Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: For how many positive integers $n \le 1000$ is \[\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor\] not divisible by $3$ ? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .)
$\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26$

Solution: First, we notice the following lemma:
$\textbf{Lemma}$ : For $N, n \in \mathbb{N}$ $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1$ if $n \mid N$ ; and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor$ if $n \nmid N.$
$\textbf{Proof}$ : Let $A = kn + r$ , with $0 \leq r < n$ . If $n \mid N$ , then $r = 0$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ $\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1.$
If $n \nmid N$ , then $1 \leq r < n$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ $\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor.$
From the lemma and the given equation, we have four possible cases: \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1 \qquad (1)\] \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (2)\] \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)\] \[\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)\]
Note that cases (2) and (3) are the cases in which the term, $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,$ is not divisible by $3$ . So we only need to count the number of $n$ 's for which cases (2) and (3) stand.
Case (2): By the lemma, we have $n \mid 1000$ and $n \nmid 999.$ Hence $n$ can be any factor of $1000$ except for $n = 1$ . Since $1000 = 2^3 * 5^3,$ there are $(3+1)(3+1) - 1 = 15$ possible values of $n$ for this case.
Case (3): By the lemma, we have $n \mid 999$ and $n \nmid 998.$ Hence $n$ can be any factor of $999$ except for $n = 1$ . Since $999 = 3^3 * 37^1,$ there are $(3+1)(1+1) - 1 = 7$ possible values of $n$ for this case.
So in total, we have total of $15+7=\boxed{22}$ possible $n$ 's.

Answer: A 22
Problem: Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$

Solution: As in the previous solution, note that we must have either $0$ or $2$ reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.
Suppose there are no reflections. Denote $90^{\circ}$ as $1$ $180^{\circ}$ as $2$ , and $270^{\circ}$ as $3$ , just for simplification purposes. We want a combination of $3$ of these that will sum to either $4$ or $8$ $0$ and $12$ are impossible since the minimum is $3$ and the max is $9$ ). $4$ can be achieved with any permutation of $(1-1-2)$ and $8$ can be achieved with any permutation of $(2-3-3)$ . This case can be done in $3+3=6$ ways.
Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have $1$ rotation left that we can do though, and the only one that will return to the original position is $2$ , which is $180^{\circ}$ AKA reflection across origin. Therefore, since all $3$ transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives $6$ ways.
$6+6=\boxed{12}$

Answer: A 12
Problem: Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$

Solution: Define $s$ as a reflection, and $r$ as a $90^{\circ}$ counterclockwise rotation. Thus, $r^4=s^2=e$ , and the five transformations can be represented as ${r, r^2, r^3, r^2s, s}$ , and $rs=sr^{-1}$
Now either $s$ doesn't appear at all or appears twice. For the former case, it's easy to see that only $r, r, r^2$ and $r^2, r^3, r^3$ will work. Both can be permuted in $3$ ways, giving $6$ ways in total.
For the latter case, note that $s$ can't appear twice, neither does $r^2s$ , else we need to get $e$ from ${r, r^2, r^3}$ , which is not possible. So $r^2s$ and $s$ must appear once each. The last transformation must be $r^2$ . A quick check shows that ${r^2, r^2s, s}$ is permutable, since $r^2s=sr^{-2}=sr^2$ (since $r^4=e$ ). This gives $6$ ways.
Thus the answer is $\boxed{12}$

Answer: A 12
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We know that $\gcd(n+57,63)=21$ and $\gcd(n-57, 120)= 60$ by the Euclidean Algorithm. Hence, let $n+57=21\alpha$ and $n-57=60 \gamma$ , where $\gcd(\alpha,3)=1$ and $\gcd(\gamma,2)=1$ . Subtracting the two equations, $38=7\alpha-20\gamma$ . Letting $\gamma = 2s+1$ , we get $58=7\alpha-40s$ . Taking modulo $40$ , we have $\alpha \equiv{14} \pmod{40}$ . We are given that $n=21\alpha -57 >1000$ , so $\alpha \geq 51$ . Notice that if $\alpha =54$ then the condition $\gcd(\alpha,3)=1$ is violated. The next possible value of $\alpha = 94$ satisfies the given condition, giving us $n=1917$
Alternatively, we could have said $\alpha = 40k+14 \equiv{0} \pmod{3}$ for $k \equiv{1} \pmod{3}$ only, so $k \equiv{0,2} \pmod{3}$ , giving us our answer. Since the problem asks for the sum of the digits of $n$ , our answer is $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: The conditions of the problem reduce to the following: $n+120 = 21k$ and $n+63 = 60\ell$ , where $\gcd(k,3) = 1$ and $\gcd(\ell,2) = 1$ . From these equations, we see that $21k - 60\ell = 57$ . Solving this Diophantine equation gives us that $k = 20a + 17$ and $\ell = 7a + 5$ . Since, $n>1000$ , we can do some bounding and get that $k > 53$ and $\ell > 17$ . Now we start bashing by plugging in numbers that satisfy these conditions: $a=4$ is the first number that works so we get $\ell = 33$ $k = 97$ . Therefore, we have $n=21(97)-120=60(33)-63=1917$ , and our answer is $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We first find that $n\equiv6\pmod{21}$ and $n\equiv57\pmod{60}$ , then we get $n=21x+6$ and $n=60y+57$ by definitions, where $x$ and $y$ are integers. It follows that $y$ must be odd, since the GCD will be $120$ instead of $60$ if $y$ is even. Also, the unit digit of $n$ must be $7$ , since the unit digit of $60y$ is always $0$ and the unit digit of $57$ is $7$ . Therefore, we find that $x$ must end in $1$ to satisfy $n$ having a unit digit of $7$ . Also, we find that $x$ must not be a multiple of $3$ or else the GCD will be $63$ . Therefore, we test values for $x$ and find that $x=91$ satisfies all these conditions. Therefore, $n=1917$ and $1+9+1+7 = \boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We are given that $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60.$ By applying the Euclidean algorithm in reverse, we have \[\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21\] and \[\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.\]
We now know that $n+183$ must be divisible by $21$ and $60,$ so it is divisible by $\text{lcm}(21, 60) = 420.$ Therefore, $n+183 = 420k$ for some integer $k.$ We know that $3 \nmid k,$ or else the first condition won't hold ( $\gcd$ will be $63$ ) and $2 \nmid k,$ or else the second condition won't hold ( $\gcd$ will be $120$ ). Since $k = 1$ gives us too small of an answer, then $k=5,$ from which $n = 1917.$ So, the answer is $1+9+1+7 = \boxed{18}.$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: $\gcd(n+63,120)=60$ tells us $n+63\equiv60\pmod {120}$ . The smallest $n+63$ that satisfies the previous condition and $n>1000$ is $1140$ , so we start from there. If $n+63=1140$ , then $n+120=1197$ . Because $\gcd(n+120,63)=21$ $n+120\equiv21\pmod {63}$ or $n+120\equiv42\pmod {63}$ . We see that $1197\equiv0\pmod {63}$ , which does not fulfill the requirement for $n+120$ , so we continue by keep on adding $120$ to $1197$ , in order to also fulfill the requirement for $n+63$ . Soon, we see that $n+120\pmod {63}$ decreases by $6$ every time we add $120$ , so we can quickly see that $n=1917$ because at that point $n+120\equiv21\pmod {63}$ . We add up all digits of $1917$ to get $\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We are able to set up the following system of congruences: \begin{align*} n &\equiv 6 \pmod {21}, \\ n &\equiv 57 \pmod {60}. \end{align*} Therefore, by definition, we are able to set-up the following system of equations: \begin{align*} n &= 21a + 6, \\ n &= 60b + 57. \end{align*} Thus, we have $21a + 6 = 60b + 57,$ from which \[7a + 2 = 20b + 19.\] We know $7a \equiv 0 \pmod {7},$ and since $7a = 20b + 17,$ therefore $20b + 17 \equiv 0 \pmod{7}.$ Simplifying this congruence further, we have \[b\equiv 3 \pmod{7}.\] Thus, by definition, $b = 7x + 3.$ Substituting this back into our original equation, we get \[n = 60(7x + 3) + 57 = 420x + 237.\] By definition, we are able to set up the following congruence: \[n \equiv 237 \pmod{420}.\] Thus, $n = 1917$ , so our answer is simply $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: First, we find $n$ . We know that it is greater than $1000$ , so we first input $n = 1000$ . From the first equation, $\gcd(63, n + 120) = 21$ , we know that if $n$ is correct, after we add $120$ to it, it should be divisible by $21$ , but not $63$ \[\frac{n + 120}{21} = \frac{1120}{21} = 53\text{ R }7.\] This does not work. To get to the nearest number divisible by $21$ , we have to add $14$ to cancel out the remainder. (Note that we don't subtract $7$ to get to $53$ $n$ is already at its lowest possible value!)
Adding $14$ to $1000$ gives us $n = 1014$ . (Note: $n$ is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
Now using the second equation, $\gcd(n + 63, 120) = 60$ , we know that if $n$ is correct, then $n+63$ is divisible by $60$ but not $120$ \[\frac{n + 63}{60} = \frac{1077}{60} = 17\text{ R }57.\] Again, this does not work. This requires some guessing and checking. We can add $21$ over and over again until $n$ is valid. This changes $n$ while also maintaining that $\frac{n + 120}{21}$ has no remainders. 
After adding $21$ once, we get $18 r 18$ . By pure luck, adding $21$ two more times gives us $19$ with no remainders. 
We now have $1077 + 21 + 21 + 21 = 1140$ . However, this number is divisible by $120$ . To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$ . Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$ . Adding $420$ again gives us $1980$ , which is valid. However, remember that this is equal to $n + 63$ , so subtracting $63$ from $1980$ gives us $1917$ , which is $n$
The sum of its digits are $1 + 9 + 1 + 7 = \boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: By the Euclidean Algorithm, we have \begin{alignat*}{8} \gcd(63,n+120)&=\hspace{1mm}&&\gcd(63,\phantom{ }\underbrace{n+120-63k_1}_{(n+120) \ \mathrm{mod} \ 63}\phantom{ })&&=21, \\  \gcd(n+63,120)&=&&\gcd(\phantom{ }\underbrace{n+63-120k_2}_{(n+63) \ \mathrm{mod} \ 120}\phantom{ },120)&&=60.  \end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$
More generally, let $t\in\{1,2\},$ so we get \begin{align*} n+120-63k_1&=21t, &\hspace{55.5mm}(1) \\ n+63-120k_2&=60. &\hspace{55.5mm}(2) \end{align*} Subtracting $(2)$ from $(1)$ and then simplifying give \begin{align*} 57-63k_1+120k_2&=21t-60 \\ 117-63k_1+120k_2&=21t \\ 39-21k_1+40k_2&=7t. \hspace{54mm}(\bigstar) \end{align*} Taking $(\bigstar)$ modulo $7$ produces \begin{align*} 4+5k_2&\equiv0\pmod{7} \\ k_2&\equiv2\pmod{7}. \end{align*} Recall that $n>1000.$ From $(2),$ it follows that \[1063-120k_2<n+63-120k_2=60,\] from which $k_2>8.$ Therefore, the possible values for $k_2$ are $9,16,23,\ldots.$
We need to check whether positive integers $k_1$ and $t$ (where $t\in\{1,2\}$ ) exist in $(1):$
Finally, the least such positive integer $n$ is $1917.$ The sum of its digits is $1+9+1+7=\boxed{18}.$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: Because we are finding value of $n$ for $n > 1000$ , let $n = 1000 + k$
Using the Euclidean Algorithm, \begin{align*} \gcd(63, n+120) &= \gcd(63, 1000 + k + 120) \\ &= \gcd(63, k + 1120 - 63 \cdot 18) \\ &= \gcd(63, k-14) \\ &= 21, \\ \gcd(n+63, 120) &= \gcd(k + 1000 + 63, 120) \\ &= \gcd(k + 1000 + 63 - 120 \cdot 9, 120) \\ &= \gcd(k-17, 120) \\ &= 60. \end{align*} So, we have \begin{align*} k &\equiv 14 \pmod{21}, \\ k &\equiv 17 \pmod{60}. \end{align*} Let $k = 21p + 14 = 60q + 17$ $7p= 20q + 1$ (by the Division Algorithm), $7p = 21q - (q - 1)$ $q - 1$ is a multiple of $7$
Let $q-1 = 7r$ $q = 7r+1$ $k = 60(7r+1) + 17 = 420r + 77$
$n = 1000 + k = 420r + 1077$ $n = 1077, 1497, 1917$
By substituting $1077$ $1497$ $1917$ into $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60$ $1077$ and $1497$ aren't valid answers, only $1917$ is. Therefore, the answer is $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We know $63 = 3 \cdot 21$ , and $\gcd(63, n + 120) = 21 \Longrightarrow n + 120 = 21a$ , where $a$ is not a multiple of $3$
Also, $120 = 2 \cdot 60$ , and $gcd(n+63, 120) = 60 \Longrightarrow n + 63 = 60b$ , where $b$ is not a multiple of $2$
Let $n+63 = 60b = x$ $n = x - 63$ $n+120 = x+57 = 21a$
Now the problem becomes $\gcd(63, x+57) =21$ and $\gcd(x, 120)=60$
Meaning $x + 57$ has to be a multiple of $21$ but not $63$ , and $x$ is a multiple of $60$ but not $120$
Using trial and error, the least values are $x = 33\cdot60 = 1980$ and $n = x-63 = 1917$
Therefore, the answer is $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: We have \begin{align*} \gcd(63, n+120) &= 21 \Longrightarrow n + 120 \equiv 0 \pmod{21}, \text{ where } 9 \nmid n + 120, \\ \gcd(n+63, 120) &= 60 \Longrightarrow n + 63 \equiv 0 \pmod{60}, \text{ where } 8 \nmid n + 63. \end{align*} So, we conclude that $n \equiv 6 \pmod{21}$ and $n \equiv 57 \pmod{60}$ , respectively.
Because the $2$ moduli $21$ and $60$ are not relatively prime, namely $\gcd{(21, 60)} = 3$ $21 = 3 \cdot 7$ , and $60 = 3 \cdot 20$ , we convert the system of $2$ linear congruences with non-coprime moduli into a system of $3$ linear congruences with coprime moduli: \begin{align*} n &\equiv 0 \pmod{3}, \\ n &\equiv 6 \pmod{7}, \\ n &\equiv 17 \pmod{20}. \end{align*} By Chinese Remainder Theorem , the general solution of system of $3$ linear congruences is \[n = 420k + 1077.\] We construct the following table: \[\begin{array}{c|c|c|c} n & 1077 & 1497 & 1917\\ \hline n + 120 & 1197 & 1617 & 2037\\ \hline n + 63 & 1140 & 1560 & 1980\\ \end{array}\] Only $n = 1917$ satisfies $9 \nmid n + 120$ and $8 \nmid n + 63$ . Therefore, the answer is $1+9+1+7=\boxed{18}$

Answer: C 18
Problem: Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution: From $\gcd(63, n + 120) = 21$ , we know that \begin{align} n + 120 &\equiv 0 \pmod{7} \Rightarrow n \equiv 6 \pmod{7} \tag{1} \\ n + 120 &\equiv 0 \pmod{3} \Rightarrow n \equiv 0 \pmod{3} \Rightarrow n \equiv 0, 3, 6 \pmod{9} \notag \end{align} Since \begin{align} n + 120 &\not\equiv 0 \pmod{9} \Rightarrow n \not\equiv 6 \pmod{9} \notag \end{align} We must have \begin{align} n &\equiv 0 \pmod{9} \tag{2} \end{align} Or \begin{align} n &\equiv 3 \pmod{9} \tag{3} \end{align} (Note that $n + 120 \equiv 0 \pmod{9}$ is impossible since then $9 | \gcd(63, n+120)$ .)
From $\gcd(n+63, 120) = 60$ , we have \begin{align} n + 63 &\equiv 0 \pmod{15} \Rightarrow n \equiv 12 \pmod {15} \tag{4} \\ n + 63 &\equiv 0 \pmod{4} \Rightarrow n \equiv 1 \pmod {4} \Rightarrow n \equiv 1, 5 \pmod{8} \notag \end{align} Since \begin{align} n + 63 &\not\equiv 0 \pmod{8} \Rightarrow n \not\equiv 1 \pmod{8} \notag \end{align} We must have \begin{align} n &\equiv 5 \pmod {8} \tag{5} \end{align} Solving equations $(1)$ $(4)$ , and $(5)$ , we get \begin{align} n &\equiv 237 \pmod{840} \tag{6} \end{align} Solving $(6)$ with $(2)$ , we have \begin{align} n &\equiv 1917 \pmod{2520} \tag{7} \end{align} Solving $(6)$ with $(3)$ , we have \begin{align} n &\equiv 237 \pmod{2520} \tag{8} \end{align} From $(7)$ , we obtain $1917$ , and from $(8)$ , we obtain $237 + 2520 = 2757$ as the smallest solutions above $1000$ . Thus, we get $1 + 9 + 1 + 7 = \boxed{18}$

Answer: C 18
Problem: What is the value of \[1-(-2)-3-(-4)-5-(-6)?\]
$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\  3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

Solution: We know that when we subtract negative numbers, $a-(-b)=a+b$
The equation becomes $1+2-3+4-5+6 = \boxed{5}$

Answer: D 5
Problem: What is the value of \[1-(-2)-3-(-4)-5-(-6)?\]
$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\  3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

Solution: Like Solution 1, we know that when we subtract $a-(-b)$ , that will equal $a+b$ as the opposite/negative of a negative is a positive. Thus, $1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6$ . We can group together a few terms to make our computation a bit simpler. $1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\boxed{5}$

Answer: D 5
Problem: Carl has $5$ cubes each having side length $1$ , and Kate has $5$ cubes each having side length $2$ . What is the total volume of these $10$ cubes?
$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45$

Solution: A cube with side length $1$ has volume $1^3=1$ , so $5$ of these will have a total volume of $5\cdot1=5$
A cube with side length $2$ has volume $2^3=8$ , so $5$ of these will have a total volume of $5\cdot8=40$
$5+40=\boxed{45}$ ~quacker88

Answer: E 45
Problem: Carl has $5$ cubes each having side length $1$ , and Kate has $5$ cubes each having side length $2$ . What is the total volume of these $10$ cubes?
$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45$

Solution: The total volume of Carl's cubes is $5$ . This is because to find the volume of a cube or a rectangular prism, you have to multiply the height by the length by the width. So in this question, it would be $1 \times 1 \times 1$ . This is equal to $1$ . Since Carl has 5 cubes, you will have to multiply $1$ by $5$ , to account for all the $5$ cubes.
Next, to find the total volume of Kate's cubes you have to do the same thing. Except, this time, the height, the width, and the length, are all $2$ , so it will be $2 \times 2 \times 2 = 8.$ Now you have to multiply by $5$ to account for all the $5$ blocks. This is $40$ . So the total volume of Kate's cubes is $40$
Lastly, to find the total of Carl's and Kate's cubes, you must add the total volume of these people. This is going to be $5+40=\boxed{45}$

Answer: E 45