Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution: Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{1524}$

Answer: B 1524
Problem: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution: Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$ . So, this case gives us $511$
Onto the second case, if there are no $0$ s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$ , there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$ ).
All in all, that gives $511 + 1022 - 9 = \boxed{1524}$ monotonous numbers.

Answer: B 1524
Problem: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution: Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$ , we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$ . When $n=1$ , there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.
The answer is:
$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n}  -  \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) -  \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{1524}$

Answer: B 1524
Problem: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution: We have 2 cases.
Case 1: Ascending order
We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ $0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$ . However, the empty set ( $\emptyset$ ) is not an integer, so we must subtract 1. Thus, $512-1=511$
Case 2: Descending order
Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$ . However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$
We have covered all the cases. We add $511$ to $1022$ , giving us $1533$ . So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers . The answer is actually $1533-9=\boxed{1524}$

Answer: B 1524
Problem: In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$ . For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{12}$

Answer: D 12
Problem: In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} \cdot \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:
1. The identity transformation. Since this doesn't change anything, there are $60$ fixed points.
2. Reflect on a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$ . Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection.
3. Rotate by $120^\circ$ counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$ $4$ , and $6$ will be the same. Similarly, the colors of circles $2$ $3$ , and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.
By Burnside's Lemma , the total number of colorings is $(1 \cdot 60+3 \cdot 4+2 \cdot 0)/(1+3+2) = \boxed{12}$

Answer: D 12
Problem: In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as
[asy] filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1));  draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); [/asy]
Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving $\frac{10-2}{2}+2=6$ arrangements in this case.
Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also $6$ arrangements in this case.
In total, we have $6+6=\boxed{12}$

Answer: D 12
Problem: In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
[asy] size(110); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution: We note that the group $G$ acting on the possible colorings is $D_3 = \{e, r, r^2, s, sr, sr^2\}$ , where $r$ is a $120^\circ$ rotation and $s$ is a reflection. In particular, the possible actions are the identity, the $120^\circ$ and $240^\circ$ rotations, and the three reflections.
We will calculate the number of colorings that are fixed under each action. Every coloring is fixed under the identity, so we count $\dfrac{6!}{3!2!1!} = 60$ fixed colorings. Note that no colorings are fixed under the rotations, since then the outer three and inner three circle must be the same color, which is impossible in our situation.
Finally, consider the reflection with a line of symmetry going through the top circle. Every fixed coloring is determined by the color of the top circle (either green or blue), and the color of the middle circles (either blue or red). Hence, there are $2\cdot 2 = 4$ colorings fixed under this reflection action. The other two actions are symmetric, so they also have $4$ fixed colorings. Hence, by Burnside's lemma, the number of unique colorings up to reflections and rotations is \[\dfrac{1}{|D_3|} (1\cdot 60 + 2\cdot 0 + 3\cdot 4) = \dfrac{1}{6}\cdot 72 = \boxed{12}.\]

Answer: D 12
Problem: Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution: We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
By inspection, $N \equiv 4 \text{ (mod 5)}$ .
The remainder when $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$ , but since $10 \equiv 1 \text{ (mod 9)}$ , we can also write this as $1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$ , which has a remainder of 0 mod 9. Solving these modular congruence using the Chinese Remainder Theorem we get the remainder to be $9 \pmod{45}$ . Therefore, the answer is $\boxed{9}$

Answer: C 9
Problem: Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution: Once we find our 2 modular congruences, we can narrow our options down to ${C}$ and ${D}$ because the remainder when $N$ is divided by $45$ should be a multiple of 9 by our modular congruence that states $N$ has a remainder of $0$ when divided by $9$ . Also, our other modular congruence states that the remainder when divided by $45$ should have a remainder of $4$ when divided by $5$ . Out of options $C$ and $D$ , only $\boxed{9}$

Answer: C 9
Problem: Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution: Realize that $10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}$ for all positive integers $k$
Apply this on the expanded form of $N$ \[N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv\] \[10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv\] \[10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{9}\]

Answer: C 9
Problem: Let $N=123456789101112\dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution: We know that $45 = 5 \cdot 9$ , so we can apply our restrictions to that. We know that the units digit must be $5$ or $0$ , and the digits must add up to a multiple of $9$ $1+2+3+4+\cdots + 44 = \frac{44 \cdot 45}{2}$ . We can quickly see this is a multiple of $9$ because $\frac{44}{2} \cdot 45 = 22 \cdot 45$ . We know $123 \ldots 4344$ is not a multiple of $5$ because the units digit isn't $5$ or $0$ . We can just subtract by 9 until we get a number whose units digit is 5 or 0.
We have $123 \ldots 4344$ is divisible by $9$ , so we can subtract by $9$ to get $123 \ldots 4335$ and we know that this is divisible by 5. So our answer is $\boxed{9}$

Answer: C 9
Problem: The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution: WLOG, let the centroid of $\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $AI = BI = CI = 2\sqrt{2}$ , so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$ , then by the Law of Cosines $AB = 2\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$ . Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$ , so the square of the area of the triangle is $\boxed{108}$

Answer: C 108
Problem: The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution: Without loss of generality, let the centroid of $\triangle ABC$ be $G = (-1,-1)$ . Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$ . Then, point $B$ must be the reflection of $C$ across the line $y=x$ , so let $B = \left(a,\frac{1}{a}\right)$ and $C=\left(\frac{1}{a},a\right)$ , where $a <-1$ . Because $G$ is the centroid, the average of the $x$ -coordinates of the vertices of the triangle is $-1$ . So we know that $a + 1/a+ 1 = -3$ . Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$ . So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$ . So $BC=2\sqrt{6}$ , and finding the square of the area gives us $\boxed{108}$

Answer: C 108
Problem: The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution: Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$ . It is known that the centroid is equidistant from the three vertices of $\triangle ABC$ . Because we have the coordinates of both $A$ and $G$ , we know that the distance from $G$ to any vertice of $\triangle ABC$ is $\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}$ . Therefore, $AG=BG=CG=2\sqrt{2}$ . It follows that from $\triangle ABG$ , where $AG=BG=2\sqrt{2}$ and $\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}$ $[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ using the formula for the area of a triangle with sine $\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)$ . Because $\triangle ACG$ and $\triangle BCG$ are congruent to $\triangle ABG$ , they also have an area of $2\sqrt{3}$ . Therefore, $[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}$ . Squaring that gives us the answer of $\boxed{108}$

Answer: C 108
Problem: The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution: Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ . Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).$
Since the centroid coordinates are the average of the vertex coordinates, we have that $\frac{x_1+x_2+x_3}{3}=1$ and $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.$
We also know that the centroid is the orthocenter in an equilateral triangle, so $CG \perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$ . From the equation $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$ . These equations are starting to resemble Vieta's:
$x_1+x_2+x_3=3$
$x_1x_2+x_1x_3+x_2x_3 = -3$
$x_1x_2x_3=-1$
$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$ . This factors as $(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,$ for the points $(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)$ . The side length is clearly $\sqrt{24}$ , so the square of the area is $\boxed{108}.$

Answer:  108
Problem: Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Let the sum of the scores of Isabella's first $6$ tests be $S$ . Since the mean of her first $7$ scores is an integer, then $S + 95 \equiv 0 \text{ (mod 7)}$ , or $S \equiv3 \text{ (mod 7)}$ . Also, $S \equiv 0 \text{ (mod 6)}$ , so by the CRT $S \equiv 24 \text{ (mod 42)}$ . We also know that $91 \cdot 6 \leq S \leq 100 \cdot 6$ , so by inspection, $S = 570$ . However, we also have that the mean of the first $5$ test scores must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$ , which implies that the $6$ th test score is $\boxed{100}$

Answer: E 100
Problem: Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: First, we find the largest sum of scores which is $100+99+98+97+96+95+94$ which equals $7(97)$ . Then we find the smallest sum of scores which is $91+92+93+94+95+96+97$ which is $7(94)$ . So the possible sums for the 7 test scores so that they provide an integer average are $7(97), 7(96), 7(95)$ and $7(94)$ which are $679, 672, 665,$ and $658$ respectively.  Now in order to get the sum of the first 6 tests, we subtract $95$ from each sum producing $584, 577, 570,$ and $563$ . Notice only $570$ is divisible by $6$ so, therefore, the sum of the first $6$ tests is $570$ . We need to find her score on the $6th$ test so we have to find which number will give us a number divisible by $5$ when subtracted from $570.$ Since $95$ is the $7th$ test score and all test scores are distinct that only leaves $\boxed{100}$

Answer: E 100
Problem: Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution: Since all of the scores are from $91 - 100$ , we can subtract 90 from all of the scores. Since the last score was a 95, the sum of the scores from the first six tests must be $3 \pmod 7$ and $0 \pmod 6$ . Trying out a few cases, the only solution possible is 30 (this is from adding numbers 1-10). The sixth test score must be $0 \pmod 5$ because $30\equiv 0\pmod5$ . The only possible test scores are $95$ and $100$ , and $95$ is already used, so the answer is $\boxed{100}$

Answer: E 100