Answer: 20.0

The least common multiple of $a$ and $b$ is $12$ , and the least common multiple of $b$ and $c$ is $15$ . What is the least possible value of the least common multiple of $a$ and $c$
$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$

Solution: We wish to find possible values of $a$ $b$ , and $c$ . By finding the greatest common factor of $12$ and $15$ , we can find that $b$ is 3. Moving on to $a$ and $c$ , in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$ $\rightarrow 4$ . Similarly, with $3$ and $c$ , we obtain $5$ . The least common multiple of $4$ and $5$ is $20 \rightarrow \boxed{20}$


Answer: 6.0

The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solution:
Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$

Solution: The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$


Answer: 6.0

Solution:
Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$

Solution: The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$


Answer: 58.0

The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$

Solution:
Let this positive integer be written as $p_1^{e_1}\cdot p_2^{e_2}$ . The number of factors of this number is therefore $(e_1+1) \cdot (e_2+1)$ , and this must equal 2021. The prime factorization of 2021 is $43 \cdot 47$ , so $e_1+1 = 43 \implies e_1=42$ and $e_2+1=47\implies e_2=46$ . To minimize this integer, we set $p_1 = 3$ and $p_2 = 2$ . Then this integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$ .
Now $m=16$ and $k=42$ so $m+k = 16 + 42 = \boxed{58}$

Solution: Recall that $6^k$ can be written as $2^k \cdot 3^k$ . Since we want the integer to have $2021$ divisors, we must have it in the form $p_1^{42} \cdot p_2^{46}$ , where $p_1$ and $p_2$ are prime numbers. Therefore, we want $p_1$ to be $3$ and $p_2$ to be $2$ . To make up the remaining $2^4$ , we multiply $2^{42} \cdot 3^{42}$ by $m$ , which is $2^4$ which is $16$ . Therefore, we have $42 + 16 = \boxed{58}$

Solution:
If a number has prime factorization $p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ , then the number of distinct positive divisors of this number is $\left( k_1 + 1 \right) \left( k_2 + 1 \right) \cdots \left( k_m + 1 \right)$
We have $2021 = 43 \cdot 47$ .
Hence, if a number $N$ has 2021 distinct positive divisors, then $N$ takes one of the following forms: $p_1^{2020}$ $p_1^{42} p_2^{46}$
Therefore, the smallest $N$ is $3^{42} 2^{46} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}$
Therefore, the answer is $\boxed{58}$


Answer: 99.0

The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?
$\textbf{(A)}\  90  \qquad \textbf{(B)}\   99  \qquad \textbf{(C)}\   100  \qquad \textbf{(D)}\   101  \qquad \textbf{(E)}\   110$

Solution: In a rectangle with dimensions $10 \times 10$ , the new rectangle would have dimensions $11 \times 9$ . The ratio of the new area to the old area is $99/100 = \boxed{99}$


Answer: 20.0

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$ . What is $b - a$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$

Solution:
Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$
$a\le 2x+3\le b$
Subtract $3$ from all of the quantities:
$a-3\le 2x\le b-3$
Divide all of the quantities by $2$
$\frac{a-3}{2}\le x\le \frac{b-3}{2}$
Since we have the range of the solutions, we can make it equal to $10$
$\frac{b-3}{2}-\frac{a-3}{2} = 10$
Multiply both sides by 2.
$(b-3) - (a-3) = 20$
Re-write without using parentheses.
$b-3-a+3 = 20$
Simplify.
$b-a = 20$
We need to find $b - a$ for the problem, so the answer is $\boxed{20}$

Solution:
Without loss of generality , let the interval of solutions be $[0, 10]$ (or any real values $[p, 10+p]$ ). Then, substitute $0$ and $10$ to $x$ . This gives $b=23$ and $a=3$ . So, the answer is $23-3=\boxed{20}$ .
~ bearjere


Answer: 10.0

The lengths of the sides of a triangle are integers, and its area is also an integer.
One side is $21$ and the perimeter is $48$ . The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

Solution:
Let $b$ and $c$ be the other two sides of the triangle.  The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.
By Heron's Formula , the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$ .  Plug in the answer choices for $b$ and write the prime factorization of the product to make sure it is a perfect square.
Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$ , so $8$ does not work.  However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$ , so $10$ works.  The third side is $17$ , and the sides satisfy the Triangle Inequality , so the answer is $\boxed{10}$


Answer: 11.0

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution:
Let $n$ $n+1$ , and $n+2$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $n + (n+1) + (n+2) = 3n+3$ . Using the fact that the length of the smallest side is $30\%$ of the perimeter, it follows that:
$n = 0.3(3n+3) \Rightarrow n = 0.9n+0.9 \Rightarrow 0.1n = 0.9 \Rightarrow n=9$ . The longest side is then $n+2 = 11$ . Thus, answer choice $\boxed{11}$ is correct.

Solution: Since the length of the shortest side is a whole number and is equal to $\frac{3}{10}$ of the perimeter, it follows that the perimeter must be a multiple of $10$ . Adding the two previous integers to each answer choice, we see that $11+10+9=30$ . Thus, answer choice $\boxed{11}$ is correct.


Answer: 893.0

The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$

Solution:
By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$ , we have that
Also,
Combining these two inequalities:
\[6.25 < n < 900\]
Thus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$


Answer: 2.0

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:
$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$

Solution: Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\boxed{2}$


Answer: 20.0

The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
[asy] size(150); draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));[/asy]
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution: If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$ . It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{20}$


Answer: 9.0

The letters $A$ $B$ $C$ and $D$ represent digits. If $\begin{tabular}{ccc}&A&B\\ +&C&A\\ \hline &D&A\end{tabular}$ and $\begin{tabular}{ccc}&A&B\\ -&C&A\\ \hline &&A\end{tabular}$ ,what digit does $D$ represent?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution:
Because $B+A=A$ $B$ must be $0$ .
Next, because $B-A=A\implies0-A=A,$ we get $A=5$ as the "0" mentioned above is actually 10 in this case.
Now we can rewrite $\begin{tabular}{ccc}&A&0\\ +&C&A\\ \hline &D&A\end{tabular}$ as $\begin{tabular}{ccc}&5&0\\ +&C&5\\ \hline &D&5\end{tabular}$ . Therefore, $D=5+C$
Finally, $A-1-C=0\implies{A=C+1}\implies{C=4}$ , So we have $D=5+C\implies{D=5+4}=\boxed{9}$


Answer: 12.0

The letters $\text{A}$ $\text{J}$ $\text{H}$ $\text{S}$ $\text{M}$ $\text{E}$ and the digits $1$ $9$ $8$ $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc}  & & AJHSME & 1989 \\  & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\  & & ........ &  \end{tabular}\]
What is the number of the line on which $\text{AJHSME 1989}$ will appear for the first time?
$\text{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24$

Solution: Every $4\text{th}$ line has $1989$ as part of it and every $6\text{th}$ line has $\text{AJHSME}$ as part of it.  In order for both to be part of line $n$ $n$ must be a multiple of $4$ and $6$ , the least of which is $\text{lcm}(4,6)=12\rightarrow \boxed{12}$


Answer: 2.0

The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$

Solution:
Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$ .
Thus, the answer is $\boxed{2}$

Solution: The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$ . Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{2}$


Answer: 1.0

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$ . What is $|r-s|$
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4}   \qquad \textbf{(D) } \frac{2}{3} \qquad   \textbf{(E) } 1$

Solution:
Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$ , and $D(B, P) = D(B',P)$ .
Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$ , and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$ .
Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$
Substituting will yield
\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ 2s &= 9 \\ s &=4.5 \\ \end{align*}
Now $|r-s| = |3.5-4.5| = \boxed{1}$

Solution:
Due to rotations preserving distance, we have that $BP = B^\prime P$ , as well as $AP = A^\prime P$ . From here, we can see that P must be on the perpendicular bisector of $\overline{BB^\prime}$ due to the property of perpendicular bisectors keeping the distance to two points constant.
From here, we proceed to find the perpendicular bisector of $\overline{BB^\prime}$ . We can see that this is just a horizontal line segment with midpoint at $(3.5, 3)$ . This means that the equation of the perpendicular bisector is $x = 3.5$
Similarly, we find the perpendicular bisector of $\overline{AA^\prime}$ . We find the slope to be $\frac{1-2}{3-1} = -\frac12$ , so our new slope will be $2$ . The midpoint of $A$ and $A^\prime$ is $(2, \frac32)$ , which we can use with our slope to get another equation of $y = 2x - \frac52$
Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of $x$ we found earlier, we find that $y=4.5$ . This means that $|r - s| = |3.5 - 4.5| = \boxed{1}$

Solution:
To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$
We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$ , and that the equation of the line $\overline{BB^\prime}$ is $y = 3$
When we solve for the perpendicular bisector of $y = -\frac{1}{2}x + \frac{5}{2}$ , we determine that it has a slope of 2, and it runs through $(2, 1.5)$ . Plugging in $1.5 = 2(2)-n$ , we get than $n = \frac{5}{2}$ . Therefore our perpendicular bisector is $y=2x-\frac{5}{2}$ . Next, we solve for the perpendicular of $y = 3$ . We know that it has an undefined slope, and it runs through $(3.5, 3)$ . We can determine that our second perpendicular bisector is $x = 3.5$
Setting the equations equal to each other, we get $2x-\frac{5}{2} = 3.5$ . Solving for x, we get that $x = \frac{9}{2}$ . Therefore, $|r - s| = |3.5 - 4.5| = \boxed{1}$

Solution:
We use the complex numbers approach to solve this problem.
Denote by $\theta$ the angle that $AB$ rotates about $P$ in the counterclockwise direction.
Thus, $A' - P = e^{i \theta} \left( A - P \right)$ and $B' - P = e^{i \theta} \left( B - P \right)$
Taking ratio of these two equations, we get \[ \frac{A' - P}{A - P} = \frac{B' - P}{B - P} . \]
By solving this equation, we get $P = \frac{7}{2} + i \frac{9}{2}$ .
Therefore, $|s-t| = \left| \frac{7}{2} - \frac{9}{2} \right| = \boxed{1}$


Answer: 13.0

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$ . What is $c$
$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution:
Writing each equation in slope-intercept form, we get $y=\frac{a}{2}x-\frac{1}{2}c$ and $y=-\frac{2}{b}x-\frac{c}{b}$ . We observe the slope of each equation is $\frac{a}{2}$ and $-\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , we see that $\frac{a}{2}=-\frac{1}{-\frac{2}{b}}$ because it is given that the two lines are perpendicular. This equation simplifies to $a=b$
Because $(1, -5)$ is a solution of both equations, we deduce $a \times 1-2 \times -5=c$ and $2 \times 1+b \times -5=-c$ . Because we know that $a=b$ , the equations reduce to $a+10=c$ and $2-5a=-c$ . Solving this system of equations, we get $c=\boxed{13}$


Answer: 665.0

The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$

Solution: It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes.  We know that there is $60$ minutes in a hour.  Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours.  Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{665}$

Solution: The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of $\boxed{665}$

Solution: 11 hours 5 min = $(11 \cdot 60) + 5 \text{min}  = 665 \text{min}$ , therefore $\boxed{665}$


Answer: 995.0

The manager of a company planned to distribute a $$50$ bonus to each employee from the company fund, but the fund contained $$5$ less than what was needed.  Instead the manager gave each employee a $$45$ bonus and kept the remaining $$95$ in the company fund.  The amount of money in the company fund before any bonuses were paid was
$\text{(A)}\ 945\text{ dollars} \qquad \text{(B)}\ 950\text{ dollars} \qquad \text{(C)}\ 955\text{ dollars} \qquad \text{(D)}\ 990\text{ dollars} \qquad \text{(E)}\ 995\text{ dollars}$

Solution:
Let $p$ be the number of people in the company, and $f$ be the amount of money in the fund.
The first sentence states that $50p = f + 5$
The second sentence states that $45p = f - 95$
Subtracing the second equation from the first, we get $5p = 100$ , leading to $p = 20$
Plugging that number into the first equation gives $50\cdot 20 = f + 5$ , leading to $f = 995$ , which is answer $\boxed{995}$

Solution:
Since the company must employ a whole number of employees, the amount of money in the fund must be $5$ dollars less than a multiple of $50$ .  Only options $A$ and $E$ satisfy that requirement.
Additionally, the number must be $95$ more than a multiple of $45$ .  Since $45 \cdot 20 = 900$ , the only number that is $95$ more than a multiple of $45$ out of options $A$ and $E$ is option $\boxed{995}$


Answer: 35.0

The marked price of a book was 30% less than the suggested retail price.  Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.  What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\mathrm{(E)\ }65\%$

Solution:
Without loss of generality, let's assume that the retail price was $100$ USD.
The marked price of the book is $30 \%$ off of $100$ which is equal to $100-100(0.3)=70.$
Half of that marked price is $0.5(70)=35.$
Therefore the percent Alice payed of the suggested retail price is $35/100=\boxed{35}.$


Answer: 22.0

The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$

Solution:
The sum of the ages of the cousins is $4$ times the mean, or $32$ .
There are an even number of cousins, so there is no single median, so $5$ must be the mean of the two in the middle.
Therefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\textbf{(D)}\ \boxed{22}$


Answer: 35.0

The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$\text{(A)}\ 19 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 40$

Solution: Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$ . The number right after $18$ also needs to be as small as possible, so it must be $19$ . This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, \boxed{35}$


Answer: 90.0

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

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

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


Answer: 90.0

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

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


Answer: 11.0

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$
$\textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12$

Solution:
We can eliminate answer choices ${\textbf{(A)}\ 5}$ and ${\textbf{(C)}\ 7}$ , because of the above statement. Now we need to test the remaining answer choices.
Case 1: $x = 6$
Mode: $6$
Median: $6$
Mean: $\frac{37}{7}$
Since the mean does not equal the median or mode, ${\textbf{(B)}\ 6}$ can also be eliminated.
Case 2: $x = 11$
Mode: $6$
Median: $6$
Mean: $6$
We are done with this problem, because we have found when $x = 11$ , the condition is satisfied. Therefore, the answer is $\boxed{11}$

Solution:
Notice that the mean of this set of numbers, in terms of $x$ , is:
$\frac{3+4+5+6+6+7+x}{7} = \frac{31+x}{7}$
Because we know that the mode must be $6$ (it can't be any of the numbers already listed, as shown above, and no matter what $x$ is, either $6$ or a new number,  it will not affect $6$ being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and $6$ equal:
\[\dfrac{31+x}{7}=6\] \[31+x=42\] \[x=\boxed{11}\]

Solution: We know the unique mode must be $6$ , so the mean must be the same number $6$ . Let's imagine a scale. $6$ exactly stands the mid-point of the scale. Numbers of $3,4,5$ represent the left side "weights" of the scale. Numbers of $6,7, x$ represent the right side "weights" of the scale. On the left side, the difference of the three "weights" between $6$ are $-3, -2, -1$ , respectively. It gives us the total difference is $-6$ . In order to allow the scale to keep balance, on the right side, the total difference must be $+6$ . Because we have already known the difference of the right side "weights" between $6$ is $0+1=1$ , partially, so the difference between $6$ and unknown $x$ must be $+6-1=+5$ . It exactly gives us the answer: $x=6+5= \boxed{11}$ . ---LarryFlora


Answer: 14.0

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$

Solution:
As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{14}$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM

Solution:
We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.
Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.
There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$
We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.
Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$
$a\textsubscript{1}$ $a\textsubscript{2}$ $a\textsubscript{3}$ $a\textsubscript{4}$ $a\textsubscript{5}$ $a\textsubscript{6}$ $a\textsubscript{7}$ $a\textsubscript{8}$ $=$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $8$ $8$ $8$ $8$ $a\textsubscript{8}$
Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{14}$ . ~By SK80, mod_x for minor edits


Answer: 14.0

Solution:
As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{14}$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM

Solution:
We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.
Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.
There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$
We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.
Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$
$a\textsubscript{1}$ $a\textsubscript{2}$ $a\textsubscript{3}$ $a\textsubscript{4}$ $a\textsubscript{5}$ $a\textsubscript{6}$ $a\textsubscript{7}$ $a\textsubscript{8}$ $=$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $8$ $8$ $8$ $8$ $a\textsubscript{8}$
Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{14}$ . ~By SK80, mod_x for minor edits


Answer: 115.0

The measure of angle $ABC$ is $50^\circ$ $\overline{AD}$ bisects angle $BAC$ , and $\overline{DC}$ bisects angle $BCA$ .  The measure of angle $ADC$ is
[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C);  label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,N); label("$50^\circ $",(9.4,8.8),SW); [/asy]
$\text{(A)}\ 90^\circ \qquad \text{(B)}\ 100^\circ \qquad \text{(C)}\ 115^\circ \qquad \text{(D)}\ 122.5^\circ \qquad \text{(E)}\ 125^\circ$

Solution:
Let $\angle CAD = \angle BAD = x$ , and let $\angle ACD = \angle BCD = y$
From $\triangle ABC$ , we know that $50 + 2x + 2y = 180$ , leading to $x + y = 65$
From $\triangle ADC$ , we know that $x + y + \angle D = 180$ .  Plugging in $x + y = 65$ , we get $\angle D = 180 - 65 = 115$ , which is answer $\boxed{115}$


Answer: 6.0

The measures of the interior angles of a convex polygon are in arithmetic progression.
If the smallest angle is $100^\circ$ , and the largest is $140^\circ$ , then the number of sides the polygon has is
$\textbf{(A) }6\qquad \textbf{(B) }8\qquad \textbf{(C) }10\qquad \textbf{(D) }11\qquad  \textbf{(E) }12$

Solution:
Let $n$ equal the number of sides the polygon has. The sum of all the interior angles of a polygon is: $180(n-2)$
The formula for an arithmetic series is $\frac{n(a_1 + a_n)}{2}$ . Set this equal to $180(n-2)$ and solve. In this case, $a_1=100$ and $a_n=140$
Our equation becomes $\frac{n(100+140)}{2} = 180(n-2) \Rightarrow 240n = 360(n-2) \Rightarrow 120n = 720$
Simplifying, we get $n = \boxed{6}$ jiang147369


Answer: 9.0

The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:
$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$

Solution:
The formula for the sum of the angles in any polygon is $180(n-2)$ . Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$ , we can find the sum of the angles.
$a_{n}=160$
$a_{1}=160-5(n-1)$
Plugging this into the formula for finding the sum of an arithmetic sequence...
$n(\frac{160+160-5(n-1)}{2})=180(n-2)$
Simplifying, we get $n^2+7n-144$
Since we want the positive solution to the quadratic, we can easily factor and find the answer is $n=\boxed{9}$


Answer: 11.0

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution:
The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$
We substitute the median for $10$ and the equation becomes $n+6=10$
Subtract both sides by 6 and we get $n=4$
$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$
The mean of those numbers is $\frac{9n+63}{9}$ which is $n+7$
Substitute $n$ for $4$ and $4+7=\boxed{11}$


Answer: 134.0

The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates.  For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate.  What was the smallest possible number of members of the committee?

Solution:
Let $v_i$ be the number of votes candidate $i$ received, and let $s=v_1+\cdots+v_{27}$ be the total number of votes cast. Our goal is to determine the smallest possible $s$
Candidate $i$ got $\frac{v_i}s$ of the votes, hence the percentage of votes they received is $\frac{100v_i}s$ . The condition in the problem statement says that $\forall i: \frac{100v_i}s + 1 \leq v_i$ .  ( $\forall$ means "for all", so this means "For all $i$ $\frac{100v_i}s + 1 \leq v_i$ is true")
Obviously, if some $v_i$ would be $0$ or $1$ , the condition would be false. Thus $\forall i: v_i\geq 2$ . We can then rewrite the above inequality as $\forall i: s\geq\frac{100v_i}{v_i-1}$
If for some $i$ we have $v_i=2$ , then from the inequality we just derived we would have $s\geq 200$
If for some $i$ we have $v_i=3$ , then $s\geq 150$
And if for some $i$ we have $v_i=4$ , then $s\geq \frac{400}3 = 133\frac13$ , and hence $s\geq 134$
Is it possible to have $s<134$ ? We just proved that to have such $s$ , all $v_i$ have to be at least $5$ . But then $s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135$ , which is a contradiction. Hence the smallest possible $s$ is at least $134$
Now consider a situation where $26$ candidates got $5$ votes each, and one candidate got $4$ votes. In this situation, the total number of votes is exactly $134$ , and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is $s=\boxed{134}$

Solution:
Let there be $N$ members of the committee.
Suppose candidate $n$ gets $a_n$ votes.
Then $a_n$ as a percentage out of $N$ is $100\frac{a_n}{N}$ . Setting up the inequality $a_n \geq 1 + 100\frac{a_n}{N}$ and simplifying, $a_n \geq \lceil(\frac{N}{N - 100})\rceil$ (the ceiling function is there because $a_n$ is an integer.
Note that if we set all $a_i$ equal to $\lceil(\frac{N}{100 - N})\rceil$ we have $N \geq 27\lceil(\frac{N}{100 - N})\rceil$ . Clearly $N = 134$ is the least such number that satisfies this inequality. Now we must show that we can find suitable $a_i$ . We can let 26 of them equal to $5$ and one of them equal to $4$ . Therefore, $N = \boxed{134}$ is the answer.
- whatRthose

Solution: Let $n$ be the total number of people in the committee, and $a_i$ be the number of votes candidate $i$ gets where $1 \leq i \leq 27$ .  The problem tells us that \[\frac{100a_i}{n} \leq a_i - 1 \implies 100a_i \leq na_i - n \implies a_i \geq \frac{n}{n-100}.\] Therefore, \[\sum^{27}_{i=1} a_i = n \geq \sum^{27}_{i=1} \frac{n}{n-100} = \frac{27n}{n-100},\] and so $n(n-127) \geq 0 \implies n \geq 127$ .  Trying $n = 127$ , we get that \[a_i \geq \frac{127}{27} \approx 4.7 \implies a_i \geq 5 \implies \sum^{27}_{a_i} a_i \geq 5 \cdot 27 = 135 \geq 127,\] a contradiction.  Bashing out a few more, we find that $\boxed{134}$ works perfectly fine.


Answer: 40.0

The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the
area of the rectangle?
$\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$

Solution:
The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle.
Let $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rhombus whose diagonals have lengths $AC=4\sqrt{5}$ and $BD=2\sqrt{5}.$ It follows that the dimensions of the rectangle are $4\sqrt{5}$ and $2\sqrt{5},$ so the area of the rectangle is $4\sqrt{5}\cdot2\sqrt{5}=\boxed{40}.$

Solution:
If a rectangle has area $K,$ then the area of the quadrilateral formed by its midpoints is $\frac{K}{2}.$
Define points $A,B,C,$ and $D$ as Solution 1 does. Since $A,B,C,$ and $D$ are the midpoints of the rectangle, the rectangle's area is $2[ABCD].$ Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $\overline{AB}\parallel\overline{CD}.$ As the parallelogram's height from $D$ to $\overline{AB}$ is $4$ and $AB=5,$ its area is $4\cdot5=20.$ Therefore, the area of the rectangle is $20\cdot2=\boxed{40}.$ ~Fruitz


Answer: 10.5

The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is
$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad  \textbf{(E) }20.5$

Solution: The answer we are looking for can be expressed as $\dfrac{100a+10b+c}{a+b+c}$ . This is equivalent to $1 + \dfrac{99a+9b}{a+b+c}$ . Because we are trying to minimize our solution, we set $c$ $9$ , so we have $1 + \dfrac{99a+9b}{a+b+9}$ . This is equal to $1 + \dfrac{9a+9b+81}{a+b+9} + \dfrac{90a-81}{a+b+9}$ , which simplifies to $10+ \dfrac{90a-81}{a+b+9}$ . Since each digit is unique, we set $b$ to $8$ , leaving us with $10 + \dfrac{90a-81}{a+17}$ . Clearly, $a$ should be minimized, so $a = 1$ and our answer is \[\boxed{10.5}.\]


Answer: 125.0

The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles , of which $s$ are squares .  The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution:
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$ . Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.
For $s$ , there are $8^2$ unit squares $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$
Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$ , and $m+n=\boxed{125}$

Solution:
First, to find the number of squares, we can look case by case by the side length of the possible squares on the checkerboard. We see that there are $8^2$ ways to place a $1$ $1$ square and $7^2$ for a $2$ $2$ square. This pattern can be easily generalized and we see that the number of squares is just $\sum^8_{i=1}{i^2}$ . This can be simplified by using the well-known formula for the sum of consecutive squares $\frac{n(n+1)(2n+1)}{6}$ to get $204$
Then, to find the number of rectangles, first note that a square falls under the definition of a rectangle. We can break up the rectangles into cases for the length x width. As we note down the cases for $1$ $1, 1$ $2 , 2$ $1, 2$ $2,...,$ we see they are respectively $8$ $8, 8$ $7, 7$ $8, 7$ $7, ...$ . We can quickly generalize this pattern to basically just ${\sum^8_{i=1}{i}}\cdot{\sum^8_{i=1}{i}}$ . This gets us ${(\frac{9\cdot8}{2})}^2,$ which is just $1296.$
Now, to calculate the ratio of $s/r,$ we divide $204$ by $1296$ to get a simplified fraction of $\frac{17}{108}.$
Thus, our answer is just $s + r = 17+108 = \boxed{125}$ ~MathWhiz35


Answer: 4.0

The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,2,3$ .  Then $A+B=$
\[\begin{tabular}{|c|c|c|}\hline 1 & &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution:
The square connected both to 1 and 2 cannot be the same as either of them, so must be 3.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 &\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The last square in the top row cannot be either 1 or 3, so it must be 2.
\[\begin{tabular}{|c|c|c|}\hline 1 & 3 & 2\\ \hline & 2 & A\\ \hline & & B\\ \hline\end{tabular}\]
The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\boxed{4}$


Answer: 7.0

The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n$

Solution:
$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$
$=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}$
$=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}$
$=\frac{\log{2000}}{\log{2000^6}}$
$=\frac{\log{2000}}{6\log{2000}}$
$=\frac{1}{6}$
Therefore, $m+n=1+6=\boxed{007}$

Solution:
Alternatively, we could've noted that, because $\frac 1{\log_a{b}} = \log_b{a}$
\begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= 2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }\\ &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ &={\log_{2000^6}{4^2 \cdot 5^3}}\\ &={\log_{2000^6}{2000}}\\ &= {\frac{1}{6}}.\end{align*}
Therefore our answer is $1 + 6 = \boxed{007}$

Solution: We know that $2 = \log_4{16}$ and $3 = \log_5{125}$ , and by base of change formula, $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$ . Lastly, notice $\log a + \log b = \log ab$ for all bases. \begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = \log_{2000^6}{16} + \log_{2000^6}{125} = \log_{2000^6}{2000} = \frac16 \implies \boxed{007}

Solution: \[\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}\] \[= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}\] \[= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5\] \[= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x\] \[\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}\] \[\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5\] \[\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}\] \[\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}\] \[\implies x = \frac{1}{6}\] \[\implies m + n = \boxed{007}\]


Answer: 46.0

The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ have this property?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$

Solution:
We take cases on the thousands digit, which must be either $1$ or $2$ :
If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \dots + 9 = 45$ possible pairs $(b, c)$ , and each pair uniquely determines the value of $d$ , so we get $45$ numbers with the given property.
If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property.
Therefore, the number of numbers with the property is $45 + 1 = \boxed{46}$

Solution:
Let's start with the case that starts with $200$ . We have only one number, which is $2002$ . If we look at the $1900s$ , we have no solutions because $1+9 = 10$ , and because we can only use digits from $1$ through $9$ , it is impossible. If we looks at the $1800s$ , we do have one solution, which is $1809$ . If we look a the $1700s$ , we have $2$ solutions, namely, $1708$ and $1719$
We can see a pattern here. The pattern is every hundred you go down, you have $1$ more solution. Therefore, we have $1+0+1+2+3+4+5+6+7+8+9$ which is = $\boxed{46}$


Answer: 2.0

The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{2}.\]


Answer: 2.0

Solution: The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{2}.\]


Answer: 1024.0

The number $2017$ is prime.  Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ .  What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$

Solution: Note that $2014\equiv -3 \mod2017$ .  We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{  if n is odd}\] and \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{  if n is even}\] Obviously, $62$ falls in the second category, so our desired value is \[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{1024}\]


Answer: 14.0

The number $25^{64}\cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is
$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35$

Solution:
Taking the root, we get $N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}$
Now, we have $N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}$
Combining the $2$ 's and $5$ 's gives us $(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}$
This is the number $2048$ with a string of sixty-four $0$ 's at the end. Thus, the sum of the digits of $N$ is $2+4+8=14\Longrightarrow\boxed{14}$


Answer: 17.0

The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Solution:
Casework by the units digit $u$ will help organize the answer.
$u=0$ gives no solutions, since no real numbers are divisible by $0$
$u=1$ has $4$ solutions, since all numbers are divisible by $1$
$u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$ ).
$u=3$ has $1$ solution: $33$ $\pm 10$ or $\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$ $\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$ , but those numbers are $3$ and $63$ , which are out of the range of the problem.
$u=4$ has $2$ solutions: $24$ and $44$ .  Adding or subtracting $10$ will kill divisibility by $4$ , since $10$ is not divisible by $4$
$u=5$ has $4$ solutions:  every number ending in $5$ is divisible by $5$
$u=6$ has $1$ solution: $36$ $\pm 10$ or $\pm 20$ will kill divisibility by $3$ , and thus kill divisibility by $6$
$u=7$ has no solutions.  The first multiples of $7$ that end in $7$ are $7$ and $77$ , but both are outside of the range of this problem.
$u=8$ has $1$ solution: $48$ $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$
$u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$
Totalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\boxed{17}$ , which is 17.


Answer: 162.0

The number $6545$ can be written as a product of a pair of positive two-digit numbers.  What is the sum of this pair of numbers?
$\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222$

Solution: The prime factorization of $6545$ is $5\cdot7\cdot11\cdot17  =385\cdot17$ , which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \[5\cdot7=35 \text{,   } 11\cdot17=187 \text{  X}\] \[5\cdot11=55 \text{,   } 7\cdot17=119 \text{  X}\] \[5\cdot17=85 \text{,   } 7\cdot11=77 \text{ }\surd\] \[85+77= \boxed{162}\]


Answer: 3.0

The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$ . Find $a_4$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution:
This problem can be approached similarly to other base number problems.
Since $120 < 695 < 720$ , divide $695$ by $120$ .  The quotient is $5$ and the remainder is $95$ , so rewrite the number as \[695 = 5 \cdot 120 + 95\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$ , so the number can be rewritten as \[695 = 5 \cdot 120 + 3 \cdot 24 + 23\] Repeat the steps to get \[695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1\] The answer is $\boxed{3}$ .  One can also stop at the second step by noting $23 < 24$


Answer: 7.0

The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$

Solution:
From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\pmod {10}.$ We now look for this one:
$9(1)=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$
The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\boxed{7}$

Solution:
This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have \[10x+3 \equiv 1 \pmod 9.\] We subtract the three, getting \[10x \equiv -2 \pmod 9.\] which simplifies to \[10x \equiv 7 \pmod 9.\] However, $9x \equiv 0 \pmod 9,$ so \[10x - 9x \equiv 7 - 0 \pmod 9\] and \[x \equiv 7 \pmod 9.\]
Let the quotient of $9$ in our modular equation be $c,$ and let our desired number be $z,$ so $x=9c+7$ and $z = 10x+3.$ We substitute these values into $z = 10x+3,$ and get \[z = 10(9c+7) + 3\] so \[z = 90c+73.\] As a result, $z \equiv 73 \pmod {90}.$
To prove generalization vigorously, we can let $a$ be the remainder when $z$ is divided by $11.$ Setting up a modular equation, we have \[90c + 73 \equiv a \pmod {11}.\] Simplifying, \[90c+7 \equiv a \pmod {11}\] If $c = 1,$ then we don't have a 2 digit number! Thus, $c=0$ and $a=\boxed{7}$

Solution:
We know that the number has to be one more than a multiple of $9$ , because of the remainder of one, and the number has to be $3$ more than a multiple of $10$ , which means that it has to end in a $3$ . Now, if we just list the first few multiples of $9$ adding one to the number we get: $10, 19, 28, 37, 46, 55, 64, 73, 82, 91$ . As we can see from these numbers,  the only one that has a three in the denominator is $73$ , thus we divide $73$ by $11$ , getting $6$ $R7$ , hence, $\boxed{7}$ .
-fn106068


Answer: 936.0

The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$

Solution:
We begin by equating the two expressions:
\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]
Squaring both sides yields:
\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]
Since $a$ $b$ , and $c$ are integers, we can match coefficients:
\begin{align*} 2ab\sqrt{6} &= 104\sqrt{6} \\  2ac\sqrt{10} &=468\sqrt{10} \\  2bc\sqrt{15} &=144\sqrt{15}\\  2a^2 + 3b^2 + 5c^2 &=2006 \end{align*}
Solving the first three equations gives: \begin{eqnarray*}ab &=& 52\\  ac &=& 234\\  bc &=& 72 \end{eqnarray*}
Multiplying these equations gives $(abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}$

Solution:
We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\sqrt{2}$ $y=\sqrt{3}$ , and $z=\sqrt{5}$ .  Since
\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]
we attempt to rewrite the radicand in this form:
\[2006+2(52xy+234xz+72yz)\]
Factoring, we see that $52=13\cdot4$ $234=13\cdot18$ , and $72=4\cdot18$ . Setting $p=13$ $q=4$ , and $r=18$ , we see that
\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]
so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$ . Square rooting gives us $13\sqrt{2}+4\sqrt{3}+18\sqrt{5}$ and our answer is $13\cdot4\cdot18=\boxed{936}$


Answer: 10.0

The number $\text{N}$ is between $9$ and $17$ .  The average of $6$ $10$ , and $\text{N}$ could be
$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution:
We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$
From what we know, we can deduce that $25<N+16<33$ , and thus \[8.\overline{3}<\frac{N+16}{3}<11\]
The only answer choice that falls in this range is choice $\boxed{10}$


Answer: 929.0

The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

Solution:
Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\leq f<1$ and $x=w+f.$
We rewrite the given equation as \[w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)\] Since $a\cdot(w+f)^2\geq0,$ it follows that $w\cdot f\geq0,$ from which $w\geq0.$
We expand and rearrange $(1)$ as \[af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(2)\] which is a quadratic with either $f$ or $w.$
For simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have \[f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)\] If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\geq1:$
Recall that $0\leq f<1,$ so $\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.$ From the discriminant, we require that $0\leq1-4a<1,$ or \[0<a\leq\frac14. \hspace{54mm}(4)\]
We consider each part of $0\leq f<1$ separately:
Now, we express $x$ in terms of $w$ and $k:$ \[x=w+f=w+wk=w(1+k).\] The sum of all solutions to the original equation is \[\sum_{w=1}^{W}w(1+k)=(1+k)\cdot\sum_{w=1}^{W}w=(1+k)\cdot\frac{W(W+1)}{2}=420. \hspace{10mm}(\bigstar)\] As $1+k<1+\frac1W,$ we conclude that $1+k$ is slightly above $1$ so that $\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(\bigstar)$ produces $k=\frac{1}{29},$ which satisfies $\frac{1}{W+1}\leq k<\frac1W,$ as required.
Finally, we solve for $a$ in $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:$ \begin{align*} \frac{1}{29}&=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a} \\ \frac{2}{29}a&=1-2a-\sqrt{1-4a} \\ \frac{60}{29}a-1&=-\sqrt{1-4a} \\ \frac{60^2}{29^2}a^2-\frac{120}{29}a+1&=1-4a \\ \frac{60^2}{29^2}a^2-\frac{4}{29}a&=0 \\ a\left(\frac{60^2}{29^2}a-\frac{4}{29}\right)&=0. \end{align*} Since $a>0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which \[a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.\] The answer is $29+900=\boxed{929}.$

Solution:
Let $x_n$ be a root in the interval $(n,n+1)$ . In this interval, $\lfloor x_n \rfloor = n$ and $\{x_n\}=x_n-n$ , so we must have $ax_n^2 = nx_n-n^2$ , i.e., $ax_n^2-nx_n+n^2=0$ . We can homogenize this equation by setting $x_n=n\zeta$ ; then $x_1=\zeta$ , and $\zeta$ is a root of $a\zeta^2-\zeta+1=0$
Suppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\ldots , N$ \[n < x_n = n\zeta < n+1\] Summing over $n\in \{1,2,\ldots , N\}$ we get \[\tfrac 12 N(N+1) < 420 = \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3)\] From the right inequality we get $27< N$ and from the left one we get $N<29$ . Thus $N=28$ . Using this in the middle equality we get $\zeta = \tfrac{30}{29}$ . Since $\zeta$ satisfies $a\zeta^2-\zeta+1=0$ , we get \[a = \zeta^{-2}(\zeta-1)= \tfrac{29^2}{30^2}\cdot \tfrac 1{29}= \tfrac{29}{900}.\] The answer is $29+900=\boxed{929}.$

Solution:
First note that $\lfloor x\rfloor \cdot \{x\}<0$ when $x<0$ while $ax^2\ge 0\forall x\in \mathbb{R}$ . Thus we only need to look at positive solutions ( $x=0$ doesn't affect the sum of the solutions).
Next, we break $\lfloor x\rfloor\cdot \{x\}$ down for each interval $[n,n+1)$ , where $n$ is a positive integer. Assume $\lfloor x\rfloor=n$ , then $\{x\}=x-n$ . This means that when $x\in [n,n+1)$ $\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2$ . Setting this equal to $ax^2$ gives \[nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}\] We're looking at the solution with the positive $x$ , which is $x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)$ . Note that if $\lfloor x\rfloor=n$ is the greatest $n$ such that $\lfloor x\rfloor \cdot \{x\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ , which is $406$ when $n=28$ , just under $420$ . Checking this gives \begin{align*} \sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)&=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420 \\ \frac{1-\sqrt{1-4a}}{2a}&=\frac{420}{406}=\frac{30}{29} \\ 29-29\sqrt{1-4a}&=60a \\ 29\sqrt{1-4a}&=29-60a \\ 29^2-4\cdot 29^2a&=29^2+3600a^2-120\cdot 29a \\ 3600a^2&=116a \\ a&=\frac{116}{3600}=\frac{29}{900} \implies \boxed{929} ~ktong


Answer: 925.0

The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$ , can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$ , and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$ , where $a > 0$ . Find the base- $10$ representation of $n$

Solution:
We have these equations: $196a+14b+c=225a+15c+b=222a+37c$ .
Taking the last two we get $3a+b=22c$ . Because $c \neq 0$ otherwise $a \ngtr 0$ , and $a \leq 5$ $c=1$
Then we know $3a+b=22$ .
Taking the first two equations we see that $29a+14c=13b$ . Combining the two gives $a=4, b=10, c=1$ . Then we see that $222 \times 4+37 \times1=\boxed{925}$

Solution: We know that $196a+14b+c=225a+15c+b=222a+37c$ . Combining the first and third equations give that $196a+14b+c=222a+37c$ , or \[7b=13a+18c\] The second and third gives $222a+37c=225a+15c+b$ , or \[22c-3a=b\] \[154c-21a=7b=13a+18c\] \[4c=a\] We can have $a=4,8,12$ , but only $a=4$ falls within the possible digits of base $6$ . Thus $a=4$ $c=1$ , and thus you can find $b$ which equals $10$ . Thus, our answer is $4\cdot225+1\cdot15+10=\boxed{925}$

Solution: We're given that $196a+14b+c=225a+15c+b=222a+37c.$ By taking the difference of the first $2$ equalities, we receive $29a+14c=13b.$ Taking $\pmod{13}$ , we receive $3a+c \equiv 0 \pmod{13}.$ We receive the following cases: $(a,c)=(4,1)$ or $(3,4).$ (Note that $(2,7)$ doesn't work since $a,c<6$ by third condition). We can just check these two, and find that $(a,b,c)=(4,10,1),$ and just plugging in $(a,c)$ into the third expression we receive $888+37=\boxed{925}.$


Answer: 417.0

The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits , any of which could be zero.  It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of possible values for $r$

Solution:
The nearest fractions to $\frac 27$ with numerator $1$ are $\frac 13, \frac 14$ ; and with numerator $2$ are $\frac 26, \frac 28 = \frac 13, \frac 14$ anyway. For $\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\frac 27 \approx .28571$ than to $\frac 13 \approx .33333$ or $\frac 14 \approx .25$
Thus $r$ can range between $\frac{\frac 14 + \frac{2}{7}}{2} \approx .267857$ and $\frac{\frac 13 + \frac{2}{7}}{2} \approx .309523$ . At $r = .2678, .3096$ , it becomes closer to the other fractions, so $.2679 \le r \le .3095$ and the number of values of $r$ is $3095 - 2679 + 1 = \boxed{417}$


Answer: 4.0

Solution:
Let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.

From the diagram, $600 = 30x$ , making $x = 20$

It follows that $20 = 5y$ , so $y = \boxed{4}$

Solution:
Another way to do this problem is to realize what makes up the bottommost number. This method doesn't work quite as well for this problem, but in a larger tree, it might be faster. Again, let the value in the empty box in the middle row be $x$ , and the value in the empty box in the top row be $y$ $y$ is the answer we're looking for.

We can write some equations:
$600 = 30x\\ 30 = 6\cdot 5\\ x = 5y$
Now we can substitute into the first equation using the two others:
$600 = (6\cdot5)(5y)\\ 600= 6\cdot5\cdot5\cdot y\\ 600=6\cdot25\cdot y\\ 600 = 150y\\ \boxed{4} = y$


Answer: 2900.0

The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$

Solution: The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$ . Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$ , so $1000 \le \frac 1x = \pi n \le 10000$ . There are $\frac{10000-1000}{\pi} \approx \boxed{2900}$ solutions for $n$ on this interval.

Solution: We know that $x$ belongs to the interval $(0.0001,0.001)$ for $\sin(1/x)$ . We see that when we plug in $x$ into $\sin(1/x)$ , the argument $(1/x)$ is always from the range $(1000, 10000)$ . Therefore, the problem simply asks for all the zeros of $\sin(x)$ with $x$ values between $(1000, 10000)$ . We know that the $x$ values of any sine graph is $\pi(n-1)$ so, we see that values of $n$ are any integer value from $320$ to $3184$ and therefore gives us an answer of approximately $2865$ which is answer $\boxed{2900}$


Answer: 36.0

The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$

Solution:
It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]
So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
\[\frac{9\pi}{\frac{\pi}{4}} = 36\]
So the answer is $\boxed{36}$


Answer: 4850.0

The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$

Solution: Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{4850}$

Solution: The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{4850}$


Answer: 28.0

The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$

Solution: We can rewrite this as $2^{32}5^{25}$ . We can also combine some of the factors to introduce factors of $10$ , whose digit count is simple to evaluate because it simply adds $0$ s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$ . We can see that this final number is $2^7$ with $25$ $0$ s annexed onto it. $2^7=128$ , which has $3$ digits, so the entire number has $25+3=28$ digits, $\boxed{28}$


Answer: 7.0

The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$

Solution: As shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\boxed{7}$ , and we are done.


Answer: 4.0

The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \   \ y=2xy\] is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad  \textbf{(E) }4$

Solution:
If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.
Case 1: $y = 0$
If $y = 0$ , then $x = x^2$ and $0 = 0$
Therefore, $x = 0$ or $x = 1$
This yields 2 solutions
Case 2: $x = \frac{1}{2}$
If $x = \frac{1}{2}$ , this means that $y = y$ , and $\frac{1}{2} = \frac{1}{4} + y^2$
Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$
This yields another 2 solutions.
$2+2 = \boxed{4}$


Answer: 3.0

The number of distinct pairs of integers $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$

Solution: We can simplify $\sqrt{1984}$ to $8\sqrt{31}$ . Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$ . The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$ . Each of these gives distinct pairs of $(x, y)$ , so there are $3$ pairs, $\boxed{3}$


Answer: 84.0

The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$ . If the populations in the years $1994$ $1995$ , and $1997$ were $39$ $60$ , and $123$ , respectively, then the population in $1996$ was
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102$

Solution:
Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.
If $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$
Translating this sentence, $(x - 39) = k(60)$
Similarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$
Since $kx = 63$ , we have $k = \frac{63}{x}$
Plugging this into the first equation, we have:
$(x - 39) = \frac{60\cdot 63}{x}$
$x - 39 = \frac{3780}{x}$
$x^2 - 39x - 3780 = 0$
$(x - 84)(x + 45) = 0$
Since $x>0$ , we must have $x=84$ , and the answer is $\boxed{84}$


Answer: 6.0

The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\frac{9}{2}$ ounces of alcohol. We want $\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\frac{9}{2}$ , to be $30\%$ , or $0.3$ , of the total amount of shaving lotion, $9+N$ ). Solving, we find that \[9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.\] So, the total amount of water we need to add is $\boxed{6}$

Solution:
The concentration of alcohol after adding $n$ ounces of water is $\frac{4.5}{9+n}$
To get a solution of 30% alcohol, we solve $\frac{4.5}{9+n}=\frac{3}{10}$
$45=27+3n$
$18=3n$
$6=n \implies \boxed{6}$


Answer: 4.0

The number of points common to the graphs of $(x-y+2)(3x+y-4)=0 \text{    and     } (x+y-2)(2x-5y+7)=0$ is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$

Solution:
By the Zero Product Property $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation.  Thus, from the first equation, $y = x+2$ or $y =-3x+4$ , and from the second equation, $y=-x+2$ or $y = \frac{2}{5}x + \frac{7}{5}$
If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation.  Since the slopes of the lines are different, none of the lines are parallel.  Thus, there are $2 \cdot 2 = \boxed{4}$ points of intersection in the two graphs.


Answer: 3.0

The number of points equidistant from a circle and two parallel tangents to the circle is:
$\text{(A) } 0\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \infty$

Solution: The distance between the two parallel tangents is the length of the circle's diameter, so the distance from a point that satisfies the conditions and the two tangents is the length of the circle's radius.  From the diagram, there are $\boxed{3}$ points that satisfies the conditions.


Answer: 2.0

The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad  \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution:
Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\frac{17}{a} \geq 2\sqrt{17}$ , with equality only if $a = \sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}$ , and finally $x \geq \sqrt{17}$
Now suppose $x > \sqrt{17}$ . Then $y - \sqrt{17} = \frac{x^{2}+17}{2x} - \sqrt{17} = (\frac{x-\sqrt{17}}{2x})(x-\sqrt{17}) < \frac{1}{2}(x-\sqrt{17})$ , so that $x > y$ . Similarly, we can get $y > z$ $z > w$ , and $w > x$ , and combining these gives $x > x$ , an obvious contradiction.
Thus we must have $x \geq \sqrt{17}$ , but $x \ngtr \sqrt{17}$ , so if $x > 0$ , the only possibility is $x = \sqrt{17}$ , and analogously from the other equations we get $x = y = z = w = \sqrt{17}$ ; indeed, by substituting, we verify that this works.
As for the other case, $x < 0$ , notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\sqrt{17}$ , so that we have $2$ solutions in total, and therefore the answer is $\boxed{2}$


Answer: 1.0

The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ 1 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 3 \qquad  \textbf{(E) }\ \text{more than 4}$

Solution: Notice that $a^0=1, a>0$ . So $2^0=1$ . So $2x^2-7x+5=0$ . Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$ . So this means that the equation has two real solutions. Therefore, select $\boxed{1}$


Answer: 2.0

Solution:
Solution by e_power_pi_times_i
Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$ . Factoring results in $(2x-5)(x-1) = 0$ , so there are $\boxed{2}$ real solutions.


Answer: 3.0

The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:
$\text{(A)  zero} \qquad \text{(B)  one} \qquad \text{(C)  two} \qquad \text{(D)  three} \qquad \text{(E)  greater than 3}$

Solution:
We know that $2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}$ .
We also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$ .
There are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$ , so the answer is $\boxed{3}$


Answer: 880.0

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:
$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$

Solution: We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\frac{5280}{6\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\boxed{880}$


Answer: 3.0

The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$

Solution:
We can write all possible triangles adding up to 12 or less \[(2, 4, 5)=11\] \[(3, 4, 5)=12\] \[(2, 3, 4)=9\]
This leaves $\boxed{3}$ scalene triangles.


Answer: 2.0

The number of sets of two or more consecutive positive integers whose sum is 100 is
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: If the first number of a group of $n$ consecutive numbers is $a$ , the $n^\text{th}$ number is $a+n-1$ .  We know that the sum of the group of numbers is $100$ , so \[\frac{n(2a+n-1)}{2} = 100\] \[2a+n-1=\frac{200}{n}\] \[2a = 1-n + \frac{200}{n}\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$ .  Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$ , so there are $\boxed{2}$ sets of two or more consecutive positive integers whose sum is $100$


Answer: 5.0

The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to
the nearest ten-thousandth of a square inch is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$

Solution: There are 5 significant digits, $1$ $1$ $0$ $2$ , and $5$ . The answer is $\boxed{5}$


Answer: 127.0

The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$

Solution: Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$ .  Solving the inequality results in $y \le 254 \frac{1}{3}$ .  From the two conditions, $y$ can be an odd number from $1$ to $253$ , so there are $127$ solutions where $x$ and $y$ are integers.  The answer is $\boxed{127}$

Solution: We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get $x_0$ $380$ and $y_0$ $1$ . then the general solution of the given diophanitine equation will be $x$ $x_0$ $3t$ and $y$ $y_0$ $2t$ . Since we need only positive integer solutions So we solve $380$ $3t$ $>$ $0$ and $1$ $2t$ $>$ $0$ to get $t$ $>$ $0$ (applying Greatest integer function) also we can clearly see that $t_{(min)}$ $=$ $0$ so,t $<$ $GIF$ $383$ $3$ ). That implies $t$ ranges from $0$ to $127$ . Hence,the correct answer is $127$ $\boxed{127}$


Answer: 8.0

The number of solutions to $\{1,~2\}\subseteq~X~\subseteq~\{1,~2,~3,~4,~5\}$ , where $X$ is a subset of $\{1,~2,~3,~4,~5\}$ is
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }8\qquad  \textbf{(E) }\text{None of these}$

Solution: $X$ has to contain $\{1,~2\}$ , so only $\{3,~4,~5\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$ . With this combinatorics in mind, the answer is simply $2^3=\boxed{8}.$ ~lopkiloinm


Answer: 8.0

The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution:
Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference.  That means the last term is $a+r(n-1)$
We can write an equation on the difference between the last and first term based on the conditions. \[a+r(n-1)-a =10.5\] \[rn-r=10.5\] Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so \[24 + r\frac{n}{2} = 30\] \[nr = 12\] Substituting the value back to a previous equation, \[12-r=10.5\] \[r=1.5\] Substituting to a previous equation again, \[1.5n-1.5=10.5\] \[n=8\] Thus, there are $\boxed{8}$ terms in the arithmetic sequence.


Answer: 5.0

The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution:
Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{5}$


Answer: 2.0

The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

Solution:
We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.

So we have $4$ possible solutions: $-2, \frac{2}{3}, 2,$ and $-\frac{4}{3}$ . Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\frac{4}{3}$ , so there are $2$ solutions, $\boxed{2}$


Answer: 2.0

The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

Solution:
We can factor the second equation to get $c(a+b)=23$ , so we see that $c$ must be a factor of $23$ , and since this is prime $c=1$ or $c=23$ . However, if $c=23$ , then $a+b=1$ , which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives
$ab+b=44$
and
$a+b=23$
From the second equation, $a=23-b$ , and substituting this into the first equation yields $b(23-b)+b=44$ , or $b^2-24b+44=0$ . Factoring this gives $(b-2)(b-22)=0$ , so $b=2$ or $b=22$ . Both of these yield integer solutions for $a$ , giving $a=21$ or $a=1$ , respectively.
Therefore, the only solutions are $(21, 2, 1)$ and $(1, 22, 1)$ , yielding $2$ solutions, $\boxed{2}$


Answer: 6.5

The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:
What is the average of the first and last numbers?
$\textbf{(A)}\ 3.5 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6.5 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$

Solution: From rule 1, the largest number, $12$ , can be second or third. From rule 2, because there are five places, the smallest number $-2$ can either be third or fourth. The median, $6$ can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are $4$ and $9$ , disregarding their order. Their average is $(4+9)/2 = \boxed{6.5}$


Answer: 85.0

The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ are randomly written on the faces of a regular octahedron so that each face contains a different number.  The probability that no two consecutive numbers, where $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution:
It is helpful to consider the cube $ABCDEFGH$ , where the vertices of the cube represent the faces of the octahedron, and the edges of the cube represent adjacent octahedral faces. Each assignment of the numbers $1,2,3,4,5,6,7$ , and $8$ to the faces of the octahedron corresponds to a permutation of $ABCDEFGH$ , and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem asks one to count octagonal circuits that can be formed by eight of these diagonals. Six of the diagonals are edges of tetrahedron $ACFH$ , six are edges of tetrahedron $DBEG$ , and four are $\textit{long}$ , joining a vertex of one tetrahedron to the diagonally opposite point from the other. Notice that each vertex belongs to exactly one long diagonal.
It follows that an octagon cannot have two successive long diagonals. Also notice that an octagonal path can jump from one tetrahedron to the other only along one of the long diagonals. It follows that an octagon must contain either 2 long diagonals separated by 3 tetrahedron edges or 4 long diagonals alternating with tetrahedron edges.
To form an octagon that contains four long diagonals, choose two opposite edges from tetrahedron $ACFH$ and two opposite edges from tetrahedron $DBEG$ . For each of the three ways to choose a pair of opposite edges from tetrahedron $ACFH$ , there are two possible ways to choose a pair of opposite edges from tetrahedron $DBEG$ . There are 6 distinct octagons of this type and $8\cdot 2$ ways to describe each of them, making 96 permutations.
To form an octagon that contains exactly two of the long diagonals, choose a three-edge path along tetrahedron $ACFH$ , which can be done in $4! = 24$ ways. Then choose a three-edge path along tetrahedron $DBEG$ which, because it must start and finish at specified vertices, can be done in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8\cdot 24\cdot 2=384$ permutations of this type.
In all, there are $96 +384 = 480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\tfrac{480}{8!}=\tfrac{1}{84}$ , and $m+n=\boxed{085}$

Solution:
Choose one face of the octahedron randomly and label it with $1$ .  There are three faces adjacent to this one, which we will call A-faces.  There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.
Clearly, the labels for the A-faces must come from the set $\{3,4,5,6,7\}$ , since these faces are all adjacent to $1$ .  There are thus $5 \cdot 4 \cdot 3 = 60$ ways to assign the labels for the A-faces.
The labels for the B-faces and C-face are the two remaining numbers from the above set, plus $2$ and $8$ .  The number on the C-face must not be consecutive to any of the numbers on the B-faces.
From here it is easiest to brute force the $10$ possibilities for the $4$ numbers on the B and C faces:
There is a total of $10$ possibilities.  There are $3!=6$ permutations (more like "rotations") of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$ .  There are $7!=5040$ ways to randomly fill in the rest of the octahedron.  So the probability is $\frac {60}{5040} = \frac {1}{84}$ . The answer is $\boxed{085}$

Solution:
Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits ( $B$ for black and $W$ for white).
Type I: $BB-WWWW-BB$ . There are $4!$ ways to arrange the black vertices and consequently two of the white vertices and $2!$ ways to arrange the other two white vertices. Since the template has a period of $8$ , there are $4!\cdot 2!\cdot 8 = 384$ circuits of type I.
Type II: $B-WW-BB-WW-B$ . There are $4!$ ways to arrange the black vertices and consequently the white vertices. Since the template has a period of $4$ , there are $4! \cdot 4 = 96$ circuits of type II.
Thus, there are $384+96=480$ circuits satisfying the given condition, out of the $8!$ possible circuits. Therefore, the desired probability is $\frac{480}{8!} = \frac{1}{84}$ . The answer is $\boxed{085}$

Solution:
As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices.
The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in $5\cdot4\cdot3=60$ ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.
The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus $\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}$ , from which the answer is $\boxed{085}$

Solution:
The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$
Case 1: Four of the vertical edges are used. $6\cdot 2=12.$
Case 2: Two of the vertical edges are used. $4\cdot 3 \cdot 2\cdot 2=48.$
So, the probability is $\frac{60}{5040}=\frac{1}{84}.$ Therefore, our answer is $\boxed{085}$

Solution:
As with some of the previous solutions, consider the cube formed by connecting the centroids of the faces on the octahedron. We choose a random vertex(hence fixing the diagram), giving us $7!$ ways as our denominator. WLOG, we color this start vertex red, and we color all $3$ vertices adjacent to it blue. We repeat this for the other vertices.
Note that there is a 1-1 correspondence between the number of valid face numberings with rotational symmetry and the number of ways to move a particle to every vertex of the cube and returning to the start vertex with only diagonal moves. We can move along either short diagonals and long diagonals.
Next, note that we can only move from the red vertices to the blue vertices and back with long diagonals(since short diagonals keep us on the color we are currently on). Thus, it is easy to check that the only possible sequences of long and short diagonals are cyclic permutations of $SSSLSSSL$ and $SLSLSLSL$
Case 1: $SSSLSSSL$ .
There are $4$ cyclic permutations, and we can traverse the entirety of the red tetrahedron in $4$ steps in $3!$ ways. Then, after the move to the blue tetrahedron from one of the non-starting red vertices, we realize that our first step cannot be to the blue vertex opposite the starting red vertex. Hence, there are $2$ possibilities for our first step. However, once we make our first move, the path is fixed. This leaves us with $4! \cdot 2$ ways.
Case 2: $SLSLSLSL$ There are $2$ cyclic permutations in this case. After we choose one of the $3$ red vertices to go to from the starting vertex and move to the blue tetrahedron, we are once again left with $2$ choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our choices are fixed. This leaves us with $2 \cdot 3 \cdot 2$ ways.
Finally, summing and dividing by $7!$ gives us $\frac{60}{7!} = \frac{1}{84} \rightarrow \boxed{085}$ , as desired. - Spacesam


Answer: 432.0

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Solution:
Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are
Because the number must have exactly two identical digits, $x\neq y$ $x\neq1$ , and $y\neq1$ . Hence, there are $3\cdot9\cdot8=216$ numbers of this form.
Now suppose that the two identical digits are not $1$ . Reasoning similarly to before, we have the following possibilities:
Again, $x\neq y$ $x\neq 1$ , and $y\neq 1$ . There are $3\cdot9\cdot8=216$ numbers of this form.
Thus the answer is $216+216=\boxed{432}$

Solution:
Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$
If we let the three distinct digits of the sequence be $a, b,$ and $c$ , with $a$ repeated twice, we can make a table with all possible sequences:
\[\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\  \end{tabular}\]
There are $6$ possible sequences.
Next, we can see how many ways we can pick $a$ $b$ , and $c$ . This is $10(9)(8) = 720$ , because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.

Solution:
We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated.
Case 1:
There are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\cdot8\cdot7=504$ numbers.
Case 2:
One of the three numbers must be $1$ , and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$ , and there are $9$ choices (any number except for $1$ ) to pick the other number that is repeated, so a total of $3\cdot9=27$ numbers.
Case 3:
We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times.
When $1$ is repeated $3$ times, then one of the digits is not $1$ . There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\cdot3=27$ numbers.
When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number.
So in Case 3 there are $27+9=36$ numbers
Case 4:
There is only $1$ number: $1111$
There are a total of $1000$ $4$ -digit numbers that begin with $1$ (from $1000$ to $1999$ ), so by complementary counting you get $1000-(504+27+36+1)=\boxed{432}$ numbers.

Solution:
Let us proceed by casework.
Case 1:
We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$ , which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\choose{2}$ $* 9 * 8 = 216$ possibilities for this case.
Case 2:
The last case will be the amount of numbers that have two identical digits thare are $1$ . There are ${3}\choose{1}$ places to pick the $1$ . For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$ .
Summing the two cases, we get $216 + 216 = \boxed{432}$


Answer: 1011.0

The numbers $16$ and $25$ are a pair of consecutive positive squares whose difference is $9$ . How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023$
$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution: Let x be the square root of the smaller of the two perfect squares. Then, $(x+1)^2 - x^2 =x^2+2x+1-x^2 = 2x+1 \le 2023$ . Thus, $x \le 1011$ . So there are $\boxed{1011}$ numbers that satisfy the equation.

Solution: The smallest number that can be expressed as the difference of a pair of consecutive positive squares is $3$ , which is $2^2-1^2$ . The largest number that can be expressed as the difference of a pair of consecutive positive squares that is less than or equal to $2023$ is $2023$ , which is $1012^2-1011^2$ . These numbers are in the form $(x+1)^2-x^2$ , which is just $2x+1$ . These numbers are just the odd numbers from 3 to 2023, so there are $[(2023-3)/2]+1=1011$ such numbers. The answer is $\boxed{1011}$


Answer: 30.0

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$
$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$

Solution: The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15$ . Solving for $a+b$ , we get $a+b=60$ . Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{30}$


Answer: 112.0

The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solution:
Let $A = \log(a)$ and $B = \log(b)$
The first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\text{th}$ term is $nB$
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$
Since the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $k$ th term is $(9k + 4)B$
Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{112}$

Solution:
If $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are in arithmetic progression , then $a^3b^7$ $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression . Therefore,
\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]
Therefore, $a^3b^7=b^{13}$ $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \Rightarrow \boxed{112}$

Solution: Given the first three terms form an arithmetic progression, we have: \[a = \log(a^3b^7)\] \[a+d = \log(a^5b^{12})\] \[a+2d = \log(a^8b^{15}).\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$ \[d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5\] \[d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.\] The desired $12$ th term in the sequence is $a+11d$ , so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$ ): \[a+11d = \log a^3b^7 + 11\log(a^2b^5)\] \[= \log a^3b^7 + \log(a^{22}b^{55})\] \[= \log a^{25}b^{62}.\] The answer must be expressed as $\log(b^n)$ , however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \[\log a^2b^5 = \log a^3b^3\] \[a^2b^5 = a^3b^3\] \[a=b^2.\] Plugging in $a=b^2$ , we have: \[a+11d = \log b^{50}b^{62}\] \[= \log b^{112} \Rightarrow \boxed{112}.\] ~ Jingwei325 $\smiley$


Answer: 2.0

The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$ . The sum of $x, y$ , and $z$ is $100$ . The number y is given by the equation $y = ax - 10$ . Then a is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$

Solution:
In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$ , and $z$ would have $5$ ). One part, after some algebra, equals $10$ , so $x$ $y$ , and $z$ are $20$ $30$ , and $50$ , respectively.
We can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$ , and then solve to get $a = \boxed{2}$


Answer: 3.0

The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime? [asy] /* Made by MRENTHUSIASM */ size(175);  void ds(pair p) { 	filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); }  ds((0.5,4.5)); ds((1.5,3.5)); ds((3.5,1.5)); ds((4.5,0.5));  add(grid(7,7,grey+linewidth(1.25)));  int adj = 1; int curUp = 2; int curLeft = 4; int curDown = 6;  label("$1$",(3.5,3.5));  for (int len = 3; len<=3; len+=2) { 	for (int i=1; i<=len-1; ++i)     		{ 			label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i));     		label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj));      		label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i));     		++curDown;     		++curLeft;     		++curUp; 		} 	++adj;     curUp = len^2 + 1;     curLeft = len^2 + len + 2;     curDown = len^2 + 2*len + 3; }  draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy] $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: We fill out the grid, as shown below:  From the four numbers that appear in the shaded squares, $\boxed{3}$ of them are prime: $19,23,$ and $47.$

Solution: Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below:  From the four numbers that appear in the shaded squares, $\boxed{3}$ of them are prime: $19,23,$ and $47.$