Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Carlos and Dick was $5$ dollars, between Dick and Elgin was $4$ dollars, and between Elgin and Ashley was $11$ dollars. How many dollars did Elgin have?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Working backwards, if $6 \le E \le 10$ , then $6 \pm 11 \le A \le 10 \pm 11$ .  Since $A$ is a positive integer, $17 \le A \le 21$
Since $17 \le A \le 21$ , we know that $17 \pm 19 \le B \le 21 \pm 19$ .  But if $B=36$ , which is the smallest possible "plus" value, then $E + A + B = 6 + 17 + 36 = 59$ , which is too much money.
Hence, $17 - 19 \le B \le 21 - 19$ .  But since $B$ must be a positive integer, that leaves only two possibilities: $B = 1$ or $B=2$ , which correspond with $E = 9$ and $E = 10$
Concentrating only on $E=9$ , we have $E=9$ leading to $A = 9 + 11 = 20$ , which leads to $B = 20 - 19 = 1$ , which leads to $C = 1 + 7 = 8$ .  Thus far we have given out $9 + 20 + 1 + 8 = 38$ dollars.  This means that Dick must have $56 - 38 = 18$ dollars.  However, the difference between Carlos and Dick is not $5$ dollars.
Thus, the right answer must be $\boxed{10}$ .  Verifying, if $E = 10$ , then $A = 10 + 11 = 21$ $B = 21 - 19 = 2$ , which leads to $C = 2 + 7 = 9$ .  Thus far, we have given out $10 + 21 + 2 + 9 = 42$ dollars, leaving $56 - 42 = 14$ dollars for Dick.  Dick does indeed have $5$ dollars more than Carlos, and $4$ dollars more than Elgin.

Answer: E 10
Problem: A rising number, such as $34689$ , is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{\text{th}}$ number in the list does not contain the digit
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution: The list starts with $12345$ .  There are $\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$ , and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.
Thus, the $71^{\text{st}}$ number is $23456$ .  There are $\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$ , and thus $20$ five digit rising numbers that begin with a $23$
Thus, the $91^{\text{st}}$ number is $24567$ .  Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\text{th}}$ number, which does not contain the digit $\boxed{5}$

Answer: B 5
Problem: Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$ $\overrightarrow{BB^\prime}$ $\overrightarrow{CC^\prime}$ , and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$ . If $AA^{\prime} = 10$ $BB^{\prime}= 8$ $CC^\prime = 18$ , and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$ , respectively, then $MN =$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Let $ABCD$ be a unit square with $A(0,0,0)$ $B(0,1,0)$ $C(1,1,0)$ , and $D(1,0,0)$ .  Assume that the rays go in the +z direction.  In this case, $A^\prime(0,0,10)$ $B^\prime(0,1,8)$ $C^\prime(1,1,18)$ , and $D^\prime(1,0,22)$ .  Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$ .  The distance $MN$ is $15 - 14 = 1$ , and the answer is $\boxed{1}$

Answer: B 1
Problem: Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: The product of two lengths with a common point brings to mind the Power of a Point Theorem
Since $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle.  Since $\angle APB = \widehat {AB} = 2 \angle ACB$ , we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$ , which is true for all inscribed angles.
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3.06,0.9); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); pair E = (0,4.5); dot(A);dot(B);dot(C);dot(P);dot(D);dot(E); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NW);label("$D$",D,NE + N);label("$P$",P,N);label("$E$", E, NW); draw(A--B--P--cycle); draw(A--C--B--cycle); draw(circle(P, 2.46)); draw(P--E);[/asy]
Since $PDB$ is a line, we have $PD + DB = PB$ , which gives $3 = DB + 2$ , or $DB = 1$
We now extend radius $PB = 3$ to diameter $EB = 6$ .  Since $EDB$ is a line, we have $ED + DB = EB$ , which gives $ED + 1 = 6$ , or $ED = 5$
Finally, we apply the power of a point theorem to point $D$ .  This states that $AD \cdot DC = DB \cdot DE$ .  Since $DB = 1$ and $DE = 5$ , the desired product is $5$ , which is $\boxed{5}$

Answer: A 5
Problem: Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Construct the angle bisector of $\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\angle EPD = \angle BCD = x,$ so $EPCB$ is cyclic. Then, $\frac{PD}{ED} = \frac{CD}{BD}$ or $\frac{2}{2x} = \frac{CD}{1}.$ Thus, $AD \cdot DC = 5x \cdot DC = 5,$ or $\boxed{5}.$

Answer: A 5
Problem: Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$ . Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$

Solution: Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$
In this case, we know that $f(x+ 4) + f(x - 4) = f(x)$ .  Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$ .  Adding those two equations gives $f(x + 8) +  f(x - 4) = 0$ after cancelling out common terms.
Again plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$ ), we find that $f(x) = -f(x + 12)$ .  Now, plugging in $x+12$ for $x$ , we get $f(x + 12) = -f(x + 24)$ .  This proves that $f(x)  = f(x + 24)$ , so there is a period of $24$ , which gives answer $\boxed{24}$ .  We now eliminate answers $A$ through $C$

Answer: D 24
Problem: Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$ . For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$

Solution: Define a super-special number to be a number whose decimal expansion only consists of $0$ 's and $1$ 's. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$ . This can be done in $8$ numbers if we take \[0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots\] Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$ .) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $0.1428\hdots$ , there must be $8$ super-special numbers, so the answer is $\boxed{8}$

Answer: B 8
Problem: For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution: If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$ , because all positive binary numbers begin with a $1$ , and if you switch digits twice, you will have a $1$ at the end.  Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.
All of these odd numbers will have an even $D(n)$ $D(n) = 0$ will be given by the numbers $1, 11, 111, 1111, 11111, 111111$ , which is a total of $6$ numbers.
We skip $D(n) = 2$ for now, and move to $D(n) = 4$ , which is easier to count.  The smallest $D(n) = 4$ happens when $n = 10101$ .  To get another number such that $D(n) = 4$ , we may extend any of the five blocks of zeros or ones by one digit.  This will form $110101, 100101, 101101, 101001, 101011$ , all of which are odd numbers that have $D(n) = 4$ .  To find seven digit numbers that have $D(n) = 4$ , we can again extend any block by one, so long as it remains less than $1100001$ or under.  There are five cases.
1)  Extending $110101$ is impossible without going over $1100001$
2)  Extending $100101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions work, giving $1000101, 1001101, 1001001, 1001011$
3)  Extending $101101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions give $1001101, 1011101, 1011001, 1011011$ .  However, $1001101$ already appeared in #2, giving only three new numbers.
4)  Extending $101001$ at the first group is impossible.  The other four extensions are $1001001, 1011001, 1010001, 1010011$ , but the first two are repeats.  Thus, there are only two new numbers.
5)  Extending $101011$ at the first group is impossible.  The other four extensions give $1001011, 1011011, 1010011, 1010111$ , but only the last number is new.
Thus, there is $1$ five digit number, $5$ six digit numbers, and $4 + 3 + 2 + 1 = 10$ seven digit numbers under $1100001$ for which $D(n) = 4$ .  That gives a total of $16$ numbers.
There smallest number for which $D(n) = 6$ is $1010101$ , which is under $98$ .  Further extensions, as well as cases where $D(n) > 6$ , are not possible.
Thus, we know that there are $6$ odd numbers that have $D(n) = 0$ , and $16$ odd numbers that have $D(n) = 4$ , and $1$ number that has $D(n) = 6$ .  The remaining odd numbers must have $D(n) = 2$ .  This means there are $49 - 6 - 16 - 1 = 26$ numbers that have $D(n) = 2$ , which is option $\boxed{26}$

Answer: C 26
Problem: For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution: For $D(n)$ to be $2$ $n$ must be in the form $1...10...01...1$ . Note that $n$ must be at least $3$ digits for $D(n)$ to be $2$
Case $1$ $n$ has $3$ digits
$n$ $101_2$ , there is only $1$ possible value for $n$ when $n$ has $3$ digits.
Case $2$ $n$ has $4$ digits
$n = 1$ _ _ $1_2$ , there could be $1$ zero or $2$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $2$ possible places. If there are $2$ zeros, there is only $1$ possible placement. Therefore, there are $2+1=3$ possible values for $n$ when $n$ has $4$ digits.
Case $3$ $n$ has $5$ digits
$n = 1$ _ _ _ $1_2$ , there could be $1$ $2$ or $3$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $3$ possible places. If there are $2$ zeros, there are $2$ possible placements. If there are $3$ zeros, there is only $1$ possible placement. Therefore, there are $3+2+1=6$ possible values for $n$ when $n$ has $5$ digits.
Case $4$ $n$ has $6$ digits
$n = 1$ _ _ _ _ $1_2$ , there could be $1$ $2$ $3$ or $4$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $4$ possible places. If there are $2$ zeros, there are $3$ possible placements. If there are $3$ zeros, there are $2$ possible placements. If there are $4$ zeros, there is only $1$ possible placement. Therefore, there are $4+3+2+1=10$ possible values for $n$ when $n$ has $6$ digits.
Case $4$ $n$ has $7$ digits
Given by the problem, $n \le 1100001_2$ . We could first count the number of $n$ that is in the form of $10$ _ _ _ _ $1_2$ , where the digits between the second and last digit are in the form of $0...01...1$
There could be $0$ $1$ $2$ $3$ or $4$ zeros. Therefore, there are $5$ possible values for $n$ when $n=10$ _ _ _ _ $1_2$
There is also the case were $n = 1100001_2$ . Therefore, there are $5+1=6$ possible values for $n$ when $n$ has $7$ digits.
Therefore, the answer is $1+3+6+10+6=\boxed{26}$

Answer: C 26
Problem: The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline  &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$

Solution: Doing the addition as is, we get $641 + 852 + 973 = 2466$ .  This number is $10$ larger than the desired sum of $2456$ .  Therefore, we must make one of the three numbers $10$ smaller.
We may either change $641 \rightarrow 631$ $852 \rightarrow 842$ , or $973 \rightarrow 963$ .  Either change results in a valid sum.  The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\boxed{7}$

Answer: D 7
Problem: Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution: If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars.  He got $6$ dollars more than this.
He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well.  Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\boxed{3}$

Answer: A 3
Problem: Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution: If Walter had done his chores for $10$ days exceptionally well, he would have earned $5 \cdot 10 = 50$ dollars.  He got $50 - 36 = 14$ dollars less than this.
He gets $2$ dollars docked from his pay if he doesn't do his chores well.  Therefore, he didn't do his chores well on $\frac{14}{2} = 7$ days.  The other $10 - 7 = 3$ days, he did them exceptionally well.  Therefore, the answer is $\boxed{3}$

Answer: A 3
Problem: Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution: Let $b$ be the number of days Walter does his chores but doesn't do them well, and let $w$ be the number of days he does his chores exceptionally well.
$b + w = 10$ since there are $10$ days Walter does chores.
$3b + 5w = 36$ since $3b$ is the amount he earns from doing his chores not well, and $5w$ is the amount he earnes from doing his chores exceptionally well, and those two sum to $36$ dollars.
Multiply the first equation by $3$ to get:
$3b + 3w = 30$
$3b + 5w = 36$
Subtract the first equation from the second equation to get:
$5w - 3w = 36 - 30$
$2w = 6$
$w = 3$
Thus, he does his chores well $3$ days, and the answer is $\boxed{3}$

Answer: A 3
Problem: $\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$

Solution: The numerator is $(3!)! = 6!$
The denominator is $3! = 6$
Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\boxed{120}$

Answer: E 120
Problem: Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$ . The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution: First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$
We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list.  If we top-load the list by making all three of the numbers greater than $9$ , the median will be the highest it can possibly be.  Thus, the maximum median is the fifth piece of data in the list, which is $8$ , giving an answer of $\boxed{8}$

Answer: D 8
Problem: A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$ , which of the following could be the age of the youngest child?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution: The bill for the three children is $9.45 - 4.95 = 4.50$ .  Since the charge is $0.45$ per year for the children, they must have $\frac{4.50}{0.45} = 10$ years among the three of them.
The twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born).  If we let the twins be $5$ years old, that leaves $10 - 2\cdot 5 = 0$ years leftover for the youngest child.  But if the twins are each $4$ years old, then the youngest child could be $10 - 2\cdot 4 = 2$ years old, which is choice $\boxed{2}$

Answer: B 2
Problem: How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

Solution: There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments.  However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment.  Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{28}$

Answer: D 28
Problem: How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

Solution: Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has $12$ edges:  Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$
A cube has $4$ spacial diagonals:  each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{28}$

Answer: D 28
Problem: A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$ .  What is the sum of the digits of $k$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$
If you doubled either of these, $k$ would not be odd.  So you must subtract $3$
If you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.
If you subtract $3$ from $108$ , you would compute $f(105)$ , which would add the $3$ back.
Thus, $f(f(f(105))) = f(f(108)) = f(54) = 27$ , and $105$ is odd.  The desired sum of the digits is $6$ , and the answer is $\boxed{6}$

Answer: B 6
Problem: Let $E(n)$ denote the sum of the even digits of $n$ . For example, $E(5681) = 6+8 = 14$ .  Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$

Solution: The problem is asking for the sum of all the even digits in the numbers $1$ to $100$ .  We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits.  This gives the list:
$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$
There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.
Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$ , and the correct answer is $\boxed{400}$

Answer: C 400