Navvye/Olympiad-Problems-CSV huggingface dataset:
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: We work backwards to solve this problem. In the $n^{th}$ test, $\sum_{i=1}^{n} i$ (mod n) = 0. In the following table, the tests already taken are in bold, the latest test is underlined. We work from the row of (mod 7), (mod 6), and (mod 5) to determine the test order by trial and error.
\[\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline  & 91 & 92 & 93 & 94 & 95 & 96 & 97 & 98 & 99 & 100 \\ \hline mod 7 & \textbf{0} & \textbf{1} & 2 & \textbf{3} & \underline{\textbf{4}} & \textbf{5} & \textbf{6} & 0 & 1 & \textbf{2} \\ \hline mod 6 & \textbf{1} & \textbf{2} & 3 & \textbf{4} & 5 & \textbf{0} & \textbf{1} & 2 & 3 & \underline{\textbf{4}} \\ \hline mod 5 & \textbf{1} & \textbf{2} & 3 & \underline{\textbf{4}} & 0 & \textbf{1} & \textbf{2} & 3 & 4 & 0 \\ \hline mod 4 & \underline{\textbf{3}} & \textbf{0} & 1 & 2 & 3 & \textbf{0} & \textbf{1} & 2 & 3 & 0 \\ \hline mod 3 & 1 & \textbf{2} & 0 & 1 & 2 & \textbf{0} & \underline{\textbf{1}} & 2 & 0 & 1 \\ \hline mod 2 & 1 & \underline{\textbf{0}} & 1 & 0 & 1 & \textbf{0} & 1 & 0 & 1 & 0 \\ \hline \end{tabular}\]
In the $4^{th}$ test, the test score could be 91 or 97.
As you can see, the test score of the $6^{th}$ test 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: Let's denote $x$ as $a_1+a_2+a_3+a_4+a_5+a_6$ , where $a_n$ is her $n$ th test score. We start by saying $\frac{x+95}{7} = [91,92,...100]$ . This results from the fact that her test averages are integers and her test scores are in the range of $91$ to $100$ . So $x+95=[637,644,...700] \implies x=[542,549,...605]$ . Next, we also see that $\frac{x}{6}=[91,92,...100]$ which becomes $x=[546,552,...600]$ . The only point of intersection of $[542,549,...605]$ and $[546,552,...600]$ is $570$ so $x=570$ . Now we can also see that $\frac{x-a_6}{5}=[91,92,...100]$ , so $x-a_6=[455,460,...500]$ , and since $x=570$ $570-a_6=[455,460,...500] \implies a_6=570-[455,460,...500]$ $a_6$ has to be in the range of $91$ and $100$ , so the only values in the set $[455,460,...500]$ that work are $470$ and $475$ which result in $a_6 = 100 \text{ or } 95$ respectively. But since her test scores are all distinct, and $95$ is already used for $a_7$ , our answer is $\boxed{100}$

Answer: E 100
Problem: What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

Solution: We can use subtraction of fractions to get \[\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{100}.\]

Answer: B 100
Problem: What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

Solution: Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{100}$

Answer: B 100
Problem: What is the value of $\dfrac{11!-10!}{9!}$
$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

Solution: We are given the equation $\frac{11!-10!}{9!}$
This is equivalent to $\frac{11(10!) - 1(10!)}{9!}$ Simplifying, we get $\frac{(11-1)(10!)}{9!}$ , which equals $10 \cdot 10$
Therefore, the answer is $10^2$ $\boxed{100}$

Answer: B 100
Problem: For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$

Answer: C 3
Problem: For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$

Answer: C 3
Problem: For every dollar Ben spent on bagels, David spent $25$ cents less. Ben paid $$12.50$ more than David. How much did they spend in the bagel store together?
$\textbf{(A)}\ $37.50 \qquad\textbf{(B)}\ $50.00\qquad\textbf{(C)}\ $87.50\qquad\textbf{(D)}\ $90.00\qquad\textbf{(E)}\ $92.50$

Solution: If Ben paid $$ 12.50$ more than David, then he paid $\frac{12.5}{.25}= $ 50.00$ . Thus, David paid $$ 37.50$ , and they spent $50.00+37.50 =$ 87.50 \implies \boxed{87.50}$

Answer: C 87.50
Problem: A rectangular box has integer side lengths in the ratio $1: 3: 4$ . Which of the following could be the volume of the box?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144$

Solution: Let the smallest side length be $x$ . Then the volume is $x \cdot 3x \cdot 4x =12x^3$ . If $x=2$ , then $12x^3 = 96 \implies \boxed{96.}$

Answer: D 96.
Problem: Star lists the whole numbers $1$ through $30$ once. Emilio copies Star's numbers, replacing each occurrence of the digit $2$ by the digit $1$ . Star adds her numbers and Emilio adds his numbers. How much larger is Star's sum than Emilio's?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110$

Solution: For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is $10\cdot 10+1\cdot 3=\boxed{103.}$

Answer: D 103.
Problem: The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$

Solution: Since $x$ is the mean, \begin{align*} x&=\frac{60+100+x+40+50+200+90}{7}\\ &=\frac{540+x}{7}. \end{align*}
Therefore, $7x=540+x$ , so $x=\boxed{90}.$

Answer: D 90
Problem: The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$

Solution: Note that $x$ must be the median so it must equal either $60$ or $90$ . You can see that the mean is also $x$ , and by intuition $x$ should be the greater one. $x=\boxed{90}.$ ~bjc

Answer: D 90
Problem: Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45$

Solution: If you started backwards you would get: \[0\Rightarrow (+40)=40 , \Rightarrow \left(\frac{1}{2}\right)=20 , \Rightarrow (+40)=60 , \Rightarrow \left(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{35}\]

Answer: C 35
Problem: Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays $40$ coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45$

Solution: If you have $x$ as the amount of money Foolish Fox started with we have $2(2(2x-40)-40)-40=0.$ Solving this we get $\boxed{35}$

Answer: C 35
Problem: A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: We are trying to find the value of $N$ such that \[1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.\] Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{9}.$

Answer: D 9
Problem: A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .
Notice that $1 + 2 + 3 \cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \cdots + 20 = 155$ and $21 + \cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract $70, 69, 68, 67, 66, 65,$ and $64, N = 63.$ Adding those two digits, we get the answer $\boxed{9}.$ - CorgiARMY

Answer: D 9
Problem: A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?
[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white);  label("$1$",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));  label("$1$",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));  label("$1$",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));  label("$1$",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));  label("$1$",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8$

Solution: Let the length of the inner rectangle be $x$
Then the area of that rectangle is $x\cdot1 = x$
The second largest rectangle has dimensions of $x+2$ and $3$ , making its area $3x+6$ . The area of the second shaded area, therefore, is $3x+6-x = 2x+6$
The largest rectangle has dimensions of $x+4$ and $5$ , making its area $5x + 20$ . The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$
The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.
Therefore, $(2x+6) - (x) = (2x+14) - (2x+6)\implies x+6 = 8\implies x =2\implies \boxed{2}$

Answer: B 2
Problem: Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution: Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat $\boxed{2}$

Answer: B 2
Problem: Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution: Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$

Answer:  2
Problem: Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution: Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$

Answer:  2
Problem: How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution: The amount of twos in our sum ranges from $0$ to $1008$ , with differences of $3$ because $2 \cdot 3 = \operatorname{lcm}(2, 3)$
The possible amount of twos is $\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{337}$

Answer: C 337
Problem: How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution: You can also see that you can rewrite the word problem into an equation $2x + 3y$ $2016$ . Therefore the question is just how many multiples of $3$ subtracted from 2016 will be an even number. We can see that $x = 1008$ $y = 0$ all the way to $x = 0$ , and $y = 672$ works, with $y$ being incremented by $2$ 's.Therefore, between $0$ and $672$ , the number of multiples of $2$ is $\boxed{337}$

Answer: C 337
Problem: How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution: We can utilize the stars-and-bars distribution technique to solve this problem. 
We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have $\binom{2016-1}{2-1}$ "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in $2015/6$ . We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0.
So, $2015/6\implies335\implies335+2=\boxed{337}$

Answer: C 337
Problem: How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution: Note that $2016 = 6 \cdot 336$ . In other words, we can write $2016$ as the sum of $336$ sixes.
In turn, we can express each $6$ as either $2 + 2 + 2$ or $3 + 3$
Therefore, we can write $2016$ as $n (2 + 2 + 2) + (336 - n) (3 + 3)$ , where $n$ is an integer between $0$ and $336$ , inclusive. Since each value of $n$ corresponds to a unique way to write the sum, we get $336 + 1 = \boxed{337}$

Answer: C 337
Problem: Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
[asy] draw(circle((0,0),3)); draw(circle((0,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((-1,sqrt(3)),1));  draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1));  draw(circle((2,0),1)); draw(circle((-2,0),1)); [/asy]
$\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi$

Solution: The big cookie has radius $3$ , since the center of the center cookie is the same as that of the large cookie.  The difference in areas of the big cookie and the seven small ones is $3^2\pi-7\pi=9\pi-7\pi=2 \pi$ .  The scrap cookie has this area, so its radius must be $\boxed{2}$

Answer: A 2
Problem: Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Solution: Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so
\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]
Multiplying both sides of the inequality by $400(5n+1)$ , we have
\[400(4n+2)<321(5n+1),\]
and by the distributive property,
\[1600n+800<1605n+321.\]
Subtracting $1600n+321$ on both sides of the inequality gives us
\[479<5n.\]
Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{12}$

Answer: A 12
Problem: Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Solution: Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$
Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$
We then cross multiply to get $80N - 400 > 79N + 79$
Finally we get $N > 479$
To achieve $N = 480$
So the sum of the digits of $N$ $\boxed{12}$

Answer: A 12
Problem: Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Solution: We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:
$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of its digits is $\boxed{12}$

Answer: A 12