2007 AMC 10B Solutions | LIVE by Po-Shen Loh

1.

Isabella’s house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?

$678$

$768$

$786$

$867$

$876$

Concepts:surface area perimeter

Difficulty rating: 720

Small Hint:

Each bedroom has four walls; find the total wall area from the perimeter and the height

Big Hint:

Multiply the room’s perimeter by its height, then subtract $60$

Solution:

The walls of one bedroom have area $2(12+10)\cdot 8 = 44\cdot 8 = 352$ square feet. Subtracting the $60$ square feet of doorways and windows leaves $352-60=292$ square feet per bedroom.

With $3$ bedrooms, the total is $3\cdot 292 = 876$ square feet.

Thus, the correct answer is E.

2.

Define the operation $\star$ by $a\star b = (a+b)b.$ What is $(3\star 5)-(5\star 3)?$

$-16$

$-8$

$0$

$8$

$16$

Concepts:custom operation order of operations

Difficulty rating: 870

Small Hint:

Compute $3\star 5$ and $5\star 3$ separately

Big Hint:

$3\star 5=(3+5)\cdot 5$ and $5\star 3=(5+3)\cdot 3$

Solution:

Since $3\star 5=(3+5)\cdot 5=40$ and $5\star 3=(5+3)\cdot 3=24,$ the difference is $40-24=16.$

Thus, the correct answer is E.

3.

A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents’ SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?

$22$

$24$

$25$

$26$

$28$

Concepts:harmonic mean rate

Difficulty rating: 980

Small Hint:

Average mileage is total distance divided by total gallons, not the average of the two rates

Big Hint:

Find the gallons used each way, then divide $240$ by their sum

Solution:

The student used $\frac{120}{30}=4$ gallons driving home and $\frac{120}{20}=6$ gallons returning, for $10$ gallons over $240$ miles.

The average is $\frac{240}{10} = 24$ miles per gallon.

Thus, the correct answer is B.

4.

The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC = 120^\circ$ and $\angle AOB = 140^\circ,$ as shown. What is the degree measure of $\angle ABC?$

$35$

$40$

$45$

$50$

$60$

Concepts:circumcircle, circumcenter, and circumradius isosceles triangle angle chasing

Difficulty rating: 1030

Small Hint:

Each radius is equal, so each of the triangles $AOB, BOC, COA$ is isosceles

Big Hint:

$\angle ABC=\angle ABO+\angle OBC,$ where each part is a base angle of an isosceles triangle

Solution:

Since $OA=OB=OC,$ triangles $AOB, BOC,$ and $COA$ are isosceles. The base angles give $\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ$ and $\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.$

Therefore $\angle ABC=20^\circ+30^\circ=50^\circ.$

Thus, the correct answer is D.

5.

In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?

All Dramps are Brafs and are Crups.

All Brafs are Crups and are Dramps.

All Arogs are Crups and are Dramps.

All Crups are Arogs and are Brafs.

All Arogs are Dramps and some Arogs may not be Crups.

Concepts:logical deduction

Difficulty rating: 1020

Small Hint:

Write each fact as an implication between the categories

Big Hint:

Follow the chain $C\Rightarrow D\Rightarrow A\Rightarrow B$

Solution:

Writing the statements as implications, being a Crup implies being a Dramp, a Dramp implies being an Arog, and an Arog implies being a Braf: $C\Rightarrow D\Rightarrow A\Rightarrow B.$

So every Crup is a Dramp, an Arog, and a Braf. The only listed statement guaranteed true is that all Crups are Arogs and Brafs.

Thus, the correct answer is D.

6.

The $2007$ AMC $10$ will be scored by awarding $6$ points for each correct response, $0$ points for each incorrect response, and $1.5$ points for each problem left unanswered. After looking over the $25$ problems, Sarah has decided to attempt the first $22$ and leave only the last $3$ unanswered. How many of the first $22$ problems must she solve correctly in order to score at least $100$ points?

$13$

$14$

$15$

$16$

$17$

Concepts:inequality linear equation

Difficulty rating: 1120

Small Hint:

The unanswered problems still contribute points

Big Hint:

Subtract the points from the $3$ blanks, then divide the remaining needed points by $6$

Solution:

The three blank problems give $3\cdot 1.5 = 4.5$ points, so Sarah needs $100-4.5=95.5$ points from the first $22.$

Since $\frac{95.5}{6}$ lies between $15$ and $16,$ she must answer at least $16$ correctly, which would give a score of $100.5.$

Thus, the correct answer is D.

7.

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A = \angle B = 90^\circ.$ What is the degree measure of $\angle E?$

$90$

$108$

$120$

$144$

$150$

Concepts:angle sum square (geometry)equilateral triangle

Difficulty rating: 1190

Small Hint:

Two right angles at adjacent vertices make part of the figure a square

Big Hint:

$ABCE$ is a square and $\triangle CDE$ is equilateral

Solution:

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ.$

The remaining sides satisfy $CD=DE=EC,$ so $\triangle CDE$ is equilateral and $\angle CED=60^\circ.$

Therefore $\angle E=\angle AEC$ $+\angle CED=90^\circ$ $+60^\circ=150^\circ.$

Thus, the correct answer is E.

8.

On the trip home from the meeting where this AMC $10$ was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0\le a\lt b\lt c\le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?

$12$

$16$

$18$

$20$

$24$

Concepts:digits parity combinations

Difficulty rating: 1290

Small Hint:

The condition $b=\dfrac{a+c}{2}$ forces $a$ and $c$ to have the same parity

Big Hint:

Count pairs $a\lt c$ among the even digits and among the odd digits; each pair fixes $b$

Solution:

Once $a$ and $c$ are chosen, $b=\dfrac{a+c}{2}$ is determined, and $a\lt b\lt c$ holds automatically. For $b$ to be an integer, $a$ and $c$ must share parity.

Choosing two even digits from $\{0,2,4,6,8\}$ gives $\binom{5}{2}=10$ pairs, and choosing two odd digits from $\{1,3,5,7,9\}$ gives another $10.$

This yields $20$ valid numbers.

Thus, the correct answer is D.

9.

A cryptographic code is designed as follows. The first time a letter appears in a given message it is replaced by the letter that is $1$ place to its right in the alphabet (assuming that the letter A is one place to the right of the letter Z). The second time this same letter appears in the given message, it is replaced by the letter that is $1+2$ places to the right, the third time it is replaced by the letter that is $1+2+3$ places to the right, and so on. For example, with this code the word “banana” becomes “cbodqg”. What letter will replace the last letter s in the message

“Lee’s sis is a Mississippi miss, Chriss!”?

$g$

$h$

$o$

$s$

$t$

Concepts:modular arithmetic triangular number

Difficulty rating: 1370

Small Hint:

Count how many times the letter s appears in the whole message

Big Hint:

The total shift is $1+2+\cdots+12,$ then reduce that modulo $26$

Solution:

The final s is the $12$ th appearance of the letter s in the message, so it is shifted $1+2+\cdots+12=\dfrac{12\cdot 13}{2}=78$ places to the right.

Since $78=3\cdot 26$ is a multiple of the alphabet length $26,$ the shift returns to the same letter, s.

Thus, the correct answer is D.

10.

Two points $B$ and $C$ are in a plane. Let $S$ be the set of all points $A$ in the plane for which $\triangle ABC$ has area $1.$ Which of the following describes $S?$

two parallel lines

a parabola

a circle

a line segment

two points

Concepts:triangle area parallel lines

Difficulty rating: 1270

Small Hint:

Fix $BC$ as the base; the area determines the height from $A$

Big Hint:

All points at a fixed distance from line $BC$ lie on two parallel lines

Solution:

Taking $BC$ as the base, the area is $\dfrac12(BC)d,$ where $d$ is the distance from $A$ to line $BC.$ The area equals $1$ exactly when $d=\dfrac{2}{BC}.$

The points at this fixed distance from line $BC$ form two lines parallel to $BC,$ one on each side.

Thus, the correct answer is A.

11.

A circle passes through the three vertices of an isosceles triangle that has two sides of length $3$ and a base of length $2.$ What is the area of this circle?

$2\pi$

$\dfrac{5}{2}\pi$

$\dfrac{81}{32}\pi$

$3\pi$

$\dfrac{7}{2}\pi$

Concepts:circumcircle, circumcenter, and circumradius triangle area circle area

Difficulty rating: 1460

Small Hint:

Use the circumradius formula $R=\dfrac{abc}{4K},$ where $K$ is the triangle’s area

Big Hint:

The sides are $3,$ $3,$ and $2;$ find the area from the base and height, then compute $R$

Solution:

The triangle has sides $3,$ $3,$ and $2.$ Its area is $\dfrac12\cdot 2\cdot\sqrt{3^2-1^2}=2\sqrt2.$

The circumradius is $R=\dfrac{abc}{4K}$ $=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2}$ $=\dfrac{9}{4\sqrt2}$ $=\dfrac{9\sqrt2}{8}.$

The area of the circle is $\pi R^2=\pi\cdot\dfrac{81\cdot 2}{64}=\dfrac{81}{32}\pi.$

Thus, the correct answer is C.

12.

Tom’s age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $\frac{T}{N}?$

$2$

$3$

$4$

$5$

$6$

Concepts:ages linear equation

Difficulty rating: 1290

Small Hint:

Express the children’s total age $N$ years ago in terms of $T$ and $N$

Big Hint:

$T-N=2(T-3N)$

Solution:

$N$ years ago Tom’s age was $T-N,$ and the sum of his three children’s ages was $T-3N.$

The condition gives $T-N=2(T-3N),$ so $T-N=2T-6N,$ which simplifies to $5N=T.$

Therefore $\frac{T}{N}=5.$

Thus, the correct answer is D.

13.

Two circles of radius $2$ are centered at $(2,0)$ and at $(0,2).$ What is the area of the intersection of the interiors of the two circles?

$\pi-2$

$\dfrac{\pi}{2}$

$\dfrac{\pi\sqrt3}{3}$

$2(\pi-2)$

$\pi$

Concepts:sector area decomposition symmetry

Difficulty rating: 1580

Small Hint:

Find the two points where the circles intersect

Big Hint:

Half the region is a quarter-circle minus an isosceles right triangle of leg $2$

Solution:

The two circles intersect at $(0,0)$ and $(2,2).$

By symmetry, half the intersection is formed by removing an isosceles right triangle of leg length $2$ from a quarter of one circle. The quarter-circle has area $\dfrac14\pi(2)^2=\pi$ and the triangle has area $\dfrac12(2)^2=2.$

Therefore the whole region has area $2(\pi-2).$

Thus, the correct answer is D.

14.

Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?

$4$

$6$

$8$

$10$

$12$

Concepts:percentage linear equation

Difficulty rating: 1330

Small Hint:

The group size does not change when two leave and two arrive

Big Hint:

The two departing girls represent $40\%-30\%$ of the group

Solution:

Two girls leave and two boys arrive, so the group size is unchanged. The two girls who left therefore represent $40\%-30\%=10\%$ of the group.

Thus the group has $2\div 0.10=20$ people, and the original number of girls was $40\%$ of $20,$ or $8.$

Thus, the correct answer is C.

15.

The angles of quadrilateral $ABCD$ satisfy $\angle A = 2\angle B = 3\angle C = 4\angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?

$125$

$144$

$153$

$173$

$180$

Concepts:angle sum ratio and proportion

Difficulty rating: 1370

Small Hint:

Write $\angle B,$ $\angle C,$ and $\angle D$ in terms of $\angle A$

Big Hint:

The four angles sum to $360^\circ$

Solution:

Let $x=\angle A.$ Then $\angle B=\dfrac{x}{2},$ $\angle C=\dfrac{x}{3},$ and $\angle D=\dfrac{x}{4}.$

The angles sum to $360^\circ,$ so $x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}=\dfrac{25x}{12}=360.$

Thus $x=\dfrac{12\cdot 360}{25}=172.8\approx 173.$

Thus, the correct answer is D.

16.

A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?

$85$

$88$

$93$

$94$

$98$

Concepts:mean ratio and proportion

Difficulty rating: 1330

Small Hint:

Take a convenient class size, such as $10$ students

Big Hint:

With one junior and nine seniors, $10\cdot 84 = 9\cdot 83 + s$

Solution:

Suppose the class has $10$ students: one junior and nine seniors. The total of all scores is $10\cdot 84=840.$

The nine seniors total $9\cdot 83=747,$ so the junior’s score is $s=840-747=93.$

Thus, the correct answer is C.

17.

Point $P$ is inside equilateral $\triangle ABC.$ Points $Q,$ $R,$ and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB},$ $\overline{BC},$ and $\overline{CA},$ respectively. Given that $PQ = 1,$ $PR = 2,$ and $PS = 3,$ what is $AB?$

$4$

$3\sqrt3$

$6$

$4\sqrt3$

$9$

Concepts:equilateral triangle area decomposition triangle area

Difficulty rating: 1640

Small Hint:

Connect $P$ to the three vertices to split $\triangle ABC$ into three smaller triangles

Big Hint:

Their areas are $\dfrac{s}{2},s,\dfrac{3s}{2}$ ; set the sum equal to $\dfrac{\sqrt3}{4}s^2$

Solution:

Let the side length be $s.$ The perpendiculars from $P$ are the heights of triangles $APB,$ $BPC,$ and $CPA,$ so their areas are $\dfrac{s}{2},$ $s,$ and $\dfrac{3s}{2}.$

Their sum equals the area of $\triangle ABC,$ which is also $\dfrac{\sqrt3}{4}s^2.$ Hence $3s=\dfrac{\sqrt3}{4}s^2.$

The positive solution is $s=4\sqrt3.$

Thus, the correct answer is D.

18.

A circle of radius $1$ is surrounded by $4$ circles of radius $r$ as shown. What is $r?$

$\sqrt2$

$1+\sqrt2$

$\sqrt6$

$3$

$2+\sqrt2$

Concepts:tangent circles square (geometry)quadratic

Difficulty rating: 1680

Small Hint:

Connect the centers of the four outer circles to form a square

Big Hint:

The square has side $2r$ and diagonal $2+2r$

Solution:

Connect the centers of the four outer circles to form a square. Adjacent outer circles are tangent, so each side has length $2r.$

The diagonal of the square passes through the center of the central circle, giving length $1+r+r+1=2+2r.$ Since a square with side $2r$ has diagonal $2r\sqrt2,$ we get $2(2r)^2=(2+2r)^2.$

Expanding gives $1+2r+r^2=2r^2,$ so $r^2-2r-1=0.$ The positive root is $r=1+\sqrt2.$

Thus, the correct answer is B.

19.

The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?

$\dfrac{1}{3}$

$\dfrac{4}{9}$

$\dfrac{1}{2}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

Concepts:basic probability parity modular arithmetic

Difficulty rating: 1490

Small Hint:

A square is shaded exactly when the two remainders have the same parity

Big Hint:

Find the probability each remainder is even, then combine the both-even and both-odd cases

Solution:

The shaded squares are those where the two remainders are both odd or both even. The first remainder is even (from the numbers $2$ and $6$ ) with probability $\dfrac13$ and odd with probability $\dfrac23.$

The second remainder is even with probability $\dfrac12$ and odd with probability $\dfrac12.$

The probability that they share parity is $\dfrac13\cdot\dfrac12+\dfrac23\cdot\dfrac12=\dfrac12.$

Thus, the correct answer is C.

20.

A set of $25$ square blocks is arranged into a $5\times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?

$100$

$125$

$600$

$2300$

$3600$

Concepts:combinations permutations arrangements with restrictions

Difficulty rating: 1700

Small Hint:

Choose which $3$ rows and which $3$ columns are used

Big Hint:

After picking $3$ rows and $3$ columns, the blocks correspond to a matching between them

Solution:

Choose $3$ of the $5$ rows in $\binom{5}{3}=10$ ways and $3$ of the $5$ columns in $\binom{5}{3}=10$ ways.

The three chosen blocks must occupy distinct rows and columns, so they form a matching between the three rows and three columns, which can be done in $3!=6$ ways.

The total is $10\cdot 10\cdot 6=600.$

Thus, the correct answer is C.

21.

Right $\triangle ABC$ has $AB=3,$ $BC=4,$ and $AC=5.$ Square $XYZW$ is inscribed in $\triangle ABC$ with $X$ and $Y$ on $\overline{AC},$ $W$ on $\overline{AB},$ and $Z$ on $\overline{BC}.$ What is the side length of the square?

$\dfrac{3}{2}$

$\dfrac{60}{37}$

$\dfrac{12}{7}$

$\dfrac{23}{13}$

$2$

Difficulty rating: 1720

Small Hint:

Find the altitude from $B$ to the hypotenuse $AC$

Big Hint:

Use the similarity between the whole triangle and the small triangle above the square

Solution:

Let $s$ be the side of the square and $h$ the altitude from $B$ to $AC.$ Then $h=\dfrac{AB\cdot BC}{AC}=\dfrac{3\cdot 4}{5}=\dfrac{12}{5}.$

The small triangle above the square is similar to $\triangle ABC$ with the square’s top side as its base, giving $\dfrac{h-s}{h}=\dfrac{s}{AC},$ so $s=\dfrac{AC\cdot h}{AC+h}.$

Substituting, $s=\dfrac{5\cdot\frac{12}{5}}{5+\frac{12}{5}}=\dfrac{12}{\frac{37}{5}}=\dfrac{60}{37}.$

Thus, the correct answer is B.

22.

A player chooses one of the numbers $1$ through $4.$ After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered $1$ through $4.$ If the number chosen appears on the bottom of exactly one die after it is rolled, then the player wins $\$1.$ If the number chosen appears on the bottom of both of the dice, then the player wins $\$2.$ If the number chosen does not appear on the bottom of either of the dice, the player loses $\$1.$ What is the expected return to the player, in dollars, for one roll of the dice?

$-\dfrac{1}{8}$

$-\dfrac{1}{16}$

$0$

$\dfrac{1}{16}$

$\dfrac{1}{8}$

Concepts:expected value binomial probability dice (probability)

Difficulty rating: 1780

Small Hint:

Each die independently shows the chosen number on the bottom with probability $\dfrac14$

Big Hint:

Weight the payoffs $-1,$ $+1,$ and $+2$ by the probabilities of $0,$ $1,$ and $2$ matches

Solution:

Each die shows the chosen number on the bottom with probability $\dfrac14.$ So the number appears $0,$ $1,$ or $2$ times with probabilities $P(0)=\dfrac{9}{16},\$ $P(1)=\dfrac{6}{16},\$ $P(2)=\dfrac{1}{16}.$

The expected return is $(-1)\cdot\dfrac{9}{16}$ $+(1)\cdot\dfrac{6}{16}$ $+(2)\cdot\dfrac{1}{16}$ $=\dfrac{-9+6+2}{16}$ $=-\dfrac{1}{16}.$

Thus, the correct answer is B.

23.

A pyramid with a square base is cut by a plane that is parallel to its base and is $2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

$2$

$2+\sqrt2$

$1+2\sqrt2$

$4$

$4+2\sqrt2$

Concepts:power scaling of length, area, and volume pyramid surface area

Difficulty rating: 1880

Small Hint:

The smaller pyramid is similar to the original

Big Hint:

The ratio of surface areas is the square of the ratio of altitudes, so $\dfrac{h}{h-2}=\sqrt2$

Solution:

Let $h$ be the altitude of the original pyramid; the smaller pyramid has altitude $h-2.$ The two pyramids are similar, so the ratio of their surface areas is the square of the ratio of their altitudes.

The smaller surface area is half the original, so $\left(\dfrac{h-2}{h}\right)^2=\dfrac12,$ giving $\dfrac{h}{h-2}=\sqrt2.$

Then $h=\sqrt2(h-2),$ so $h(\sqrt2-1)=2\sqrt2$ and $h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2.$

Thus, the correct answer is E.

24.

Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base- $10$ representation consists of only $4$ ’s and $9$ ’s, with at least one of each. What are the last four digits of $n?$

$4444$

$4494$

$4944$

$9444$

$9944$

Concepts:divisibility digits optimization

Difficulty rating: 1980

Small Hint:

Divisibility by $9$ constrains the digit sum; divisibility by $4$ constrains the last two digits

Big Hint:

The digit sum forces at least nine $4$ ’s, and the last two digits must both be $4$

Solution:

Since $n$ is divisible by $9,$ its digit sum is a multiple of $9.$ With $k$ fours and $m$ nines, the digit sum is $4k+9m,$ so $4k$ is divisible by $9,$ forcing $k$ to be divisible by $9.$ Thus $k\ge 9,$ and with at least one $9,$ the number has at least ten digits.

For divisibility by $4,$ the last two digits must form a multiple of $4,$ and among $44,$ $49,$ $94,$ and $99$ only $44$ works, so $n$ ends in $44.$

The smallest such ten-digit number places the single $9$ in the lowest available position, giving $4{,}444{,}444{,}944.$ Its last four digits are $4944.$

Thus, the correct answer is C.

25.

How many pairs of positive integers $(a,b)$ are there such that $a$ and $b$ have no common factors greater than $1$ and

$\frac{a}{b}+\frac{14b}{9a}$

is an integer?

$4$

$6$

$9$

$12$

infinitely many

Concepts:divisibility greatest common divisor Diophantine Equation

Difficulty rating: 2170

Small Hint:

Combine into the single fraction $\dfrac{9a^2+14b^2}{9ab}$ and use $\gcd(a,b)=1$

Big Hint:

Deduce that $14$ is divisible by $a$ and $9$ is divisible by $b,$ then test the finitely many pairs

Solution:

Combining, the expression is $\dfrac{9a^2+14b^2}{9ab}.$ For this to be an integer, $a$ must divide $9a^2+14b^2,$ hence $14b^2$ is divisible by $a.$ Since $\gcd(a,b)=1,$ we get that $14$ is divisible by $a.$ Similarly, $9a^2$ is divisible by $b,$ so $9$ is divisible by $b.$

So $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking these, only $b=3$ makes the expression an integer for each allowed $a.$

The valid pairs are $(1,3),$ $(2,3),$ $(7,3),$ and $(14,3),$ for a total of $4.$

Thus, the correct answer is A.