1974 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $x\ne0$ or $4$ and $y\ne0$ or $6,$ then $\frac2x+\frac3y=\frac12$ is equivalent to

$4x+3y=xy$

$y=\frac{4x}{6-y}$

$\frac{x}{2}+\frac{y}{3}=2$

$\frac{4y}{y-6}=x$

none of these

Concepts:rational equation factoring

Difficulty rating: 1330

Small Hint:

Clear all three denominators at once

Big Hint:

After multiplying by $2xy,$ isolate the terms containing $x$ and factor

Solution:

Multiplying by $2xy$ gives $4y+6x=xy,$ so $4y=x(y-6).$ Since $y\ne6,$ $x=\frac{4y}{y-6}.$

Therefore, the correct answer is D.

2.

Let $x_1$ and $x_2$ be such that $x_1\ne x_2$ and $3x_i^2-hx_i=b,$ $i=1,$ $2.$ Then $x_1+x_2$ equals

$-\frac h3$

$\frac h3$

$\frac b3$

$2b$

$-\frac b3$

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1410

Small Hint:

Both given numbers satisfy the same quadratic equation

Big Hint:

Move $b$ to the left and use the sum-of-roots formula

Solution:

The distinct numbers $x_1,x_2$ are the two roots of $3x^2-hx-b=0.$ By Vieta’s formulas, their sum is $-\frac{(-h)}{3}=\frac{h}{3}.$

Therefore, the correct answer is B.

3.

The coefficient of $x^7$ in the polynomial expansion of $(1+2x-x^2)^4$ is

$-8$

$12$

$6$

$-12$

none of these

Concepts:binomial theorem polynomial

Difficulty rating: 1830

Small Hint:

A degree- $7$ term must select degrees $2,$ $2,$ $2,$ $1$ from the four factors

Big Hint:

Choose which factor contributes $2x$ ; the other three contribute $-x^2$

Solution:

The only way four selected terms can have total degree $7$ is to choose $2x$ once and $-x^2$ three times. There are $4$ choices for the first factor, so the coefficient is $4(2)(-1)^3=-8.$

Therefore, the correct answer is A.

4.

What is the remainder when $x^{51}+51$ is divided by $x+1?$

$0$

$1$

$49$

$50$

$51$

Concepts:polynomial substitution exponent

Difficulty rating: 1410

Small Hint:

For division by $x+1,$ evaluate the polynomial at $-1$

Big Hint:

The exponent $51$ is odd

Solution:

By the polynomial remainder theorem, the remainder is $(-1)^{51}+51=-1+51=50.$

Therefore, the correct answer is D.

5.

Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E,$ if $\angle BAD=92^\circ$ and $\angle ADC=68^\circ,$ find $\angle EBC.$

$66^\circ$

$68^\circ$

$70^\circ$

$88^\circ$

$92^\circ$

Concepts:cyclic quadrilateral angle chasing

Difficulty rating: 1460

Small Hint:

Opposite angles of a cyclic quadrilateral are supplementary

Big Hint:

The desired exterior angle is also supplementary to $\angle ABC$

Solution:

Because $ABCD$ is cyclic, $\angle ABC+\angle ADC=180^\circ.$ Since $BE$ extends $BA,$ $\angle EBC+\angle ABC=180^\circ$ as well. Hence $\angle EBC=\angle ADC=68^\circ.$ The given $92^\circ$ angle is unnecessary.

Therefore, the correct answer is B.

6.

For positive real numbers $x$ and $y$ define $x*y=\frac{x\cdot y}{x+y};$ then

“ $*$ ” is commutative but not associative

“ $*$ ” is associative but not commutative

“ $*$ ” is neither commutative nor associative

“ $*$ ” is commutative and associative

none of these

Concepts:custom operation fraction

Difficulty rating: 1740

Small Hint:

The formula is visibly unchanged when $x$ and $y$ are interchanged

Big Hint:

Rewrite the operation through its reciprocal: $\frac1{x*y}=\frac1x+\frac1y$

Solution:

Symmetry in $x,$ $y$ makes the operation commutative. Also, $\frac1{x*y}=\frac1x+\frac1y.$ Thus both $(x*y)*z$ and $x*(y*z)$ have reciprocal $\frac1x+\frac1y+\frac1z,$ so they are equal. The operation is associative too.

Therefore, the correct answer is D.

7.

A town’s population increased by $1{,}200$ people, and then this new population decreased by $11\%.$ The town now had $32$ less people than it did before the $1{,}200$ increase. What is the original population?

$1{,}200$

$11{,}200$

$9{,}968$

$10{,}000$

none of these

Concepts:percentage linear equation

Difficulty rating: 1310

Small Hint:

Let $P$ be the original population and apply the decrease to $P+1200$

Big Hint:

The final population is both $0.89(P+1200)$ and $P-32$

Solution:

Let the original population be $P.$ Then $0.89(P+1200)=P-32.$ Therefore $0.11P=1100,$ giving $P=10{,}000.$

Therefore, the correct answer is D.

8.

What is the smallest prime number dividing the sum $3^{11}+5^{13}?$

$2$

$3$

$5$

$3^{11}+5^{13}$

none of these

Concepts:parity prime

Difficulty rating: 920

Small Hint:

Both bases are odd

Big Hint:

The sum of two odd integers is divisible by the smallest prime

Solution:

Both $3^{11}$ and $5^{13}$ are odd, so their sum is even. Hence it is divisible by $2,$ the smallest prime.

Therefore, the correct answer is A.

9.

The integers greater than one are arranged in five columns as follows:

	$2$	$3$	$4$	$5$
$9$	$8$	$7$	$6$
	$10$	$11$	$12$	$13$
$17$	$16$	$15$	$14$
	$\vdots$	$\vdots$	$\vdots$	$\vdots$

(Four consecutive integers appear in each row; in the first, third and other odd numbered rows, the integers appear in the last four columns and increase from left to right; in the second, fourth and other even numbered rows, the integers appear in the first four columns and increase from right to left.)

In which column will the number $1{,}000$ fall?

first

second

third

fourth

fifth

Concepts:pattern recognition modular arithmetic

Difficulty rating: 1430

Small Hint:

Look at where the displayed multiples of $8$ occur

Big Hint:

Each two-row cycle contains eight consecutive integers

Solution:

Each pair of rows is a repeating block of eight integers. In every such block, its multiple of $8$ lies in the second column, as do $8$ and $16.$ Since $1000=125\cdot8,$ it also lies in the second column.

Therefore, the correct answer is B.

10.

What is the smallest integral value of $k$ such that $2x(kx-4)-x^2+6=0$ has no real roots?

$-1$

$2$

$3$

$4$

$5$

Concepts:quadratic inequality

Difficulty rating: 1650

Small Hint:

First rewrite the equation in standard quadratic form

Big Hint:

No real roots means the discriminant is negative

Solution:

The equation is $(2k-1)x^2-8x+6=0.$ Its discriminant is $64-24(2k-1)=88-48k.$ This is negative exactly when $k\gt\frac{11}{6},$ so the least integer possible is $2.$

Therefore, the correct answer is B.

11.

If $(a,b)$ and $(c,d)$ are two points on the line whose equation is $y=mx+k,$ then the distance between $(a,b)$ and $(c,d),$ in terms of $a,$ $c,$ and $m,$ is

$|a-c|\sqrt{1+m^2}$

$|a+c|\sqrt{1+m^2}$

$\frac{|a-c|}{\sqrt{1+m^2}}$

$|a-c|(1+m^2)$

$|a-c||m|$

Concepts:coordinate geometry distance formula

Difficulty rating: 1390

Small Hint:

Use the line equation to express $b-d$ in terms of $a-c$

Big Hint:

Substitute $b-d=m(a-c)$ into the distance formula

Solution:

Since both points are on the line, $b-d=m(a-c).$ Their distance is therefore $\begin{aligned} \ell&=\sqrt{(a-c)^2+(b-d)^2}\\ &=\sqrt{(a-c)^2(1+m^2)}\\ &=|a-c|\sqrt{1+m^2}. \end{aligned}$

Therefore, the correct answer is A.

12.

If $g(x)=1-x^2$ and $f(g(x))=\frac{1-x^2}{x^2}$ when $x\ne0,$ then $f(\frac{1}{2})$ equals

$\frac{3}{4}$

$1$

$3$

$\frac{\sqrt2}{2}$

$\sqrt2$

Concepts:function substitution

Difficulty rating: 1620

Small Hint:

Choose a nonzero $x$ for which $g(x)=\frac{1}{2}$

Big Hint:

The equation $1-x^2=\frac{1}{2}$ determines $x^2$ directly

Solution:

If $g(x)=\frac{1}{2},$ then $x^2=\frac{1}{2},$ and either corresponding $x$ is nonzero. Hence $f(\frac{1}{2})=\frac{1-x^2}{x^2} =\frac{\frac{1}{2}}{\frac{1}{2}}=1.$

Therefore, the correct answer is B.

13.

Which of the following is equivalent to “If $P$ is true then $Q$ is false”?

“ $P$ is true or $Q$ is false.”

“If $Q$ is false then $P$ is true.”

“If $P$ is false then $Q$ is true.”

“If $Q$ is true then $P$ is false.”

“If $Q$ is true then $P$ is true.”

Concepts:logical deduction

Difficulty rating: 1450

Small Hint:

An implication is equivalent to its contrapositive

Big Hint:

Reverse the implication and negate both “ $P$ true” and “ $Q$ false”

Solution:

The contrapositive of “ $P$ true implies $Q$ false” is “ $Q$ not false implies $P$ not true.” This is exactly “If $Q$ is true then $P$ is false.”

Therefore, the correct answer is D.

14.

Which statement is correct?

If $x\lt0,$ then $x^2\gt x.$

If $x^2\gt0,$ then $x\gt0.$

If $x^2\gt x,$ then $x\gt0.$

If $x^2\gt x,$ then $x\lt0.$

If $x\lt1,$ then $x^2\lt x.$

Concepts:inequality counterexample

Difficulty rating: 1160

Small Hint:

For a negative $x,$ compare $x^2$ first with zero

Big Hint:

Simple test values such as $-1,$ $\frac12,$ and $2$ reject the other implications

Solution:

If $x\lt0,$ then $x^2\gt0\gt x,$ so choice A always holds. The values $x=-1,$ $x=-1,$ $x=2,$ and $x=-1$ respectively disprove B, C, D, and E.

Therefore, the correct answer is A.

15.

If $x\lt-2$ then $\bigl|1-|1+x|\bigr|$ equals

$2+x$

$-2-x$

$x$

$-x$

$-2$

Concepts:absolute value inequality

Difficulty rating: 1430

Small Hint:

Resolve the inner absolute value first using $1+x\lt-1$

Big Hint:

After replacing $|1+x|$ by $-1-x,$ determine the sign of $2+x$

Solution:

Since $x\lt-2,$ $1+x\lt0,$ so $|1+x|=-1-x.$ Thus $\bigl|1-|1+x|\bigr|$ becomes $|1-(-1-x)|,$ which is $|2+x|=-2-x$ because $2+x\lt0.$

Therefore, the correct answer is B.

16.

A circle of radius $r$ is inscribed in a right isosceles triangle, and a circle of radius $R$ is circumscribed about the triangle. Then $\frac{R}{r}$ equals

$1+\sqrt2$

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

$\frac{\sqrt2-1}{2}$

$\frac{1+\sqrt2}{2}$

$2(2-\sqrt2)$

Concepts:special right triangle incircle, incenter, and inradius circumcircle, circumcenter, and circumradius

Difficulty rating: 1670

Small Hint:

Let each leg have length $a$ and express both radii in terms of $a$

Big Hint:

For a right triangle, the circumradius is half the hypotenuse; the inradius is $\frac{a+a-a\sqrt2}{2}$

Solution:

With both legs of length $a$ and hypotenuse $a\sqrt2,$ $\begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2}. \end{aligned}$ Therefore $\begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2. \end{aligned}$

Therefore, the correct answer is A.

17.

If $i^2=-1,$ then $(1+i)^{20}-(1-i)^{20}$ equals

$-1024$

$-1024i$

$0$

$1024$

$1024i$

Concepts:complex number exponent

Difficulty rating: 1540

Small Hint:

Compute the fourth power of each conjugate base

Big Hint:

Both $(1+i)^4$ and $(1-i)^4$ equal $-4$

Solution:

Since $(1+i)^2=2i,$ we have $(1+i)^4=-4.$ Likewise $(1-i)^2=-2i$ and $(1-i)^4=-4.$ Hence each twentieth power equals $(-4)^5=-1024,$ and their difference is $0.$

Therefore, the correct answer is C.

18.

If $\log_8 3=p$ and $\log_3 5=q,$ then, in terms of $p$ and $q,$ $\log_{10}5$ equals

$pq$

$\frac{3p+q}{5}$

$\frac{1+3pq}{p+q}$

$\frac{3pq}{1+3pq}$

$p^2+q^2$

Concepts:logarithm

Difficulty rating: 1780

Small Hint:

Convert every logarithm to base $2$

Big Hint:

The givens imply $\log_2 3=3p$ and $\log_2 5=3pq$

Solution:

From $p=\log_8 3,$ $\log_2 3=3p.$ Therefore $\log_2 5=(\log_2 3)(\log_3 5)=3pq.$ Since $\log_2 10=1+\log_2 5=1+3pq,$ $\log_{10}5=\frac{\log_2 5}{\log_2 10} =\frac{3pq}{1+3pq}.$

Therefore, the correct answer is D.

19.

In the adjoining figure $ABCD$ is a square and $CMN$ is an equilateral triangle. If the area of $ABCD$ is one square inch, then the area of $CMN$ in square inches is

$2\sqrt3-3$

$1-\frac{\sqrt3}{3}$

$\frac{\sqrt3}{4}$

$\frac{\sqrt2}{3}$

$4-2\sqrt3$

Concepts:equilateral triangle Pythagorean Theorem triangle area

Difficulty rating: 2070

Small Hint:

The square has side $1;$ set $DM=NB=x$ by comparing the equal sides $CM$ and $CN$

Big Hint:

Equate $CM^2=1+x^2$ with $MN^2=2(1-x)^2$

Solution:

The square has side $1.$ Since $CM=CN,$ symmetry of their squared lengths gives $DM=NB=x.$ Also, $\begin{aligned} CM^2&=1+x^2,\\ MN^2&=2(1-x)^2. \end{aligned}$ Equating these because $CMN$ is equilateral gives $x^2-4x+1=0.$ The root in $(0,1)$ is $x=2-\sqrt3.$

The triangle’s side has square $1+x^2,$ so its area is $\begin{aligned} A_\triangle&=\frac{\sqrt3}{4}(1+x^2)\\ &=\frac{\sqrt3}{4} \bigl(1+(2-\sqrt3)^2\bigr)\\ &=2\sqrt3-3. \end{aligned}$

Therefore, the correct answer is A.

20.

Let $\begin{aligned} T={}&\frac1{3-\sqrt8} -\frac1{\sqrt8-\sqrt7}\\ &+\frac1{\sqrt7-\sqrt6} -\frac1{\sqrt6-\sqrt5}\\ &+\frac1{\sqrt5-2}; \end{aligned}$ then

$T\lt1$

$T=1$

$1\lt T\lt2$

$T\gt2$

$\displaystyle T=\frac1{\substack{(3-\sqrt8)(\sqrt8-\sqrt7)(\sqrt7-\sqrt6)\\ {}\cdot(\sqrt6-\sqrt5)(\sqrt5-2)}}$

Concepts:rationalizing denominator telescoping

Difficulty rating: 1740

Small Hint:

Rationalize each denominator separately

Big Hint:

Every denominator is a difference of square roots whose squares differ by $1$

Solution:

Rationalizing the five terms gives $\begin{aligned} T={}&(3+\sqrt8)-(\sqrt8+\sqrt7)\\ &+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)\\ &+(\sqrt5+2)=5. \end{aligned}$ Thus $T\gt2.$

Therefore, the correct answer is D.

21.

In a geometric series of positive terms the difference between the fifth and fourth terms is $576,$ and the difference between the second and first terms is $9.$ What is the sum of the first five terms of this series?

$1061$

$1023$

$1024$

$768$

none of these

Concepts:geometric sequence summation

Difficulty rating: 1900

Small Hint:

Write the first term as $a$ and the common ratio as $r$

Big Hint:

Divide $ar^3(r-1)=576$ by $a(r-1)=9$

Solution:

Let the first term be $a$ and the ratio be $r.$ The conditions give $\begin{aligned} ar^3(r-1)&=576,\\ a(r-1)&=9. \end{aligned}$ Their quotient gives $r^3=64,$ so $r=4.$ Then $3a=9,$ hence $a=3.$ The sum is $\begin{aligned} S&=3(1+4+16+64+256)\\ &=3\cdot341\\ &=1023. \end{aligned}$

Therefore, the correct answer is B.

22.

The minimum value of $\sin\frac A2-\sqrt3\cos\frac A2$ is attained when $A$ is

$-180^\circ$

$60^\circ$

$120^\circ$

$0^\circ$

none of these

Concepts:trigonometric identity trigonometry

Difficulty rating: 1780

Small Hint:

Combine the sine and cosine into one shifted sine with amplitude $2$

Big Hint:

Use $\sin u-\sqrt3\cos u=2\sin(u-60^\circ)$

Solution:

Let $u=\frac{A}{2}.$ Then $\sin u-\sqrt3\cos u$ equals $2\sin(u-60^\circ).$ Its minimum occurs when $u-60^\circ=270^\circ+360^\circ n,$ so $A=660^\circ+720^\circ n$ for an integer $n.$ None of the first four choices has this form.

Therefore, the correct answer is E.

23.

In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r,$ with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T''$ as point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is

$\frac{25}{6}$

$6$

$\frac{25}{4}$

a number other than $\frac{25}{6},$ $6,$ or $\frac{25}{4}$

not determinable from the given information

Concepts:tangent line right triangle similarity

Difficulty rating: 2070

Small Hint:

The center lies on each angle bisector formed by two tangents from the same external point

Big Hint:

Show that $\angle POQ=90^\circ$ and use the altitude-to-hypotenuse theorem at $T''$

Solution:

The radii to the tangency points make congruent right triangles from each external point, so $OP$ and $OQ$ bisect the two tangent angles. Because the other two tangents are parallel, these half-angles add to $90^\circ;$ hence $\angle POQ=90^\circ.$

Thus $OT''=r$ is the altitude to the hypotenuse $PQ.$ Equal tangent segments give $PT''=PT=4$ and $QT''=QT'=9.$ The altitude theorem yields $r^2=(PT'')(QT'')=4\cdot9=36,$ so $r=6.$

Therefore, the correct answer is B.

24.

A fair die is rolled six times. The probability of rolling at least a five at least five times is

$\frac{13}{729}$

$\frac{12}{729}$

$\frac{2}{729}$

$\frac{3}{729}$

none of these

Concepts:binomial probability dice (probability)

Difficulty rating: 1900

Small Hint:

A roll is a success with probability $\frac{2}{6}=\frac{1}{3}$

Big Hint:

Add the disjoint cases of exactly five successes and exactly six successes

Solution:

Let a success be a $5$ or $6,$ so its probability is $\frac{1}{3}.$ The desired probability is $\begin{aligned} &\binom65\left(\frac13\right)^5 \left(\frac23\right) +\left(\frac13\right)^6\\ &\qquad=\frac{12}{729}+\frac1{729}\\ &\qquad=\frac{13}{729}. \end{aligned}$

Therefore, the correct answer is A.

25.

In parallelogram $ABCD$ of the accompanying diagram, line $DP$ is drawn bisecting $BC$ at $N$ and meeting $AB$ (extended) at $P.$ From vertex $C,$ line $CQ$ is drawn bisecting side $AD$ at $M$ and meeting $AB$ (extended) at $Q.$ Lines $DP$ and $CQ$ meet at $O.$ If the area of parallelogram $ABCD$ is $k,$ then the area of triangle $QPO$ is equal to

$k$

$\frac{6k}{5}$

$\frac{9k}{8}$

$\frac{5k}{4}$

$2k$

Concepts:parallelogram coordinate geometry triangle area

Difficulty rating: 2100

Small Hint:

Use $A=(0,0),$ $B=(1,0),$ $D=(u,v),$ and $C=(u+1,v)$

Big Hint:

The midpoint lines meet the baseline at $Q=(-1,0)$ and $P=(2,0)$ ; then find $O$ ’s height

Solution:

Choose coordinates $A=(0,0),$ $B=(1,0),$ $D=(u,v),$ and $C=(u+1,v),$ so $k=v.$ The midpoints are $\begin{aligned} M&=(\frac{u}{2},\frac{v}{2}),\\ N&=(1+\frac{u}{2},\frac{v}{2}). \end{aligned}$ Extending $CM$ and $DN$ to $y=0$ gives $Q=(-1,0)$ and $P=(2,0).$

Parameterizing $DP$ and $CQ$ shows their intersection is $O=(\frac{3u}{4}+\frac{1}{2},\frac{3v}{4}).$ Thus $QP=3$ and the height from $O$ is $\frac{3v}{4},$ so $\begin{aligned} [QPO]&=\frac12(3)\left(\frac{3v}{4}\right)\\ &=\frac{9v}{8}=\frac{9k}{8}. \end{aligned}$

Therefore, the correct answer is C.

26.

The number of distinct positive integral divisors of $(30)^4$ excluding $1$ and $(30)^4$ is

$100$

$125$

$123$

$30$

none of these

Concepts:prime factorization factor counting

Difficulty rating: 1790

Small Hint:

Factor $30^4$ into powers of $2,$ $3,$ and $5$

Big Hint:

Choose each prime exponent independently from $0$ through $4,$ then remove two divisors

Solution:

Because $30^4=2^4\cdot3^4\cdot5^4,$ a divisor independently chooses each of three exponents from $0,$ $1,$ $2,$ $3,$ $4.$ There are $5^3=125$ divisors. Excluding $1$ and $30^4$ leaves $125-2=123.$

Therefore, the correct answer is C.

27.

If $f(x)=3x+2$ for all real $x,$ then the statement:

“ $|f(x)+4|\lt a$ whenever $|x+2|\lt b$ and $a\gt0$ and $b\gt0$ ”

is true when

$b\le \frac{a}{3}$

$b\gt \frac{a}{3}$

$a\le \frac{b}{3}$

$a\gt \frac{b}{3}$

The statement is never true.

Concepts:absolute value function inequality

Difficulty rating: 1670

Small Hint:

Factor $f(x)+4$ in terms of $x+2$

Big Hint:

The condition becomes $3|x+2|\lt a$ whenever $|x+2|\lt b$

Solution:

We have $|f(x)+4|=|3x+6|=3|x+2|.$ If $|x+2|\lt b$ and $b\le \frac{a}{3},$ then $3|x+2|\lt3b\le a,$ as required. If $b\gt \frac{a}{3},$ one can choose $|x+2|$ strictly between $\frac{a}{3}$ and $b,$ so the statement fails. Thus the exact condition is $b\le \frac{a}{3}.$

Therefore, the correct answer is A.

28.

Which of the following is satisfied by all numbers $x$ of the form $x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}},$ where $a_1$ is $0$ or $2,$ $a_2$ is $0$ or $2,$ $\ldots,$ $a_{25}$ is $0$ or $2?$

$0\le x\lt\frac{1}{3}$

$\frac{1}{3}\le x\lt\frac{2}{3}$

$\frac{2}{3}\le x\lt1$

$0\le x\lt\frac{1}{3}$ or $\frac{2}{3}\le x\lt1$

$\frac{1}{2}\le x\le\frac{3}{4}$

Concepts:number base geometric sequence bounding to limit cases

Difficulty rating: 2090

Small Hint:

Separate the cases $a_1=0$ and $a_1=2$

Big Hint:

Bound the remaining terms by the infinite geometric tail $\sum_{n=2}^{\infty}\frac{2}{3^n}$

Solution:

If $a_1=0,$ then $0\le x\le\sum_{n=2}^{25}\frac2{3^n} \lt\sum_{n=2}^{\infty}\frac2{3^n}=\frac13.$ If $a_1=2,$ then $x\ge\frac{2}{3},$ while $x\le\sum_{n=1}^{25}\frac2{3^n} \lt\sum_{n=1}^{\infty}\frac2{3^n}=1.$ Hence every such $x$ lies in one of the two stated outer thirds.

Therefore, the correct answer is D.

29.

For $p=1,$ $2,$ $\ldots,$ $10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1;$ then $S_1+S_2+\cdots+S_{10}$ is

$80{,}000$

$80{,}200$

$80{,}400$

$80{,}600$

$80{,}800$

Concepts:arithmetic sequence summation

Difficulty rating: 1740

Small Hint:

Use the $40$ -term arithmetic-series formula for a fixed $p$

Big Hint:

Simplify $S_p=20\bigl(2p+39(2p-1)\bigr)$ before summing over $p$

Solution:

For each $p,$ $\begin{aligned} S_p&=\frac{40}{2} \bigl(2p+39(2p-1)\bigr)\\ &=1600p-780. \end{aligned}$ Since $1+2+\cdots+10=55,$ therefore $\begin{aligned} \sum_{p=1}^{10}S_p &=1600(55)-7800\\ &=80{,}200. \end{aligned}$

Therefore, the correct answer is B.

30.

A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If $R$ is the ratio of the lesser part to the greater part, then the value of $R^{\left[R^{\left(R^2+R^{-1}\right)}+R^{-1}\right]}+R^{-1}$ is

$2$

$2R$

$R^{-1}$

$2+R^{-1}$

$2+R$

Concepts:quadratic exponent substitution

Difficulty rating: 2100

Small Hint:

Translate the division condition into $R=\frac{1}{1+R}$

Big Hint:

From $R^2+R=1,$ derive $R^{-1}=R+1$ and simplify from the innermost exponent outward

Solution:

Scale the greater part to $1,$ making the lesser part $R.$ The condition gives $R=\frac{1}{1+R},$ so $R^2+R=1,\qquad R^{-1}=R+1.$ Consequently $R^2+R^{-1}=R^2+R+1=2.$ Working outward, the exponent of the outer $R$ becomes $\begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2. \end{aligned}$ The whole expression is therefore $R^2+R^{-1}=2.$

Therefore, the correct answer is A.