and $r=\frac{5\sqrt{22}}{2}.$

Therefore, the correct answer is D.

20.

A box contains $2$ pennies, $4$ nickels and $6$ dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of the coins drawn is at least $50$ cents?

$\frac{37}{924}$

$\frac{91}{924}$

$\frac{127}{924}$

$\frac{132}{924}$

none of these

Answer: C

Concepts:combinations sampling without replacement casework

Difficulty rating: 2100

Small Hint:

There are $\binom{12}{6}$ equally likely sets of six coins

Big Hint:

Classify favorable selections by whether they contain six, five, or four dimes

Solution:

There are $\binom{12}{6}=924$ selections. A value of at least $50$ cents occurs with six dimes; with five dimes and any one other coin; or with four dimes and two nickels. The favorable count is $\begin{aligned} &\binom66\\ &\quad+\binom65\binom61\\ &\quad+\binom64\binom42\\ &=1+36+90\\ &=127. \end{aligned}$ Thus the probability is $\frac{127}{924}.$

Therefore, the correct answer is C.

21.

In triangle $ABC,$ $\angle CBA=72^\circ,$ $E$ is the midpoint of side $AC,$ and $D$ is a point on side $BC$ such that $2BD=DC;$ $AD$ and $BE$ intersect at $F.$ The ratio of the area of $\triangle BDF$ to the area of quadrilateral $FDCE$ is

$\frac15$

$\frac14$

$\frac13$

$\frac25$

none of these

Answer: A

Concepts:area ratio midpoint ratio and proportion

Difficulty rating: 2040

Small Hint:

Area ratios are unchanged by an affine transformation, so convenient coordinates may be used

Big Hint:

Find the ratio $AF:FD$ , then compare the small triangles with $\triangle ABC$

Solution:

Scale the total area of $\triangle ABC$ to $1.$ Since $BD:DC=1:2,$ we have $[ABD]=\frac{1}{3}$ and $[ACD]=\frac{2}{3}.$ The midpoint and trisection conditions give $AF:FD=3:1.$ Hence $[BDF]=\frac14[ABD]=\frac1{12}.$ Also $\begin{aligned} [AEF] &=\frac{AE}{AC}\frac{AF}{AD}[ACD]\\ &=\frac12\cdot\frac34\cdot\frac23\\ &=\frac14. \end{aligned}$ Therefore $[FDCE]=\frac{2}{3}-\frac{1}{4}=\frac{5}{12},$ and the requested ratio is $\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5}.$

Therefore, the correct answer is A.

22.

For each real number $x,$ let $f(x)$ be the minimum of the numbers $4x+1,$ $x+2,$ and $-2x+4.$ Then the maximum value of $f(x)$ is

$\frac13$

$\frac12$

$\frac23$

$\frac52$

$\frac83$

Answer: E

Concepts:function optimization linear equation

Difficulty rating: 1980

Small Hint:

Sketch the three lines and follow their lower envelope

Big Hint:

The maximum occurs where the active increasing line meets the decreasing line

Solution:

The lower envelope is $4x+1$ until it meets $x+2$ at $x=\frac{1}{3},$ then is $x+2$ until that line meets $-2x+4$ at $x=\frac{2}{3},$ and thereafter is $-2x+4.$ Thus its maximum occurs at $x=\frac{2}{3}$ and equals $\frac23+2=\frac83.$

Therefore, the correct answer is E.

23.

Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x,$ where $x$ is a real number such that $0\lt x\lt\frac\pi2.$ The length of the hypotenuse is

$\frac43$

$\frac32$

$\frac{3\sqrt5}{5}$

$\frac{2\sqrt5}{3}$

not uniquely determined by the given information

Answer: C

Concepts:right triangle coordinate geometry Pythagorean Theorem

Difficulty rating: 2100

Small Hint:

Put the hypotenuse on the $x$ -axis and assign coordinates to the right-angle vertex

Big Hint:

Add the squares of the two trisector-segment lengths so the unknown horizontal position cancels

Solution:

Let the hypotenuse have endpoints $(0,0)$ and $(h,0),$ and let the right-angle vertex be $(u,v).$ The right-angle condition gives $u^2+v^2=hu.$ The squared distances to $(\frac{h}{3},0)$ and $(\frac{2h}{3},0)$ sum to $\begin{aligned} &(u-\frac{h}{3})^2+v^2\\ &\quad+(u-\frac{2h}{3})^2+v^2 =\frac{5h^2}{9}. \end{aligned}$ This also equals $\sin^2x+\cos^2x=1.$ Hence $h=\frac{3}{\sqrt5}=\frac{3\sqrt5}{5}.$

Therefore, the correct answer is C.

24.

For some real number $r,$ the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2.$ Which of the following numbers is closest to $r?$

$1.22$

$1.32$

$1.42$

$1.52$

$1.62$

Answer: D

Concepts:polynomial factoring estimation

Difficulty rating: 2040

Small Hint:

A repeated root appears as a squared linear factor

Big Hint:

Factor the cubic completely and then compare the repeated root with the listed decimals

Solution:

The polynomial factors as $\begin{aligned} &8x^3-4x^2-42x+45\\ &\qquad=(2x-3)^2(2x+5). \end{aligned}$ Thus the repeated root is $r=\frac{3}{2}=1.5,$ and the nearest listed number is $1.52.$

Therefore, the correct answer is D.

25.

In the nondecreasing sequence of odd integers $\begin{aligned} (a_1,a_2,a_3,\ldots) ={}&(1,3,3,3,\\ &5,5,5,5,5,\ldots) \end{aligned}$ each positive odd integer $k$ appears $k$ times. It is a fact that there are integers $b,$ $c,$ and $d$ such that, for all positive integers $n,$ $a_n=b\left\lfloor\sqrt{n+c}\right\rfloor+d,$ where $\lfloor x\rfloor$ denotes the largest integer not exceeding $x.$ The sum $b+c+d$ equals

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:floor and ceiling functions sum of first n odd numbers pattern recognition

Difficulty rating: 2200

Small Hint:

The first $m$ positive odd integers have sum $m^2$

Big Hint:

Express the least $m$ with $m^2\ge n$ using a floor of $\sqrt{n-1}$

Solution:

The final occurrence of $2m-1$ is in position $1+3+\cdots+(2m-1)=m^2.$ Therefore $\begin{aligned} a_n&=2\left\lceil\sqrt n\right\rceil-1\\ &=2\left\lfloor\sqrt{n-1}\right\rfloor+1. \end{aligned}$ Thus $b=2,$ $c=-1,$ and $d=1,$ so $b+c+d=2.$

Therefore, the correct answer is C.

26.

Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others. A tetrahedron, each of whose edges has length $s,$ is circumscribed around the balls. Then $s$ equals

$4\sqrt2$

$4\sqrt3$

$2\sqrt6$

$1+2\sqrt6$

$2+2\sqrt6$

Answer: E

Concepts:3D geometry incircle, incenter, and inradius similarity

Difficulty rating: 2160

Small Hint:

The four ball centers form a regular tetrahedron of edge $2$

Big Hint:

The outer faces are parallel to the corresponding center-tetrahedron faces and one radius farther away

Solution:

The ball centers form a regular tetrahedron of edge $2.$ Its inradius is $d=\frac{2\sqrt6}{12}=\frac{\sqrt6}{6}.$ Each face of the circumscribed tetrahedron is parallel to the corresponding face of this center tetrahedron and lies one unit farther from the common center. By similarity, $\frac{s}{2}=\frac{d+1}{d}.$ Hence $s=2+\frac{2}{d}=2+2\sqrt6.$

Therefore, the correct answer is E.

27.

The sum $\sqrt[3]{5+2\sqrt{13}} +\sqrt[3]{5-2\sqrt{13}}$ equals

$\frac32$

$\frac{\sqrt[3]{65}}4$

$\frac{1+\sqrt[6]{13}}2$

$\sqrt[3]2$

none of these

Answer: E

Concepts:radical sum and difference of cubes algebraic manipulation

Difficulty rating: 2200

Small Hint:

Call the two cube roots $u$ and $v$ , and compute $uv$

Big Hint:

Use $(u+v)^3=u^3+v^3+3uv(u+v)$

Solution:

Let the two real cube roots be $u$ and $v.$ Then $\begin{aligned} uv&=\sqrt[3]{25-52}=-3,\\ u^3+v^3&=10. \end{aligned}$ If $t=u+v,$ then $t^3=10-9t,$ so $\begin{gathered} t^3+9t-10\\ =(t-1)(t^2+t+10)\\ =0. \end{gathered}$ The only real solution is $t=1,$ which is not listed.

Therefore, the correct answer is E.

28.

The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals

$17$

$20$

$21$

$64$

$65$

Answer: C

Concepts:polynomial roots of unity modular exponentiation

Difficulty rating: 2100

Small Hint:

Evaluate the polynomial at a nonreal cube root of unity $\omega$

Big Hint:

Use $1+\omega=-\omega^2$ and consider $n$ modulo $3$

Solution:

Let $\omega^2+\omega+1=0.$ Since $1+\omega=-\omega^2,$ the polynomial evaluated at $\omega$ is $\omega^{2n}+1+\omega^{4n}.$ This is $0$ unless $\omega^{2n}=1,$ in which case it is $3.$ The latter occurs exactly when $n$ is divisible by $3.$ Among the choices only $21$ is divisible by $3,$ so that is the value for which divisibility fails.

Therefore, the correct answer is C.

29.

How many ordered triples $(x,y,z)$ of integers satisfy the system of equations below? $\begin{aligned} x^2-3xy+2y^2-z^2&=31,\\ -x^2+6yz+2z^2&=44,\\ x^2+xy+8z^2&=100. \end{aligned}$

$0$

$1$

$2$

a finite number greater than two

infinitely many

Answer: A

Concepts:Diophantine Equation modular arithmetic perfect square

Difficulty rating: 2100

Small Hint:

Add all three equations before trying to solve for the variables

Big Hint:

Rewrite the resulting quadratic form as a sum of two squares and reduce modulo $4$

Solution:

Adding the three equations gives $\begin{aligned} x^2-2xy+2y^2 &{}+6yz+9z^2\\ &=175, \end{aligned}$ or $(x-y)^2+(y+3z)^2=175.$ A square is congruent to $0$ or $1\pmod4,$ so a sum of two squares cannot be congruent to $3\pmod4.$ But $175\equiv3\pmod4,$ a contradiction. Thus there are no integer triples.

Therefore, the correct answer is A.

30.

A six digit number (base $10$ ) is squarish if it satisfies the following conditions:

(i) none of its digits is zero;

(ii) it is a perfect square; and

(iii) the first two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.

How many squarish numbers are there?

$0$

$2$

$3$

$8$

$9$

Answer: B

Concepts:digits perfect square systematic listing

Difficulty rating: 2160

Small Hint:

List the two-digit perfect squares having no zero digit

Big Hint:

Use the first pair to restrict the square root to a short interval, then filter by the final pair

Solution:

Each two-digit block must belong to $\{16,25,36,49,64,81\}.$ Restricting the square root by the first block and retaining only squares with an allowed final block leaves the following possible middle blocks:

first block	possible middle blocks
$16$	$32,$ $40,$ $48,$ $56,$ $64,$ $72$
$25$	$40,$ $50,$ $60,$ $70,$ $80,$ $90$
$36$	$48,$ $60,$ $72,$ $84,$ $96$
$49$	$56,$ $70,$ $84,$ $98$
$64$	$64,$ $80,$ $96$
$81$	$72,$ $90$

Only the middle block $64$ is allowed, producing $\begin{aligned} 166464&=408^2,\\ 646416&=804^2. \end{aligned}$ Thus there are $2$ squarish numbers.

Therefore, the correct answer is B.