Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$ , then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers  ( $4$ $5$ $6$ ), then Tadd must say the next four numbers ( $7$ $8$ $9$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution: Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||}   \hline  Round & Tadd & Todd & Tucker \\ [0.5ex]   \hline\hline  1 & 1 & 2-3 & 4-6 \\   \hline  2 & 7-10 & 11-15 & 16-21 \\  \hline  3 & 22-28 & 29-36 & 37-45 \\  \hline  4 & 46-55 & 56-66 & 67-78 \\ [1ex]   \hline \end{tabular}$
Tadd says $1$ number in round 1, $4$ numbers in round 2, $7$ numbers in round 3, and in general $3n - 2$ numbers in round n. At the end of round n, the number of numbers Tadd has said so far is $1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n-1)}{2}$ , by the sum of arithmetic series formula.
We find that $\dfrac{37(110)}{2}=2035$ , so Tadd says his 2035th number at the end of his turn in round 37. That also means that Tadd says his 2019th number in round 37. At the end of Tadd's turn in round 37, the children have, in total, completed $36+36+37=109$ turns. In general, at the end of turn $n$ , the nth triangular number is said, or $\dfrac{n(n+1)}{2}$ . So at the end of turn 109 (or the end of Tadd's turn in round 37), Tadd says the number $\dfrac{109(110)}{2}=5995$ . Recalling that this was the 2035th number said by Tadd, so the 2019th number he said was $5995-16=5979$
Thus, the answer is $\boxed{5979}$

Answer: C 5979
Problem: Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$ , then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers  ( $4$ $5$ $6$ ), then Tadd must say the next four numbers ( $7$ $8$ $9$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution: Firstly, as in Solution 1, we list how many numbers Tadd says, Todd says, and Tucker says in each round.
Tadd: $1, 4, 7, 10, 13 \cdots$
Todd: $2, 5, 8, 11, 14 \cdots$
Tucker: $3, 6, 9, 12, 15 \cdots$
We can find a general formula for the number of numbers each of the kids say after the $n$ th round. For Tadd, we can either use the arithmetic series sum formula (like in Solution 1) or standard summation results to get $\sum_{i=1}^n 3n-2=-2n+3\sum_{i=1}^n n=-2n+\frac{3n(n+1)}{2}=\frac{3n^2-n}{2}$
Now, to find the number of rotations Tadd and his siblings go through before Tadd says his $2019$ th number, we know the inequality $\frac{3n^2-n}{2}<2019$ must be satisfied, and testing numbers gives the maximum integer value of $n$ as $36$
The next main insight, in order to simplify the computation process, is to notice that the $2019$ th number Tadd says is simply the number of numbers Todd and Tucker say plus the $2019$ Tadd says, which will be the answer since Tadd goes first.
Carrying out the calculation thus becomes quite simple:
\[\left(\sum_{i=1}^{36} 3n+\sum_{i=1}^{36} 3n-1\right)+2019=\left(\sum_{i=1}^{36} 6n-1\right)+2019=(5+11+17...+215)+2019=\frac{36(220)}{2}+2019\]
At this point, we can note that the last digit of the answer is $9$ , which gives $\boxed{5979}$ . (Completing the calculation will confirm the answer, if you have time.)

Answer: C 5979
Problem: Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$ , then Todd must say the next two numbers ( $2$ and $3$ ), then Tucker must say the next three numbers  ( $4$ $5$ $6$ ), then Tadd must say the next four numbers ( $7$ $8$ $9$ $10$ ), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$ th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$

Solution: Think of a turn as adding a block or an interval of numbers to each person. Here are the first couple "blocks" that Tadd has (a block will be denoted in interval notation): [ $1$ ], [ $7$ $10$ ], [ $22$ $28$ ], [ $46$ $55$ ]. From simple inspection, we can notice two things.
$1$ ) The start points of each interval is increasing arithmetically by $9$
$2$ ) The length of each "block" is increasing by $3$
Since each block length is increasing by 3, then we want $1$ $4$ $7$ +.. $3n-2$ $2019$ or $n(3n-1)/2 = 2019$ , where $n$ is the amount of blocks we have. In a short amount of time, you will see that $n$ is not an integer from that equation, but thats perfectly okay since we can just find an $n$ that gets it close to 2019, then we can work from there. From inspection, you will see that a good n value that does that exact job is $37$
We can then deduce that the start point of the $37$ th block will be $1+(6+15+24..+321) = 5887$ if we refer to fact $1$ . We can easily find the length of 37th block by $36*3+1 = 109$ if we refer to fact $2$ . Remember that the length increases by $3$ for each block. So we then can deduce that our endpoint for the 37th block is $5887+109-1 = 5995$ . The endpoint of the $37$ th block will be the $2035$ th number said because we can plug $37$ into $n(3n-1)/2$ , which tells us how many numbers Tadd has said where $n$ is the number of blocks Tadd has. From there we know that the $2019$ th number is in $37$ th block, so that means we can take our endpoint and subtract $2035-2019 = 16$ from the endpoint of the $37$ th block, which is $5995$ , so we get $5995-16 = 5979$ or $\boxed{5979}$ .
~~triggod

Answer: C 5979
Problem: Let $p$ $q$ , and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$ . It is given that there exist real numbers $A$ $B$ , and $C$ such that \[\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}\] for all $s\not\in\{p,q,r\}$ . What is $\tfrac1A+\tfrac1B+\tfrac1C$
$\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247$

Solution: Multiplying both sides by $(s-p)(s-q)(s-r)$ yields \[1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)\] As this is a polynomial identity, and it is true for infinitely many $s$ , it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$ ). This means we can plug in $s = p$ to find that $\frac1A = (p-q)(p-r)$ .  Similarly, we can find $\frac1B = (q-p)(q-r)$ and $\frac1C = (r-p)(r-q)$ .  Summing them up, we get that \[\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr\] We can express $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr)$ , and by Vieta's Formulas, we know that this expression is equal to $324$ . Vieta's also gives $pq + qr + pr = 80$ (which we also used to find $p^2+q^2+r^2$ ), so the answer is $324 -80 = \boxed{244}$

Answer: B 244
Problem: Let $p$ $q$ , and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$ . It is given that there exist real numbers $A$ $B$ , and $C$ such that \[\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}\] for all $s\not\in\{p,q,r\}$ . What is $\tfrac1A+\tfrac1B+\tfrac1C$
$\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247$

Solution: Solution 1 uses a trick from Calculus that seemingly contradicts the restriction $s\not\in\{p,q,r\}$ . I am going to provide a solution with pure elementary algebra. \[A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)=1\] \[s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0\] \[\begin{cases} A+B+C=0 & (1)\\ Aq+Ar+Bp+Br+Cp+Cq=0 & (2)\\ Aqr+Bpr+Cpq=1 & (3) \end{cases}\] From $(1)$ we get $A=-(B+C)$ $B=-(A+C)$ $C=-(A+B)$ , substituting them in $(2)$ , we get $Ap + Bq + Cr=0$ $(4)$
$(4)- (1) \cdot r$ $A(p-r)+B(q-r)=0$ $(5)$
$(3) - (1) \cdot pq$ $Aq(r-p)+Bp(r-q)=1$ $(6)$
$(6) + (5) \cdot p$ $A(r-p)(q-p)=1$
$A = \frac{1}{(r-p)(q-p)}$ , by symmetry, $B = \frac{1}{(r-q)(p-q)}$ $C = \frac{1}{(q-r)(p-r)}$
The rest is similar to solution 1, we get $\boxed{244}$

Answer: B 244
Problem: For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution: The main insight is that
\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ . Thus,
\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$ , or in other words, if $\frac{(n-1)!}{n}$ , is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem . There are $15$ primes between $1$ and $50$ , inclusive, so there are $15 + 1 = 16$ terms for which
\[\frac{(n^2-1)!}{(n!)^{n}}\]
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$ , as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{34}$

Answer: D 34
Problem: For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution: We can use the P-Adic Valuation (more info could be found here: Mathematicial notation ) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :
\[v_p (n!)= \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.
We also know that , $v_p (m^n) = n \cdot v_p (m)$ .
Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :
\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.
If we plug in $n=p$ , by Legendre's we get two equations:
\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty  \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :
\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.
Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:
\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\]
(as $v_p(p^4!) = p^3 + p^2 + p + 1$ and $p^4$ contains 4 factors of $p$ ) and
\[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:
\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality : \[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]
Therefore, there are 16 values that don't work and $50-16 = \boxed{34}$ values that work.

Answer: D 34
Problem: For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution: Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So we get $50-15-1=\boxed{34}$ and that happens to be right.

Answer:  34
Problem: Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?
$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$

Solution: Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{27}$

Answer: E 27
Problem: In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution: $60\%$ of seniors do not play a musical instrument. If we denote $x$ as the number of seniors, then \[\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500\]
\[\frac{3}{5}x + 150 - \frac{3}{10}x = 234\] \[\frac{3}{10}x = 84\] \[x = 84\cdot\frac{10}{3} = 280\]
Thus there are $500-x = 220$ non-seniors. Since 70% of the non-seniors play a musical instrument, $220 \cdot \frac{7}{10} = \boxed{154}$

Answer: B 154
Problem: In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution: Let $x$ be the number of seniors, and $y$ be the number of non-seniors. Then \[\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234\]
Multiplying both sides by $10$ gives us \[6x + 3y = 2340\]
Also, $x + y = 500$ because there are 500 students in total.
Solving these system of equations give us $x = 280$ $y = 220$
Since $70\%$ of the non-seniors play a musical instrument, the answer is simply $70\%$ of $220$ , which gives us $\boxed{154}$

Answer: B 154
Problem: In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution: We can clearly deduce that $70\%$ of the non-seniors do play an instrument, but, since the total percentage of instrument players is $46.8\%$ , the non-senior population is quite low. By intuition, we can therefore see that the answer is around $\text{B}$ or $\text{C}$ . Testing both of these gives us the answer $\boxed{154}$

Answer: B 154
Problem: In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution: We know that $40\%$ of the seniors play a musical instrument, and $30\%$ of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is \[46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234\] We can also conclude that $60\%$ of the seniors do not play an instrument, $70\%$ of the non seniors do play an instrument, and $500-234 = 266$ people do play an instrument.
We can now set up the following equations, where $s$ is the number of seniors and $n$ is the number of non-seniors: \[0.3n + 0.6s = 234\] \[0.7n + 0.4s = 266\] By elimination, we get $1.5n$ to be equal to $330$ . This means that $n = \frac{330}{1.5} = 220$ .
The answer is $70$ percent of $220$ . This is equal to \[0.7*220 = 7*22 = 154\] Therefore, the answer is $\boxed{154}$

Answer: B 154
Problem: All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?
$\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

Solution: If all lines satisfy the condition, then we can just plug in values for $a$ $b$ , and $c$ that form an arithmetic progression. Let's use $a=1$ $b=2$ $c=3$ , and $a=1$ $b=3$ $c=5$ . Then the two lines we get are: \[x+2y=3\] \[x+3y=5\] Use elimination to deduce \[y = 2\] and plug this into one of the previous line equations. We get \[x+4 = 3 \Rightarrow x=-1\] Thus the common point is $\boxed{1,2}$

Answer: A 1,2
Problem: All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?
$\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

Solution: We know that $a$ $b$ , and $c$ form an arithmetic progression, so if the common difference is $d$ , we can say $a,b,c = a, a+d, a+2d.$ Now we have $ax+ (a+d)y = a+2d$ , and expanding gives $ax + ay + dy = a + 2d.$ Factoring gives $a(x+y-1)+d(y-2) = 0$ . Since this must always be true (regardless of the values of $a$ and $d$ ), we must have $x+y-1 = 0$ and $y-2 = 0$ , so $x,y = -1, 2,$ and the common point is $\boxed{1,2}$

Answer: A 1,2
Problem: There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution: \[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$ ). The answer is therefore $1 + 9 = \boxed{10}$

Answer: C 10
Problem: There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution: Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$

Answer: C 10
Problem: There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution: Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$

Answer: C 10
Problem: There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution: Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \boxed{10}$

Answer: C 10
Problem: There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution: Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{10}$

Answer: C 10
Problem: Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution: If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$

Answer: B 21
Problem: Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution: We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{21}$

Answer: B 21
Problem: Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution: We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$

Answer: B 21