1995 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Kim earned scores of $87,$ $83$ and $88$ on her first three mathematics examinations. If Kim receives a score of $90$ on the fourth exam, then her average will

remain the same

increase by $1$

increase by $2$

increase by $3$

increase by $4$

Concepts:arithmetic mean

Difficulty rating: 690

Small Hint:

Compare $90$ with the average of the first three scores

Big Hint:

Find the old and new totals before dividing by the number of exams

Solution:

The first three scores total $87+83+88=258,$ so their average is $\frac{258}{3}=86.$ With the fourth score, the total is $348,$ and the new average is $\frac{348}{4}=87.$ It increases by $1,$ so the correct answer is B.

2.

If $\sqrt{2+\sqrt{x}}=3,$ then $x=$

$1$

$\sqrt7$

$7$

$49$

$121$

Concepts:radical equations

Difficulty rating: 890

Small Hint:

Square both sides once to isolate $\sqrt{x}$

Big Hint:

After isolating the inner radical, square a second time

Solution:

Squaring gives $2+\sqrt{x}=9,$ so $\sqrt{x}=7.$ Squaring again gives $x=49.$ Thus the correct answer is D.

3.

The total in-store price for an appliance is $\$99.99.$ A television commercial advertises the same product for three easy payments of $\$29.98$ and a one-time shipping and handling charge of $\$9.98.$ How much is saved by buying the appliance from the television advertiser?

$6$ cents

$7$ cents

$8$ cents

$9$ cents

$10$ cents

Concepts:decimal arithmetic unit prices

Difficulty rating: 800

Small Hint:

Add all three payments and the one-time charge

Big Hint:

Subtract the advertised total from the in-store price and convert dollars to cents

Solution:

The advertised total is $3(29.98)+9.98=99.92$ dollars. The savings is $99.99-99.92=0.07$ dollars, or $7$ cents. Thus the correct answer is B.

4.

If $M$ is $30\%$ of $Q,$ $Q$ is $20\%$ of $P,$ and $N$ is $50\%$ of $P,$ then $\frac{M}{N}=$

$\frac3{250}$

$\frac3{25}$

$1$

$\frac65$

$\frac43$

Concepts:percentages ratios

Difficulty rating: 960

Small Hint:

Express both $M$ and $N$ as multiples of $P$

Big Hint:

Convert the percentages to fractions before forming $\frac{M}{N}$

Solution:

We have $M=(0.30)(0.20)P=0.06P$ and $N=0.50P.$ Hence $\frac{M}{N}=\frac{0.06}{0.50}=\frac{3}{25}.$ Thus the correct answer is B.

5.

A rectangular field is $300$ feet wide and $400$ feet long. Random sampling indicates that there are, on the average, three ants per square inch throughout the field. [ $12$ inches = $1$ foot.] Of the following, the number that most closely approximates the number of ants in the field is

$500$ thousand

$5$ million

$50$ million

$500$ million

$5$ billion

Concepts:area conversion estimation

Difficulty rating: 1070

Small Hint:

First find the field’s area in square feet

Big Hint:

One square foot contains $12^2$ square inches

Solution:

The area is $300\cdot400=120{,}000$ square feet, or $120{,}000\cdot144=17{,}280{,}000$ square inches. The estimate is $3(17{,}280{,}000)=51{,}840{,}000,$ closest to $50$ million. Thus the correct answer is C.

6.

The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked $x?$

$A$

$B$

$C$

$D$

$E$

Concepts:cube nets spatial reasoning

Difficulty rating: 960

Small Hint:

Hold face $A$ fixed and fold each neighboring square by $90^\circ$

Big Hint:

Track the outward normal direction of each face as you move through the net

Solution:

Hold $A$ as the front face. Folding makes $x$ the left face and $B$ the top face. The face $C,$ attached to the right of $B,$ then folds to the right face, opposite $x.$ Thus the correct answer is C.

7.

The radius of Earth at the equator is approximately $4000$ miles. Suppose a jet flies once around Earth at a speed of $500$ miles per hour relative to Earth. If the flight path is a negligible height above the equator, then, among the following choices, the best estimate of the number of hours of flight is

$8$

$25$

$50$

$75$

$100$

Concepts:circumference rate

Difficulty rating: 1220

Small Hint:

Approximate the equator’s length with $2\pi r$

Big Hint:

Divide the trip distance by $500$ and use a simple estimate for $\pi$

Solution:

The equator is approximately $2\pi(4000)=8000\pi$ miles long. The flight time is $\frac{8000\pi}{500}=16\pi\approx50$ hours. Thus the correct answer is C.

8.

In triangle $ABC,$ $\angle C=90^\circ,$ $AC=6$ and $BC=8.$ Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC},$ respectively, and $\angle BED=90^\circ.$ If $DE=4,$ then $BD=$

$5$

$\frac{16}3$

$\frac{20}3$

$\frac{15}2$

$8$

Difficulty rating: 1290

Small Hint:

The two right angles show that $\overline{DE}\parallel\overline{AC}$

Big Hint:

Use similarity between $\triangle BDE$ and $\triangle BAC$

Solution:

The large right triangle has $AB=\sqrt{6^2+8^2}=10.$ Since $\overline{DE}\parallel\overline{AC},$ triangles $BDE$ and $BAC$ are similar, with scale factor $\frac{DE}{AC}=\frac{4}{6}=\frac{2}{3}.$ Therefore $BD=(\frac{2}{3})(10)=\frac{20}{3}.$ Thus the correct answer is C.

9.

Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is

$10$

$12$

$14$

$16$

$18$

Concepts:geometric counting symmetry

Difficulty rating: 1380

Small Hint:

Separate the triangles by size before counting

Big Hint:

Count one orientation and use the square’s rotational symmetry

Solution:

There are $8$ smallest triangles, each bounded by a half-side, a half-diagonal, and a half-midline. There are $4$ triangles whose base is a full side and vertex is the center, and $4$ half-square triangles cut off by a diagonal. The total is $8+4+4=16.$ Thus the correct answer is D.

10.

The area of the triangle bounded by the lines $y=x,$ $y=-x$ and $y=6$ is

$12$

$12\sqrt2$

$24$

$24\sqrt2$

$36$

Concepts:coordinate geometry triangle area

Difficulty rating: 1100

Small Hint:

Find where each slanted line meets $y=6$

Big Hint:

Use the horizontal segment on $y=6$ as the triangle’s base

Solution:

The vertices are $(0,0),$ $(6,6),$ and $(-6,6).$ The horizontal base has length $12,$ and the height is $6,$ so the area is $\frac12(12)(6)=36.$ Thus the correct answer is E.

11.

How many base $10$ four-digit numbers, $N=\underline{a\,b\,c\,d},$ satisfy all three of the following conditions?

(i) $4{,}000\le N\lt6{,}000;$

(ii) $N$ is a multiple of $5;$

(iii) $3\le b\lt c\le6.$

$10$

$18$

$24$

$36$

$48$

Concepts:counting divisibility

Difficulty rating: 1600

Small Hint:

Determine the possible digits $a$ and $d$ from conditions (i) and (ii)

Big Hint:

Choose two distinct digits from $3,$ $4,$ $5,$ $6$ for $b\lt c$

Solution:

The thousands digit $a$ is $4$ or $5,$ and divisibility by $5$ makes $d$ either $0$ or $5.$ The pair $(b,c)$ is obtained by choosing two of $3,$ $4,$ $5,$ $6$ in increasing order, giving $\binom42=6$ pairs. Thus the count is $2\cdot2\cdot6=24,$ and the correct answer is C.

12.

Let $f$ be a linear function with the properties that $f(1)\le f(2),$ $f(3)\ge f(4),$ and $f(5)=5.$ Which of the following statements is true?

$f(0)\lt0$

$f(0)=0$

$f(1)\lt f(0)\lt f(-1)$

$f(0)=5$

$f(0)\gt5$

Concepts:linear functions monotonicity

Difficulty rating: 1600

Small Hint:

Translate each inequality into a condition on the slope

Big Hint:

The two slope conditions together determine the entire linear function

Solution:

From $f(1)\le f(2),$ the slope is nonnegative. From $f(3)\ge f(4),$ it is nonpositive. Thus the slope is $0,$ so $f$ is constant. Since $f(5)=5,$ we have $f(0)=5.$ Thus the correct answer is D.

13.

The addition below is incorrect. The display can be made correct by changing one digit $d,$ wherever it occurs, to another digit $e.$ Find the sum of $d$ and $e.$

$\begin{array}{rrrrrrr}&7&4&2&5&8&6\\+&8&2&9&4&3&0\\\hline1&2&1&2&0&1&6\end{array}$

$4$

$6$

$8$

$10$

more than $10$

Concepts:column addition digits

Difficulty rating: 1630

Small Hint:

Add from right to left and track each carry

Big Hint:

Remember that the same replacement must be made at every occurrence of $d$

Solution:

The units, tens, and hundreds columns work with carries $0,$ $1,$ $1.$ In the thousands column, replacing every $2$ by $6$ gives $6+9+1=16,$ so the result digit is also $6.$ The next columns then give $4+6+1=11$ and $7+8+1=16,$ producing the corrected equation $746586+869430=1616016.$ Thus $d+e=2+6=8,$ and the correct answer is C.

14.

If $f(x)=ax^4-bx^2+x+5$ and $f(-3)=2,$ then $f(3)=$

$-5$

$-2$

$1$

$3$

$8$

Concepts:even and odd powers functions

Difficulty rating: 1020

Small Hint:

The terms with even powers are unchanged when $x$ is replaced by $-x$

Big Hint:

Compare $f(3)$ and $f(-3)$ without solving for $a$ or $b$

Solution:

The even-power terms and constant are the same at $3$ and $-3,$ while the linear term changes from $-3$ to $3.$ Hence $f(3)-f(-3)=6,$ so $f(3)=2+6=8.$ Thus the correct answer is E.

15.

Five points on a circle are numbered $1,$ $2,$ $3,$ $4,$ and $5$ in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point $5,$ after $1995$ jumps it will be on point

$1$

$2$

$3$

$4$

$5$

Concepts:cycles modular arithmetic

Difficulty rating: 1470

Small Hint:

Write down the landing points for the first few jumps

Big Hint:

Once a point repeats, reduce the remaining number of jumps modulo the cycle length

Solution:

Starting at $5,$ the landing points are $1,2,4,1,2,4,\ldots$ Thus the cycle $1,2,4$ has length $3.$ Since $1995$ is divisible by $3,$ the $1995$ th landing point is $4.$ Thus the correct answer is D.

16.

Anita attends a baseball game in Atlanta and estimates that there are $50{,}000$ fans in attendance. Bob attends a baseball game in Boston and estimates that there are $60{,}000$ fans in attendance. A league official who knows the actual numbers attending the two games notes that:

i. The actual attendance in Atlanta is within $10\%$ of Anita’s estimate.

ii. Bob’s estimate is within $10\%$ of the actual attendance in Boston.

To the nearest $1{,}000,$ the largest possible difference between the numbers attending the two games is

$10{,}000$

$11{,}000$

$20{,}000$

$21{,}000$

$22{,}000$

Concepts:percent error inequalities

Difficulty rating: 1740

Small Hint:

In the two statements, the $10\%$ is taken of different quantities

Big Hint:

Find both attendance intervals, then maximize the larger endpoint minus the smaller endpoint

Solution:

Atlanta’s actual attendance $A$ satisfies $45{,}000\le A\le55{,}000.$ For Boston’s actual attendance $B,$ the condition is $|60{,}000-B|\le0.1B,$ so $\frac{60{,}000}{1.1}\le B\le\frac{60{,}000}{0.9}.$ The largest difference is therefore $\frac{60{,}000}{0.9}-45{,}000 =21{,}666.\overline6,$ which rounds to $22{,}000.$ Thus the correct answer is E.

17.

Given regular pentagon $ABCDE,$ a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A.$ The number of degrees in minor arc $\overset{\frown}{AD}$ is

$72$

$108$

$120$

$135$

$144$

Concepts:regular polygons tangents central angles

Difficulty rating: 1780

Small Hint:

The radii to $A$ and $D$ are perpendicular to the tangent sides

Big Hint:

Use the $72^\circ$ exterior angle of a regular pentagon to compare $\overline{AB}$ and $\overline{DC}$

Solution:

Let $O$ be the circle’s center. The direction changes by $72^\circ$ at each vertex of the pentagon, so the acute angle between lines $AB$ and $DC$ is $36^\circ.$ The radii to their tangency points are perpendicular to these lines; for the circle shown, the minor central angle $\angle AOD$ is the supplementary angle $180^\circ-36^\circ=144^\circ.$ Therefore minor arc $\overset{\frown}{AD}$ measures $144^\circ,$ and the correct answer is E.

18.

Two rays with common endpoint $O$ form a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB=1.$ The maximum possible length of $OB$ is

$1$

$\frac{1+\sqrt3}{\sqrt2}$

$\sqrt3$

$2$

$\frac4{\sqrt3}$

Concepts:law of cosines discriminant

Difficulty rating: 1700

Small Hint:

Apply the Law of Cosines to $\triangle AOB$

Big Hint:

Treat the resulting relation as a quadratic in $OA$ and require a real solution

Solution:

Let $OA=x$ and $OB=y.$ The Law of Cosines gives $1=x^2+y^2-\sqrt3xy.$ As a quadratic in $x,$ this has discriminant $3y^2-4(y^2-1)=4-y^2.$ A real $x$ requires $y\le2,$ and equality is attained when $x=\sqrt3.$ Hence the maximum is $2,$ and the correct answer is D.

19.

Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ as shown with $\overline{DE}\perp\overline{BC}.$ The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is

$\frac16$

$\frac14$

$\frac13$

$\frac25$

$\frac12$

Concepts:equilateral triangles coordinates area ratios

Difficulty rating: 1740

Small Hint:

Let the outer triangle have side $1$ and the inner triangle have side $s$

Big Hint:

Use the $60^\circ$ side slopes together with the fact that $\overline{DE}$ is vertical

Solution:

Let $ABC$ have side $1,$ with $B=(0,0)$ and $C=(1,0).$ If the inner side is $s,$ write $D=(d,0)$ and $E=(d,s).$ Since $E$ lies on $AC,$ $s=\sqrt3(1-d).$ The third inner vertex is $F=(d-\frac{\sqrt3}{2}s,\frac12s),$ and $F$ lies on $AB,$ so $\frac{s}{2}=\sqrt3\left(d-\frac{\sqrt3}{2}s\right),$ giving $d=\frac{2s}{\sqrt3}.$ Substitution yields $s=\frac{1}{\sqrt3}.$ Areas of equilateral triangles scale as the square of their sides, so the ratio is $s^2=\frac{1}{3}.$ Thus the correct answer is C.

20.

If $a,$ $b$ and $c$ are three (not necessarily different) numbers chosen randomly and with replacement from the set $\{1,2,3,4,5\},$ the probability that $ab+c$ is even is

$\frac25$

$\frac{59}{125}$

$\frac12$

$\frac{64}{125}$

$\frac35$

Concepts:parity probability

Difficulty rating: 1810

Small Hint:

The sum is even when $ab$ and $c$ have the same parity

Big Hint:

The product $ab$ is odd only when both selected factors are odd

Solution:

There are $3$ odd and $2$ even choices. The product $ab$ is odd with probability $\frac{9}{25}$ and even with probability $\frac{16}{25}.$ Therefore $\begin{aligned} P(ab+c\text{ even}) &=\frac9{25}\cdot\frac35\\ &\quad{}+\frac{16}{25}\cdot\frac25\\ &=\frac{59}{125}. \end{aligned}$ Thus the correct answer is B.

21.

Two nonadjacent vertices of a rectangle are $(4,3)$ and $(-4,-3),$ and the coordinates of the other two vertices are integers. The number of such rectangles is

$1$

$2$

$3$

$4$

$5$

Concepts:rectangles lattice points diagonals

Difficulty rating: 2280

Small Hint:

A rectangle’s diagonals have equal length and the same midpoint

Big Hint:

Enumerate integer vectors of length $5$ from the origin, identifying antipodal pairs

Solution:

The given diagonal has midpoint $(0,0)$ and half-length $5.$ The other diagonal must therefore have endpoints $(u,v)$ and $(-u,-v)$ with $u^2+v^2=25.$ Conversely, two equal diagonals with the same midpoint form a rectangle. There are $12$ integer points on this circle: $\begin{gathered} (\pm5,0),(0,\pm5),\\ (\pm3,\pm4),(\pm4,\pm3). \end{gathered}$ Antipodal points determine the same diagonal, giving $6$ possibilities. One is the given diagonal itself, which is degenerate, so $6-1=5$ rectangles remain. Thus the correct answer is E.

22.

A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13,$ $19,$ $20,$ $25$ and $31,$ although this is not necessarily their order around the pentagon. The area of the pentagon is

$459$

$600$

$680$

$720$

$745$

Concepts:rectangles Pythagorean theorem area subtraction

Difficulty rating: 1910

Small Hint:

Two pentagon sides are the original rectangle’s full side lengths

Big Hint:

Look for two differences among the lengths that form the legs of a right triangle with a third listed length

Solution:

The full rectangle sides must be longer than the two remnants on those same sides. The only assignment whose two differences and remaining cut side form a right triangle is the rectangle $25$ by $31,$ with remnants $20$ and $19.$ The removed corner then has legs $25-20=5$ and $31-19=12,$ whose hypotenuse is the listed side $13.$ Thus the pentagon’s area is $\begin{aligned} 25\cdot31-\frac12(5)(12) &=775-30\\ &=745. \end{aligned}$ Hence the correct answer is E.

23.

The sides of a triangle have lengths $11,$ $15,$ and $k,$ where $k$ is an integer. For how many values of $k$ is the triangle obtuse?

$5$

$7$

$12$

$13$

$14$

Concepts:triangle inequality obtuse triangles

Difficulty rating: 2060

Small Hint:

First use the triangle inequality to bound the integer $k$

Big Hint:

Test the obtuse inequality separately when $15$ is longest and when $k$ is longest

Solution:

The triangle inequality gives $5\le k\le25.$ For $k\le15,$ the longest side is $15,$ and the triangle is obtuse when $15^2\gt11^2+k^2,$ which gives $5\le k\le10,$ or $6$ values. For $k\gt15,$ it is obtuse when $k^2\gt11^2+15^2=346,$ giving $19\le k\le25,$ or $7$ values. The total is $6+7=13,$ so the correct answer is D.

24.

There exist positive integers $A,$ $B,$ and $C,$ with no common factor greater than $1,$ such that $A\log_{200}5+B\log_{200}2=C.$ What is $A+B+C?$

$6$

$7$

$8$

$9$

$10$

Concepts:logarithms prime factorization

Difficulty rating: 1900

Small Hint:

Combine the left side into one logarithm

Big Hint:

Use $200=2^3\cdot5^2$ and compare prime exponents

Solution:

Combining logarithms gives $\log_{200}(5^A2^B)=C,$ so $5^A2^B=200^C=5^{2C}2^{3C}.$ Hence $A=2C$ and $B=3C.$ The relatively prime positive triple is $(A,B,C)=(2,3,1),$ whose sum is $6.$ Thus the correct answer is A.

25.

A list of five positive integers has mean $12$ and range $18.$ The mode and median are both $8.$ How many different values are possible for the second largest element of the list?

$4$

$6$

$8$

$10$

$12$

Concepts:mean median mode integer cases

Difficulty rating: 2070

Small Hint:

Sort the five integers; the median and mode force at least two entries to equal $8$

Big Hint:

Let the smallest entry be $m$ and use the range and total sum to express the second largest entry

Solution:

Write the sorted list as $m,8,8,d,m+18.$ (If the fourth entry were $8,$ the sum and ordering conditions would be impossible.) Since the total is $5(12)=60,$ $m+8+8+d+(m+18)=60,$ so $d=26-2m.$ The conditions $8\le d\le m+18$ and $m\le8$ give $3\le m\le8.$ These six values yield $d=20,$ $18,$ $16,$ $14,$ $12,$ $10,$ all valid. Thus there are $6$ possibilities, and the correct answer is B.

26.

In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O,$ $\overline{AB}\perp\overline{CD},$ and chord $\overline{DF}$ intersects $\overline{AB}$ at $E.$ If $DE=6$ and $EF=2,$ then the area of the circle is

$23\pi$

$\frac{47}2\pi$

$24\pi$

$\frac{49}2\pi$

$25\pi$

Concepts:intersecting chords coordinates circles

Difficulty rating: 2170

Small Hint:

Intersecting chords gives $AE\cdot EB=DE\cdot EF$

Big Hint:

Place $O$ at the origin and express $F$ using the fact that $E$ divides $\overline{DF}$ in a $3:1$ ratio

Solution:

Let the radius be $r$ and $OE=x.$ Intersecting chords gives $(r+x)(r-x)=6\cdot2,$ so $r^2-x^2=12.$ Put $O=(0,0),$ $D=(0,-r),$ and $E=(x,0).$ Since $DE:EF=3:1,$ $\begin{aligned} F&=D+\frac43(E-D)\\ &=\left(\frac{4x}{3},\frac r3\right). \end{aligned}$ Because $F$ lies on the circle, $\frac{16x^2}{9}+\frac{r^2}{9}=r^2,$ so $r^2=2x^2.$ Therefore $x^2=12$ and $r^2=24.$ The area is $24\pi,$ so the correct answer is C.

27.

Consider the triangular array of numbers with $0,$ $1,$ $2,$ $3,$ $\ldots$ along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows $1$ through $6$ are shown.

$\begin{array}{cccccc} &&0&&&\\ &&1&1&&\\ &2&2&2&&\\ &3&4&4&3&\\ 4&7&8&7&4&\\ 5&11&15&15&11&5 \end{array}$

Let $f(n)$ denote the sum of the numbers in row $n.$ What is the remainder when $f(100)$ is divided by $100?$

$12$

$30$

$50$

$62$

$74$

Concepts:recurrences modular arithmetic

Difficulty rating: 2170

Small Hint:

Relate a row’s sum to the previous row’s sum by counting how often each old entry contributes

Big Hint:

Solve the resulting recurrence, then compute the power of $2$ modulo $100$

Solution:

Every previous entry contributes to two entries of the next row, and the two new boundary entries contribute an additional $2.$ Thus $f(n)=2f(n-1)+2,$ with $f(1)=0.$ Hence $f(n)=2^n-2.$ Since $2^{20}\equiv76\pmod{100}$ and $76^2\equiv76\pmod{100},$ we have $2^{100}\equiv76\pmod{100}.$ Therefore $f(100)\equiv74\pmod{100},$ and the correct answer is E.

28.

Two parallel chords in a circle have lengths $10$ and $14,$ and the distance between them is $6.$ The chord parallel to these chords and midway between them is of length $\sqrt a,$ where $a$ is

$144$

$156$

$168$

$176$

$184$

Concepts:circle chords coordinate geometry

Difficulty rating: 2290

Small Hint:

If a chord is at signed distance $y$ from the center, its half-length satisfies $h^2+y^2=r^2$

Big Hint:

Let the signed distances of the $14$ - and $10$ -chords differ by $6$ , and subtract their equations

Solution:

Let the signed distances from the center to the $14$ - and $10$ -chords be $u$ and $v,$ with $u-v=6.$ Then $r^2-u^2=7^2,\qquad r^2-v^2=5^2.$ Hence $v^2-u^2=24,$ so $(v-u)(v+u)=24.$ Since $v-u=-6,$ we get $u+v=-4,$ and therefore $u=1$ and $v=-5.$ The midway chord is at signed distance $-2,$ while $r^2=49+1=50.$ Its squared length is $4(50-4)=184,$ so $a=184.$ Thus the correct answer is E.

29.

For how many three-element sets of positive integers $\{a,b,c\}$ is it true that $a\cdot b\cdot c=2310?$

$32$

$36$

$40$

$43$

$45$

Concepts:prime factorization set partitions

Difficulty rating: 2280

Small Hint:

Factor $2310$ into distinct primes and assign each prime to one of the factors

Big Hint:

Count separately the cases in which one factor is $1$ and in which all three factors exceed $1$

Solution:

Since $2310=2\cdot3\cdot5\cdot7\cdot11,$ each prime belongs to exactly one of the three factors. If all factors exceed $1,$ their unordered prime groups form a partition of five objects into three nonempty blocks, counted by $S(5,3)=\frac{3^5-3\cdot2^5+3}{6}=25.$ If one factor is $1,$ the primes are partitioned into two nonempty blocks, giving $S(5,2)=2^4-1=15.$ The factors are distinct because their disjoint prime sets differ. Thus the total is $25+15=40,$ and the correct answer is C.

30.

A large cube is formed by stacking $27$ unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is

$16$

$17$

$18$

$19$

$20$

Concepts:three-dimensional coordinates generating functions

Difficulty rating: 2330

Small Hint:

Model the large cube as $[0,3]^3$ and write the plane through its center perpendicular to the diagonal

Big Hint:

Index each unit cube by its minimum value of $x+y+z$ and determine which index sums allow an intersection

Solution:

Take the large cube as $[0,3]^3.$ The plane is $x+y+z=\frac{9}{2}.$ A unit cube with lower corner $(i,j,k),$ where $i,j,k\in\{0,1,2\},$ contains values of $x+y+z$ from $s=i+j+k$ to $s+3.$ It intersects the plane exactly when $s=2,$ $s=3,$ or $s=4.$ The coefficients of $x^2,$ $x^3,$ $x^4$ in $(1+x+x^2)^3$ are $6,$ $7,$ $6.$ Thus the plane intersects $6+7+6=19$ unit cubes, and the correct answer is D.