Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The ratio of boys to girls in Mr. Brown's math class is $2:3$ .  If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$

Solution: Let the number of boys be $2x$ .  It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]
The difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\boxed{6}$

Answer: D 6
Problem: If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$

Solution: If the average score of the first six is $84$ , then the sum of those six scores is $6\times 84=504$
The average score of the first seven is $85$ , so the sum of the seven is $7\times 85=595$
Taking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\boxed{91}$

Answer: D 91
Problem: If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$

Solution: Let's remove the condition that the average of the first seven tests is $85$ , and say the 7th test score was a $0$ . Then, the average of the first seven tests would be \[\frac{6\times 84}{7}=72\]
If we increase the seventh test score by $7$ , the average will increase by $\frac{7}{7}=1$ . We need the average to increase by $85-72=13$ , so the seventh test score is $7\times 13=91$ more than $0$ , which is clearly $91$ .  This is choice $\boxed{91}$

Answer: D 91
Problem: If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$

Solution: Let the width be $w$ and the length be $l$ . Then, the original perimeter is $2(w+l)$
After the increase, the new width and new length are $1.1w$ and $1.1l$ , so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$
Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}
$\boxed{10}$

Answer: B 10
Problem: If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$

Solution: Assume WLOG that the rectangle is a square with length $10$ and width $10$ .  Thus, the square has a perimeter of $40$ .  Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11.  Thus, the new square's perimeter is $44$ , and because $44$ is $110\%$ of $40$ , our answer is $\boxed{10}$

Answer: B 10
Problem: Mr. Green receives a $10\%$ raise every year.  His salary after four such raises has gone up by what percent?
$\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\%$

Solution: Assume his salary is originally $100$ dollars. Then, in the next year, he would have $110$ dollars, and in the next, he would have $121$ dollars. The next year he would have $133.1$ dollars and in the final year, he would have $146.41$ . As the total increase is greater than $45\%$ , the answer is $\boxed{45}$

Answer: E 45
Problem: King Middle School has $1200$ students.  Each student takes $5$ classes a day.  Each teacher teaches $4$ classes.  Each class has $30$ students and $1$ teacher.  How many teachers are there at King Middle School?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$

Solution: If each student has $5$ classes, and there are $1200$ students, then they have a total of $5\times 1200=6000$ classes among them.
Each class has $30$ students, so there must be $\frac{6000}{30}=200$ classes. Each class has $1$ teacher, so the teachers have a total of $200$ classes among them.
Each teacher teaches $4$ classes, so if there are $t$ teachers, they have $4t$ classes among them. This was found to be $200$ , so \[4t=200\Rightarrow t=50\]
This is answer choice $\boxed{50}$

Answer: E 50
Problem: In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same.  The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution: Let the number in the top circle be $a$ and then $b$ $c$ $d$ $e$ , and $f$ , going in clockwise order. Then, we have \[S=a+b+c\] \[S=c+d+e\] \[S=e+f+a\]
Adding these equations together, we get
\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}
where the last step comes from the fact that since $a$ $b$ $c$ $d$ $e$ , and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$
The left hand side is divisible by $3$ and $75$ is divisible by $3$ , so $a+c+e$ must be divisible by $3$ .  The largest possible value of $a+c+e$ is then $15+14+13=42$ , and the corresponding value of $S$ is $\frac{75+42}{3}=39$ , which is choice $\boxed{39}$

Answer: D 39
Problem: In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same.  The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$

Solution: To make the sum the greatest, put the three largest numbers $(13,14$ and $15)$ in the corners. Then, balance the sides by putting the least integer $(10)$ between the greatest sum $(14$ and $15)$ . Then put the next least integer $(11)$ between the next greatest sum ( $13 +15$ ). Fill in the last integer $(12)$ and you can see that the sum of any three numbers on a side is (for example) $14 +10 + 15 = 39$ $\boxed{39}$ . 
-by goldenn

Answer: D 39
Problem: Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\  \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]

Solution: Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (both are even numbers). Thus, if one of these is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is $\boxed{3}$

Answer: A 3
Problem: Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$

Solution: Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$
No prime number ends in $4$ , since all numbers that end in $4$ are divisible by $2$ .  Thus, we are only looking for numbers that end in $9$
Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$
Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$
Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$ , since all numbers are odd), we get:
$29, 49, 89.$
Noting that $49$ is not prime, we have only $29$ and $89$ , which give a sum of $118$ , so the answer is $\boxed{118}$

Answer: A 118
Problem: The marked price of a book was 30% less than the suggested retail price.  Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.  What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\mathrm{(E)\ }65\%$

Solution: Without loss of generality, let's assume that the retail price was $100$ USD.
The marked price of the book is $30 \%$ off of $100$ which is equal to $100-100(0.3)=70.$
Half of that marked price is $0.5(70)=35.$
Therefore the percent Alice payed of the suggested retail price is $35/100=\boxed{35}.$

Answer: C 35
Problem: What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$

Solution: $2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$ , a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$ , thus making the answer $\boxed{7}$

Answer: D 7
Problem: What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\  2 \qquad \textbf{(B)}\  3 \qquad \textbf{(C)}\  4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\  6$

Solution: The sum of the interior angles of a hexagon is $720$ degrees.  In a convex polygon, each angle must be strictly less than $180$ degrees.
Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than $90\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.
Four acute angles and two obtuse angles can only sum to less than $90\cdot 4 + 180\cdot 2 = 720$ degrees.  This is a strict inequality, so these angles could not form a hexagon.  (The limiting figure would be four right angles and two straight angles, which would really be a square with two "extra" points on two sides to form the straight angles.)
Three acute angles and three obtuse angles work.  For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\cdot 3 + 160\cdot 3 = 720$ degrees, which is a genuine hexagon.  Thus, the answer is $\boxed{3}$

Answer: B 3
Problem: At the end of $1994$ , Walter was half as old as his grandmother. The sum of the years in which they were born was $3838$ . How old will Walter be at the end of $1999$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\  49\qquad \textbf{(C)}\  53\qquad \textbf{(D)}\  55\qquad \textbf{(E)}\ 101$

Solution: In $1994$ , if Water is $x$ years old, then Walter's grandmother is $2x$ years old.
This means that Walter was born in $1994 - x$ , and Walter's grandmother was born in $1994 - 2x$
The sum of those years is $3838$ , so we have:
$1994 - x + 1994 - 2x = 3838$
$3988 - 3x = 3838$
$x = 50$
If Walter is $50$ years old in $1994$ , then he will be $55$ years old in $1999$ , thus giving answer $\boxed{55}$

Answer: D 55
Problem: The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$ . The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$ . If it costs $137.94  to label all the lockers, how many lockers are there at the school?
$\textbf{(A)}\ 2001 \qquad  \textbf{(B)}\ 2010 \qquad  \textbf{(C)}\ 2100 \qquad  \textbf{(D)}\ 2726 \qquad  \textbf{(E)}\ 6897$

Solution: Since all answers are over $2000$ , work backwards and find the cost of the first $1999$ lockers.  The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\cdot 90 = 3.60$ .  Lockers $100$ through $999$ cost $0.06\cdot 900 = 54.00$ , and lockers $1000$ through $1999$ inclusive cost $0.08\cdot 1000 = 80.00$
This gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$ .  There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers.  These lockers are $2000$ and $2001$ , leading to answer $\boxed{2001}$

Answer: A 2001
Problem: What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$ , each with leading coefficient 1?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution: The intersections of the two polynomials, $p(x)$ and $q(x)$ , are precisely the roots of the equation $p(x)=q(x) \rightarrow p(x) - q(x) = 0$ . Since the leading coefficients of both polynomials are $1$ , the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\boxed{3}$

Answer: C 3
Problem: Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad  \textbf{(B)}\ 8 \qquad  \textbf{(C)}\ 9 \qquad  \textbf{(D)}\ 10 \qquad  \textbf{(E)}\ 11$

Solution: Alina and Tina must sing more than $4$ , but less than $7$ , songs.  Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.
However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$ .  Thus, $7 + 4 + a + t$ must be divisible by $3$ , and $a+t - 1$ must be divisible by $3$
Since $10 \le a+t \le 12$ by the bounds of $5 \le a, t \le 6$ , we have $9 \le a + t - 1 \le 11$ .  Because the middle number must be a multiple of $3$ , we set $a + t - 1 = 9$ , leading to $a + t = 10$ and $a = t = 5$
So, the four girls sang $4, 5, 5$ and $7$ songs in trios.  That means there were $\frac{4 + 5 + 5 + 7}{3} = 7$ songs sung, and the answer is $\boxed{7}$

Answer: A 7
Problem: Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad  \textbf{(B)}\ 0.2 \qquad  \textbf{(C)}\ 0.3 \qquad  \textbf{(D)}\ 0.4 \qquad  \textbf{(E)}\ 0.5$

Solution: $(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1$ , so $\sec x + \tan x = \boxed{0.5}$

Answer: E 0.5
Problem: Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad  \textbf{(B)}\ 0.2 \qquad  \textbf{(C)}\ 0.3 \qquad  \textbf{(D)}\ 0.4 \qquad  \textbf{(E)}\ 0.5$

Solution: Note that $\sec x - \tan x = (1-\sin x)/\cos x$ and $\sec x + \tan x = (1+\sin x)/\cos x$ . Let $(1+\sin x)/\cos x = y$ . Multiplying, we get $(1-\sin^{2}x)/\cos^{2}x = 1$ .Then, $2y = 1$ $\sec x + \tan x =  \boxed{0.5}$ . ~songmath20  
Edited 5.1.2023

Answer: E 0.5
Problem: Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$

Solution: According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$
From the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$
The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$ , from which $Q(99)=-1$ . This means that $99$ is a root of the polynomial $Q(x)+1$ . In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$
Substituting this into the original formula for $P(x)$ we get \[P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =\] \[= S(x)(x-99)(x-19) - (x-19) + 99\]
Therefore when $P(x)$ is divided by $(x-19)(x-99)$ , the remainder is $\boxed{118}$

Answer:  118
Problem: Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$

Solution: Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear. We can write $P(x)$ in the form \[P(x) = Q(x)(x-19)(x-99) + cx+d\] where $cx+d$ is the remainder.
By the Remainder Theorem, plugging in $19$ and $99$ gives us a system of equations. \[99c+d = 19\] \[19c+d = 99\]
Solving gives us $c=-1$ and $d = 118$ , thus, our answer is $\boxed{118}$

Answer: C 118
Problem: The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms.  Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$

Solution: Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$
By definition, $a_3=m$
Next, $a_4$ is the mean of $m-x$ $m+x$ and $m$ , which is again $m$
Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$
It follows that $m=a_9=99$ . From $19=a_1=m-x$ we get that $x=80$ . And thus $a_2 = m+x = \boxed{179}$

Answer: E 179
Problem: The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms.  Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$

Solution: Let $a_1=a$ and $a_2=b$ . Then, $a_3=\frac{a+b}{2}$ $a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},$ and so on.
Since $a_3=a_4$ $a_n=a_3$ for all $n\geq3.$
Hence, $a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$
Subtracting $a_1$ from $198,$ we get $b=a_2=\boxed{179}.$

Answer: E 179
Problem: The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$ . Find $a+c$
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$

Solution: Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:
Obviously, the maximum of the first graph is achieved when $x=a$ , and its value is $-0+b=b$ . Similarly, the minimum of the other graph is $(c,d)$ . Therefore the two remaining vertices of the area between the graphs are $(a,b)$ and $(c,d)$
As the area has four right angles, it is a rectangle. Without actually computing $a$ and $c$ we can therefore conclude that $a+c=2+8=\boxed{10}$

Answer:  10
Problem: There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ .  Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

Solution: Multiply out the $7!$ to get
\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]
By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$ , so $a_7 = 2$ . Then we move $a_7$ to the left and divide through by $7$ to obtain
\[\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\]
We then repeat this procedure $\pmod{6}$ , from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$ , and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem ). The answer is $9 \Longrightarrow \boxed{9}$

Answer: B 9
Problem: There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ .  Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

Solution: We start by multiplying both sides by $7!$ , and we get: \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] After doing some guess and check, we find that the answer is $\boxed{9}$

Answer: B 9
Problem: There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ .  Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

Solution: Let's clear up the fractions: \[\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\] \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] Notice that if we divide everything by $7$ then we would have: \[514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7\] Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$ , so $a_7=2$
Similarly, if we divide everything by $6$ , then we would have: \[85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6\] Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$ , so $a_6=4$
The pattern repeats itself, so in the end we have $a_2=1$ $a_3=1$ $a_4=1$ $a_5=0$ $a_6=4$ $a_7=2$ . So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{9}$

Answer: B 9