1965 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The number of real values of $x$ satisfying the equation $2^{2x^2-7x+5}=1$ is:

$0$

$1$

$2$

$4$

more than $4$

Concepts:exponent quadratic factoring

Difficulty rating: 1080

Small Hint:

A positive base other than $1$ has value $1$ only at exponent $0$

Big Hint:

Solve $2x^2-7x+5=0$

Solution:

Since the base is $2,$ the exponent must be zero. Factoring gives $\begin{gathered} 2x^2-7x+5=0,\\ (2x-5)(x-1)=0. \end{gathered}$ Thus $x=\frac{5}{2}$ or $x=1,$ giving two real values.

Therefore, the correct answer is C.

2.

A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side is:

$1:1$

$1:6$

$1:\pi$

$3:\pi$

$6:\pi$

Concepts:regular polygon arc circumference ratio and proportion

Difficulty rating: 1520

Small Hint:

Each side of an inscribed regular hexagon equals the circle’s radius

Big Hint:

The shorter arc is one sixth of the circumference

Solution:

If the radius is $r,$ the hexagon side is $r.$ Its intercepted minor arc has length $\frac{2\pi r}{6}=\frac{\pi r}{3}.$ The ratio is $r:(\frac{\pi r}{3})=3:\pi.$

Therefore, the correct answer is D.

3.

The expression $81^{-(2^{-2})}$ has the same value as:

$\dfrac1{81}$

$\dfrac13$

$3$

$81$

$81^4$

Concepts:exponent radical order of operations

Difficulty rating: 1350

Small Hint:

Evaluate the exponent $2^{-2}$ first

Big Hint:

Write $81$ as $3^4$

Solution:

Since $2^{-2}=\frac{1}{4},$ $\begin{aligned} 81^{-(2^{-2})} &=81^{-\frac{1}{4}}\\ &=(3^4)^{-\frac{1}{4}}=\frac13. \end{aligned}$

Therefore, the correct answer is B.

4.

Line $l_2$ intersects line $l_1$ and line $l_3$ is parallel to $l_1.$ The three lines are distinct and lie in a plane. The number of points equidistant from all three lines is:

$0$

$1$

$2$

$4$

$8$

Concepts:parallel lines angle bisector counting intersections

Difficulty rating: 1530

Small Hint:

Points equidistant from the two parallel lines lie on their midway parallel

Big Hint:

Intersect that midway line with the two angle bisectors of $l_1$ and $l_2$

Solution:

Points equidistant from parallel lines $l_1$ and $l_3$ lie on the unique line midway between them. Points equidistant from intersecting lines $l_1$ and $l_2$ lie on either of their two angle bisectors. Each angle bisector meets the midway parallel once, giving two points.

Therefore, the correct answer is C.

5.

When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:

$15$

$45$

$114$

$135$

$150$

Concepts:repeating decimal fraction

Difficulty rating: 1260

Small Hint:

Multiply the decimal by $100$ and subtract the original number

Big Hint:

Simplify $\frac{36}{99}$

Solution:

Let $z=0.363636\ldots.$ Then $100z-z=36,$ so $z=\frac{36}{99}=\frac{4}{11}.$ The requested sum is $4+11=15.$

Therefore, the correct answer is A.

6.

If $10^{\log_{10}9}=8x+5,$ then $x$ equals:

$0$

$\dfrac12$

$\dfrac58$

$\dfrac98$

$\dfrac{2\log_{10}3-5}{8}$

Concepts:logarithm linear equation

Difficulty rating: 1280

Small Hint:

Use $10^{\log_{10}9}=9$

Big Hint:

Solve the resulting linear equation $9=8x+5$

Solution:

The exponential and logarithm cancel, so $10^{\log_{10}9}=9.$ Thus $8x+5=9,$ giving $x=\frac{1}{2}.$

Therefore, the correct answer is B.

7.

The sum of the reciprocals of the roots of the equation $ax^2+bx+c=0$ is:

$\dfrac1a+\dfrac1b$

$-\dfrac cb$

$\dfrac bc$

$-\dfrac ab$

$-\dfrac bc$

Concepts:quadratic Vieta’s Formulas algebraic manipulation

Difficulty rating: 1510

Small Hint:

If the roots are $r,s,$ then $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}$

Big Hint:

Use $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$

Solution:

By Vieta’s formulas, $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}.$ Therefore $\begin{aligned} \frac1r+\frac1s &=\frac{r+s}{rs}\\ &=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac bc. \end{aligned}$

Thus, the correct answer is E.

8.

One side of a given triangle is $18$ inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle. The length of this segment, in inches, is:

$6\sqrt6$

$9\sqrt2$

$12$

$6\sqrt3$

$9$

Difficulty rating: 1580

Small Hint:

The smaller triangle above the segment has two-thirds of the original area

Big Hint:

Linear scale factors are square roots of area scale factors

Solution:

The trapezoid occupies one third of the area, so the smaller similar triangle occupies two thirds. Its linear scale factor is $\sqrt{\frac{2}{3}}.$ Hence the parallel segment has length $18\sqrt{\frac23}=6\sqrt6.$

Therefore, the correct answer is A.

9.

The vertex of the parabola $y=x^2-8x+c$ will be a point on the $x$ -axis if the value of $c$ is:

$-16$

$-4$

$4$

$8$

$16$

Concepts:parabola completing the square

Difficulty rating: 1360

Small Hint:

Complete the square in $x^2-8x+c$

Big Hint:

The vertex’s $y$ -coordinate must equal $0$

Solution:

Completing the square gives $y=(x-4)^2+c-16.$ The vertex is $(4,c-16),$ so it lies on the $x$ -axis when $c=16.$

Therefore, the correct answer is E.

10.

The statement $x^2-x-6\lt0$ is equivalent to the statement:

$-2\lt x\lt3$

$x\gt-2$

$x\lt3$

$x\gt3$ and $x\lt-2$

$x\gt3$ or $x\lt-2$

Concepts:quadratic factoring inequality

Difficulty rating: 1470

Small Hint:

Factor $x^2-x-6$

Big Hint:

A positive-leading quadratic is negative between its two roots

Solution:

We have $x^2-x-6=(x-3)(x+2).$ This product is negative exactly between its roots, so $-2\lt x\lt3.$

Therefore, the correct answer is A.

11.

Consider the statements:

$\begin{aligned} \mathrm{I:}\quad &(\sqrt{-4})(\sqrt{-16})\\ &=\sqrt{(-4)(-16)}, \end{aligned}$

$\mathrm{II:}\quad \sqrt{(-4)(-16)}=\sqrt{64},$

and

$\mathrm{III:}\quad \sqrt{64}=8.$

Of these the following are incorrect:

none

$\mathrm{I}$ only

$\mathrm{II}$ only

$\mathrm{III}$ only

$\mathrm{I}$ and $\mathrm{III}$ only

Concepts:radical complex number

Difficulty rating: 1500

Small Hint:

The product rule for square roots requires nonnegative real radicands

Big Hint:

Evaluate statements II and III directly

Solution:

Statement I improperly applies $\sqrt a\sqrt b=\sqrt{ab}$ to negative radicands; over the reals its left side is undefined, and with principal complex roots it equals $-8,$ not $8.$ Statements II and III both correctly simplify to $8.$ Thus only I is incorrect.

Therefore, the correct answer is B.

12.

A rhombus is inscribed in triangle $ABC$ in such a way that one of its vertices is $A$ and two of its sides lie along $AB$ and $AC.$ If $AC=6$ inches, $AB=12$ inches, and $BC=8$ inches, the side of the rhombus, in inches, is:

$2$

$3$

$3\dfrac12$

$4$

$5$

Concepts:rhombus similarity ratio and proportion

Difficulty rating: 1720

Small Hint:

Let the rhombus side be $s$ , with adjacent vertices $\frac{s}{12}$ of the way along $AB$ and $\frac{s}{6}$ along $AC$

Big Hint:

The opposite rhombus vertex lies on $BC$ , so its two affine coefficients sum to $1$

Solution:

Use $A$ as the origin and let the vectors to $B,C$ be $u,v.$ If the rhombus side is $s,$ its opposite vertex is $(\frac{s}{12})u+(\frac{s}{6})v.$ A point lies on $BC$ when these coefficients sum to $1.$ Thus $\frac{s}{12}+\frac{s}{6}=1,$ giving $s=4.$

Therefore, the correct answer is D.

13.

Let $n$ be the number of number-pairs $(x,y)$ which satisfy $5y-3x=15$ and $x^2+y^2\leq16.$ Then $n$ is:

$0$

$1$

$2$

more than two, but finite

greater than any finite number

Concepts:circle distance formula inequality

Difficulty rating: 1610

Small Hint:

The inequality describes the disk of radius $4$ centered at the origin

Big Hint:

Compare the line’s distance $\frac{15}{\sqrt{34}}$ from the origin with $4$

Solution:

The line’s distance from the origin is $\frac{15}{\sqrt{(-3)^2+5^2}}=\frac{15}{\sqrt{34}}\lt4.$ It therefore crosses the interior of the disk $x^2+y^2\leq16$ in a segment containing infinitely many points.

Thus, the correct answer is E.

14.

The sum of the numerical coefficients in the complete expansion of $(x^2-2xy+y^2)^7$ is:

$0$

$7$

$14$

$128$

$128^2$

Concepts:polynomial substitution

Difficulty rating: 1470

Small Hint:

Set both variables equal to $1$ to obtain the coefficient sum

Big Hint:

The expression inside the seventh power becomes $1-2+1$

Solution:

Setting $x=y=1$ makes every monomial equal $1,$ so the value is the sum of the numerical coefficients. It is $(1-2+1)^7=0.$

Therefore, the correct answer is A.

15.

The symbol $25_b$ represents a two-digit number in the base $b.$ If the number $52_b$ is double the number $25_b,$ then $b$ is:

$7$

$8$

$9$

$11$

$12$

Concepts:number base place value linear equation

Difficulty rating: 1500

Small Hint:

Translate $25_b$ as $2b+5$ and $52_b$ as $5b+2$

Big Hint:

Solve $5b+2=2(2b+5)$

Solution:

In ordinary notation, $25_b=2b+5$ and $52_b=5b+2.$ Thus $5b+2=2(2b+5),$ so $b=8.$

Therefore, the correct answer is B.

16.

Let line $AC$ be perpendicular to line $CE.$ Connect $A$ to the midpoint $D$ of $CE,$ and connect $E$ to the midpoint $B$ of $AC.$ If $AD$ and $EB$ intersect in point $F,$ and $BC=CD=15$ inches, then the area of triangle $DFE,$ in square inches, is:

$50$

$50\sqrt2$

$75$

$\dfrac{15}{2}\sqrt{105}$

$100$

Concepts:median (geometry)centroid coordinate geometry triangle area

Difficulty rating: 1830

Small Hint:

Since $B,D$ are midpoints and $BC=CD=15,$ both perpendicular legs have length $30$

Big Hint:

Use coordinates $C=(0,0), A=(0,30)$ and $E=(30,0)$ ; the two medians meet at the centroid

Solution:

Put $C=(0,0),\ A=(0,30),$ and $E=(30,0).$ Then $B=(0,15)$ and $D=(15,0).$ Lines $AD$ and $EB$ are medians of triangle $ACE,$ so they meet at the centroid $F=(10,10).$ Segment $DE$ has length $15$ and $F$ is $10$ units above its line. Therefore $[DFE]=\frac12(15)(10)=75.$

Thus, the correct answer is C.

17.

Given the true statement: The picnic on Sunday will not be held only if the weather is not fair. We can then conclude that:

If the picnic is held, Sunday’s weather is undoubtedly fair.

If the picnic is not held, Sunday’s weather is possibly unfair.

If it is not fair Sunday, the picnic will not be held.

If it is fair Sunday, the picnic may be held.

If it is fair Sunday, the picnic will be held.

Concepts:logical deduction

Difficulty rating: 1650

Small Hint:

Translate “ $P$ only if $Q$ ” as $P\to Q$

Big Hint:

Take the contrapositive of “not held implies not fair”

Solution:

The statement says: if the picnic is not held, then the weather is not fair. Its contrapositive is: if the weather is fair, then the picnic will be held.

Therefore, the correct answer is E.

18.

If $1-y$ is used as an approximation to the value of $\dfrac1{1+y},$ $|y|\lt1,$ the ratio of the error made to the correct value is:

$y$

$y^2$

$\dfrac1{1+y}$

$\dfrac y{1+y}$

$\dfrac{y^2}{1+y}$

Concepts:estimation algebraic manipulation

Difficulty rating: 1580

Small Hint:

Subtract the approximation $1-y$ from the correct value

Big Hint:

Divide that error by $\frac{1}{1+y}$

Solution:

The error is $\frac1{1+y}-(1-y)=\frac{y^2}{1+y}.$ Dividing by the correct value $\frac{1}{1+y}$ gives the ratio $y^2.$

Therefore, the correct answer is B.

19.

If $x^4+4x^3+6px^2+4qx+r$ is exactly divisible by $x^3+3x^2+9x+3,$ the value of $(p+q)r$ is:

$-18$

$12$

$15$

$27$

$45$

Concepts:polynomial divisibility algebraic manipulation

Difficulty rating: 1830

Small Hint:

The quotient must be monic and linear

Big Hint:

Subtract $(x+1)(x^3+3x^2+9x+3)$ and compare coefficients

Solution:

The leading terms force quotient $x+1.$ Expanding, $\begin{aligned} &(x+1)(x^3+3x^2+9x+3)\\ &\quad=x^4+4x^3+12x^2+12x+3. \end{aligned}$ Thus $6p=12,\ 4q=12,\ r=3,$ so $p=2,\ q=3$ and $(p+q)r=15.$

Therefore, the correct answer is C.

20.

For every $n$ the sum $S_n$ of $n$ terms of an arithmetic progression is $2n+3n^2.$ The $r$ th term is:

$3r^2$

$3r^2+2r$

$6r-1$

$5r+5$

$6r+2$

Concepts:arithmetic sequence summation algebraic manipulation

Difficulty rating: 1360

Small Hint:

The $r$ th term equals $S_r-S_{r-1}$

Big Hint:

Compute the difference of the two quadratic expressions

Solution:

The $r$ th term is the difference of consecutive partial sums. Substitution and simplification give $S_r-S_{r-1}=6r-1.$

Therefore, the correct answer is C.

21.

It is possible to choose $x\gt\dfrac23$ in such a way that the value of

$\log_{10}(x^2+3)-2\log_{10}x$

is:

negative

zero

one

smaller than any positive number that might be specified

greater than any positive number that might be specified

Concepts:logarithm bounding to limit cases

Difficulty rating: 1860

Small Hint:

Combine the logarithms into $\log_{10}(1+\frac{3}{x^2})$

Big Hint:

Let $x$ grow without bound

Solution:

The expression is $\begin{gathered} \log_{10}\left(\frac{x^2+3}{x^2}\right)\\ =\log_{10}\left(1+\frac3{x^2}\right). \end{gathered}$ It is always positive, but tends to $0$ as $x$ grows. Therefore it can be made smaller than any specified positive number.

Thus, the correct answer is D.

22.

If $a_2\ne0$ and $r,$ $s$ are the roots of $a_0+a_1x+a_2x^2=0,$ then the equality

$\begin{aligned} a_0+a_1x+a_2x^2 &=a_0\left(1-\frac xr\right)\\ &\quad\cdot\left(1-\frac xs\right) \end{aligned}$

holds:

for all values of $x,$ $a_0\ne0$

for all values of $x$

only when $x=0$

only when $x=r$ or $x=s$

only when $x=r$ or $x=s,$ $a_0\ne0$

Concepts:quadratic Vieta’s Formulas factoring

Difficulty rating: 2000

Small Hint:

The right side requires both roots to be nonzero

Big Hint:

Use $rs=\frac{a_0}{a_2}$ to compare the leading coefficients

Solution:

If $a_0\ne0,$ then $rs=\frac{a_0}{a_2}\ne0$ and $\begin{gathered} a_0\left(1-\frac xr\right) \left(1-\frac xs\right)\\ =\frac{a_0}{rs}(x-r)(x-s)\\ =a_2(x-r)(x-s), \end{gathered}$ which is the original polynomial for every $x.$ If $a_0=0,$ at least one denominator on the right is zero.

Therefore, the correct answer is A.

23.

If we write $|x^2-4|\lt N$ for all $x$ such that $|x-2|\lt0.01,$ the smallest value we can use for $N$ is:

$0.0301$

$0.0349$

$0.0399$

$0.0401$

$0.0499$

Concepts:absolute value inequality bounding to limit cases

Difficulty rating: 1910

Small Hint:

Factor $|x^2-4|=|x-2||x+2|$

Big Hint:

The larger one-sided bound occurs as $x$ approaches $2.01$

Solution:

For $1.99\lt x\lt2.01,$ $\begin{gathered} |x^2-4|=|x-2||x+2|\\ \lt(0.01)(4.01)=0.0401. \end{gathered}$ Values with $x$ arbitrarily close to $2.01$ make the expression arbitrarily close to $0.0401,$ so no smaller bound works.

Therefore, the correct answer is D.

24.

Given the sequence $10^{\frac{1}{11}},$ $10^{\frac{2}{11}},$ $10^{\frac{3}{11}},$ $\ldots,$ $10^{\frac{n}{11}},$ the smallest value of $n$ such that the product of the first $n$ members of this sequence exceeds $100{,}000$ is:

$7$

$8$

$9$

$10$

$11$

Concepts:exponent summation inequality

Difficulty rating: 1530

Small Hint:

Add the exponents in the product

Big Hint:

Require $\frac{n(n+1)}{22}\gt5$

Solution:

The product is $10^{\frac{1+2+\cdots+n}{11}} =10^{\frac{n(n+1)}{22}}.$ To exceed $100{,}000=10^5,$ we need $n(n+1)\gt110.$ At $n=10$ equality holds, while $n=11$ exceeds it.

Therefore, the correct answer is E.

25.

Let $ABCD$ be a quadrilateral with $AB$ extended to $E$ so that $AB=BE.$ Lines $AC$ and $CE$ are drawn to form angle $ACE.$ For this angle to be a right angle it is necessary that quadrilateral $ABCD$ have:

all angles equal

all sides equal

two pairs of equal sides

one pair of equal sides

one pair of equal angles

Concepts:right triangle midpoint circumcircle, circumcenter, and circumradius

Difficulty rating: 1830

Small Hint:

Because $AB=BE,$ point $B$ is the midpoint of $AE$

Big Hint:

In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices

Solution:

If $\angle ACE=90^\circ,$ then $AE$ is the hypotenuse of right triangle $ACE.$ Its midpoint $B$ is equidistant from $A,C,E.$ Thus $AB=BC,$ so quadrilateral $ABCD$ necessarily has at least one pair of equal sides. No condition involving $D$ follows.

Therefore, the correct answer is D.

26.

For the numbers $a,$ $b,$ $c,$ $d,$ $e$ define $m$ to be the arithmetic mean of all five numbers; $k$ to be the arithmetic mean of $a$ and $b;$ $l$ to be the arithmetic mean of $c,$ $d,$ and $e;$ and $p$ to be the arithmetic mean of $k$ and $l.$ Then, no matter how $a,$ $b,$ $c,$ $d,$ $e$ are chosen, we shall always have:

$m=p$

$m\geq p$

$m\gt p$

$m\lt p$

none of these

Concepts:mean weighted mean counterexample

Difficulty rating: 1610

Small Hint:

Express $m$ in terms of the subgroup means $k$ and $l$

Big Hint:

Compare $m=\frac{2k+3l}{5}$ with $p=\frac{k+l}{2}$

Solution:

We have $m=\frac{2k+3l}{5}$ and $p=\frac{k+l}{2},$ so $m-p=\frac{l-k}{10}.$ This difference may be positive, zero, or negative depending on the chosen numbers. None of the first four relations always holds.

Therefore, the correct answer is E.

27.

When $y^2+my+2$ is divided by $y-1$ the quotient is $f(y)$ and the remainder is $R_1.$ When $y^2+my+2$ is divided by $y+1$ the quotient is $g(y)$ and the remainder is $R_2.$ If $R_1=R_2,$ then $m$ is:

$0$

$1$

$2$

$-1$

an undetermined constant

Concepts:polynomial substitution linear equation

Difficulty rating: 1470

Small Hint:

Use the remainder theorem at $y=1$ and $y=-1$

Big Hint:

Set $m+3=3-m$

Solution:

The remainder theorem gives $R_1=1+m+2=m+3$ and $R_2=1-m+2=3-m.$ Equality implies $m=0.$

Therefore, the correct answer is A.

28.

An escalator (moving staircase) of $n$ uniform steps visible at all times descends at constant speed. Two boys, $A$ and $Z,$ walk down the escalator steadily as it moves, $A$ negotiating twice as many escalator steps per minute as $Z.$ $A$ reaches the bottom after taking $27$ steps while $Z$ reaches the bottom after taking $18$ steps. Then $n$ is:

$63$

$54$

$45$

$36$

$30$

Concepts:relative speed system of equations

Difficulty rating: 1880

Small Hint:

Let $Z$ ’s stepping rate be $v$ and the escalator rate be $e$

Big Hint:

Equate $27+\frac{27e}{2v}$ and $18+\frac{18e}{v}$

Solution:

Let $Z$ take $v$ steps per minute, so $A$ takes $2v,$ and let the escalator contribute $e$ steps per minute. Their travel times are $\frac{27}{2v}$ and $\frac{18}{v}.$ Thus $n=27+\frac{27e}{2v} =18+\frac{18e}{v}.$ This gives $\frac{e}{v}=2,$ and then $n=18+18(2)=54.$

Therefore, the correct answer is B.

29.

Of $28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only. No student takes English only or History only, and six students take Mathematics and History, but no English. The number taking English and History only is five times the number taking all three subjects. If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:

$5$

$6$

$7$

$8$

$9$

Concepts:Venn Diagram basic counting Diophantine Equation

Difficulty rating: 1830

Small Hint:

Let $t$ be the all-three count and $x$ both the Math-only and Math-English-only count

Big Hint:

The total becomes $2x+6+6t=28$

Solution:

Let $t$ be the number taking all three and $x$ the number taking Mathematics and English only, which also equals the Mathematics-only count. The disjoint-region total is $x+x+6+5t+t=28,$ so $x=11-3t.$ Since $t$ is positive and even, $t=2$ is the only value giving a nonnegative listed count, and $x=5.$

Therefore, the correct answer is A.

30.

Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ in $D.$ At $D$ a tangent is drawn cutting leg $CA$ in $F.$ This information is not sufficient to prove that:

$DF$ bisects $CA$

$DF$ bisects $\angle CDA$

$DF=FA$

$\angle A=\angle BCD$

$\angle CFD=2\angle A$

Concepts:tangent line right triangle angle bisector theorem counterexample

Difficulty rating: 2290

Small Hint:

Use coordinates $C=(0,0), A=(a,0), B=(0,b)$

Big Hint:

The tangent meets $CA$ at its midpoint; test the angle-bisector claim with the angle-bisector theorem

Solution:

Put $C=(0,0),\ A=(a,0),$ and $B=(0,b).$ The second intersection of $AB$ with the circle of diameter $BC$ is $D=\left(\frac{ab^2}{a^2+b^2}, \frac{a^2b}{a^2+b^2}\right).$ The tangent at $D$ meets $CA$ at $F=(\frac{a}{2},0).$ Hence $DF$ bisects $CA;$ also $FC=FD=FA$ by equal tangents from $F,$ which supplies choices C and E, and the circle/right-triangle angles supply choice D.

If $DF$ bisected $\angle CDA,$ the angle-bisector theorem in triangle $CDA$ would require $\frac{CF}{FA}=\frac{CD}{DA}.$ The left side is $1,$ while $\frac{CD}{DA}=\frac{ab}{a^2}=\frac ba,$ which need not be $1.$ Thus choice B is not provable.

Therefore, the correct answer is B.

31.

The number of real values of $x$ satisfying the equality $(\log_a x)(\log_b x)=\log_a b,$ where $a$ and $b$ are positive constants different from $1,$ is:

$0$

$1$

$2$

an integer greater than $2$

not finite

Concepts:logarithm quadratic algebraic manipulation

Difficulty rating: 2000

Small Hint:

Write every logarithm with natural logs

Big Hint:

The equation reduces to $(\ln x)^2=(\ln b)^2$

Solution:

Change of base gives $\frac{\ln x}{\ln a}\frac{\ln x}{\ln b} =\frac{\ln b}{\ln a}.$ Since the denominators are nonzero, $(\ln x)^2=(\ln b)^2.$ Thus $x=b$ or $x=\frac{1}{b}.$ These are distinct because $b\ne1,$ so there are two values.

Therefore, the correct answer is C.

32.

An article costing $C$ dollars is sold for $\$100$ at a loss of $x$ percent of the selling price. It is then resold at a profit of $x$ percent of the new selling price $S'.$ If the difference between $S'$ and $C$ is $1\dfrac19$ dollars, then $x$ is:

undetermined

$\dfrac{80}{9}$

$10$

$\dfrac{95}{9}$

$\dfrac{100}{9}$

Concepts:percentage money quadratic

Difficulty rating: 1950

Small Hint:

The first loss makes $C=100+x$

Big Hint:

The resale condition gives $S'=\frac{10000}{100-x}$

Solution:

A loss of $x\%$ of the $\$100$ selling price means $C=100+x.$ On resale, $S'-100=(\frac{x}{100})S',$ so $S'=\frac{10000}{100-x}.$ Therefore $\begin{gathered} \frac{10000}{100-x}-(100+x)\\ =\frac{x^2}{100-x} =\frac{10}{9}. \end{gathered}$ Thus $9x^2+10x-1000=0,$ whose positive root is $x=10.$

Therefore, the correct answer is C.

33.

If the number $15!,$ that is, $15\cdot14\cdot13\cdots1,$ ends with $k$ zeros when given to the base $12$ and ends with $h$ zeros when given to the base $10,$ then $k+h$ equals:

$5$

$6$

$7$

$8$

$9$

Concepts:factorial trailing zeros prime factorization number base

Difficulty rating: 2000

Small Hint:

Count the powers of $2,$ $3,$ and $5$ in $15!$

Big Hint:

A base- $12$ zero uses $2^2\cdot3$ , while a base- $10$ zero uses $2\cdot5$

Solution:

The prime valuations are $\begin{gathered} v_2(15!)=11,\\ v_3(15!)=6,\\ v_5(15!)=3. \end{gathered}$ Hence $k=\min(\lfloor\frac{11}{2}\rfloor,6)=5$ in base $12,$ and $h=\min(11,3)=3$ in base $10.$ Thus $k+h=8.$

Therefore, the correct answer is D.

34.

For $x\geq0,$ the smallest value of $\dfrac{4x^2+8x+13}{6(1+x)}$ is:

$1$

$2$

$\dfrac{25}{12}$

$\dfrac{13}{6}$

$\dfrac{34}{5}$

Concepts:optimization AM-GM Inequality substitution

Difficulty rating: 1950

Small Hint:

Set $t=x+1\geq1$

Big Hint:

Rewrite the expression as $\frac23t+\frac{3}{2t}$

Solution:

Let $t=x+1\geq1.$ Since $4x^2+8x+13=4t^2+9,$ the expression is $\frac23t+\frac{3}{2t}.$ By AM-GM this is at least $2\sqrt{(\frac{2}{3})(\frac{3}{2})}=2,$ with equality at $t=\frac{3}{2},$ which is allowed.

Therefore, the correct answer is B.

35.

The length of a rectangle is $5$ inches and its width is less than $4$ inches. The rectangle is folded so that two diagonally opposite vertices coincide. If the length of the crease is $\sqrt6,$ then the width is:

$\sqrt2$

$\sqrt3$

$2$

$\sqrt5$

$\sqrt{\frac{11}{2}}$

Concepts:paper folding perpendicular bisector coordinate geometry Pythagorean Theorem

Difficulty rating: 2290

Small Hint:

The crease is the perpendicular bisector of a rectangle diagonal

Big Hint:

If the width is $w\lt4,$ its intersections with the long sides differ horizontally by $\frac{w^2}{5}$

Solution:

Place the rectangle at $(0,0),$ $(5,0),$ $(5,w),$ and $(0,w).$ The crease sending $(0,0)$ to $(5,w)$ is $5x+wy=\frac{25+w^2}{2}.$ Because $w\lt4,$ it meets the two horizontal sides. Between those intersections the vertical change is $w$ and the horizontal change is $\frac{w^2}{5}.$ Thus $6=w^2+\frac{w^4}{25}.$ Setting $z=w^2$ gives $z^2+25z-150=0,$ so $z=5$ and $w=\sqrt5.$

Therefore, the correct answer is D.

36.

Given distinct straight lines $OA$ and $OB.$ From a point in $OA$ a perpendicular is drawn to $OB;$ from the foot of this perpendicular a line is drawn perpendicular to $OA.$ From the foot of this second perpendicular a line is drawn perpendicular to $OB;$ and so on indefinitely. The lengths of the first and second perpendiculars are $a$ and $b,$ respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:

$\dfrac b{a-b}$

$\dfrac a{a-b}$

$\dfrac{ab}{a-b}$

$\dfrac{b^2}{a-b}$

$\dfrac{a^2}{a-b}$

Concepts:geometric sequence similarity summation

Difficulty rating: 2190

Small Hint:

Successive right triangles formed by the two fixed lines are similar

Big Hint:

The perpendicular lengths form a geometric sequence with first term $a$ and ratio $\frac{b}{a}$

Solution:

Each new right triangle has the same acute angle, so the perpendicular lengths form a geometric sequence. Since the first two lengths are $a,b,$ the common ratio is $\frac{b}{a}.$ Convergence implies $0\lt \frac{b}{a}\lt1,$ and the sum is $\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}.$

Therefore, the correct answer is E.

37.

Point $E$ is selected on side $AB$ of triangle $ABC$ in such a way that $AE:EB=1:3,$ and point $D$ is selected on side $BC$ so that $CD:DB=1:2.$ The point of intersection of $AD$ and $CE$ is $F.$ Then

$\frac{EF}{FC}+\frac{AF}{FD}$

is:

$\dfrac45$

$\dfrac54$

$\dfrac32$

$2$

$\dfrac52$

Concepts:mass points ratio and proportion

Difficulty rating: 2170

Small Hint:

Assign masses $m_A=3,$ $m_B=1,$ and $m_C=2$

Big Hint:

Use the combined masses at $E$ and $D$ to read the two cevian ratios

Solution:

The side ratios are represented by masses $m_A=3,$ $m_B=1,$ and $m_C=2.$ Then $m_E=m_A+m_B=4$ and $m_D=m_B+m_C=3.$ Along $CE,$ $\frac{EF}{FC}=\frac{m_C}{m_E}=\frac12,$ while along $AD,$ $\frac{AF}{FD}=\frac{m_D}{m_A}=1.$ Their sum is $\frac{3}{2}.$

Therefore, the correct answer is C.

38.

$A$ takes $m$ times as long to do a piece of work as $B$ and $C$ together; $B$ takes $n$ times as long as $C$ and $A$ together; and $C$ takes $x$ times as long as $A$ and $B$ together. Then $x,$ in terms of $m$ and $n,$ is:

$\dfrac{2mn}{m+n}$

$\dfrac1{2(m+n)}$

$\dfrac1{m+n-mn}$

$\dfrac{1-mn}{m+n+2mn}$

$\dfrac{m+n+2}{mn-1}$

Concepts:rate system of equations algebraic manipulation

Difficulty rating: 2170

Small Hint:

Let the work rates of $A,$ $B,$ and $C$ be $a,$ $b,$ and $c$

Big Hint:

Translate the first two conditions as $b+c=ma$ and $c+a=nb$

Solution:

Let the work rates be $a,$ $b,$ and $c.$ The time statements give $\begin{gathered} b+c=ma,\\ c+a=nb,\\ a+b=xc. \end{gathered}$ From the first two, $\frac ba=\frac{m+1}{n+1}$ and $\frac ca=\frac{mn-1}{n+1}.$ Hence $x=\frac{a+b}{c} =\frac{m+n+2}{mn-1}.$

Therefore, the correct answer is E.

39.

A foreman noticed an inspector checking a $3$ -inch hole with a $2$ -inch plug and a $1$ -inch plug and suggested that two more gauges be inserted to be sure that the fit was snug. If the new gauges are alike, then the diameter $d$ of each, to the nearest hundredth of an inch, is:

$0.87$

$0.86$

$0.83$

$0.75$

$0.71$

Concepts:tangent circles coordinate geometry estimation

Difficulty rating: 2410

Small Hint:

Use radii $\frac{3}{2},$ $1,$ and $\frac{1}{2},$ and let a new gauge have radius $r$

Big Hint:

If its center is $(u,v)$ , subtract its three tangency-distance equations

Solution:

Place the hole’s center at the origin. The $2$ -inch and $1$ -inch plug centers are $(0,-\frac{1}{2})$ and $(0,1).$ Let a new gauge have radius $r$ and center $(u,v).$ Tangency to the two plugs gives $\begin{gathered} u^2+(v+\frac{1}{2})^2=(1+r)^2,\\ u^2+(v-1)^2=(\frac{1}{2}+r)^2, \end{gathered}$ while internal tangency to the hole gives $u^2+v^2=(\frac{3}{2}-r)^2.$ Subtracting pairs of equations yields $v=\frac12+\frac r3$ and $v=\frac32-2r.$ Thus $r=\frac{3}{7},$ so $d=2r=\frac{6}{7}\approx0.86.$

Therefore, the correct answer is B.

40.

Let $n$ be the number of integer values of $x$ such that $P=x^4+6x^3+11x^2+3x+31$ is the square of an integer. Then $n$ is:

$4$

$3$

$2$

$1$

$0$

Concepts:perfect square polynomial bounding to limit cases systematic listing

Difficulty rating: 2710

Small Hint:

Rewrite $P=(x^2+3x+1)^2-3(x-10)$

Big Hint:

Distinct integer squares $U^2$ and $V^2$ differ by at least $2\max(|U|,|V|)-1$

Solution:

Let $N=x^2+3x+1.$ Then $P=N^2-3(x-10).$ At $x=10,\ P=N^2=131^2,$ so one value works.

If $x\gt10$ and $P=y^2,$ then $N^2-y^2=3(x-10).$ Distinct integer squares differing from $N^2$ differ by at least $2|N|-1,$ which is already greater than $3(x-10),$ a contradiction. If $x\lt10,$ the analogous bound $\begin{gathered} 3(10-x)=y^2-N^2\\ \geq2|y|-1\gt2|N|-1 \end{gathered}$ can hold only for $-6\leq x\leq2.$ Direct substitution for these nine integers gives no square. Hence $x=10$ is the unique solution and $n=1.$

Therefore, the correct answer is D.