Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$

Solution: Because $P$ $Q$ $R$ , and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational. The greatest value of $PQ$ happens when $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. Another possible pair could be $(-4,3)$ and $(5,0)$ . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from $(-4,3)$ to $(4,-3)$ . The distance between $(3,-4)$ and $(-4,3)$ is greater than the distance between $(5,0)$ and $(4,-3)$ . Therefore, the segment from $(3,-4)$ to $(-4,3)$ is the longest attainable (the other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$ $(3, 4)$ and $(-4, -3)$ $(-3, 4)$ and $(4, -3)$ . The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3)$ and vice versa. Using the distance formula, we get that $PQ$ is $\sqrt{98}$ and that $RS$ is $\sqrt{2}.$ $\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{7}$

Answer: D 7
Problem: Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$

Solution: We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using $7$ (though $5 \sqrt{2}$ actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let $RS$ be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of $RS$ is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$ . Assuming that $\frac{PQ}{RS} = 7$ , we plug in $RS = \sqrt{2}$ and solve for $PQ$ $PQ=7\sqrt{2}$ . Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of $x$ and $y$ does $\sqrt{x^{2}+y^{2}}=7\sqrt2$ ? We see that this can easily be made into a $45-45-90$ triangle. But, instead of substituting $y=x$ into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a $45-45-90$ triangle, the two $45$ degree sides have side length $s$ , then the hypotenuse is $s\sqrt2$ . Using this, we can see that $s=7$ , and since our equation does in fact yield a sensible solution, we can be assured that our answer is $\boxed{7}$

Answer: D 7
Problem: Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$

Solution: By inspection, when $R$ is at $(3, 4)$ and $S$ is at $(4, 3),$ it makes $RS$ as small as possible with a distance of $\sqrt{2}$ . The greatest possible length of $PQ$ arises when $P$ is at $(-3, 4)$ and $Q$ is at $(4, -3).$ Using the distance formula, we find that $PQ$ has a length of $7\sqrt{2}.$ The requested fraction is then $\dfrac{PQ}{RS} = \dfrac{7\sqrt{2}}{\sqrt{2}} = \boxed{7}$

Answer: D 7
Problem: Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$ , since if she gets to her turn again, she is back where she started with probability of winning $P$ . The chance she wins on her first turn is $\frac{1}{3}$ . The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$ . Multiplying gives us $\frac{2}{5}$ . Thus, \[P = \frac{1}{3} + \frac{2}{5}P\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5=\boxed{4}$

Answer: D 4
Problem: Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} \cdots$ This can be represented as an infinite geometric series: \[P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5 = \boxed{4}.$

Answer: D 4
Problem: Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We can solve this by using 'casework,' the cases being:
Case 1: Amelia wins on her first turn.
Case 2 Amelia wins on her second turn.
and so on.
The probability of her winning on her first turn is $\dfrac13$ . The probability of all the other cases is determined by the probability that Amelia and Blaine all lose until Amelia's turn on which she is supposed to win. So, the total probability of Amelia winning is: \[\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)\cdot\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)^2\cdot\dfrac{1}{3}+\cdots.\] Factoring out $\dfrac13$ we get a geometric series: \[\dfrac{1}{3}\left(1+\dfrac{2}{5}+\left(\dfrac{2}{5}\right)^2+\cdots\right) = \dfrac{1}{3}\cdot\dfrac{1}{3/5} = \boxed{59}.\]

Answer:  59
Problem: Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution: Let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.
We can split this problem up into two cases:
$\textbf{Case 1: }$ A sits on an edge seat.
Since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of $2 \cdot 2 \cdot 2 \cdot 2 = 16$
$\textbf{Case 2: }$ A does not sit in an edge seat.
Still, the only two people that can sit next to A are either D or E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are $3 \cdot 2 \cdot 2 = 12$ seatings in this case.
Adding up all the seatings, we have $16+12 = \boxed{28}$

Answer: C 28
Problem: Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution: Label the seats (from left to right) $1$ through $5$ . The number of ways to seat Derek and Eric in the five seats with no restrictions is $5 \cdot 4=20$ . The number of ways to seat Derek and Eric such that they sit next to each other is $8$ (we can treat Derek and Eric as a "block". There are four ways to seat this "block", and two ways to permute Derek and Eric, for a total of $4\cdot 2=8$ ), so the number of ways such that Derek and Eric don't sit next to each other is $20-8=12$ . Note that once Derek and Eric are seated, we can divide into three cases.
The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us $0$ ways.
Another possible case is if Derek and Eric sit in seats $2$ and $4$ in some order. There are $2$ possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are $3!=6$ ways to do this. So the second case gives us $2 \cdot 6=12$ total ways.
The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are $12-2-2=8$ ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us $8 \cdot 2=16$ ways.
So in total there are $12+16=28$ ways. Our answer is $\boxed{28}$ Minor $\LaTeX$ edits by fasterthanlight

Answer: C 28
Problem: Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution: We start with complementary counting. After all, it's much easier to count the cases where some of these restraints are true, than when they aren't.
PIE:
Let's count the total number of cases where one of these is true:
When Alice is with Bob: $2\cdot4!=48$
When Alice is with Carla: $2\cdot4!=48$
When Derek is with Eric: $2\cdot4!=48$
Then, we count the cases where two of these are true.
Alice is next to Carla, and Alice is also next to Bob.
There are two ways to rearrange Alice, Bob, and Carla so that this is true:
BAC and CAB. $2\cdot3!=12$
Alice is next to Carla, and Derek is also next to Eric. $2\cdot2\cdot3!=24$
Alice is next to Bob, and Derek is also next to Eric. $2\cdot2\cdot3!=24$
Finally, we count the cases where all three of these are true: $2\cdot2\cdot2=8$
We add up the cases where one of these are true: $48\cdot3=144$
Subtract the cases where two of these are true: $144-60=84$
And finally add back the cases where three of these are true: $84+8=92$
Thus, our answer is $5!-92=28$ , or $\boxed{28}$

Answer: C 28
Problem: Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution: To find the number of ways, we do casework.
Case 1: Alice sits in the first seat (leftmost)
Since Alice refuses to sit with Bob and Carla, then the seat on her immediate right must be Derek or Eric. The middle seat must be Bob or Carla (because Derek and Eric refuse to sit together). The seat to the right of the middle seat could be whoever is left over from Derek and Eric, or whoever is left together from Bob and Carla. The last seat only has one person left. There are $4$ ways to permute Bob, Carla, Derek, and Eric, and $2$ ways to pick who goes in the seat to the right of the middle seat, for $4\cdot2=8$ seating arrangements here.
Case 2: Alice sits in the second seat
Derek and Eric must be on both sides of Alice because otherwise, we would have to put Bob or Carla next to Alice which is forbidden. Then Bob and Carla take the remaining two seats. There are $4$ ways to permute and $4$ arrangements here.
Case 3: Alice sits in the middle seat
Once again, Derek and Eric must be on both sides of Alice, and Bob and Carla need to take the two remaining seats. There are also $4$ ways to permute and $4$ arrangements here.
Case 4: Alice sits in the fourth seat
By symmetry, this is the same as case 2. There are $4$ arrangements.
Case 5: Alice sits in the last seat (rightmost)
By symmetry, this is the same as case 1. There are $8$ arrangements.
Adding up the cases, there are $8+4+4+4+8=28$ total seating arrangements $\Longrightarrow \boxed{28}$

Answer: C 28
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: Note that $n \equiv S(n) \pmod{9}$ . This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$ . Thus, if $S(n) = 1274$ , then $n \equiv 5 \pmod{9}$ , and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that satisfies $n+1 \equiv 6 \pmod{9}$ is $\boxed{1239}$

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: One divisibility rule that we can use for this problem is that a multiple of $9$ will always have its digits sum to a multiple of $9$ . We can find out that the least number of digits the number $n$ has is $142$ , with $141$ $9$ 's and $1$ $5$ , assuming the rule above. No matter what arrangement or different digits we use, the divisibility rule stays the same. To make the problem simpler, we can just use the $141$ $9$ 's and $1$ $5$ . By randomly mixing the digits up, we are likely to get: $9999$ ... $9995999$ ... $9999$ . By adding $1$ to this number, we get: $9999$ ... $9996000$ ... $0000$ .
Knowing that $n+1$ is divisible by $9$ when $6$ , we can subtract $6$ from every available choice, and see if the number is divisible by $9$ afterwards.
After subtracting $6$ from every number, we can conclude that $1233$ (originally $1239$ ) is the only number divisible by $9$ . 
So our answer is $\boxed{1239}$

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: The number $n$ can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.
If $A$ is correct, then $n$ must be some number $99999999...9$ , because when we add one to $99999999...9$ we get $10000000...00$ . Thus, if $1$ is the correct answer, then the equation $9x=1274$ must have an integer solution (i.e. $1274$ must be divisible by $9$ ). But since it does not, $1$ is not the correct answer.
If $B$ is correct, then $n$ must be some number $29999999...9$ , because when we add one to $29999999...9$ , we get $30000000...00$ . Thus, if $3$ is the correct answer, then the equation $2+9x=1274$ must have an integer solution. But since it does not, $3$ is not the correct answer.
Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if $S(n+1)=N$ , then $n$ must be a number whose initial digits sum to $N-1$ , and whose other, terminating digits, are all $9$ . Thus, we can evaluate the three final possibilities by seeing if the equation $(N-1)+9x=1274$ has an integer solution.
The equation does not have an integer solution for $N=12$ , so $C$ is not correct. However, the equation does have an integer solution for $N=1239$ $x=4$ ), so $\boxed{1239}$ is the answer.

Answer: D 1239
Problem: Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution: If adding $1$ to $n$ does not carry any of its digits, then $S(n+1)=S(n)+1$ (ex: $25+1=26$ . Sum of digits $7 \rightarrow 8$ ). But since no answer choice is $1275$ , that means $n$ has some amount of $9$ 's from right to left.
When $n+1$ , some $9$ 's will bump to 0, not affected its $\pmod 9$ . But the first non-9 digit (from right to left) will be bumped up by 1. So $S(n) + 1 \pmod {9} \equiv S(n+1) \pmod{9}$ . For example, $34999+1=35000$ , and the sum of digits $7+27 \rightarrow 8+0$
Since $S(n) \equiv 5 \pmod{9}$ , that means $S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that meets this requirement is $\boxed{1239}.$

Answer: D 1239
Problem: How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Solution: We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$ , so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$ , which simplifies to $2300$ .
Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\dbinom53$ is $10$ , and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$ .
We can also count the ones with $4$ on a diagonal. That is $\dbinom43$ , which is 4, and there are $4$ of those diagonals, so that results in $16$ .
We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$ .
We can also count the ones with a slope of $\frac12$ $2$ $-\frac12$ , or $-2$ , with $3$ points in each. Note that there are $3$ such lines, for each slope, present in the grid. In total, this results in $12$ .
Finally, we subtract all the ones in a line from $2300$ , so we have $2300-120-16-4-12=\boxed{2148}$

Answer: B 2148
Problem: How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Solution: There are $5 \times 5 = 25$ total points in all. So, there are $\dbinom{25}3 = 2300$ ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line.
There are $10 \times 10 = 100$ cases where the 3 points chosen make up a vertical or horizontal line.
There are $2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40$ cases where the 3 points all land on the diagonals of the square.
There are $3 \times 4=12$ ways where the 3 points make the a slope of $\frac{1}{2}$ $-\frac{1}{2}$ $2$ , and $-2$
Hence, there are $100+40+12=152$ cases where the chosen 3 points make a line. The answer would be $2300-152=\boxed{2148}$

Answer: B 2148
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: $f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x+p)\]
where $-p\in\mathbb{R}$ is the fourth root of $f(x)$ . 
(Using $(x+p) = (x-r))$ instead of $(x-r)$ makes the following computations less messy.)
Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x+p)\\ &=x^4+(a+p)x^3+(1+ap)x^2+(10+p)x+10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a+p&=1\\ 1+ap=b\\ 10+p&=100\\ 10p&=c.\\ \end{align*}
Let's solve for $a,b,c,$ and $p$ . Since $10+p=100$ $p=90$
Since $a+p=1$ $a=-89$
(Solution 1.1 branches from here and takes a shortcut.)
$c=(10)(90)=900$
Then, since $b=1+ap$ $b=-8009$ . Thus,
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
(Solution 1.2 branches from here and takes another shortcut)
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007}

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: A faster ending to Solution 1 is as follows.
\begin{align*} f(1)&=(1+p)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &= (7)(13)(11)(-7) = (1001)(-7) \\ &=\boxed{7007}

Answer: C 7007
Problem: For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution: Also a faster ending to Solution 1 is as follows.
To find $f(1)$ we just need to find the sum of the coefficients which is $1 + 1 - 8009 + 100 + 900= \boxed{7007}.$

Answer: C 7007