Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

Solution: We start by simplifying the recursion: $x_{n+1} = x_n + \frac{4-x_n}{x_n+6}$ .
While $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \[x_{n} - x_{n+1} = \frac{x_n-4}{x_n+6} = \frac{f_n}{10 + f_n} \approx \frac{f_n}{10}.\] Now notice that we want to find $m$ , such that $x_m \leq 4 + \frac{1}{2^{20}}$ , and $x_0 = 4 + \frac{1}{2^0}$ . 
We are going to find an approximate number of steps to go from $4 + \frac{1}{2^i}$ to $4 + \frac{1}{2^{i+1}}$ . 
If $x_j = 4 + \frac{1}{2^i}$ , then $f_j = \frac{1}{2^i}$ and if $x_k = 4 + \frac{1}{2^{i+1}}$ , then $f_k = \frac{1}{2^{i+1}}$ . Then \[x_j - x_k = \frac{1}{2^{i+1}} \approx \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10}.\] Furthermore, \[\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \frac{1}{2^{i+1}}.\] Therefore, \[(k-j) \cdot \frac{\frac{1}{2^i}}{10} > \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10} > (k-j) \cdot \frac{\frac{1}{2^{i+1}}}{10},\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \frac{1}{2^0}$ to $f_m = \frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\boxed{81,242}$

Answer: C 81,242
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).
We know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equivalent to $0110$ . Thus, our sequence must be of the forms $0\ldots010$ or $0\ldots0110$ . In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$ . In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$ . Therefore, it must be the case that $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions).
It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ $f(4)=1$ since the only possible valid sequence is $0110$ $f(5)=1$ since the only possible valid sequence is $01010$
The recursive sequence is then as follows:
\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]
So, our answer is $\boxed{65}$

Answer: C 65
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: After any particular $0$ , the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$ , i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$ s and $3$ s to get $18$ . There are a number of ways to do this:
Case 1 : nine $2$ s - there is only $1$ way to arrange them.
Case 2 : two $3$ s and six $2$ s - there are ${8\choose2} = 28$ ways to arrange them.
Case 3 : four $3$ s and three $2$ s - there are ${7\choose4} = 35$ ways to arrange them.
Case 4 : six $3$ s - there is only $1$ way to arrange them.
Summing the four cases gives $1+28+35+1=\boxed{65}$

Answer: C 65
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:
Case 1 : Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ . 
Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.
Case 2 : There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.
Case 3 : Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.
Case 4 : No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$
Adding these four cases, we have $1+28+35+1=\boxed{65}$ as our final answer.

Answer: C 65
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: Any valid sequence must start with a $0$ . We can then think of constructing a sequence as adding groups of terms to this $0$ , each ending in $0$ . (This is always possible because every valid string ends in $0$ .) For example, we can represent the string $01011010110110$ as: $0-10-110-10-110-110$ .
To not have any consecutive 0s, we must have at least one $1$ before the next $0$ . However, we cannot have three or more $1$ s before the next $0$ because we cannot have three consecutive $1$ s. Consequently, we can only have one or two $1$ s.
So we can have the groups: $10$ and $110$
After the initial $0$ , we have $18$ digits left to fill in the string. Let the number of $10$ blocks be $x$ , and $110$ be $y$ . Then $x$ and $y$ must satisfy $2x+3y=18$ . We recognize this as a Diophantine equation. Taking $\pmod{2}$ yields $y=0 \pmod{2}$ . Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$ $(6, 2)$ $(3, 4)$ , and $(0, 6)$ . We now handle each of these cases separately.
$(9, 0)$ : Only one arrangement, namely all $10$ s.
$(6, 2)$ : We have 6 groups of $11$ , and $2$ groups of $110$ . This has $\binom{6+2}{2}=28$ cases.
$(3, 4)$ : This means we have 3 groups of $10$ , and 4 groups of $110$ . This has $\binom{3+4}{3}=35$ cases.
$(0, 6)$ : Only one arrangement, namely all $110$
Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{65}$ .
~Math4Life2020

Answer: C 65
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: Suppose the number of $0$ s is $n$ . We can construct the sequence in two steps:
Step 1: put $n-1$ of $1$ s between the $0$ s;
Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$ s in the $n-1$ spots where there is a $1$ . There are $\binom{n-1}{20-2n}$ ways of doing this.
Now we find the possible values of $n$
First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$ s);
And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$ s).
Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{65}.\]

Answer: C 65
Problem: How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution: For a valid sequence of length $n$ , the sequence must be in the form of $01xx...xx10$ . By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of:
So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$
We will calculate $f(19)$ by Dynamic Programming
$f(3) = 1$
$f(4) = 1$
$f(5) = 1$
$f(6) = 2$
$f(7) = 2$
$f(8) = 3$
$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$
$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$
$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$
$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$
$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$
$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$
$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$
$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$
$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$
$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$
$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{65}$

Answer: C 65
Problem: A unit of blood expires after $10!=10\cdot 9 \cdot 8 \cdots 1$ seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?
$\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}$

Solution: There are $60 \cdot 60 \cdot 24 = 86400$ seconds in a day, which means that Yasin's blood expires in $10! \div 86400 = 42$ days. Since there are $31$ days in January (consult a calendar), then $42-31+1$ (Jan 1 doesn't count) is $12$ days into February, so $\boxed{12}$

Answer: E 12
Problem: How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution: We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{24}$ ways to choose the classes.

Answer: E 24
Problem: How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution: Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.
Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$ . Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.
Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$ . Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$ . Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{24}$

Answer: E 24
Problem: How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution: We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s: \[10101\] Now we need to place 1 additional 0. There are 4 places to put it: \[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\] It can be placed in any 1 of the underscores.
Since there are $3!=6$ ways to order the math classes, the answer is $6\cdot 4=\boxed{24}$

Answer: E 24
Problem: How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution: Let $M$ represent a math class and $N$ represent a non-math class. We have _N_N_N_, where the spaces represent the possible spots the $M’s$ could go in. There are $4\choose 3$ ways to choose the spots for the math classes and $3!=6$ ways to order the classes. Hence, the answer is $4 \cdot 6 = \boxed{24}$ .
~azc1027

Answer: E 24
Problem: Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$
$\textbf{(A) }   (0,4)   \qquad        \textbf{(B) }   (4,5)   \qquad    \textbf{(C) }   (4,6)   \qquad   \textbf{(D) }  (5,6)  \qquad  \textbf{(E) }   (5,\infty)$

Solution: For each of the false statements, we identify its corresponding true statement. Note that:
We construct the following table: \[\begin{array}{c||c|c} & &  \\ [-2.5ex] \textbf{Hiker} & \textbf{False Statement} & \textbf{True Statement} \\ [0.5ex] \hline & & \\ [-2ex] \textbf{Alice} & [6,\infty) & [0,6) \\  & & \\ [-2.25ex] \textbf{Bob} & [0,5] & (5,\infty)  \\ & & \\ [-2.25ex] \textbf{Charlie} & [0,4] & (4,\infty) \end{array}\] Taking the intersection of the true statements, we have \[[0,6)\cap(5,\infty)\cap(4,\infty)=(5,6)\cap(4,\infty)=\boxed{5,6}.\] ~MRENTHUSIASM

Answer: D 5,6
Problem: Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least $6$ miles away," Bob replied, "We are at most $5$ miles away." Charlie then remarked, "Actually the nearest town is at most $4$ miles away." It turned out that none of the three statements were true. Let $d$ be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of $d$
$\textbf{(A) }   (0,4)   \qquad        \textbf{(B) }   (4,5)   \qquad    \textbf{(C) }   (4,6)   \qquad   \textbf{(D) }  (5,6)  \qquad  \textbf{(E) }   (5,\infty)$

Solution: Think of the distances as if they are on a number line. Alice claims that $d > 6$ , Bob says $d < 5$ , while Charlie thinks $d < 4$ . This means that all possible numbers less than $5$ and greater than $6$ are included. However, since the three statements are actually false, the distance to the nearest town is one of the numbers not covered. Therefore, the answer is $\boxed{5,6}$

Answer: D 5,6
Problem: Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution: If $65\%$ of the votes were likes, then $35\%$ of the votes were dislikes. $65\%-35\%=30\%$ , so $90$ votes is $30\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $\boxed{300}$

Answer: B 300
Problem: Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution: Let's consider that Sangho has received 100 votes. This means he has received 65 upvotes and 35 downvotes. Part of these upvotes and downvotes cancel out, so Sangho is now left with a total of 30 upvotes, or a score increase of 30. In order for his score to be 90, Sangho must receive three sets of 100 votes. Therefore, the answer is $\boxed{300}$

Answer: B 300
Problem: Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution: Let $x$ be the amount of votes cast, $35\%$ of $x$ would be dislikes and $65\%$ of $x$ would be likes. Since a like earns the video 1 point and a dislike takes 1 point away from the video, the amount of points Sangho's video will have in terms of $x$ is $65\%-35\%=30\%$ of $x$ , or $\frac{3}{10}$ of $x$ . Because Songho's video had a score of 90 points, $\frac{3x}{10} = 90$ . After we solve for $x$ , we get $x = \boxed{300}$

Answer: B 300
Problem: Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution: Let x be the amount of like votes and y be the amount of dislike votes. Then $x-y=90$ . Since $65\%$ of the total votes are like votes, $0.65(x+y)=x$ . then $.65y = .35x$ or $13/7$ y = $x$ . Plugging that in to the original equation, $6/7 y$ $90$ and $y =105$ Then $x=195$ and the total number of votes is $105+195 = 300$ . The answer is $\boxed{300}$

Answer: B 300
Problem: Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution: Obviously, $90/(65\%-35\%) = 90/30\% = 300$ . The answer is $\boxed{300}$

Answer: B 300
Problem: For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution: Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need:
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$

Answer: E 9
Problem: For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution: Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$ . We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$ . For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$ . Also keep in mind that the expression will be an integer for $n=0$ , which gives us a total of $\boxed{9}$ for $n.$

Answer: E 9
Problem: For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution: The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$
The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.
In total, there are $\boxed{9}$ values for $n.$

Answer: E 9
Problem: Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) }   0   \qquad        \textbf{(B) }   1   \qquad    \textbf{(C) }   2   \qquad   \textbf{(D) }  3  \qquad  \textbf{(E) }   4$

Solution: Let $x$ be the number of $5$ -cent coins that Joe has. Therefore, he must have $(x+3) \ 10$ -cent coins and $(23-(x+3)-x) \ 25$ -cent coins. Since the total value of his collection is $320$ cents, we can write \begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. \end{align*} Joe has six $5$ -cent coins, nine $10$ -cent coins, and eight $25$ -cent coins. Thus, our answer is $8-6 = \boxed{2}.$

Answer: C 2