Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$ , where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution: First, $x=2,1$ are trivial solutions
We assume from the shape of a parabola and the nature of the floor function that any additional roots will be near 2 and 1
We can now test values for $\lfloor{x}\rfloor$
$\lfloor{x}\rfloor=0$
We have $0-3x+2=0$ . Solving, we have $x=\frac{2}{3}$ . We see that $\lfloor{\frac{2}{3}}\rfloor=0$ , so this solution is valid
$\lfloor{x}\rfloor=-1$
We have $1-3x+2=0$ . Solving, we have $x=1$ $\lfloor{1}\rfloor\neq-1$ , so this is not valid. We assume there are no more solutions in the negative direction and move on to $\lfloor{x}\rfloor=3$
$\lfloor{x}\rfloor=3$
We have $9-3x+2=0$ . Solving, we have $x=\frac{11}{3}$ . We see that $\lfloor{\frac{11}{3}}\rfloor=3$ , so this solution is valid
$\lfloor{x}\rfloor=4$
We have $16-3x+2=0$ . Solving, we have $x=6$ $\lfloor{6}\rfloor\neq4$ , so this is not valid. We assume there are no more solutions.
Our final answer is $\boxed{4}$

Answer: B 4
Problem: How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$ , where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution: Denote $a = \lfloor x \rfloor$ .
Denote $b = x - \lfloor x \rfloor$ .
Thus, $b \in \left[ 0 , 1 \right)$
The equation given in this problem can be written as \[ a^2 - 3 \left( a + b \right) + 2 = 0 . \]
Thus, \begin{align*} 3 b & = a^2 - 3 a + 2 . \end{align*}
Because $b \in \left[ 0 , 1 \right)$ , we have $3 b \in \left[ 0 , 3 \right)$ .
Thus, \[ a^2 - 3 a + 2 = 0, 1, \mbox{ or } 2 . \]
If $a^2-3a+2=0$ $(a-2)(a-1)=0$ so $a$ can be $1, 2$
If $a^2-3a+2=1$ $a^2-3a+1=0$ which we find has no integer solutions after finding the discriminant.
If $a^2-3a+2=2$ $a^2-3a=0$ -> $a(a-3)=0$ so $a$ can also be $0, 3$
Therefore, $a = 1$ , 2, 0, 3.
Therefore, the number of solutions is $\boxed{4}$

Answer: B 4
Problem: How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$ , where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution: $x=1, 2$ are trivial solutions.
Let $x=n+f$ for some integer $n$ and some number $f$ such that $-1<f<1$ \[\lfloor{x}\rfloor^2-3x+2= \lfloor{n+f}\rfloor^2-3(n+f)+2=n^2+-3(n+f)+2.\] So now we have \[n^2-3(n+f)+2 = 0,\] which we can rewrite as \[n(n-3)=3f-2.\] Since $n$ is an integer, $n(n-3)$ is an integer, so $3f-2$ is an integer. Since $-1<f<1$ , the only possible values of $f$ are $\frac{1}{3}$ $\frac{2}{3}$ $-\frac{1}{3}$ , and $-\frac{2}{3}$ . Plugging in each value, we find that the only value of $f$ that produces integer solutions for $n$ is $f=\frac{2}{3}$ . If $f=\frac{2}{3}$ $n=0$ or $n=3$ . Hence, there is a total of 4 possible solutions, so the answer is $\boxed{4}$ . 
~azc1027

Answer: B 4
Problem: How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$ , where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution: We rewrite the equation as ${\lfloor x\rfloor}^2-3\lfloor x\rfloor-3\{x\}+2=0$ , where $\{x\}$ is the fractional part of $x$
Denote $\lfloor x\rfloor = x_1$ and $\{x\} = x_2.$ Thus \[{x_1}^2-3{x_1}-3{x_2}+2=0.\]
By definition, $0\leq x_2\leq 1$ . We then have ${x_1}^2-3{x_1}+2=3{x_2}$ and therefore $0\leq {x_1}^2-3{x_1}+2\leq 3$
Solving, we have $\left[\frac{3-\sqrt{13}}{2},1\right]\cup \left[2,\frac{3+\sqrt{13}}{2}\right]$ . But since $x_1$ is an integer, we have $x_1$ can only be $0,1,2,$ or $3$
Testing, we see these values of $x_1$ work, and therefore the answer is just $\boxed{4}$

Answer: B 4
Problem: How many distinct values of $x$ satisfy $\lfloor{x}\rfloor^2-3x+2=0$ , where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$

Solution: We know that for integer values of x, the graph is just $x^2-3x+2$ . From the interval $[x, x+1]$ , the square stays the same, so the graph has a line segment that goes down by 3 and right by 2. This is very easy to graph, so we see that there are 4 solutions. Or, we notice that only $x = 0, 1, 2, 3$ results in a $x^2-3x+2$ in the interval $[0, 3]$ .That is $\boxed{4}$ solutions.

Answer: B 4
Problem: An arithmetic sequence of positive integers has $\text{n} \ge 3$ terms, initial term $a$ , and common difference $d > 1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote was $222$ . What is $a + d + n$
$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution: Since one of the terms was either $1$ more or $1$ less than it should have been, the sum should have been $222-1=221$ or $222+1=223.$
The formula for an arithmetic series is $an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(2a+d(n-1)\right).$ This can quickly be rederived by noticing that the sequence goes $a,a+d,a+2d,a+3d,\dots,a+(n-1)d$ , and grouping terms.
We know that $\dfrac n2(2a+d(n-1))=221$ or $223$ . Let us now show that $223$ is not possible.
If $\dfrac n2(2a+d(n-1))=223$ , we can simplify this to be $n(a+d(n-1))=223\cdot2.$ Since every expression in here should be an integer, we know that either $n=2$ and $a+d(n-1)=223$ or $n=223$ and $a+d(n-1)=2.$ The latter is not possible, since $n\ge3,d>1,$ and $a>0.$ The former is also impossible, as $n\ge3.$ Thus, $\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221$
We can factor $221$ as $13\cdot17$ . Using similar reasoning, we see that $221\cdot2$ can not be paired as $2$ and $221$ , but rather must be paired as $13$ and $17$ with a factor of $2$ somewhere.
Let us first try $n=13.$ Our equation simplifies to $2a+12d=34\implies a+6d=17.$ We know that $d>1,$ so we try the smallest possible value: $d=2.$ This would give us $a=17-2\cdot6=17-12=5.$ (Indeed, this is the only possible $d$ .)
There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer $a+d+n=5+2+13=\boxed{20.}$

Answer: B 20.
Problem: An arithmetic sequence of positive integers has $\text{n} \ge 3$ terms, initial term $a$ , and common difference $d > 1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote was $222$ . What is $a + d + n$
$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution: There are $n$ terms, the $x$ th term is $a+(x-1)d$ , summation is $an+dn(n-1)/2=n(a+\frac{d(n-1)}{2})$
The summation of the set is $222 \pm 1 = 221,223$ . First, $221$ : its only possible factors are $1,13,17,221$ , and as said by the problem, $n\ge3$ , so $n$ must be $13,17,$ or $221$ . Let's start with $n=13$ . Then, $a+6d=17$ , and this means $a=5$ $d=2$ . Summing gives $13+5+2=\boxed{20}$ . We don't need to test any more cases, since the problem writes that all $a+d+n$ are the same.

Answer: B 20
Problem: An arithmetic sequence of positive integers has $\text{n} \ge 3$ terms, initial term $a$ , and common difference $d > 1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote was $222$ . What is $a + d + n$
$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution: We must have the sum of terms of the arithmetic sequence is $222\pm 1$ , which is $221$ or $223$
Since we have $223$ is prime, it cannot be the sum of the arithmetic sequence.
We see that $221$ is just $13\times 17$
We can write any arithmetic sequence with an odd amount of terms like this: $b-md,\cdots ,b-2d,b-d,b,b+d,b+2d,\cdots b+md$ where b is the middle term and d is the common difference.
By the sum of an arithmetic sequence, we have $13b=221$ and $17b=221$ and therefore $(b,n)$ $=$ $(13,17)$ or $(17,13)$
Then $a+d+n=b-md+d+n=30-(m-1)d$
We must have that m is either $\frac{17-1}{2}$ or $\frac{13-1}{2}$ , so m is either $6$ or $8$
So $a+d+n=30-5d$ or $30-7d$
Taking $\mod 7$ , we have no answer choices that give $2$ , and then taking $\mod 5$ gives the only answer that works is $20$
Therefore we have $a+d+n=\boxed{20.}$

Answer: B 20.
Problem: An arithmetic sequence of positive integers has $\text{n} \ge 3$ terms, initial term $a$ , and common difference $d > 1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote was $222$ . What is $a + d + n$
$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution: The formula for the sum of an arithmetic sequence is $n(\frac{a_1+a_n}{2})$ , where $a$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. Let $a$ be the first term, $d$ be the common difference, and $n$ be the number of terms of Carl's sequence. Since the sum the sequence is $1$ less or $1$ more than $222$ , we have \[n(\frac{a+d(n-1)}{2})\pm{1} = 222\] \[n(\frac{a+d(n-1)}{2}) = 222\pm{1}.\] The right-hand side is either $221$ or $223$ . We know that it has to be divisible by $n$ so we can find the factors of $221$ and $223$ . Checking all the primes less than $15$ , we find that $223$ is prime and $221=13\cdot17$
Because $n\ge3$ , the sum must be $221$ and the only possible values of $n$ are $13$ and $17$ . We can test both cases.
Case 1: $n=17$ Substituting for $n$ gives us $a+8d=17$ . Since the sequence consists of only positive integers, $d$ is an integer. We know that $d>1$ but if $d\ge2$ , then $8d>13$ . Hence, this case is not possible.
Case 2: $n=13$ Substituting for $n$ gives us $a+6d=17$ . Using the same logic from case 1, we get $3>d>1$ , so $d=2$ . Solving for $a$ , we get $a=5$ . Therefore, $a+d+n=5+2+13=20$ , so the answer is $\boxed{20.}$ .
~azc1027

Answer: B 20.
Problem: What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ $0\le v\le1,$ and $0\le w\le1$
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$

Solution: [asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5);   pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1);    //draw(A--B--D--C--cycle);    draw(A--B); label("1",midpoint(A--B),W); label("2",midpoint(D--B),N); draw(A--C,dashed); draw(B--D); draw(C--D, dashed);  draw(A--AA); label("5",midpoint(A--AA),W); draw(B--BB,dashed); draw(C--CC,dashed); draw(D--DD); label("5",midpoint(D--DD),E); label("1",midpoint(CC--DD),E); label("2",midpoint(AA--CC),S);  // Dotted vertices dot(A); dot(B); dot(C); dot(D);    dot(AA); dot(BB); dot(CC); dot(DD);  draw(AA--BB,dashed); draw(AA--CC); draw(BB--DD,dashed); draw(CC--DD);  label("(0,0)",AA,W); label("(-3,4)",A,SW); label("(-1,5)",D,E); label("(2,1)",DD,NE); [/asy] Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$ . We first fix $u$ and $v$ to $0$ , and graph $(-3w,4w)$ from $0\le w\le1$ . When $w$ is $0$ , we have the point $(0,0)$ , and when $w$ is $1$ , we have the point $(-3,4)$ . We see that since this is a directly proportional function, we can just connect the dots like this:
[asy] 	import graph; 	Label f; size(5cm); 	unitsize(0.7cm);  	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	[/asy]
Now, when we vary $u$ from $0$ to $2$ , this line is translated to the right $2$ units:
[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4)); 	[/asy]
We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$ , so we shade in the region:
[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	[/asy]
We can also shift this quadrilateral one unit up, because of $v$ . Thus, this is our figure:
[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((2,0)--(-1,4));  	filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); 	filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray);  draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed); 	[/asy]
[asy] 	import graph; 	Label f;  	unitsize(0.7cm); size(5cm); 	xaxis(-5,5,Ticks(f, 5.0, 1.0)); 	yaxis(-5,5,Ticks(f, 5.0, 1.0));  	draw((0,0)--(-3,4)); 	draw((1,0)--(-2,4));  	filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray); 	[/asy]
The length of the boundary is simply $1+2+5+1+2+5$ $5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$ .). This equals $\boxed{16.}$

Answer: E 16.
Problem: What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ $0\le v\le1,$ and $0\le w\le1$
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$

Solution: We can find the "boundary points" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$ . We now test each one.
Case 1: $u = 0, v = 0, w = 0 \implies (0, 0)$
Case 2: $u = 0, v = 0, w = 1 \implies (-3, 4)$
Case 3: $u = 0, v = 1, w = 0 \implies (0, 1)$
Case 4: $u = 0, v = 1, w = 1 \implies (-3, 5)$
Case 5: $u = 1, v = 0, w = 0 \implies (2, 0)$
Case 6: $u = 1, v = 0, w = 1 \implies (-1, 4)$
Case 7: $u = 1, v = 1, w = 0 \implies (2, 1)$
Case 8: $u = 1, v = 1, w = 1 \implies (-1, 5)$
When graphed on a coordinate plane, the points appear as follows.
[asy]  import graph; import geometry; Label f;  size(5cm); unitsize(0.7cm);  xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0));  pair A = (0, 0); dot (A); pair B = (-3, 4); dot (B); pair C = (0, 1); dot (C); pair D = (-3, 5); dot (D);  pair E = (2, 0); dot (E); pair F = (-1, 4); dot (F); pair G = (2, 1); dot (G); pair H = (-1, 5); dot (H);  [/asy]
Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.
[asy]  import graph; import geometry; Label f;  size(5cm); unitsize(0.7cm);  xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0));  draw((0,0)--(-3,4)--(-3,5)--(-1,5)--(2,1)--(2,0)--cycle, gray);  [/asy]
Each of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total area is $2 + 1 + 5 + 2 + 1 + 5 = \boxed{16}$

Answer: E 16
Problem: Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution: Mike's speed is $\frac{15}{57}=\frac{5}{19}$ laps per minute.
In the first $27$ minutes, he completed approximately $\frac{5}{19}\cdot27\approx\frac{1}{4}\cdot28=\boxed{7}$ laps.

Answer: B 7
Problem: Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution: Mike runs $1$ lap in $\frac{57}{15}=\frac{19}{5}$ minutes. So, in $27$ minutes, Mike ran about $\frac{27}{\frac{19}{5}} \approx \boxed{7}$ laps.

Answer: B 7
Problem: Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution: Mike's rate is \[\frac{15}{57}=\frac{x}{27},\] where $x$ is the number of laps he can complete in $27$ minutes. 
If you cross multiply, $57x = 405$
So, $x = \frac{405}{57} \approx \boxed{7}$

Answer: B 7
Problem: Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution: Note that $27$ minutes is a little bit less than half of $57$ minutes. Mike will therefore run a little bit less than $15/2=7.5$ laps, which is about $\boxed{7}$

Answer: B 7
Problem: Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution: Note that $57$ minutes is almost equal to $1$ hour. Running $15$ laps in $1$ hour is running approximately $1$ lap every $4$ minutes. This means that in $27$ minutes, Mike will run approximately $\frac{27}{4}$ laps. This is very close to $\frac{28}{4} = \boxed{7}$

Answer: B 7
Problem: The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution: Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$

Answer: E 5
Problem: The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution: Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$
So, the answer is $|x-y|=|42-47|=\boxed{5}.$

Answer: E 5
Problem: The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution: In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer: E 5
Problem: In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?
$\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}$

Solution: The formula for fuel efficiency is \[\frac{\text{Distance}}{\text{Gas Consumption}}.\] Note that $1$ mile equals $\frac 1m$ kilometers. We have \[\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.\] Therefore, the answer is $\boxed{100}.$

Answer: E 100
Problem: In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?
$\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}$

Solution: Since it can be a bit odd to think of "liters per $100$ km", this statement's numerical value is equivalent to $100$ km per $1$ liter:
$1$ km  requires $l$ liters, so the numerator is simply $l$ . Since $l$ liters is $1$ gallon, and $x$ miles is $1$ gallon, we have $1\text{ liter} = \frac{x}{l}$
Therefore, the requested expression is \[100\cdot\frac{m}{(\frac{x}{l})} = \boxed{100}.\] -Benedict T (countmath1)

Answer: E 100
Problem: The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solution: Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$

Answer: B 6
Problem: The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solution: The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$

Answer: B 6
Problem: A data set consists of $6$ (not distinct) positive integers: $1$ $7$ $5$ $2$ $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$

Solution: First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$

Answer: D 36