Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$ $B$ , and $C$ are adjacent to vertex $D$ . The perpendicular distance from $D$ to the plane containing $A$ $B$ , and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]
$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$

Solution: Let $x$ be the desired distance. Recall that the volume of a pyramid is given by $\frac{1}{3}\cdot h \cdot B$ , where $B$ is the area of the base and $h$ is the height. Consider pyramid $ABCD$ . Letting $ABC$ be the base, the volume of $ABCD$ is given by $\frac{1}{3} \cdot x \cdot [ABC]$ , but if we let $BCD$ be the base, the volume is given by $\frac{1}{3} \cdot [BCD]\cdot [AD] = \frac{1}{3} \cdot [\frac{1}{2} \cdot 4 \cdot 4] \cdot 3 = 8$ . Clearly, these two volumes must be equal, so we get the equation $\frac{1}{3}\cdot x \cdot[ABC]=8$ . Thus, to find $x$ , we just need to find $[ABC]$
By the Pythagorean Theorem, $AB=\sqrt{AD^2+DB^2}=5$ $AC=\sqrt{AD^2+DC^2}=5$ $BC=\sqrt{BD^2+DC^2}=4\sqrt{2}$
The altitude to $BC$ in triangle $ABC$ has length $\sqrt{AC^2-\frac{BC}{2}^2}=\sqrt{17}$ , so $[ABC]=\frac{1}{2}\cdot 4\sqrt{2} \cdot \sqrt{17} = 2\sqrt{34}$ . Then $x=\frac{24}{[ABC]}=\frac{24}{2\sqrt{34}}=\frac{6\sqrt{34}}{17}$ or about $2.1$ . The answer is $\boxed{2.1}$

Answer: C 2.1
Problem: If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution: Working with the second part of the problem first, we know that $3n$ has $30$ divisors.  We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.
$30 \rightarrow p^{29}$
$2 \cdot 15 \rightarrow pq^{14}$
$3\cdot 10 \rightarrow p^2q^9$
$5\cdot 6 \rightarrow p^4q^5$
$2\cdot 3\cdot 5 \rightarrow pq^2r^4$
The variables $p, q, r$ are different prime factors, and one of them must be $3$ .  We now try to count the factors of $2n$ , to see which prime factorization is correct and has $28$ factors.
In the first case, $p=3$ is the only possibility.  This gives $2n = 2\cdot p^{28}$ , which has $2\cdot {29}$ factors, which is way too many.
In the second case, $p=3$ gives $2n = 2q^{14}$ .  If $q=2$ , then there are $16$ factors, while if $q\neq 2$ , there are $2\cdot 15 = 30$ factors.
In the second case, $q=3$ gives $2n = 2p3^{13}$ .  If $p=2$ , then there are $3\cdot 13$ factors, while if $p\neq 2$ , there are $2\cdot 2 \cdot 13$ factors.
In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$ .  If $q=2$ , then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$ , there are $2\cdot 2\cdot 10$ factors.
In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$ .  If $p=2$ , then there are $4\cdot 9$ factors, while if $p \neq 2$ , there are $2\cdot 3\cdot 9$ factors.
In the fourth case, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$ .  If $q=2$ , then there are $7\cdot 4= 28$ factors.  This is the factorization we want.
Thus, $3n = 3^4 \cdot 2^5$ , which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$ , which has $4\cdot 7 = 28$ factors.
In this case, $6n = 3^4\cdot 2^6$ , which has $5\cdot 7 = 35$ factors, and the answer is $\boxed{35}$

Answer: C 35
Problem: If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution: Because $2n$ has $28$ factors and $3n$ has $30$ factors, we should rewrite the number $n = 2^{e_1}3^{e_2}... p_n^{e_n}$ As the formula for the number of divisors for such a number gives: $(e_1+1)(e_2+1)... (e_n+1)$ We plug in the variations we need to make for the cases $2n$ and $3n$ $2n$ has $(e_1+2)(e_2+1)(e_3+1)... (e_n+1) = 28$ $3n$ has $(e_1+1)(e_2+2)(e_3+1)...(e_n+1) = 30$
If we take the  top and divide by the bottom, we get the following equation: $\frac{(e_1+2)(e_2+1)}{(e_1+1)(e_2+2)} = \frac{14}{15}$ . Letting $e_1=x$ and $e_2 = y$ for convenience and expanding this out gives us: $xy-13x+16y+2=0$
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into: $(x+16)(y-13) +2 + 208 = 0$ or $(x+16)(y-13) = - 210$
As we want to be dealing with rather reasonable numbers for $x$ and $y$ , we try to make the $x+16$ term the slightly larger term and the $y-13$ term the slightly smaller term. This effect is achieved when $x+16 = 21$ and $y-13 = -10$ . Therefore, $x = 5, y = 3$ . We get that this already satisfies the requirements for the number we are looking for, and we take $(5+2)(3+2) = \boxed{35}$

Answer:  35
Problem: If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution: Let $\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,$ be the prime factorization of $\, n$ . Then the number of positive divisors of $\, n \,$ is $\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \;$ . In view of the given information, we have \[28 = (e_1 + 2)(e_2 + 1)P\] and \[30 = (e_1 + 1)(e_2 + 2)P,\] where $\, P = (e_3 + 1)(e_4 + 1) \cdots \;$ . Subtracting the first equation from the second, we obtain $\, 2 = (e_1 - e_2)P, \,$ so either $\, e_1 - e_2 = 1 \,$ and $\, P = 2, \,$ or $\, e_1 - e_2 = 2 \,$ and $\, P = 1$ . The first case yields $\, 14 = (e_1 + 2)e_1 \,$ and $\, (e_1 + 1)^2 = 15$ ; since $\, e_1 \,$ is a nonnegative integer, this is impossible. In the second case, $\, e_2 = e_1 - 2 \,$ and $\, 30 = (e_1 + 1)e_1, \,$ from which we find $\, e_1 = 5 \,$ and $\, e_2 = 3$ . Thus $\, n = 2^5 3^3, \,$ so $\, 6n = 2^6 3^4 \,$ has $\, (6+1)(4+1) = \boxed{35} \,$ positive divisors.

Answer:  35
Problem: A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5.  The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers.  Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409$

Solution: In hexagon $ABCDEF$ , let $AB=BC=CD=3$ and let $DE=EF=FA=5$ . Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$ . Similarly, $\angle CBE =\angle CFE=60^{\circ}$ . Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$ $Q$ that of $\overline{BE}$ and $\overline{AD}$ , and $R$ that of $\overline{CF}$ and $\overline{AD}$ . Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); real angleUnit = 15; draw(Circle(origin,1)); pair D = dir(22.5); pair C = dir(3*angleUnit + degrees(D)); pair B = dir(3*angleUnit + degrees(C)); pair A = dir(3*angleUnit + degrees(B)); pair F = dir(5*angleUnit + degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); draw(A--D^^B--E^^C--F); label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); [/asy]
Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$ . Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \[\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.\] It follows that \[\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$

Answer:  409
Problem: A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5.  The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers.  Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409$

Solution: All angle measures are in degrees.
Let the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ .  Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ .  We look for $AD$
Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\angle BCA=\angle BAC$ .  Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ $\widehat{AB}$ means arc AB).  Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$
Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$ .  Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$ .  Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$ $\angle CDE=120$ as well.
Now I focus on triangle $FAB$ .  By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$ , so $BF=7$ .  Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$ , we can now use the Law of Sines to get: \[\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.\]
Now I focus on triangle $AFD$ $\angle AFD=3\phi$ and $\angle ADF=\theta$ , and we are given that $AF=5$ , so \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.\] We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$ , but we need to find $\sin{3\phi}$ .  Using various identities, we see \begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*} Returning to finding $AD$ , we remember \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.\] Plugging in and solving, we see $AD=\frac{360}{49}$ . Thus, the answer is $360 + 49 = 409$ , which is answer choice $\boxed{409}$

Answer: E 409
Problem: A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5.  The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers.  Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409$

Solution: Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$ . Major arc $\overarc{AB}$ $=240^{\circ}$ $\angle ACB = 120^{\circ}$
By the Law of Cosine, $AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
$\frac{\angle AOB}{2} = 60^{\circ}$ , therefore, $r = \frac{\frac{AB}{2}}{\sin 60^{\circ}} = \frac{\frac{7}{2}}{ \frac{\sqrt{3}}{2} } = \frac{7\sqrt{3}}{3}$
$\sin \frac{\theta}{2} = \frac{\frac32}{r} = \frac{3}{2r} = \frac{3}{2 \cdot \frac{7\sqrt{3}}{3}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \theta}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \theta}{2} = 3 \cdot \sin \frac{\theta}{2} - 4 \cdot \sin(\frac{ \theta}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$

Answer: E 409
Problem: A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5.  The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers.  Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409$

Solution: Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$
$\sin \frac{\alpha}{2} = \frac{\frac32}{r}$ $\sin \frac{\alpha}{2} = \frac{3}{2r}$
$\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}$ $\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}$
$\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35$ $5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})$
$5 \cdot \sin \frac{\alpha}{2} = 3 ( \sin 60^{\circ} \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \cos 60^{\circ}) = 3 ( \frac{\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2} - \frac12 \cdot \sin \frac{\alpha}{2})$
$13 \cdot \sin \frac{\alpha}{2} = \frac{3\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2}$
Let $\sin \frac{\alpha}{2} = a$ $\cos \frac{\alpha}{2} = \sqrt{1-a^2}$ $13a = \frac{3\sqrt{3}}{2} \cdot  \sqrt{1-a^2}$
$169a^2 = 27-27a^2$ $196a^2=27$ $\sin \frac{\alpha}{2} = a = \sqrt{\frac{27}{196}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \alpha}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \alpha}{2} = 3 \cdot \sin \frac{\alpha}{2} - 4 \cdot \sin(\frac{ \alpha}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$

Answer: E 409
Problem: A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5.  The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers.  Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409$

Solution: Note that major arc $\overarc{AE}$ is two thirds of the circumference, therefore, $\angle AFE = 120^{\circ}$
By the Law of Cosine, $AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
By the Ptolemy's theorem of quadrilateral $ABDE$ $AD \cdot BE = AB \cdot DE + BD \cdot AE$ $AD = BE$ $AD^2= 3 \cdot 5 + 7^2 = 64$ $AD = 8$
By the Ptolemy's theorem of quadrilateral $ABCD$ $AC \cdot BD = BC \cdot AD + AB \cdot CD$ $7AC = 3 \cdot 8 + 3 \cdot 5 = 39$ $AC = \frac{39}{7}$
By the Ptolemy's theorem of quadrilateral $ABCF$ $AC \cdot BF = AB \cdot CF + BC \cdot AF$ $AC = BF$ $(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3$ $CF = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$

Answer: E 409
Problem: The addition below is incorrect. The display can be made correct by changing one digit $d$ , wherever it occurs, to another digit $e$ . Find the sum of $d$ and $e$
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$

Solution: If we change $0$ , the units column would  be incorrect.
If we change $1$ , then the leading $1$ in the sum would be incorrect.
However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.
Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$ .  This does not match up to the bottom.
Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$ .  This is the number on the bottom if the $2$ s were changed to $6$ s.
Thus $d=2$ and $e=6$ . so $d+e= 8 \boxed{8}$

Answer: C 8
Problem: If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$

Solution: Substituting $x = -3$ , we get \[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\] But $f(-3) = 2$ , so $81a - 9b + 2 = 2$ , which means $81a - 9b = 0$ . Then \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]

Answer:  8
Problem: pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$ , although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$

Solution: Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$ , to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{745}$

Answer: E 745
Problem: A list of five positive integers has mean $12$ and range $18$ . The mode and median are both $8$ . How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$

Solution: Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$ , at least two of the five numbers must be $8$ . The last number we denote as $b$
Then their average is $\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26$ . Clearly $a \le 8$ . Also we have $b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a$ . Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$ ; listing shows that all such values work. The answer is $\boxed{6}$

Answer: B 6
Problem: For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$

Solution: $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$ , we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$ , we count the exact same case when we put those prime factors which were in $b$ into $c$
Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{40}.$

Answer: C 40
Problem: For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$

Solution: The prime factorization of $2310$ is $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.$ Therefore, we have the equation \[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\] where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$ s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.
Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]

Answer:  40
Problem: A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$

Solution: Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$ . The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$
The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$ . Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$
There are six cubes such that $x+y+z=2$ : permutations of $(1,1,0)$ and $(2,0,0)$ Symmetrically, there are six cubes such that $x+y+z=4$ Finally, there are seven cubes such that $x+y+z=3$ : permutations of $(2,1,0)$ and the central cube $(1,1,1)$
That gives a total of $\boxed{19}$ intersected cubes.

Answer:  19