1982 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

When the polynomial $x^3-2$ is divided by the polynomial $x^2-2,$ the remainder is

$2$

$-2$

$-2x-2$

$2x+2$

$2x-2$

Answer: E

Concepts:polynomial algebraic manipulation

Difficulty rating: 1110

Small Hint:

Work modulo $x^2-2$

Big Hint:

Replace $x^2$ by $2$ in $x^3-2$

Solution:

Modulo $x^2-2,$ we have $x^2\equiv2,$ so $x^3-2\equiv2x-2.$ This linear polynomial is the remainder.

Therefore, the correct answer is E.

2.

If a number eight times as large as $x$ is increased by two, then one fourth of the result equals

$2x+\frac12$

$x+\frac12$

$2x+2$

$2x+4$

$2x+16$

Answer: A

Concepts:order of operations algebraic manipulation

Difficulty rating: 900

Small Hint:

Translate the first two operations before dividing

Big Hint:

Simplify $\frac{8x+2}{4}$ term by term

Solution:

Eight times $x,$ increased by two, is $8x+2.$ One fourth is $\frac{8x+2}{4}=2x+\frac12.$

Therefore, the correct answer is A.

3.

Evaluate $(x^x)^{(x^x)}$ at $x=2.$

$16$

$64$

$256$

$1024$

$65{,}536$

Answer: C

Concepts:exponent substitution

Difficulty rating: 1020

Small Hint:

Evaluate the inner $x^x$ first

Big Hint:

The same value becomes both the base and exponent

Solution:

At $x=2,$ $x^x=2^2=4.$ Hence $(x^x)^{(x^x)}=4^4=256.$

Therefore, the correct answer is C.

4.

The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, measured in square centimeters. The radius of the semicircle, measured in centimeters, is

$\pi$

$\frac2\pi$

$1$

$\frac12$

$\frac4\pi+2$

Answer: E

Concepts:circumference perimeter area algebraic manipulation

Difficulty rating: 1290

Small Hint:

Include the diameter in the perimeter of the region

Big Hint:

Set $\pi r+2r$ equal to $\frac{\pi r^2}{2}$

Solution:

For $r\gt0,$ equality of perimeter and area gives $\pi r+2r=\frac{\pi r^2}{2}.$ Dividing by $r$ and solving yields $r=\frac{2(\pi+2)}{\pi}=2+\frac{4}{\pi}.$

Therefore, the correct answer is E.

5.

Two positive numbers $x$ and $y$ are in the ratio $a:b,$ where $0\lt a\lt b.$ If $x+y=c,$ then the smaller of $x$ and $y$ is

$\frac{ac}{b}$

$\frac{bc-ac}{b}$

$\frac{ac}{a+b}$

$\frac{bc}{a+b}$

$\frac{ac}{b-a}$

Answer: C

Concepts:ratio and proportion

Difficulty rating: 1100

Small Hint:

Write the numbers as $ka$ and $kb$

Big Hint:

Use their sum to determine the common scale factor

Solution:

Write $x=ka$ and $y=kb.$ Since $a\lt b,$ $x$ is smaller. From $k(a+b)=c,$ $k=\frac{c}{a+b},$ so $x=\frac{ac}{a+b}.$

Therefore, the correct answer is C.

6.

The sum of all but one of the interior angles of a convex polygon equals $2570^\circ.$ The remaining angle is

$90^\circ$

$105^\circ$

$120^\circ$

$130^\circ$

$144^\circ$

Answer: D

Concepts:angle sum algebraic manipulation

Difficulty rating: 1360

Small Hint:

A polygon’s total interior angle sum is a multiple of $180^\circ$

Big Hint:

The missing convex angle must place the total strictly between $2570^\circ$ and $2750^\circ$

Solution:

The total must be $(n-2)180^\circ$ and the missing angle lies between $0^\circ$ and $180^\circ.$ The only multiple of $180^\circ$ between $2570^\circ$ and $2750^\circ$ is $2700^\circ,$ leaving $2700^\circ-2570^\circ=130^\circ.$

Therefore, the correct answer is D.

7.

If the operation $x*y$ is defined by $x*y=(x+1)(y+1)-1,$ then which one of the following is false?

$x*y=y*x$ for all real $x$ and $y.$

$x*(y+z)$ equals $(x*y)+(x*z)$ for all real $x,$ $y,$ and $z.$

$(x-1)*(x+1)$ equals $(x*x)-1$ for all real $x.$

$x*0=x$ for all real $x.$

$x*(y*z)=(x*y)*z$ for all real $x,$ $y,$ and $z.$

Answer: B

Concepts:custom operation distributive property counterexample

Difficulty rating: 1390

Small Hint:

First simplify the operation to $x+y+xy$

Big Hint:

Test the proposed distributive law with $x=1$ and $y=z=0$

Solution:

Since $x*y=x+y+xy,$ the operation is commutative, has identity $0,$ and becomes ordinary multiplication after adding $1.$ Thus it is associative, and direct expansion also verifies C. Statement B fails when $x=1$ and $y=z=0:$ its left side is $1*0=1,$ but its right side is $(1*0)+(1*0)=2.$ Hence B is false.

Therefore, the correct answer is B.

8.

By definition $r!=r(r-1)\cdots1$ and $\binom jk=\frac{j!}{k!(j-k)!},$ where $r,$ $j,$ $k$ are positive integers and $k\lt j.$ If $\binom n1,$ $\binom n2,$ $\binom n3$ form an arithmetic progression with $n\gt3,$ then $n$ equals

$5$

$7$

$9$

$11$

$12$

Answer: B

Concepts:factorial combinations arithmetic sequence quadratic

Difficulty rating: 1730

Small Hint:

The middle term of an arithmetic progression is the average of its neighbors

Big Hint:

Substitute the factorial formulas into $2\binom n2=\binom n1+\binom n3$

Solution:

The progression condition is $2\binom n2=\binom n1+\binom n3.$ Substitution and cancellation give $n^2-9n+14=0,$ or $(n-2)(n-7)=0.$ Since $n\gt3,$ $n=7.$

Therefore, the correct answer is B.

9.

A vertical line divides the triangle with vertices $(0,0),$ $(1,1)$ and $(9,1)$ in the $xy$ -plane into two regions of equal area. The equation of the line is $x=$

$2.5$

$3.0$

$3.5$

$4.0$

$4.5$

Answer: B

Concepts:coordinate geometry area calculus

Difficulty rating: 1630

Small Hint:

For $1\le x\le9,$ the triangle lies between $y=\frac{x}{9}$ and $y=1$

Big Hint:

Set the accumulated area to one half of the triangle’s total area

Solution:

The triangle has area $4.$ The portion with $0\le x\le1$ has area $\int_0^1\left(x-\frac x9\right)\,dx=\frac49.$ For a cut $x=t\ge1,$ the area to its left is therefore $\frac49+\int_1^t(1-\frac{x}{9})\,dx.$ Setting this equal to $2$ gives $t-\frac{t^2}{18}=\frac{5}{2},$ whose relevant solution is $t=3.$

Therefore, the correct answer is B.

10.

In the adjoining diagram, $BO$ bisects $\angle CBA,$ $CO$ bisects $\angle ACB,$ and $MN$ is parallel to $BC.$ If $AB=12,$ $BC=24,$ and $AC=18,$ then the perimeter of $\triangle AMN$ is

$30$

$33$

$36$

$39$

$42$

Answer: A

Difficulty rating: 1910

Small Hint:

Because $O$ is the incenter, its distance from $BC$ is the inradius

Big Hint:

Compare the altitude of $\triangle AMN$ with the altitude of $\triangle ABC$

Solution:

The sides $12,$ $18,$ $24$ have semiperimeter $27$ and area $\sqrt{27\cdot15\cdot9\cdot3}=27\sqrt{15},$ so the inradius is $r=\sqrt{15}.$ Taking $BC$ as base, the altitude is $h=\frac{2(27\sqrt{15})}{24}=\frac{9\sqrt{15}}{4}.$ Thus the similarity scale from $\triangle ABC$ to $\triangle AMN$ is $1-\frac{r}{h}=1-\frac{4}{9}=\frac{5}{9}.$ Its perimeter is $(\frac{5}{9})(12+18+24)=30.$

Therefore, the correct answer is A.

11.

How many integers with four different digits are there between $1{,}000$ and $9{,}999$ such that the absolute value of the difference between the first digit and the last digit is $2?$

$672$

$784$

$840$

$896$

$1{,}008$

Answer: C

Concepts:digits basic counting casework

Difficulty rating: 1710

Small Hint:

First count the allowed ordered pairs of first and last digits

Big Hint:

Once those digits are fixed, choose two distinct middle digits from the remaining eight

Solution:

For leading digits $1,\ldots,9,$ the number of possible last digits differing by $2$ is $1+1+2+2+2$ plus $2+2+1+1,$ or $14.$ In addition, leading digit $2$ may end in $0,$ giving $15$ endpoint pairs. The middle digits can then be chosen in $8\cdot7$ ordered ways. Thus the count is $15\cdot56=840.$

Therefore, the correct answer is C.

12.

Let $f(x)=ax^7+bx^3+cx-5,$ where $a,$ $b,$ and $c$ are constants. If $f(-7)=7,$ then $f(7)$ equals

$-17$

$-7$

$14$

$21$

not uniquely determined

Answer: A

Concepts:symmetry (algebra)polynomial function

Difficulty rating: 1360

Small Hint:

Separate the constant term from the odd-powered terms

Big Hint:

If $g(x)=f(x)+5,$ compare $g(7)$ and $g(-7)$

Solution:

Let $g(x)=ax^7+bx^3+cx,$ which is odd. Since $f(-7)=g(-7)-5=7,$ $g(-7)=12.$ Therefore $g(7)=-12,$ and $f(7)=-12-5=-17.$

Therefore, the correct answer is A.

13.

If $a\gt1,$ $b\gt1$ and $p=\frac{\log_b(\log_ba)}{\log_ba},$ then $a^p$ equals

$1$

$b$

$\log_ab$

$\log_ba$

$a^{\log_ba}$

Answer: D

Concepts:logarithm exponent

Difficulty rating: 1730

Small Hint:

Let $t=\log_ba,$ so $a=b^t$

Big Hint:

Rewrite $p$ as $\frac{\log_bt}{t}$

Solution:

Set $t=\log_ba,$ so $a=b^t$ and $p=\frac{\log_bt}{t}.$ Then $a^p=(b^t)^{\frac{\log_bt}{t}},$ which equals $b^{\log_bt}=t.$ Therefore $a^p=\log_ba.$

Therefore, the correct answer is D.

14.

In the adjoining figure, points $B$ and $C$ lie on line segment $AD,$ and $AB,$ $BC,$ and $CD$ are diameters of circles $O,$ $N,$ and $P,$ respectively. Circles $O,$ $N,$ and $P$ all have radius $15,$ and the line $AG$ is tangent to circle $P$ at $G.$ If $AG$ intersects circle $N$ at points $E$ and $F,$ then chord $EF$ has length

$20$

$15\sqrt2$

$24$

$25$

none of these

Answer: C

Concepts:circle tangent line chord similarity

Difficulty rating: 2090

Small Hint:

Use right triangle $APG$ to find the distance from $N$ to line $AG$

Big Hint:

A chord at distance $d$ from a circle’s center has length $2\sqrt{r^2-d^2}$

Solution:

Here $AP=75,$ $PG=15,$ and $AG\perp PG.$ The distance from $N$ to $AG$ scales with $\frac{AN}{AP}=\frac{45}{75},$ so it is $15(\frac{45}{75})=9.$ Therefore the chord in the radius- $15$ circle has length $2\sqrt{15^2-9^2}=2\sqrt{144}=24.$

Therefore, the correct answer is C.

15.

Let $\lfloor z\rfloor$ denote the greatest integer not exceeding $z.$ Let $x$ and $y$ satisfy the simultaneous equations $\begin{aligned} y&=2\lfloor x\rfloor+3,\\ y&=3\lfloor x-2\rfloor+5. \end{aligned}$ If $x$ is not an integer, then $x+y$ is

an integer

between $4$ and $5$

between $-4$ and $4$

between $15$ and $16$

$16.5$

Answer: D

Concepts:floor and ceiling functions system of equations inequality

Difficulty rating: 1660

Small Hint:

For nonintegral $x,$ relate $\lfloor x-2\rfloor$ to $\lfloor x\rfloor$

Big Hint:

Let $n=\lfloor x\rfloor$ and equate the two formulas for $y$

Solution:

For nonintegral $x,$ let $n=\lfloor x\rfloor;$ then $\lfloor x-2\rfloor=n-2.$ Thus $2n+3=3(n-2)+5,$ so $n=4$ and $y=11.$ Since $4\lt x\lt5,$ we have $15\lt x+y\lt16.$

Therefore, the correct answer is D.

16.

In the adjoining figure, a wooden cube has edges of length $3$ meters. Square holes of side one meter, centered in each face, are cut through to the opposite face. The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside, in square meters, is

$54$

$72$

$76$

$84$

$86$

Answer: B

Concepts:surface area cube geometry 3D geometry

Difficulty rating: 1860

Small Hint:

Start with the outer faces after removing their central squares

Big Hint:

For each of the three tunnels, count the four interior walls not removed by the crossing tunnels

Solution:

The six outer faces contribute $6(3^2-1^2)=48.$ Each of the three length- $3$ square tunnels has four inner walls, but the central unit segment of every wall is removed by a perpendicular tunnel, leaving area $2$ per wall. Thus the inside contributes $3\cdot4\cdot2=24.$ The total is $48+24=72.$

Therefore, the correct answer is B.

17.

How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^x+3=0?$

$0$

$1$

$2$

$3$

$4$

Answer: C

Concepts:exponent substitution quadratic

Difficulty rating: 1590

Small Hint:

Substitute $t=3^x,$ noting that $t\gt0$

Big Hint:

The equation becomes a quadratic in $t$

Solution:

Let $t=3^x\gt0.$ Then $9t^2-28t+3=0,$ whose roots are $t=3$ and $t=\frac{1}{9}.$ Both are positive and each corresponds to one real $x,$ so there are $2$ solutions.

Therefore, the correct answer is C.

18.

In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ.$ Find the cosine of $\angle BHD.$

$\frac{\sqrt3}{6}$

$\frac{\sqrt2}{6}$

$\frac{\sqrt6}{3}$

$\frac{\sqrt6}{4}$

$\frac{\sqrt6-\sqrt2}{4}$

Answer: D

Concepts:rectangular prism vector 3D geometry trigonometry

Difficulty rating: 2230

Small Hint:

Assign coordinates at $H$ along the three mutually perpendicular edges

Big Hint:

Translate the two given angles into relationships among the three edge lengths, then use a dot product

Solution:

Let $H=(0,0,0),$ $F=(u,0,0),$ $G=(0,v,0),$ and $C=(0,0,w).$ Then $B=(u,0,w)$ and $D=(0,v,w).$ The $60^\circ$ condition gives $w=\sqrt3u,$ while the $45^\circ$ condition gives $v=w.$ Hence $\begin{aligned} \cos\angle BHD &=\frac{w^2} {\sqrt{u^2+w^2}\sqrt{2w^2}}\\ &=\frac{\sqrt6}{4}. \end{aligned}$

Therefore, the correct answer is D.

19.

Let $f(x)=|x-2|+|x-4|$ $-|2x-6|,$ for $2\le x\le8.$ The sum of the largest and smallest values of $f(x)$ is

$1$

$2$

$4$

$6$

none of these

Answer: B

Concepts:absolute value function optimization casework

Difficulty rating: 1630

Small Hint:

Break the interval at the zeros $2,$ $3,$ and $4$

Big Hint:

Simplify $f(x)$ on each resulting interval

Solution:

On $[2,3],$ $f(x)=2x-4;$ on $[3,4],$ $f(x)=8-2x;$ and on $[4,8],$ $f(x)=0.$ Thus the minimum is $0$ and the maximum is $2,$ whose sum is $2.$

Therefore, the correct answer is B.

20.

The number of pairs of positive integers $(x,y)$ which satisfy the equation $x^2+y^2=x^3$ is

$0$

$1$

$2$

not finite

none of these

Answer: D

Concepts:Diophantine Equation perfect square

Difficulty rating: 1940

Small Hint:

Rearrange as $y^2=x^2(x-1)$

Big Hint:

Choose $x-1$ to be a perfect square

Solution:

For every positive integer $k,$ take $x=k^2+1$ and $y=k(k^2+1)=kx.$ Then $y^2=k^2x^2=x^2(x-1),$ so $x^2+y^2=x^3.$ This supplies infinitely many positive-integer pairs.

Therefore, the correct answer is D.

21.

In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ.$ Median $CM$ is perpendicular to median $BN,$ and side $BC=s.$ The length of $BN$ is

$s\sqrt2$

$\frac32s\sqrt2$

$2s\sqrt2$

$\frac{s\sqrt5}{2}$

$\frac{s\sqrt6}{2}$

Answer: E

Concepts:median (geometry)coordinate geometry vector right triangle

Difficulty rating: 2150

Small Hint:

Place $C=(0,0),$ $B=(0,s),$ and $A=(t,0)$

Big Hint:

Use a dot product for the directions of the two medians

Solution:

Set $C=(0,0),$ $B=(0,s),$ and $A=(t,0).$ Then $M=(\frac{t}{2},\frac{s}{2})$ and $N=(\frac{t}{2},0).$ Perpendicularity gives $(\frac{t}{2},\frac{s}{2})\cdot(\frac{t}{2},-s)=0,$ hence $t^2=2s^2.$ Therefore $\begin{aligned} BN&=\sqrt{(\frac{t}{2})^2+s^2}\\ &=\sqrt{\frac{s^2}{2}+s^2}\\ &=\frac{s\sqrt6}{2}. \end{aligned}$

Therefore, the correct answer is E.

22.

In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at a point $P$ between the walls. Resting against one wall at $Q,$ a distance $k$ above the ground, the ladder makes a $45^\circ$ angle with the ground. Resting against the other wall at $R,$ a distance $h$ above the ground, the ladder makes a $75^\circ$ angle with the ground. The width $w$ is equal to

$a$

$RQ$

$k$

$\frac{h+k}{2}$

$h$

Answer: E

Concepts:trigonometry right triangle trigonometric identity

Difficulty rating: 1910

Small Hint:

Express the two horizontal portions of the alley using cosines

Big Hint:

Compare $\cos75^\circ+\cos45^\circ$ with $\sin75^\circ$

Solution:

The horizontal distances from $P$ to the walls are $a\cos75^\circ$ and $a\cos45^\circ,$ so $w=a(\cos75^\circ+\cos45^\circ).$ The sum-to-product identity gives $\cos75^\circ+\cos45^\circ=\sin75^\circ.$ Since $h=a\sin75^\circ,$ $w=h.$

Therefore, the correct answer is E.

23.

The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is

$\frac34$

$\frac7{10}$

$\frac23$

$\frac9{14}$

none of these

Answer: A

Concepts:law of sines law of cosines trigonometric identity

Difficulty rating: 2150

Small Hint:

Let the smallest and largest angles be $\theta$ and $2\theta$

Big Hint:

Use the law of sines to compare the shortest and longest consecutive sides

Solution:

Let the consecutive sides be $n,$ $n+1,$ $n+2,$ opposite angles $\theta,$ $\phi,$ $2\theta.$ By the law of sines, $\frac{n+2}{n}=\frac{\sin2\theta}{\sin\theta}=2\cos\theta.$ The law of cosines at the smallest angle also gives $\cos\theta=\frac{n^2+6n+5}{2(n+1)(n+2)}.$ Equating these expressions yields $n^2-3n-4=0,$ so $n=4.$ Hence the sides are $4,$ $5,$ $6$ and $\cos\theta=\frac{6}{2\cdot4}=\frac{3}{4}.$

Therefore, the correct answer is A.

24.

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If $AG=2,$ $GF=13,$ $FC=1$ and $HJ=7,$ then $DE$ equals

$2\sqrt{22}$

$7\sqrt3$

$9$

$10$

$13$

Answer: A

Concepts:power of a point circle equilateral triangle

Difficulty rating: 2310

Small Hint:

All three sides of the triangle have length $16$

Big Hint:

Apply power of a point from $A,$ $B,$ and $C$ to relate the two base segments and $DE$

Solution:

First, power of $A$ gives $AH(AH+7)=AG\cdot AF.$ The latter product is $2\cdot15=30,$ so $AH=3,$ $BJ=6,$ and $BH=13.$ Let $BD=u,$ $DE=x,$ $EC=v.$ From the secants at $B$ and $C,$ $\begin{aligned} u(u+x)&=BH\cdot BJ\\ &=13\cdot6=78,\\ v(v+x)&=CF\cdot CG\\ &=1\cdot14=14. \end{aligned}$ Also $u+x+v=16.$ Subtracting the power equations and using the sum gives $u-v=4.$ Hence $u=10-\frac{x}{2}$ and $v=6-\frac{x}{2}.$ Substitution into $v(v+x)=14$ gives $x^2=88,$ so $DE=2\sqrt{22}.$

Therefore, the correct answer is A.

25.

The adjoining figure is a map of part of a city: the small rectangles are blocks and the spaces in between are streets. Each morning a student walks from intersection $A$ to intersection $B,$ always walking along streets shown, always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\frac12$ (independent of all other choices) whether to go east or south. Find the probability that, on any given morning, he walks through intersection $C.$

$\frac{11}{32}$

$\frac12$

$\frac47$

$\frac{21}{32}$

$\frac34$

Answer: D

Concepts:basic probability lattice paths binomial probability

Difficulty rating: 2090

Small Hint:

Reaching $C$ means making the third eastward move before the fourth southward move

Big Hint:

Condition on the number $i=0,1,2,3$ of south moves made before the third east move

Solution:

If $i$ south moves occur before the third east move, the final step to $C$ is east and the preceding $2+i$ steps contain two east moves. Thus $\begin{aligned} \Pr(C) &=\sum_{i=0}^3\binom{2+i}{2}2^{-(3+i)}\\ &=\frac18+\frac3{16}+\frac6{32}\\ &\quad+\frac{10}{64}\\ &=\frac{21}{32}. \end{aligned}$

Therefore, the correct answer is D.

26.

If the base $8$ representation of a perfect square is $ab3c,$ where $a\ne0,$ then $c$ is

$0$

$1$

$3$

$4$

not uniquely determined

Answer: B

Concepts:modular arithmetic perfect square number base quadratic residue

Difficulty rating: 1860

Small Hint:

Only the last two base- $8$ digits matter modulo $64$

Big Hint:

List the quadratic residues modulo $64$ that lie between $30_8$ and $37_8$

Solution:

The last two octal digits $3c_8$ represent $24+c\pmod{64}.$ Squares modulo $64$ in the range $24,\ldots,31$ include only $25.$ Thus $24+c=25,$ so $c=1.$

Therefore, the correct answer is B.

27.

Suppose $z=a+bi$ is a solution of the polynomial equation $\begin{aligned} c_4z^4+ic_3z^3+c_2z^2\\ {}+ic_1z+c_0=0, \end{aligned}$ where $c_0,$ $c_1,$ $c_2,$ $c_3,$ $c_4,$ $a,$ and $b$ are real constants and $i^2=-1.$ Which one of the following must also be a solution?

$-a-bi$

$a-bi$

$-a+bi$

$b+ai$

none of these

Answer: C

Concepts:complex number symmetry (algebra)polynomial

Difficulty rating: 2090

Small Hint:

Conjugate the entire equation

Big Hint:

Compare the conjugated equation at $\bar z$ with the original polynomial evaluated at $-\bar z$

Solution:

Conjugating the equation changes each $i$ to $-i$ and $z$ to $\bar z.$ Because the odd-powered terms also change sign when the input is negated, this conjugated equation is precisely the original polynomial evaluated at $-\bar z.$ Thus $-\bar z=-a+bi$ must be a root.

Therefore, the correct answer is C.

28.

A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac7{17}.$ What number was erased?

$6$

$7$

$8$

$9$

can not be determined

Answer: B

Concepts:mean arithmetic sequence divisibility

Difficulty rating: 2040

Small Hint:

If the last written number is $n,$ the remaining count is $n-1$

Big Hint:

Use the bounds $1\le e\le n$ on the erased value to narrow the possible values of $n$

Solution:

The new average $\frac{602}{17}$ is below the original average $\frac{n+1}{2}$ but differs from it by less than $\frac{n}{n-1},$ forcing $n=69$ or $70.$ Since the remaining sum $(n-1)(\frac{602}{17})$ must be integral, $n-1$ is divisible by $17,$ so $n=69.$ The erased number is $\begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7. \end{aligned}$

Therefore, the correct answer is B.

29.

Let $x,$ $y,$ and $z$ be three positive real numbers whose sum is $1.$ If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is

$\frac1{32}$

$\frac1{36}$

$\frac4{125}$

$\frac1{127}$

none of these

Answer: A

Concepts:optimization inequality extremal argument

Difficulty rating: 2340

Small Hint:

Order the variables $x\le y\le z$ ; then the active constraint is $z\le2x$

Big Hint:

A minimum occurs on the boundary $z=2x,$ where $y=1-3x$

Solution:

Order $x\le y\le z.$ At a minimum the spread is maximal, so $z=2x$ and $y=1-3x.$ The ordering requires $\frac{1}{5}\le x\le\frac{1}{4}.$ Thus $xyz=2x^2(1-3x).$ Its only interior critical point is a maximum, so compare endpoints: the values are $\frac{4}{125}$ and $\frac{1}{32},$ respectively. The minimum is $\frac{1}{32}.$

Therefore, the correct answer is A.

30.

Find the units digit in the decimal expansion of $(15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}.$

$0$

$2$

$5$

$9$

none of these

Answer: D

Concepts:recursion modular arithmetic units digit

Difficulty rating: 2400

Small Hint:

Pair $\alpha=15+\sqrt{220}$ with its conjugate $\beta=15-\sqrt{220}$

Big Hint:

The integers $S_n=\alpha^n+\beta^n$ satisfy a short recurrence, while $0\lt\beta\lt1$

Solution:

Let $\alpha=15+\sqrt{220}$ and $\beta=15-\sqrt{220},$ so $0\lt\beta\lt1.$ The integers $S_n=\alpha^n+\beta^n$ satisfy $S_n=30S_{n-1}-5S_{n-2},$ hence $S_n\equiv0\pmod{10}$ for every $n\ge1.$ Therefore $\begin{aligned} \alpha^{19}+\alpha^{82} &=S_{19}+S_{82}\\ &\quad-(\beta^{19}+\beta^{82}) \end{aligned}$ is a multiple of $10$ minus a positive number less than $1.$ Its integer part therefore ends in $9.$

Therefore, the correct answer is D.