Answer: 3.0

Solution:
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:  Now, it becomes clear that there are $\boxed{3}$ intersection points.

Solution:
$x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$ $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
$2y-3$ is negative so $|2y-3| = 3-2y = 1$ $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$
$\textbf{Case 2:}$ $y = 1$ :
It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$ . We already have this solution from Case 2 as $y = 1$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$ $y = \frac{1}{2}$ and $x = \frac{3}{2}$ .
Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$ , and the answer is $\boxed{3}$

Solution:
Note that $||x| - |y||$ can take on one of four values: $x + y$ $x - y$ $-x + y$ $-x -y$ . So we have 4 cases:
Case 1: ||x| - |y|| = x+y \[x+3y=3\] \[x+y=1\]
Subtracting:
$2y=2 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$
Case 2: ||x| - |y|| = x-y \[x+3y=3\] \[x-y=1\]
Subtacting:
$4y=2 \Rightarrow y=\dfrac{1}{2}$ an $x=\dfrac{3}{2}$
$\text{Result: } \left(\dfrac{3}{2},\dfrac{1}{2}\right)$
Case 3: ||x| - |y|| = -x+y \[x+3y=3\] \[-x+y=1\]
Adding:
$4y=4 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$ . Since this is the same solution as we got in Case 1, we can not count this (otherwise, we would have overcounted).
Case 4: ||x| - |y|| = -x-y \[x+3y=3\] \[-x-y=1\]
Adding:
$2y=4 \Rightarrow y=2$ and $x=-3$
$\text{Result: } (-3,2)$
Answer
We solved each case by elimination (either adding the two equations or subtracting), to obtain three solutions: $(0, 1)$ $(-3,2)$ $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
Our answer is $\boxed{3}$

Solution: Just as in solution $2$ , we derive the equation $||3-3y|-|y||=1$ . If we remove the absolute values, the equation collapses into four different possible values. $3-2y$ $3-4y$ $2y-3$ , and $4y-3$ , each equal to either $1$ or $-1$ . Remember that if $P-Q=a$ , then $Q-P=-a$ . Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$ , and $3-4y$ and $4y-3$ . Find the values of $y$ when $3-2y=1$ $3-2y=-1$ $3-4y=1$ and $3-4y=-1$ . We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$ , so the answer is $\boxed{3}$

Solution:
Just as in solution $2$ , we derive the equation $x=3-3y$ . Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$ . Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$ . From here we know that $3y-3y^2$ is either positive or negative.
When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$ .
When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$ . Clearly, there are $3$ different pairs of values and that gives us $\boxed{3}$

Solution: Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$ ) and the other is linear. Subtract to get $\boxed{3}$


Answer: 4.0

How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: We have $ab = c$ $bc = a$ , and $ca = b$ , so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0$ , and as $a$ $b$ , and $c$ are all non-zero, we cannot have $abc = 0$ , so we must have $abc = 1$ . Now substituting $bc = a$ gives $a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1$ . If $a = 1$ , then the system becomes $b = c, bc = 1, c = b$ , so either $b = c = 1$ or $b = c = -1$ , giving $2$ solutions. If $a = -1$ , the system becomes $-b = c, bc = -1, -c = b$ , so $-b = c = 1$ or $b = -c = 1$ , giving another $2$ solutions. Thus the total number of solutions is $2 + 2 = 4$ , which is answer $\boxed{4}$


Answer: 15.0

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$

Solution: We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,h)=1\] \[\max(c,i)=2\] \[\max(d,g)=2\] \[\max(e,h)=2\] and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$ . We also know that since $\max(b,h)=1$ that $e=2$ . So now some equations have become useless to us...let's take them out. \[\max(b,h)=1\] \[\max(d,g)=2\] are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$ . Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$ . Thus our answer is $5\times 3=\boxed{15}.$

Solution:
It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$ .
So $x=8,24$
$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}(y,z)=900$ and $72\nmid 900$
So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.
$25\mid z$ because $z$ must source all powers of $5$ $z\in\{25,50,75,100,150,300\}$ $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.
By different sourcing of powers of $2$ and $3$
\[(8,9):z=300\] \[(8,18):z=300\] \[(8,36):z=75,150,300\] \[(24,9):z=100,300\] \[(24,18):z=100,300\] \[(24,36):z=25,50,75,100,150,300\]
$z=100$ is "enabled" by $x$ sourcing the power of $3$ $z=75,150$ is uncovered by $y$ sourcing all powers of $2$ . And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.
Counting the cases, $1+1+3+2+2+6=\boxed{15}.$

Solution:
As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.

Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$

Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$

Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$

This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\]


Answer: 15.0

Solution: We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,h)=1\] \[\max(c,i)=2\] \[\max(d,g)=2\] \[\max(e,h)=2\] and $c=f=0$ since $\text{lcm}(x,y)$ isn't a multiple of 5. Since $\max(d,g)=2$ we know that $a=3$ . We also know that since $\max(b,h)=1$ that $e=2$ . So now some equations have become useless to us...let's take them out. \[\max(b,h)=1\] \[\max(d,g)=2\] are the only two important ones left. We do casework on each now. If $\max(b,h)=1$ then $(b,h)=(1,0),(0,1)$ or $(1,1)$ . Similarly if $\max(d,g)=2$ then $(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)$ . Thus our answer is $5\times 3=\boxed{15}.$

Solution:
It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\text{lcm}(x,y)=72$ so $8\mid x$ or $9\mid x$ or both. But $9\nmid x$ because $\text{lcm}(x,z)=600$ and $9\nmid 600$ .
So $x=8,24$
$y$ can be $9,18,36$ in both cases of $x$ but NOT $72$ because $\text{lcm}(y,z)=900$ and $72\nmid 900$
So there are six sets of $x,y$ and we will list all possible values of $z$ based on those.
$25\mid z$ because $z$ must source all powers of $5$ $z\in\{25,50,75,100,150,300\}$ $z\ne\{200,225\}$ because of $\text{lcm}$ restrictions.
By different sourcing of powers of $2$ and $3$
\[(8,9):z=300\] \[(8,18):z=300\] \[(8,36):z=75,150,300\] \[(24,9):z=100,300\] \[(24,18):z=100,300\] \[(24,36):z=25,50,75,100,150,300\]
$z=100$ is "enabled" by $x$ sourcing the power of $3$ $z=75,150$ is uncovered by $y$ sourcing all powers of $2$ . And $z=25,50$ is uncovered by $x$ and $y$ both at full power capacity.
Counting the cases, $1+1+3+2+2+6=\boxed{15}.$

Solution:
As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.

Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$

Analyzing for powers of $2$ , it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$ . Turning towards the vertices $y$ and $z$ , we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$

Using the same logic as we did for powers of $2$ , it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$ , we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$
The final and last case is the powers of $5$

This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$ , we get the answer: \[5\cdot3\cdot1=\boxed{15}\]


Answer: 12.0

How many ordered triples of integers $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$

Solution:
WLOG , let $a \ge 0$ , and let $a \ge b$ .  We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \ge 0$ and $a_0 > b_0$ , we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original problem.
Furthermore, we assert that these four solutions are distinct.  We can say that $a > b$ , since if $a=b$ , we have $c = 19 - 2a$ for the first equation and either $c = 97 - a^2$ or $c = a^2 - 97$ for the second equation.  Equating $19 - 2a = 97 - a^2$ gives no integer solution, while equating $19 - 2a = a^2 - 97$ also gives no integer solution.
Thus, we can now assume WLOG that $a \ge 0$ and $a > b$ , and each pair of $(a_0,b_0)$ that we get will generate four unique solutions: $(a_0, b_0), (b_0, a_0), (-a_0, -b_0), (-b_0, -a_0)$
We now divide the problem into $c \ge 0$ and $c < 0$
If $c \ge 0$ , we have $|a + b| + c = 19$ and $ab + c = 97$
Solving both equations for $c$ and equating them, we get that $ab - |a + b| = 78$ .  Splitting these up, we find that either $ab - a - b = 78$ or $ab + a+ b = 78$ .  Factoring both with SFFT gives $(a-1)(b-1) = 79$ or $(a +1)(b+1) = 79$ .  We factor with the restrctions that $a \ge 0$ and $a > b$ .  Since $79$ is prime, we have:
$a - 1 = 79$ and $b - 1 = 1$ , which leads to $(80,2)$
$a + 1 = 79$ and $b + 1 = 1$ , which leads to $(78, 0)$
Each of those solutions could generate $3$ more solutions, giving a total of $8$ potential solutions.  However, in each the first set of four solutions, we have $|a + b| = 82$ , which from the original first equation $|a + b| + c = 19$ gives $c = 19 - 82$ , which contradicts our initial assumption that $c\ge 0$ .  Similarly, for the second set of four solutions, we have $|a + b| = 78$ , which leads to $c = 19 - 78$ , also contradicting $c \ge 0$
If $c < 0$ , we have $|a + b| + c = 19$ and $ab - c = 97$ .  We note that $19 - c$ must be positive whenever $c$ is negative, and thus $|a + b| = a + b$
Solving both equations for $c$ and using SFFT as above gives $(a+1)(b+1) = 117$ .  Since $117 = 3^2\cdot 13$ , we factor with the restriction that $a\ge 0$ and $a > b$ .  Thus, we can let $a+1  \in \{117, 39, 13\}$ , which means $a \in \{116, 38, 12\}$ .  These give corresponding $b+1\in \{1, 3, 9\}$ , which leads to corresponding $b \in \{0, 2, 8\}$ .  Combining the solutions, we have $(a,b) = (116, 0), (38, 2), (12, 8)$
Each of these three solutions permutes, negates, and permute-negates into $4$ solutions as described in the start of the solution, for a total of $12$ solutions.
Checking our solutions to ensure $c < 0$ , we find in the first set of four solutions, $|a + b| = 116$ , and thus $c = 19 - 116$ , which is indeed negative.
In the second set of four solutions, $|a + b| = 40$ , which leads to $c = 19 - 40$ , which is also negative.
Finally, in the third set of four solutoins, $|a + b| = 20$ , which leads to $c = 19 - 20$ , which is negative.
Thus, there are $12$ ordered triples, and the answer is $\boxed{12}$


Answer: 2.0

How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad  \textbf{(E) }\text{more than }4$

Solution:
Solution by e_power_pi_times_i
If $m+n=mn$ $mn-m-n = (m-1)(n-1)-1 = 0$ . Then $(m-1)(n-1) = 1$ , and $(m,n) = (2,2) or (0,0)$ . The answer is $\boxed{2}$


Answer: 18.0

Solution: We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$ .  The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$ $\overline{CD}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{CD}$ $\overline{EF}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{GH}$ , and $\overline{CD}\text{ and }\overline{EF}$ .  These are $6$ pairs total.  We can do the same for the lines in the same direction as $\overline{AE}$ and $\overline{AD}$ .  This means there are $6\cdot 3=\boxed{18}$ total pairs of parallel lines.

Solution: Look at any edge, let's say $\overline{AB}$ . There are three ways we can pair $\overline{AB}$ with another edge. $\overline{AB}\text{ and }\overline{EF}$ $\overline{AB}\text{ and }\overline{HG}$ , and $\overline{AB}\text{ and }\overline{DC}$ . There are 12 edges on a cube. 3 times 12 is 36. We have to divide by 2 because every pair is counted twice, so $\frac{36}{2}$ is $\boxed{18}$ total pairs of parallel lines.

Solution: We can use the feature of 3-Dimension in a cube to solve the problem systematically. For example, in the 3-D of the cube, $\overline{AB}$ $\overline{BC}$ , and $\overline{BF}$ have $4$ different parallel edges respectively. So it gives us the total pairs of parallel lines are $\binom{4}{2}\cdot3 =\boxed{18}$


Answer: 4.0

How many pairs of positive integers $(a,b)$ are there such that $a$ and $b$ have no common factors greater than $1$ and:
\[\frac{a}{b} + \frac{14b}{9a}\]
is an integer?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many}$

Solution:
Let $\frac{a}{b} = x$ and $\frac{14b}{9a} = y.$ Then for rational numbers $x,y$ we have $xy = \frac{14}{9}$ and $x+y \in Z.$ One can prove that $x$ and $y$ must both have the same denominator in order for their sum to be an integer. This denominator is $3$ meaning $b=3$ and $a$ is any factor of $14.$ This yields $\boxed{4}$ solutions.
~ab2024


Answer: 58.0

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution: We compute $2^8+1=257$ . We're all familiar with what $6^3$ is, namely $216$ , which is too small. The smallest cube greater than it is $7^3=343$ $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$ , which therefore will clearly be the largest cube less than $2^{18}+1$ . So, the required number of cubes is $64-7+1= \boxed{58}$

Solution: First, $2^8+1=257$ . Then, $2^{18}+1=262145$ . Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ends with $64$ . Now, by counting how many numbers are between these, we find the answer to be $\boxed{58}$

Solution: First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number $-$ the lowest number $+ 1$ . People will forget the $+1$ so the only possibilities are C and E. We can clearly see that C is too small so our answer is $\boxed{58}$

Solution: There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$ , which is over $2^8+1=257$ . We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$ , which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gives us 64 and $64^3 = 262,144$ , which is perfect for the $2^{18}+1=262,145$ . But we are not done. Since it is inclusive, we must add 1 to this solution, which give us $\boxed{58}$


Answer: 672.0

How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$

Solution: We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$ . We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$ . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$ . To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{672}$

Solution:
(Explanation of 1)
Factorials up to 9 in product lead to prime factorization $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$ So the number of pairs possible is { $2^{0}$ $,$ $2^{2}$ $,$ ... $2^{30}$ }{ $3^{0}$ $,$ $3^{2}$ $,$ ... $3^{12}$ }{ $5^{0}$ $,$ $5^{2}$ $,$ ... $5^{4}$ }{ $7^{0}$ $,$ $7^{2}$
Which leads to resulting number of pairs = $16*7*3*2$ $=$ $672$ $\Rightarrow\boxed{672}$


Answer: 2.0

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution:
Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$ . We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$ .  That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$ $\frac{-1+i\sqrt{3}}{2}$ $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$
In order $r$ to be a root of $P$ $re^{2\pi i / 3}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$ $re^{i\frac{2\pi}{3}}$ $re^{i\frac{4\pi}{3}}$ . However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$ , if $w$ is a root, then $we^{2\pi i / 3}$ and $we^{4\pi i / 3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$ $re^{2\pi i / 3}$ , or $re^{4\pi i / 3}$ . Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p$ . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$ , meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$ . In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{2\pi i / 3}$ and $re^{4 \pi i / 3}$ being conjugates and since $m+n+p = 5$ , (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{2}$

Solution:
Let $x_1=r$ , then \[x_2=\frac{-1+i\sqrt{3}}{2} r,\] \[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,\] \[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,\] which means $x_4$ is the same as $x_1$
Now we have 3 different roots of the polynomial, $x_1$ $x_2$ , and $x_3$ . Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root $x_4=p$ which is different from the three roots we already know, then there must be two other roots, \[x_5=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =\left( \frac{-1-i\sqrt{3}}{2} \right) p,\] \[x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 p=p,\] different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from $x_1$ $x_2$ , and $x_3$
The polynomial then can be written like $f(x)=(x-x_1)^m (x-x_2)^n (x-x_3)^q$ , where $m$ $n$ , and $q$ are non-negative integers and $m+n+q=5$ . Since $a$ $b$ $c$ and $d$ are real numbers, then $n$ must be equal to $q$ . Therefore $(m,n,q)$ can only be $(1,2,2)$ or $(3,1,1)$ , so the answer is $\boxed{2}$


Answer: 6.0

How many positive cubes divide $3! \cdot 5! \cdot 7!$
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6$

Solution:
$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$
Therefore, a perfect cube that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}$ where $a$ $b$ $c$ , and $d$ are nonnegative multiples of $3$ that are less than or equal to $8, 5, 2$ and $1,$ respectively.
So:
$a\in\{0,3,6\}$ $3$ possibilities)
$b\in\{0,3\}$ $2$ possibilities)
$c\in\{0\}$ $1$ possibility)
$d\in\{0\}$ $1$ possibility)
So the number of perfect cubes that divide $3! \cdot 5! \cdot 7!$ is $3\cdot2\cdot1\cdot1 = \boxed{6}$

Solution:
$3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)$
In the expression, we notice that there are 3 $3's$ , 3 $2's$ , and 3 $1's$ . This gives us our first 3 cubes: $3^3$ $2^3$ , and $1^3$
However, we can also multiply smaller numbers in the expression to make bigger expressions. For example, $(2 \cdot 2) \cdot 4 \cdot 4=4 \cdot 4 \cdot 4 = 4^3$ (one 2 comes from the $3!$ , and the other from the $5!$ ). Using this method, we find:
$(3 \cdot 2) \cdot (3 \cdot 2) \cdot 6 = 6^3$
and
$(3 \cdot 4) \cdot (3 \cdot 4) \cdot (2 \cdot 6) = 12^3$
So, we have 6 cubes total: $1^3 ,2^3, 3^3, 4^3, 6^3,$ and $12^3$ for a total of $\boxed{6}$ cubes


Answer: 7.0

How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

Solution: Any even multiple of $3$ is a multiple of $6$ , so we need to find multiples of $6$ that are perfect squares and less than $2020$ . Any solution that we want will be in the form $(6n)^2$ , where $n$ is a positive integer. The smallest possible value is at $n=1$ , and the largest is at $n=7$ (where the expression equals $1764$ ). Therefore, there are a total of $\boxed{7}$ possible numbers.-PCChess

Solution: A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$ , where $3^2$ is the multiple or $3$ and $2^2$ makes it even. Simplifying we have $36^2 \cdot x^2$ . We can divide $2020$ by $36$ (floor) and get $56$ see the result. We can then see that there are $7$ different values for $x$ . It can't be larger or else $x^2 > 56$ . And thus $\boxed{7}$

Solution:
It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$ $12^2$ $18^2$ $24^2$ $30^2$ $36^2$ , and $42^2$ . Since $48^2=2196 > 2020$ , there are $\boxed{7}$ valid answers.
~airbus-a321, November 2023


Answer: 42.0

How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: We can first find the prime factorization of $23,232$ , which is $2^6\cdot3^1\cdot11^2$ . Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{42}$

Solution: Observe that $69696$ $264^2$ , so this is $\frac{1}{3}$ of $264^2$ which is $88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3$ , which has $3 \cdot 7 \cdot 2 = 42$ factors. The answer is $\boxed{42}$


Answer: 3.0

How many positive factors of 36 are also multiples of 4?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution:
The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$
The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$
Only $4, 12$ and $36$ appear on both lists, so the answer is $3$ , which is option $\boxed{3}$

Solution: $36 = 4^1 \cdot 3^2$ . All possible factors of $36$ will be here, except for ones divisible by $2$ and not by $4$ $(1+1)\cdot (2+1) = 6$ . Subtract factors not divisible by $4$ , which are $1$ $3^1$ , and $3^2$ $6-3=3$ , which is $\boxed{3}$

Solution:
Divide $36$ by $4$ , and the remaining factors, when multiplied by $4$ , will be factors of $36$
$36 \div 4 = 9$ , which has $3$ factors, giving us option $\boxed{3}$


Answer: 54.0

How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

Solution:
The prime factorization of 2004 is $2^2\cdot 3\cdot 167$ .  Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$
We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization.  For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$
A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$ .  Thus we need to find how many $(a,b,c)$ satisfy
We can think of this as partitioning the exponents to $a+1,$ $b+1,$ and $c+1$ .  So let's partition the 2's first.  There are two 2's so this is equivalent to partitioning two items in three containers.  We can do this in ${4 \choose 2} = 6$ ways.  We can partition the 3 in three ways and likewise we can partition the 167 in three ways.  So we have $6\cdot 3\cdot 3 = \boxed{54}$ as our answer.

Solution:
Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 * 2^2 * 3$
167, 2, 2, 3
4, 3, 167
12, 167
4, 501
2, 1002
2, 3, 334
2, 2, 501*
6, 2, 167
3, 668
6, 334
2004*
To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of $2004^{2004}$ is $2^{4008} * 3^{2004} * 167^{2004}.$ Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star.
For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to.
For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.
Therefore we have $8 * 6$ normal multiples and $3 *2$ "half" multiples. Sum to get $\boxed{54}$ as our answer.


Answer: 37.0

How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution:
Prime factorizing $201^9$ , we get $3^9\cdot67^9$ .
A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$ . This yields $5\cdot5 = 25$ perfect squares.
Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0, 3, 6$ , or $9$ for both $3$ and $67$ , which yields $4\cdot4 = 16$ perfect cubes, for a total of $25+16 = 41$
Subtracting the overcounted powers of $6$ $3^0\cdot67^0$ $3^0\cdot67^6$ $3^6\cdot67^0$ , and $3^6\cdot67^6$ ), we get $41-4 = \boxed{37}$

Solution:
Observe that $201 = 67 \cdot 3$ . Now divide into cases:
Case 1 : The factor is $3^n$ . Then we can have $n = 2$ $3$ $4$ $6$ $8$ , or $9$
Case 2 : The factor is $67^n$ . This is the same as Case 1.
Case 3 : The factor is some combination of $3$ s and $67$ s.
This would be easy if we could just have any combination, as that would simply give $6 \cdot 6$ . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for $n$
$n = 2$ is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
$n = 3$ is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of $3$
$n = 4$ is a "square".
$n = 6$ is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.
$n = 8$ is a "square"
$n = 9$ is a "cube".
Now let's consider subcases:
Subcase 1 : The squares are with each other.
Since we have $3$ square terms, and they would pair with $3$ other square terms, we get $3 \cdot 3 = 9$ possibilities.
Subcase 2 : The cubes are with each other.
Since we have $2$ cube terms, and they would pair with $2$ other cube terms, we get $2 \cdot 2 = 4$ possibilities.
Subcase 3 : A number pairs with $n=6$
Since any number can pair with $n=6$ (as it gives both a square and a cube), there would be $6$ possibilities. Remember however that there can be two different bases ( $3$ and $67$ ), and they would produce different results. Thus, there are in fact $6 \cdot 2 = 12$ possibilities.
Finally, summing the cases gives $6+6+9+4+12 = \boxed{37}$

Solution: We first prime factorize $201^9 = 3^9 \cdot 67^9$ . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of $3$ in the exponent. The largest square for $3$ can be $3^8$ , so their must be $\dfrac {8}{2} = 4$ ways. The largest cube is $3^9$ , so there must be $\dfrac{9}{3} = 3$ . Minus one $3^6$ due to overlapping and we get $4 + 3 -1 = 6$ ways for $3$ to be a cube/square. We can see that this same thing happens for $67^9$ due to the same exponent. Adding $0$ as a case, we have our answer; $6 \cdot 6 + 1 = \boxed{37}$

Solution: Notice that $201=3 \cdot 67$ . We factorize $201^9$ to get $3^9 \cdot 67^9$ . We then list perfect squares and cubes. $3^2$ $3^4$ $3^6$ $3^8$ $3^3$ $3^6$ $3^9$ $67^2$ $67^4$ $67^6$ $67^8$ $67^3$ $67^6$ $67^9$ . Notice that the powers of $6$ overlap. We must not forget $1$ though. Of course, all of these factors already work. This gives us $15-2=3$ . Next, we count the perfect squares. Since there are $4$ options we have $4 \cdot 4=16$ . We do the same for the perfect cubes except with 3 options this time, and we have $3 \cdot 3=9$ . However, we accidentally overcounted $3^6 \cdot 67^6$ . We add our answers and subtract $1$ to get $13+16+9-1 = \boxed{37}$


Answer: 7.0

How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution: Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{7}$ factors have more than $3$ factors.

Solution: As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ $2$ , or $3$ factors (namely $1$ $2$ $4$ $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{7}$

Solution: Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5\cdot2$ and $b = 101\cdot2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{7}$


Answer: 784.0

How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$

Solution:
Observation: We see that there is a pattern with $10^k \pmod{1001}$ \[10^0 \equiv 1 \pmod{1001}\] \[10^1 \equiv 10 \pmod{1001}\] \[10^2 \equiv 100 \pmod{1001}\] \[10^3 \equiv -1 \pmod{1001}\] \[10^4 \equiv -10 \pmod{1001}\] \[10^5 \equiv -100 \pmod{1001}\] \[10^6 \equiv 1 \pmod{1001}\] \[10^7 \equiv 10 \pmod{1001}\] \[10^8 \equiv 100 \pmod{1001}\]
So, this pattern repeats every 6.
Also, $10^j-10^i \equiv 0 \pmod{1001}$ , so $10^j \equiv 10^i \pmod{1001}$ , and thus, \[j \equiv i \pmod{6}\] . Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$

Solution: Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$ ; therefore, the order of 10 modulo $7,11$ , and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$

Solution:
$10^j - 10^i \equiv 0 \pmod{1001} \iff 10^{j - i} - 1 \equiv 0 \pmod{1001} \iff 10^{j - i} \equiv 1 \pmod{1001} \iff j \equiv i \pmod 6$ . If $j \equiv i \equiv n \pmod 6$ for $n = 0, 1, 2, 3$ , there are $17$ choices for each value of $n$ , yielding $4 \cdot \dbinom{17}{2} = 544$ . However, if $n = 4, 5$ , there are only $16$ choices, giving us $2 \cdot \dbinom{16}{2} = 240$ . So, our final answer is $544 + 240 = \boxed{784}$ .
~Puck_0


Answer: 412.0

How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$

Solution:
First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$
Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$
For ${\lfloor x\rfloor}=0$ $N=1$ as long as $x \neq 0$ . This gives us $1$ value of $N$
For ${\lfloor x\rfloor}=1$ $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$
Therefore, $N=1$ . However, we got $N=1$ in case 1 so it got counted twice.
For ${\lfloor x\rfloor}=2$ $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$
This gives us $3^2-2^2=5$ $N$ 's
For ${\lfloor x\rfloor}=3$ $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$
This gives us $4^3-3^3=37$ $N$ 's
For ${\lfloor x\rfloor}=4$ $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$
This gives us $5^4-4^4=369$ $N$ 's
Since $x$ must be less than $5$ , we can stop here and the answer is $1+5+37+369= \boxed{412}$ possible values for $N$

Solution: For a positive integer $k$ , we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$ . Then $x=\sqrt[k]{N}$ , and because $k \le x < k+1$ , we have $k^k \le x^k < (k+1)^k$ , and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$ . The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$ , and this is equal to $(k+1)^k-k^k$ . If $k>4$ , the value of $x^k$ exceeds $1000$ , so we only need to consider $k \le 4$ . The requested number of values of $N$ is the same as the number of values of $x^k$ , which is $\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}$


Answer: 4.0

How many positive integers $b$ have the property that $\log_{b} 729$ is a positive integer?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }$

Solution: If $\log_{b} 729 = n$ , then $b^n = 729$ . Since $729 = 3^6$ $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{4}$


Answer: 48.0

How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

Solution:
We set up the following equation as the problem states:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\]
Breaking each number into its prime factorization, we see that the equation becomes
\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.\]
We can now determine the prime factorization of $n$ . We know that its prime factors belong to the set $\{2, 3, 5, 7\}$ , as no factor of $10!$ has $11$ in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.
There can be anywhere between $3$ and $8$ $2$ 's and $1$ to $4$ $3$ 's. However, since $n$ is a multiple of $5$ , and we multiply the $\text{gcd}$ by $5$ , there can only be $3$ $5$ 's in $n$ 's prime factorization. Finally, there can either $0$ or $1$ $7$ 's.
Thus, we can multiply the total possibilities of $n$ 's factorization to determine the number of integers $n$ which satisfy the equation, giving us $6 \times 4 \times 1 \times 2 = \boxed{48}$ . ~ciceronii

Solution:
Like the Solution 1, we start from the equation:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\] Assume $\text{lcm}{(5!, n)}=k\cdot5!$ , with some integer $k$ . It follows that $k\cdot 4!=\text{gcd}{(10!, n)}$ . It means that $n$ has a divisor $4!$ . Since $n$ is a multiple of $5$ $n$ has a divisor $5!$ . Thus, $\text{lcm}{(5!, n)}=n=k\cdot5!$ . The equation can be changed as \[k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}\] \[k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}\] We can see that $k$ is also a multiple of $5$ , with a form of $5\cdot m$ . Substituting it in the above equation, we have \[m=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, m)}\] Similarly, $m$ is a multiple of $5$ , with a form of $5\cdot s$ . We have \[s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}\] The equation holds, if $s$ is a divisor of $2^5\cdot3^3\cdot7$ , which has $(5+1)(3+1)(1+1)=\boxed{48}$ divisors. ~Linty Huang

Solution:
As in the previous solutions, we start with
\[\text{lcm}(5!,n) = 5\text{gcd}(10!,n)\]
From this we have that $\text{lcm}(5!,n) \,|\, 5\text{gcd}(10!,n)$ , and in particular, $n \,|\, 5\text{gcd}(10!,n)$ . However, $\text{gcd}(10!,n)\, |\, n$ , so we must have $\text{gcd}(10!,n) = n$ or $\text{gcd}(10!,n) = n/5$ . If $\text{gcd}(10!,n) = n$ , then we have $\text{lcm}(5!,n) = 5n$ ; because $5\, |\, 5!$ , this implies that 5 does not divide $n$ , so we must have $\text{gcd}(10!,n) = n/5$
Now we have $\text{lcm}(5!,n) = n$ , implying that $5!\, |\, n$ , and $n/5\, |\, 10!$ . Writing out prime factorizations, this gives us
\[2^3 \cdot 3 \cdot 5 \,|\, n\] \[n \,|\, 2^8 \cdot 3^4 \cdot 5^2 \cdot 7\]
So $n$ can have 3, 4, 5, 6, 7, or 8 factors of 2; 1, 2, 3, or 4 factors of two; and 0 or 1 factors of 7. Note that $\text{gcd}(2^8 \cdot 3^4 \cdot 5^2 \cdot 7,n) = n/5$ implies that $n$ has 2 factors of 5. Thus, there are $6 \cdot 4 \cdot 2 = 48$ possible choices for $n$ , and our answer is $\boxed{48}$


Answer: 6.0

How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

Solution:
We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = 400$ and $n = 2500$ , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation, which we abbreviate as $L=R$
For $n = 330$ $L=19$ but $18^2 < 330 < 19^2$ so $R=18$
For $n = 400$ $L=20$ and $R=20$
For $n = 470$ $L=21$ $R=21$
For $n = 540$ $L=22$ but $540 > 23^2$ so $R=23$
Now we move to $n = 2500$
For $n = 2430$ $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$
For $n = 2360$ $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$
For $n = 2290$ $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$
For $n = 2220$ $L=46$ but $47^2 < 2220$ so $R=47$
For $n = 2500$ $L=50$ and $R=50$
For $n = 2570$ $L=51$ but $2570 < 51^2$ so $R=50$
Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{6}$

Solution:
This is my first solution here, so please forgive me for any errors.
We are given that \[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\]
$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ . As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+50$ for $k\in\mathbb{Z}$
Therefore, \[\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor\]
Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$
Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$
Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$
$\sqrt{155}$ is larger than $12$ and smaller than $13$ , so instead, we can say $k\leq 6$ or $k\geq 32$
Combining this with $5\leq k\leq 35$ , we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$ , meaning that our answer is $\boxed{6}$ .
-Solution By Qqqwerw

Solution: We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separately gives us two inequalities: \[n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50\] \[n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}\] Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is $2+4 = \boxed{6}$

Solution:
Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get \[j+14 \leq \sqrt{70j-20} < j+15\] We can square the inequality to get \[196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}\] The left inequality simplifies to $(j-36)(j-6) \leq 0$ , which yields \[6 \le j \le 36.\] The right inequality simplifies to $(j-20)^2 - 155 > 0$ , which yields \[j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32\]
Solving $j < 8$ , and $6 \le j \le 36$ , we get $6 \le j < 8$ , for $2$ values $j\in \{6, 7\}$
Solving $j >32$ , and $6 \le j \le 36$ , we get $32 < j \le 36$ , for $4$ values $k\in \{33, \ldots , 36\}$
Thus, our answer is $2 + 4 = \boxed{6}$

Solution:
Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le  70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which yields \[20 \le k \le 50.\] The right inequality simplifies to $(k-34)^2 > 155$ , which yields \[k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46\] Solving $k < 22$ , and $20 \le k \le 50$ , we get $20 \le k < 22$ , for $2$ values $k\in \{20, 21\}$
Solving $k >46$ , and $20 \le k \le 50$ , we get $46 < k \le 50$ , for $4$ values $k\in \{47, \ldots , 50\}$
Thus, our answer is $2 + 4 = \boxed{6}$

Solution:
If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not square, then we can say that, for a positive integer $k$ , we have
\begin{align*}
k^2 < &n < (k+1)^2 \\
k^2 + 1000 < &n + 1000  = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\
k^2 + 1000 < &70k < (k+1)^2 + 1000.
\end{align*}
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$ . To solve the first one, we have
\begin{align*}
k^2 - 70k + 1000 &< 0 \\
(k-20)(k-50) &< 0\\
\end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.
The second inequality yields
\begin{align*}
70k &< k^2 + 2k + 1 + 1000 \\
0 &< k^2 -68k + 1001.
\end{align*}
This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).
Notation wise, we need all integers $k$ such that
\[k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)\] or \[k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).\]
For the first one, since our uppoer bound is a little less than $22$ , the $k$ that works is $21$ . For the second, our lower bound is a little more than $46$ , so the $k$ that work are $47, 48,$ and $49$
$\boxed{6}$ total solutions for $n$ , since each value of $k$ corresponds to exactly one value of $n$


Answer: 6.0

Solution:
We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = 400$ and $n = 2500$ , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation, which we abbreviate as $L=R$
For $n = 330$ $L=19$ but $18^2 < 330 < 19^2$ so $R=18$
For $n = 400$ $L=20$ and $R=20$
For $n = 470$ $L=21$ $R=21$
For $n = 540$ $L=22$ but $540 > 23^2$ so $R=23$
Now we move to $n = 2500$
For $n = 2430$ $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$
For $n = 2360$ $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$
For $n = 2290$ $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$
For $n = 2220$ $L=46$ but $47^2 < 2220$ so $R=47$
For $n = 2500$ $L=50$ and $R=50$
For $n = 2570$ $L=51$ but $2570 < 51^2$ so $R=50$
Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{6}$

Solution:
This is my first solution here, so please forgive me for any errors.
We are given that \[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\]
$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ . As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+50$ for $k\in\mathbb{Z}$
Therefore, \[\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor\]
Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$
Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$
Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$
$\sqrt{155}$ is larger than $12$ and smaller than $13$ , so instead, we can say $k\leq 6$ or $k\geq 32$
Combining this with $5\leq k\leq 35$ , we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$ , meaning that our answer is $\boxed{6}$ .
-Solution By Qqqwerw

Solution: We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separately gives us two inequalities: \[n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50\] \[n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}\] Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is $2+4 = \boxed{6}$

Solution:
Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get \[j+14 \leq \sqrt{70j-20} < j+15\] We can square the inequality to get \[196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}\] The left inequality simplifies to $(j-36)(j-6) \leq 0$ , which yields \[6 \le j \le 36.\] The right inequality simplifies to $(j-20)^2 - 155 > 0$ , which yields \[j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32\]
Solving $j < 8$ , and $6 \le j \le 36$ , we get $6 \le j < 8$ , for $2$ values $j\in \{6, 7\}$
Solving $j >32$ , and $6 \le j \le 36$ , we get $32 < j \le 36$ , for $4$ values $k\in \{33, \ldots , 36\}$
Thus, our answer is $2 + 4 = \boxed{6}$

Solution:
Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le  70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which yields \[20 \le k \le 50.\] The right inequality simplifies to $(k-34)^2 > 155$ , which yields \[k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46\] Solving $k < 22$ , and $20 \le k \le 50$ , we get $20 \le k < 22$ , for $2$ values $k\in \{20, 21\}$
Solving $k >46$ , and $20 \le k \le 50$ , we get $46 < k \le 50$ , for $4$ values $k\in \{47, \ldots , 50\}$
Thus, our answer is $2 + 4 = \boxed{6}$

Solution:
If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not square, then we can say that, for a positive integer $k$ , we have
\begin{align*}
k^2 < &n < (k+1)^2 \\
k^2 + 1000 < &n + 1000  = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\
k^2 + 1000 < &70k < (k+1)^2 + 1000.
\end{align*}
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$ . To solve the first one, we have
\begin{align*}
k^2 - 70k + 1000 &< 0 \\
(k-20)(k-50) &< 0\\
\end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.
The second inequality yields
\begin{align*}
70k &< k^2 + 2k + 1 + 1000 \\
0 &< k^2 -68k + 1001.
\end{align*}
This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).
Notation wise, we need all integers $k$ such that
\[k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)\] or \[k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).\]
For the first one, since our uppoer bound is a little less than $22$ , the $k$ that works is $21$ . For the second, our lower bound is a little more than $46$ , so the $k$ that work are $47, 48,$ and $49$
$\boxed{6}$ total solutions for $n$ , since each value of $k$ corresponds to exactly one value of $n$


Answer: 125.0

How many positive integers $n$ satisfy the following condition:
$(130n)^{50} > n^{100} > 2^{200}\ ?$
$\textbf{(A) } 0\qquad \textbf{(B) } 7\qquad \textbf{(C) } 12\qquad \textbf{(D) } 65\qquad \textbf{(E) } 125$

Solution:
We're given $(130n)^{50} > n^{100} > 2^{200}$ , so
$\sqrt[50]{(130n)^{50}} > \sqrt[50]{n^{100}} > \sqrt[50]{2^{200}}$ (because all terms are positive) and thus
$130n > n^2 > 2^4$
$130n > n^2 > 16$
Solving each part separately:
$n^2 > 16 \Longrightarrow n > 4$
$130n > n^2 \Longrightarrow 130 > n$
So $4 < n < 130$
Therefore the answer is the number of positive integers over the interval $(4,130)$ which is $129-5+1 = \boxed{125}$

Solution:
We're given $\left(130n\right)^{50}>n^{100}>2^{200}$
Alternatively to solution 1, first deal with the first half: $\left(130n\right)^{50}>\left(n^{2}\right)^{50}$ . Because the exponents are equal, we can ignore them and solve for $n$ $130n>n^{2}$ , or $n<130$
The second half: $n^{100}>2^{200}$ , or $n^{100}>4^{100}$ , which means $n>4$
Therefore $4<n<130$ and $n=\left\{5,\ 6,\ 7,\ \cdots ,\ 128,\ 129\right\}$ which contains the same number of elements as $\left\{1,\ 2,\ 3,\ \cdots ,\ 124,\ 125\right\}$ which clearly contains $125$ values or choice $\boxed{125}$


Answer: 9.0

How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is $28$ .”
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Solution:
Let $m$ and $n$ be positive integers such that $m>n$ and $m+n=28.$ It follows that $m=2n+d$ for some positive integer $d.$ We wish to find the number of possible values for $d.$
By substitution, we have $(2n+d)+n=28,$ from which $d=28-3n.$ Note that $n=1,2,3,\ldots,9$ each generate a positive integer for $d,$ so there are $\boxed{9}$ possible values for $d.$


Answer: 109.0

How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?

Solution:
Suppose $n$ is such an integer . Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$
In the first case, the three proper divisors of $n$ are $1$ $p$ and $q$ .  Thus, we need to pick two prime numbers less than $50$ . There are fifteen of these ( $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$ ) so there are ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type.
In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$ .  Thus we need to pick a prime number whose square is less than $50$ .  There are four of these ( $2, 3, 5,$ and $7$ ) and so four numbers of the second type.
Thus there are $105+4=\boxed{109}$ integers that meet the given conditions.


Answer: 1.0

How many positive integers less than $1000$ are $6$ times the sum of their digits?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$

Solution:
The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Out of the numbers $1$ to $162$ the one with the largest sum of digits is $99$ , and the sum is $9+9=18$ . Hence the sum of digits will be at most $18$
Also, each number with this property is divisible by $6$ , therefore it is divisible by $3$ , and thus also its sum of digits is divisible by $3$ . Thus, the number is divisible by $18$
We only have six possibilities left for the sum of the digits: $3$ $6$ $9$ $12$ $15$ , and $18$ , but since the number is divisible by $18$ , the digits can only add to $9$ or $18$ . This leads to the integers $18$ $36$ $54$ $72$ $90$ , and $108$ being possibilities. We can check to see that $\boxed{1}$ solution: the number $54$ is the only solution that satisfies the conditions in the problem.

Solution: The sum of the digits is at most $9+9+9=27$ . Therefore the number is at most $6\cdot 27 = 162$ . Since the number is $6$ times the sum of its digits, it must be divisible by $6$ , therefore also by $3$ , therefore the sum of its digits must be divisible by $3$ . With this in mind we can conclude that the number must be divisible by $18$ , not just by $6$ . Since the number is divisible by $18$ , it is also divisible by $9$ , therefore the sum of its digits is divisible by $9$ , therefore the number is divisible by $54$ , which leaves us with $54$ $108$ and $162$ . Only $54$ is $6$ times its digits, hence the answer is $\boxed{1}$


Answer: 927.0

How many positive integers less than 10,000 have at most two different digits

Solution:
First, let's count numbers with only a single digit.  We have nine of these for each length, and four lengths, so 36 total numbers.
Now, let's count those with two distinct digits.  We handle the cases "0 included" and "0 not included" separately.
There are ${9 \choose 2}$ ways to choose two digits, $A$ and $B$ .  Given two digits, there are $2^n - 2$ ways to arrange them in an $n$ -digit number, for a total of $(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22$ such numbers (or we can list them: $AB, BA, AAB, ABA, BAA, ABB, BAB, BBA, AAAB, AABA, ABAA,$ $BAAA, AABB, ABAB, BAAB, ABBA, BABA, BBAA, ABBB, BABB, BBAB, BBBA$ ).  Thus, we have ${9 \choose 2} \cdot 22 = 36\cdot22 = 792$ numbers of this form.
Now, suppose 0 is one of our digits.  We have nine choices for the other digit.  For each choice, we have $2^{n - 1} - 1$ $n$ -digit numbers we can form, for a total of $(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11$ such numbers (or we can list them: $A0, A00, A0A, AA0, A000, AA00, A0A0, A00A, AAA0, AA0A, A0AA$ ).  This gives us $9\cdot 11 = 99$ numbers of this form.
Thus, in total, we have $36 + 792 + 99 = \boxed{927}$ such numbers.

Solution:
We use casework on the number of digits for this problem.
If the number has a single digit, namely the number $n \in [1,9],$ we can clearly all such $n$ work.
If the number has two digits, or the number $n \in [10,99]$ we can clearly see all such $n$ work.
If the number $n$ has three digits, there are a total of $900$ three digit numbers, and there are $9 \cdot 9 \cdot 8$ numbers that have all distinct digits so there are $900 - 9 \cdot 9 \cdot 8$ total three digit numbers that work.
If the number $n$ has four digits, then we have a total of $9 +\binom{9}{2}\left(\binom{4}{1}+\binom{4}{2} +\binom{4}{3}\right) + 9 \cdot \left(\binom{3}{1}+\binom{3}{2} +\binom{3}{3}\right)$ so in total we get $\boxed{927}$ numbers that work.


Answer: 83.0

How many positive perfect squares less than $10^6$ are multiples of $24$

Solution:
The prime factorization of $24$ is $2^3\cdot3$ . Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$ .
This means that each square is in the form $(12c)^2$ , where $12 c$ is a positive integer less than $\sqrt{10^6}$ . There are $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}$ solutions.

Solution: The perfect squares divisible by $24$ are all multiples of $12$ $12^2$ $24^2$ $36^2$ $48^2$ , etc... Since they all have to be less than $10^6$ , or $1000^2$ , the closest multiple of $12$ to $1000$ is $996$ $12*83$ ), so we know that this is the last term in the sequence. Therefore, we know that there are $\boxed{083}$ perfect squares divisible by $24$ that are less than $10^6$


Answer: 8.0

How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$

Solution: Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$

Solution: Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{8}$ perfect squares less than 2023.

Solution: Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{8}$ positive perfect squares no greater than $80$

Solution: We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{8}$ solutions.


Answer: 8.0

Solution: Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$

Solution: Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{8}$ perfect squares less than 2023.

Solution: Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{8}$ positive perfect squares no greater than $80$

Solution: We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{8}$ solutions.


Answer: 5.0

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: Looking at the values, we notice that $11-7=4$ $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a multiple of $11$ $9$ , and $6$ . The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where $k$ is a positive integer.  This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$ , which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values, so our answer is $\boxed{5}$

Solution: Let us create the equations: $6x+2 = 9y+5 = 11z+7$ , and we know $100 \leq 11z+7 <1000$ , it gives us $9 \leq z \leq 90$ , which is the range of the value of z. Because of $6x+2=11z+7$ , then $6x=11z+5=6z+5(z+1)$ , so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$ , then $9y=11z+2=9z+2(z+1)$ , so $(z+1)$ must also be a mutiple of $9$ . Hence, the value of $(z+1)$ must be a common multiple of $6$ and $9$ , which means multiples of $18 (LCM \text{ of }\ 6, 9)$ . So, let's say $z+1 = 18p$ ; then, $9 \leq z = 18p-1 \leq 90$ ,  so $1 \leq p \leq 91/18\ or \ 1 \leq p \leq 5$ . Thus, the answer is $\boxed{5}$

Solution: By the Chinese Remainder Theorem , we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{5}.$

Solution: We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent to their LCM. Thus, $x+4 \equiv 0 \mod 198,$ since 198 is the LCM of $6, 9,$ and $11.$ Since these numbers have to be three digits, they can only be $194, 392, 590, 788,$ and $986.$ This gives us the answer of $\boxed{5}.$

Solution: Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$ . Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$ . Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$ . From this, we can figure out that $N$ can be 5 different three digit numbers -- $194, 392, 590, 788,$ and $986$ $\therefore$ , the answer is $\boxed{5}.$


Answer: 12.0

How many positive two-digit integers are factors of $2^{24}-1$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$
~ pi_is_3.14

Solution:
Repeating difference of squares
$2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$
$2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$
$2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$
The sum of cubes formula gives us:
$2^{12}+1=(2^4+1)(2^8-2^4+1)$
$2^{12}+1 = 17\cdot241$
A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$ , since multiplying by $241$ will make any factor too large.
Multiplying $17$ by $3$ or $5$ will give a two-digit factor; $17$ itself will also work. The next smallest factor, $7$ , gives a three-digit number.  Thus, there are $3$ factors that are multiples of $17$
Multiplying $13$ by $3$ $5$ , or $7$ will also give a two-digit factor, as well as $13$ itself. Higher numbers will not work, giving $4$ additional factors.
Multiply $7$ by $3$ $5$ , or $3^2$ for a two-digit factor. There are no more factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors.
Multiply $5$ by $3$ or $3^2$ for a two-digit factor. All higher factors have been counted already, so there are $2$ more factors.
Thus, the total number of factors is $3+4+3+2=\boxed{12}$


Answer: 6.0

How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)
$\text{(A)} \ 4 \qquad  \text{(B)} \ 5 \qquad  \text{(C)} \ 6 \qquad  \text{(D)} \ 7 \qquad  \text{(E)} \ 8$

Solution: List out all numbers that have 7 as the ones digit less than 100: ${7, 17, 27, 37, 47, 57, 67, 77, 87, 97}$ . Only $7, 17,37, 47,67,$ and $97$ are prime. Thus, it is $\boxed{6}$ . -slackroadia


Answer: 4.0

How many quadratic polynomials with real coefficients are there such that the set of roots equals the set of coefficients? (For clarification: If the polynomial is $ax^2+bx+c,a\neq 0,$ and the roots are $r$ and $s,$ then the requirement is that $\{a,b,c\}=\{r,s\}$ .)
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } \text{infinitely many}$

Solution:
Firstly, if $r=s$ , then $a=b=c$ , so the equation becomes $ax^2 + ax + a = 0 \Rightarrow x^2 + x + 1=0$ , which has no real roots.
Hence there are three cases we need to consider:
Case 1 $a=b=r$ and $c=s \neq r$ :
The equation becomes $ax^2+ax+c=0$ , and by Vieta's Formulas, we have $a+c=-1$ and $ac = \frac{c}{a}$ . This second equation becomes $(a^2-1)c=0$ . Hence one possibility is $c=0$ , in which case $a=-1$ , giving the equation $-x^2 - x = 0$ , which has roots $0$ and ${-}1$ . This gives one valid solution. On the other hand, if $c\neq 0$ , then $a^2-1=0$ , so $a = \pm 1$ . If $a=1$ , we have $c=-2$ , and the equation is $x^2 + x - 2 = 0$ , which clearly works, giving a second valid solution. If $a=-1$ , then we have $c=0$ , which has already been considered, so this possibility gives no further valid solutions.
Case 2 $a=c=r$ $b=s \neq r$ :
The equation becomes $ax^2 + bx + a = 0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = 1$ . These equations reduce to $a^3 + a + 1 = 0$ . By sketching a graph of this function, we see that there is exactly one real root. (Alternatively, note that as it is of odd degree, there is at least one real root, and by differentiation, it has no stationary points, so there is at most one real root. Combining these gives exactly one real root.) This gives a third valid solution.
Case 3 $a=r$ $b=c=s \neq r$ :
The equation becomes $ax^2 + bx+b=0$ , so by Vieta's Formulas, we have $a+b = -\frac{b}{a}$ and $ab = \frac{b}{a}$ .
Observe that $b \neq 0$ , as if it were $0$ , the equation would just have one real root, $0$ , so this would not give a valid solution. Thus, taking the second equation and dividing both sides by $b$ , we deduce have $a=\frac{1}{a}$ , so $a=\pm 1$ . If $a=1$ , we have $1+b=-b$ , giving $b=-\frac{1}{2}$ , so the equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$ , which is a fourth valid solution. If $a=-1$ , we have $1+b=b$ , which is a contradiction, so this case gives no further valid solutions.
Hence the total number of valid solutions is $\boxed{4}$


Answer: 2.0

How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^{x}+3=0$
$\text {(A)} 0 \qquad  \text {(B)} 1 \qquad  \text {(C)} 2 \qquad  \text {(D)} 3 \qquad  \text {(E)} 4$

Solution: Let $a = 3^x$ . Then the preceding equation can be expressed as the quadratic, \[9a^2-28a+3 = 0\] Solving the quadratic yields the roots $3$ and $1/9$ . Setting these equal to $3^x$ , we can immediately see that there are $\boxed{2}$ real values of $x$ that satisfy the equation.


Answer: 159.0

How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$

Solution:
The range of the sine function is $-1 \le y \le 1$ . It is periodic (in this problem) with a period of $\frac{2}{5}$
Thus, $-1 \le \frac{1}{5} \log_2 x \le 1$ , and $-5 \le \log_2 x \le 5$ . The solutions for $x$ occur in the domain of $\frac{1}{32} \le x \le 32$ . When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$ ) of the sine curve and another curve that is $< 1$ , so there are $\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$ , there is exactly $1$ touching point between the two functions: $\left(\frac{1}{5},0\right)$ . When $y < 0$ or $x < 1$ , we can count $4$ more solutions. The solution is $154 + 1 + 4 = \boxed{159}$

Solution: Notice that the equation is satisfied twice for every sine period (which is $\frac{2}{5}$ ), except in the sole case when the two equations equate to $0$ . In that case, the equation is satisfied twice but only at the one instance when $y=0$ . Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: $32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed{159}$


Answer: 2.0

How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet?  For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution:
The first thing one would want to do is place a possible letter that works and then stem off of it.  For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it.  Unfortunately, after that step, we can't do too much, since:
$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed because of the $cd$
We get the same problem if we start with a $d$ , since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$
If we start with a $b$ , the next letter would have to be a $d$ , and since we can put an $a$ next to it and then a $c$ after that, this configuration works.  The same approach applies if we start with a $c$
So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$ , giving us:
\[1 + 1 = \boxed{2}.\]

Solution:
Case 1: the first letter is A
Subcase 1: the second letter is C
The next letter must either be B or D, both of which do not satisfy the conditions.
Subcase 2: the second letter is D
The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.
Case 2: the first letter is B
The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.
Case 3: the first letter is C
The next letter is forced to be A, the third letter is D and the last letter is B. This works.
Case 4: the first letter is D
Subcase 1: the second letter is A
The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other.
Subcase 2: the second letter is B
The third letter cannot be A or C, so this doesn't work.
Summing the cases, there are two that work: $BDAC$ and $CADB$ $\Longrightarrow \boxed{2}$


Answer: 11.0

Solution:
The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$
$3+5+3=\boxed{11}$


Answer: 6.0

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

Solution:
By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$
This means that $a^2=2b+1$
We know that $a,b>0$ and that $b<100$
We also know that $a^2$ is odd and thus $a$ is odd, since the right side of the equation is odd. $2b$ is even. $2b+1$ is odd.
So $a=1,3,5,7,9,11,13$ , but if $a=1$ , then $b=0$ . Thus $a\neq1.$
$a=3,5,7,9,11,13$
The answer is $\boxed{6}$


Answer: 65.0

How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution:
Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).
We know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equivalent to $0110$ . Thus, our sequence must be of the forms $0\ldots010$ or $0\ldots0110$ . In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$ . In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$ . Therefore, it must be the case that $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions).
It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ $f(4)=1$ since the only possible valid sequence is $0110$ $f(5)=1$ since the only possible valid sequence is $01010$
The recursive sequence is then as follows:
\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]
So, our answer is $\boxed{65}$

Solution:
After any particular $0$ , the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$ , i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$ s and $3$ s to get $18$ . There are a number of ways to do this:
Case 1 : nine $2$ s - there is only $1$ way to arrange them.
Case 2 : two $3$ s and six $2$ s - there are ${8\choose2} = 28$ ways to arrange them.
Case 3 : four $3$ s and three $2$ s - there are ${7\choose4} = 35$ ways to arrange them.
Case 4 : six $3$ s - there is only $1$ way to arrange them.
Summing the four cases gives $1+28+35+1=\boxed{65}$

Solution:
We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:
Case 1 : Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ .
Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.
Case 2 : There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.
Case 3 : Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.
Case 4 : No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$
Adding these four cases, we have $1+28+35+1=\boxed{65}$ as our final answer.

Solution:
Any valid sequence must start with a $0$ . We can then think of constructing a sequence as adding groups of terms to this $0$ , each ending in $0$ . (This is always possible because every valid string ends in $0$ .) For example, we can represent the string $01011010110110$ as: $0-10-110-10-110-110$ .
To not have any consecutive 0s, we must have at least one $1$ before the next $0$ . However, we cannot have three or more $1$ s before the next $0$ because we cannot have three consecutive $1$ s. Consequently, we can only have one or two $1$ s.
So we can have the groups: $10$ and $110$
After the initial $0$ , we have $18$ digits left to fill in the string. Let the number of $10$ blocks be $x$ , and $110$ be $y$ . Then $x$ and $y$ must satisfy $2x+3y=18$ . We recognize this as a Diophantine equation. Taking $\pmod{2}$ yields $y=0 \pmod{2}$ . Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$ $(6, 2)$ $(3, 4)$ , and $(0, 6)$ . We now handle each of these cases separately.
$(9, 0)$ : Only one arrangement, namely all $10$ s.
$(6, 2)$ : We have 6 groups of $11$ , and $2$ groups of $110$ . This has $\binom{6+2}{2}=28$ cases.
$(3, 4)$ : This means we have 3 groups of $10$ , and 4 groups of $110$ . This has $\binom{3+4}{3}=35$ cases.
$(0, 6)$ : Only one arrangement, namely all $110$
Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{65}$ .
~Math4Life2020

Solution:
Suppose the number of $0$ s is $n$ . We can construct the sequence in two steps:
Step 1: put $n-1$ of $1$ s between the $0$ s;
Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$ s in the $n-1$ spots where there is a $1$ . There are $\binom{n-1}{20-2n}$ ways of doing this.
Now we find the possible values of $n$
First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$ s);
And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$ s).
Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{65}.\]

Solution:
For a valid sequence of length $n$ , the sequence must be in the form of $01xx...xx10$ . By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of:
So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$
We will calculate $f(19)$ by Dynamic Programming
$f(3) = 1$
$f(4) = 1$
$f(5) = 1$
$f(6) = 2$
$f(7) = 2$
$f(8) = 3$
$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$
$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$
$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$
$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$
$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$
$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$
$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$
$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$
$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$
$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$
$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{65}$


Answer: 65.0

Solution:
Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).
We know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equivalent to $0110$ . Thus, our sequence must be of the forms $0\ldots010$ or $0\ldots0110$ . In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$ . In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$ . Therefore, it must be the case that $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions).
It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ $f(4)=1$ since the only possible valid sequence is $0110$ $f(5)=1$ since the only possible valid sequence is $01010$
The recursive sequence is then as follows:
\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]
So, our answer is $\boxed{65}$

Solution:
After any particular $0$ , the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$ , i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$ s and $3$ s to get $18$ . There are a number of ways to do this:
Case 1 : nine $2$ s - there is only $1$ way to arrange them.
Case 2 : two $3$ s and six $2$ s - there are ${8\choose2} = 28$ ways to arrange them.
Case 3 : four $3$ s and three $2$ s - there are ${7\choose4} = 35$ ways to arrange them.
Case 4 : six $3$ s - there is only $1$ way to arrange them.
Summing the four cases gives $1+28+35+1=\boxed{65}$

Solution:
We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:
Case 1 : Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ .
Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.
Case 2 : There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.
Case 3 : Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.
Case 4 : No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$
Adding these four cases, we have $1+28+35+1=\boxed{65}$ as our final answer.

Solution:
Any valid sequence must start with a $0$ . We can then think of constructing a sequence as adding groups of terms to this $0$ , each ending in $0$ . (This is always possible because every valid string ends in $0$ .) For example, we can represent the string $01011010110110$ as: $0-10-110-10-110-110$ .
To not have any consecutive 0s, we must have at least one $1$ before the next $0$ . However, we cannot have three or more $1$ s before the next $0$ because we cannot have three consecutive $1$ s. Consequently, we can only have one or two $1$ s.
So we can have the groups: $10$ and $110$
After the initial $0$ , we have $18$ digits left to fill in the string. Let the number of $10$ blocks be $x$ , and $110$ be $y$ . Then $x$ and $y$ must satisfy $2x+3y=18$ . We recognize this as a Diophantine equation. Taking $\pmod{2}$ yields $y=0 \pmod{2}$ . Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$ $(6, 2)$ $(3, 4)$ , and $(0, 6)$ . We now handle each of these cases separately.
$(9, 0)$ : Only one arrangement, namely all $10$ s.
$(6, 2)$ : We have 6 groups of $11$ , and $2$ groups of $110$ . This has $\binom{6+2}{2}=28$ cases.
$(3, 4)$ : This means we have 3 groups of $10$ , and 4 groups of $110$ . This has $\binom{3+4}{3}=35$ cases.
$(0, 6)$ : Only one arrangement, namely all $110$
Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{65}$ .
~Math4Life2020

Solution:
Suppose the number of $0$ s is $n$ . We can construct the sequence in two steps:
Step 1: put $n-1$ of $1$ s between the $0$ s;
Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$ s in the $n-1$ spots where there is a $1$ . There are $\binom{n-1}{20-2n}$ ways of doing this.
Now we find the possible values of $n$
First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$ s);
And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$ s).
Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{65}.\]

Solution:
For a valid sequence of length $n$ , the sequence must be in the form of $01xx...xx10$ . By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of:
So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$
We will calculate $f(19)$ by Dynamic Programming
$f(3) = 1$
$f(4) = 1$
$f(5) = 1$
$f(6) = 2$
$f(7) = 2$
$f(8) = 3$
$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$
$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$
$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$
$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$
$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$
$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$
$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$
$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$
$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$
$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$
$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{65}$


Answer: 380.0

How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$

Solution:
There are $\binom{20}{2}$ selections; however, we count these twice, therefore
$2\cdot\binom{20}{2} = \boxed{380}$ . The wording of the question implies D, not E.


Answer: 382.0

Solution: Consider the 20 term sequence of $0$ 's and $1$ 's.  Keeping all other terms 1, a sequence of $k>0$ consecutive 0's can be placed in $21-k$ locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are $20+19+\cdots+1=\binom{21}{2}$ strings with consecutive zeros. The same argument shows there are $\binom{21}{2}$ strings with consecutive 1's. This yields $2\binom{21}{2}$ strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases $01111...$ $00111...$ $000111...$ , ..., $000...0001$ (of which there are 19) as well as the cases $10000...$ $11000...$ $111000...$ , ..., $111...110$ (of which there are 19 as well). This yields $2\binom{21}{2}-2\cdot19=\boxed{382}$

Solution: First, we think of ways to make all the $1$ 's consecutive. If there are no consecutive $1$ 's, there are $\binom{20}{0}$ ways to order them. If there is one consecutive $1$ , there are $\binom{20}{1}$ ways to order them. If there are two consecutive $1$ 's, then there are $\binom{19}{1}$ ways to order them (We treat the two $1$ 's like a block, and then order that block with 18 other $0$ 's). Continuing in this fashion, there are $\binom{20}{0} + \binom{20}{1} + \binom{19}{1} + \cdots + \binom{1}{1} = 1 + 20 + 19 + \cdots + 2 + 1 = 210 + 1 = 211$ ways to order consecutive $1$ 's. From symmetry, there are also $211$ ways to order the $0$ 's. Now, from PIE, we subtract out the cases where both the $1$ 's and the $0$ 's are consecutive. We do this because when counting the ways to order the $1$ 's, we counted all of these cases once. Then, we did so again when ordering the $0$ 's. So, to only have all of these cases once, we must subtract them. If $1$ is the leftmost digit, then there are $20$ cases where all the $1$ 's and $0$ 's are consecutive (we basically are choosing how many $1$ 's are consecutive, and there are $20$ possibilities. All other digits become $0$ , which are automatically consecutive since the $1$ 's are consecutive. There are also $20$ cases when $0$ is the left-most digit. Thus, there are a total of $211 + 211 - 20 - 20 = \boxed{382}$ as acceptable answers.


Answer: 3.0

How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5$

Solution:
Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a $\frac{1}{2}$ . The only possibility is $\frac{15}{2}$ , from which we get:
Thus, the correct answer is $\boxed{3}.$

Solution:
Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] \[n^2 + n(2a - 1) - 30 = 0\]
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now $a$ cannot be 15 as we need 2 terms. So a can only be less the 15.
Trying all the values of a from 1 to 14 we observe that $a = 4$ $a = 7$ and $a = 1$ provide the only real solutions to the above equation.The three possibilites of a and n are.
\[(a,n) = (4,3),(7, 2),(1, 6)\]
The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into $n^2 + n(2a - 1) - 30 = 0$
\[n^2 + 7n - 30 = 0\] \[n^2 + 13n - 30 = 0\] \[n^2 - n - 30 = 0\]
Since there are 3 possibilities the answer is $\boxed{3}.$


Answer: 3.0

Solution:
Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a $\frac{1}{2}$ . The only possibility is $\frac{15}{2}$ , from which we get:
Thus, the correct answer is $\boxed{3}.$

Solution:
Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] \[n^2 + n(2a - 1) - 30 = 0\]
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now $a$ cannot be 15 as we need 2 terms. So a can only be less the 15.
Trying all the values of a from 1 to 14 we observe that $a = 4$ $a = 7$ and $a = 1$ provide the only real solutions to the above equation.The three possibilites of a and n are.
\[(a,n) = (4,3),(7, 2),(1, 6)\]
The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into $n^2 + n(2a - 1) - 30 = 0$
\[n^2 + 7n - 30 = 0\] \[n^2 + 13n - 30 = 0\] \[n^2 - n - 30 = 0\]
Since there are 3 possibilities the answer is $\boxed{3}.$