Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

Solution: Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt));  path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0)--(2,-2)--(0,-2)--cycle; path f4=(-3,-3)--(-3,-1)--(-3,1)--(-3,3)--(-1,3)--(1,3)--(3,3)--(3,1)--(3,-1)--(3,-3)--(1,-3)--(-1,-3)--cycle;  draw((-3,-3)--(3,3)); draw((-3,3)--(3,-3)); draw(f1,mred); draw(f2,mblue); draw(f3,mred); draw(f4,mblue); [/asy]
In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square. 
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$

Answer:  761
Problem: The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

Solution: When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761}

Answer:  761
Problem: The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

Solution: The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$ . The first finite differences go $4, 8, 12, 16, \dots$ . The second finite differences go $4, 4, 4, \dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we already know $d_1 = 1$ $d_2 = 3$ , and $d_3 = 13$ . Thus, \[a + b + c = d_1 = 1\] \[4a + 2b + c = d_2 = 3\] \[9a + 3b + c = d_3 = 13\] Solving this system for $a$ $b$ , and $c$ gives $a = 2$ $b = -2$ $c = 1$ . Finally, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{761}$

Answer: E 761
Problem: Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution: From $|a-b|=2$ we get that $a=b\pm 2$
Similarly, $b=c\pm3$ and $c=d\pm4$
Substitution gives $a=d\pm 4\pm 3\pm 2$ . This gives $|a-d|=|\pm 4\pm 3\pm 2|$ . There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$
$4+3+2=9$
$4+3-2=5$
$4-3+2=3$
$-4+3+2=1$
$4-3-2=-1$
$-4+3-2=-3$
$-4-3+2=-5$
$-4-3-2=-9$
Therefore, the only possible values of $|a-d|$ are $9$ $5$ $3$ , and $1$ . Their sum is $\boxed{18}$

Answer: D 18
Problem: Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution: If we add the same constant to all of $a$ $b$ $c$ , and $d$ , we will not change any of the differences. Hence we can assume that $a=0$
From $|a-b|=2$ we get that $|b|=2$ , hence $b\in\{-2,2\}$
If we multiply all four numbers by $-1$ , we will not change any of the differences. (This is due to the fact that we are calculating $|d|$ at the end ~Williamgolly) WLOG we can assume that $b=2$
From $|b-c|=3$ we get that $c\in\{-1,5\}$
From $|c-d|=4$ we get that $d\in\{-5,1,3,9\}$
Hence $|a-d|=|d|\in\{1,3,5,9\}$ , and the sum of possible values is $1+3+5+9 = \boxed{18}$

Answer: D 18
Problem: Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution: Let \[\begin{cases} |a-b| = 2, \\ |b-c| = 3, \\ |c-d| = 4, \\ |d-a| = X. \\ \end{cases}\] Note that we have \[\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies\] \[\pm X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies X \pm 2 \pm 3 \pm 4 = 0,\] from which it follows that \[X = \pm 4 \pm 3 \pm 2.\]
Note that $X = |a-d|$ must be positive however, the only arrangements of $+$ and $-$ signs on the RHS which make $X$ positive are \[(4, 3, 2) \ \ \implies \ \ X = 9\] \[(4, 3, -2) \ \ \implies \ \ X = 5\] \[(4, -3, 2) \ \ \implies \ \ X = 3\] \[(-4, 3, 2) \ \ \implies \ \ X = 1.\] (There are no cases with $2$ or more negative as $4-3-2<0.$
Thus, the answer is \[1+3+5+9 = \boxed{18}.\]

Answer: D 18
Problem: Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution: Let $a=0$
Then $\begin{cases} b=2 \\ b=-2 \\ \end{cases}$
Thus $\begin{cases} c=5 \\ c=-1 \\ c=1 \\ c=-5 \\ \end{cases}$
Therefore $\begin{cases} d=9 \\ d=1 \\ d=3 \\ d=-5 \\ d=5 \\ d=-3 \\ d=-1 \\ d=-9 \\ \end{cases}$
So $|a-d|=1, 9, 3, 5$ and the sum is $\boxed{18}$

Answer: D 18
Problem: At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$

Solution: Out of the soccer players, $40\%$ swim. As the soccer players are $60\%$ of the whole, the swimming soccer players are $0.4 \cdot 0.6 = 0.24 = 24\%$ of all children.
The non-swimming soccer players then form $60\% - 24\% = 36\%$ of all the children.
Out of all the children, $30\%$ swim. We know that $24\%$ of all the children swim and play soccer, hence $30\%-24\% = 6\%$ of all the children swim and don't play soccer.
Finally, we know that $70\%$ of all the children are non-swimmers. And as $36\%$ of all the children do not swim but play soccer, $70\% - 36\% = 34\%$ of all the children do not engage in any activity.
A quick summary of what we found out:
Now we can compute the answer. Out of all children, $70\%$ are non-swimmers, and again out of all children $36\%$ are non-swimmers that play soccer. Hence the percent of non-swimmers that play soccer is $\frac{36}{70} \approx 51\% \Rightarrow \boxed{51}$

Answer: D 51
Problem: At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$

Solution: Let us set the total number of children as $100$ . So $60$ children play soccer, $30$ swim, and $0.4\times60=24$ play soccer and swim.
Thus, $60-24=36$ children only play soccer.
So our numerator is $36$
Our denominator is simply $100-\text{Swimmers}=100-30=70$
And so we get $\frac{36}{70}$ which is roughly $51.4\% \Rightarrow \boxed{51}$

Answer: D 51
Problem: At Jefferson Summer Camp, $60\%$ of the children play soccer, $30\%$ of the children swim, and $40\%$ of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
$\mathrm{(A)}\ 30\% \qquad \mathrm{(B)}\ 40\% \qquad \mathrm{(C)}\ 49\% \qquad \mathrm{(D)}\ 51\% \qquad \mathrm{(E)}\ 70\%$

Solution: WLOG, let the total number of students be $100$ . Draw a venn diagram with 2 circles encompassing these 4 regions:
Non-soccer players, non-swimmers: 34 people
Soccer players, non-swimmers: 36 people
Soccer players, swimmers: 24 people
Non-soccer players, swimmers: 6 people.
Hence the answer is $\frac{36}{70}=\frac{18}{35}$ . We know this is a little bit larger than $\frac 12$ because $\frac{17.5}{35}=\frac 12$ $\boxed{51}$

Answer: D 51
Problem: Circle $A$ has radius $100$ . Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$ . The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?
$\mathrm{(A)}\ 4\ \qquad \mathrm{(B)}\ 8\ \qquad \mathrm{(C)}\ 9\ \qquad \mathrm{(D)}\ 50\ \qquad \mathrm{(E)}\ 90\ \qquad$

Solution: The circumference of circle $A$ is $200\pi$ , and the circumference of circle $B$ with radius $r$ is $2r\pi$ . Since circle $B$ makes a complete revolution and ends up on the same point , the circumference of $A$ must be a multiple of the circumference of $B$ , therefore the quotient must be an integer.
Thus, $\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}$
Therefore $r$ must then be a factor of $100$ , excluding $100$ because the problem says that $r<100$ $100\: =\: 2^2\; \cdot \; 5^2$ . Therefore $100$ has $(2+1)\; \cdot \; (2+1)\;$ factors*. But you need to subtract $1$ from $9$ , in order to exclude $100$ . Therefore the answer is $\boxed{8}$

Answer:  8
Problem: Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$

Solution: Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1\mathrm km/min = 60\mathrm km/h$ .) This solves to $v_A=45$ and $v_L=15$
As the distance decreases at a rate of $1$ kilometer per minute, after $5$ minutes the distance between them will be $20-5=15$ kilometers.
From this point on, only Lauren will be riding her bike. As there are $15$ kilometers remaining and $v_L=15$ , she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $5+60 = \boxed{65}$

Answer:  65
Problem: Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$

Solution: Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is $\frac{3}{4} \textbf{km/min}$ , and Lauren's $\frac{1}{4} \textbf{km/min}$ . Therefore, after 5 minutes, Andrea will have biked $\frac{3}{4} \cdot 5 = \frac{15}{4}km$
In all, Lauren will have to bike $20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}km$ . Because her speed is $\frac{1}{4} \textbf{km/min}$ , the time elapsed will be $\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{65}$

Answer: D 65
Problem: Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
$\mathrm{(A)}\ 20 \qquad \mathrm{(B)}\ 30 \qquad \mathrm{(C)}\ 55 \qquad \mathrm{(D)}\ 65 \qquad \mathrm{(E)}\ 80$

Solution: Since the distance between them decreases at a rate of $1$ kilometer per minute when they are both biking, their combined speed is $1$ kilometer per minute. Andrea travels three times as fast as Lauren, so they travel at speeds of $\frac{3}{4}$ kilometers per minute and $\frac{1}{4}$ kilometers per minute, respectively.
After $5$ minutes, the distance between them will have decreased by $5$ kilometers, so they will be $20-5 = 15$ kilometers apart when Andrea stops. Then, Lauren will take $\frac{15}{\frac{1}{4}} = 15*4=60$ more minutes to reach Andrea.
They started to bike $5$ minutes before Andrea stopped, so the total time Lauren passed from the time they started biking to the time Lauren reached Andrea is $60+5=65$ minutes. Hence, the answer is $\boxed{65}$ . ~azc1027

Answer: B 65
Problem: Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ .  Diagonals $AC$ and $BD$ intersect at $E$ $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas.  What is $AE$
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$

Solution: The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\boxed{6}$

Answer: A 6
Problem: Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ .  Diagonals $AC$ and $BD$ intersect at $E$ $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas.  What is $AE$
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$

Solution: Using the fact that $[AED] = [BEC]$ and the fact that $\triangle AEB \sim \triangle EDC$ (which should be trivial given the two equal triangles) we have that
\[\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}\]
We know that $DC=EC,$ so we have
\[\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}\]
Thus
\[\frac{AE}{EC} = \frac{3}{4}\]
But $EC = 14 - AE$ so we have
\[\frac{AE}{14 - AE} = \frac{3}{4}\]
Simplifying gives $AE = \boxed{6}.$

Answer:  6
Problem: Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$

Solution: We will try to use symmetry as much as possible.
Pick the first vertex $A$ , its choice clearly does not influence anything.
Pick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.
In the first case, there are $2$ faces that contain the edge $AB$ . In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$ , the entire triangle $ABC$ will be on a face. On the other hand, if $C$ is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good $C$ is $2/6 = 1/3$
In the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$ . Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$
In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$
Summing up all cases, the resulting probability is: \[\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{47}\]

Answer:  47
Problem: Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$

Solution: There are $\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
There are $\binom{4}{3}=4$ ways to choose three points from the vertices of a single face. Since there are six faces, $4 \times 6 = 24$
Thus, the probability of what we don't want is $\frac{24}{56} = \frac{3}{7}$ . Using complementary probability,
\[1- \frac 37 = \boxed{47}\]

Answer:  47
Problem: For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ .  Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ .  What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution: The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$
This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ $N(4) = \boxed{7}$

Answer:  7
Problem: For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ .  Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ .  What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution: Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\boxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ $N(k)$ must be less than or equal to $7$

Answer: B 7
Problem: For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ .  Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ .  What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution: Similar to the other solutions, notice that $I_k$ can be written as $10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6$ . Factoring out $2^6$ we see that
$I_k = 2^6(2^{k-4}5^{k+2}+1)$
Notice that for $k < 4$ $2^{k-4}$ will not be an integer, and will "steal" some $2$ 's from the $2^6$ . We don't want this to happen, since we want to maximize the exponent of $2$ . We start by considering $k = 4$ . Then
$I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)$
$5^6$ is an odd number; more specifically, it ends in $25$ (all powers of $5$ after $5^1$ end in $25$ ). Therefore the value in the parentheses will be some large number that ends in $26$ . Considering the rules of divisibility, we find that $5^6+1$ is even, so it is divisible by $2$ . Now our exponent of $2$ is at $7$ . But the divisibility rule for $4$ is the last $2$ digits of the number must be divisible by $4$ . We know the last digits: $26$ , which is not divisible by $4$ . Therefore $5^6 + 1$ is divisible by $2$ , but not $4$ . Testing more values of $k$ , we find that for $k \ge 5$ , the last digit becomes $1$ , which means it is not even divisible by $2$ . Therefore the highest possible exponent of $2$ that we can reach is $7 \Rightarrow \boxed{7}$

Answer: B 7
Problem: For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ .  Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ .  What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution: Let $m=k+2$ $v_2(10^m+2^6)=6$ if $m>6$ and $v_2(10^m+2^6)=m$ if $m<6$ . 
However, if $m=6$ , then $v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)$ . By LTE, $v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3$ . Since $v_2(5^6-1)=3$ $v_2(5^6+1)$ must equal $1$ . So, the answer is $6+1=7 \Rightarrow \boxed{7}$

Answer: B 7
Problem: Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution: If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.
If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{2}$ bagels.

Answer: B 2
Problem: Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$

Solution: Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
We can multiply both by $12$ , getting $\dfrac{4-3}{6-4} = \boxed{12}$

Answer:  12
Problem: Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$
$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$

Solution: We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\cdot4=12$ . 
Next, we multiply both the numerator and denominator by $12$ $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{12}$ .
-sosiaops

Answer: C 12
Problem: Paula the painter had just enough paint for 30 identically sized rooms.  Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms.  How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 25$

Solution: Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$

Answer: C 15