Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution: We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$
Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (second one), because the problem only ask's for the power of $5$ . We get $4$
Next, we can look at the multiples of 5 in $98!$ $98/5 = 19$ so there is 19 multiples of 5. We get $19$
But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$ . How many multiples of $25$ are between $1$ and $98$ $3$ $25,50,75,$ and that's it. We get $3$
Finally, we add all of the numbers (powers of $5$ ) up. That is $4 + 19 + 3$ , which is just $26$
So the answer is $26$ . Which is answer choice D $\boxed{26}$

Answer: D 26
Problem: Suppose $a$ $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$
$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$

Solution: There are $2$ cases to consider:
Case $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]
Case $2$ $2$ of $a$ $b$ , and $c$ are negative and the other is positive. WLOG, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]
Note these are the only valid cases, for neither $3$ negatives nor $3$ positives would work as they cannot sum up to $0$ . In both cases, we get that the given expression equals $\boxed{0}$

Answer: A 0
Problem: Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution: It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour.  By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\] We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{25}.\]

Answer: C 25
Problem: Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

Solution: We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor.\]
To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days, which is \[\left \lfloor \frac{365}{12} \right \rfloor +  \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor.\]
We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor.$ Therefore, our answer is \[365 - \left( \left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor \right) +  \left( \left \lfloor \frac{365}{12} \right \rfloor +  \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{146}.\]

Answer: D 146
Problem: Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

Solution: For any randomly chosen day, there is a $\frac{2}{3}$ chance the first child does not call her, a $\frac{3}{4}$ chance the second child does not call her and a $\frac{4}{5}$ chance the third child does not call her. So, in a randomly chosen day, there is a $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$ chance no child calls her.
In particular, this "unrigorous" reasoning becomes rigorous, by linearity of expectation, when we take a sample of $360$ days; over the first $360$ days, $360 \times \frac{2}{5} = 144$ days will go without a phone call. Now, we simply check the remaining days; on day $361$ and $362$ , nobody calls her; on day $363$ , the child that calls every three days will call her; on day $364$ , the child that calls every four days will call her; on day $365$ , the child that calls every five days will call her. Thus, we add two more days to $144$ , to get our answer of $\boxed{146}$

Answer: D 146
Problem: 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 $660 + 5 = \boxed{665}$

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

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

Answer: C 665
Problem: In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution: Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{12}$

Answer: A 12
Problem: In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution: A triangle with the same height and base as a rectangle is half of the rectangle's area.  This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area.  Therefore, we get $\frac{48}{4} =\boxed{12}$

Answer: A 12
Problem: In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution: We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{12}$

Answer: A 12
Problem: In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution: A triangle is half of a rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer $\boxed{12}$

Answer: A 12
Problem: In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Solution: We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. We add 24 and 12 to get 36. Then, we calculate the area of rectangle ABCD, which is 48. We subtract 36 from 48, resulting in $\boxed{12}$

Answer: A 12
Problem: Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$

Solution: Let $r$ be the remaining student's score.  We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ .  We can use basic algebra to solve for $r$ \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{40}$

Answer: A 40
Problem: Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$

Solution: Since $90$ is $20$ more than $70$ , and $80$ is $10$ more than $70$ , for $70$ to be the average, the other number must be $30$ less than $70$ , or $\boxed{40}$

Answer: A 40
Problem: When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$

Solution: When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes per mile. Therefore, it takes him $\boxed{10}$ minutes longer to walk a mile now compared to when he was a boy.

Answer: B 10
Problem: 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, $73-11(6)=73-66=\boxed{7}$

Answer: E 7
Problem: 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: 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 $a=\boxed{7}$

Answer: E 7
Problem: 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: 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: E 7
Problem: The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$

Solution: We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{4}$

Answer: B 4
Problem: The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$

Solution: To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$ , and $3$ names with length $4$ , the $10$ th name would have $4$ letters. Thus our answer is $\boxed{4}$

Answer: B 4
Problem: Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

Solution: We can group each subtracting pair together: \[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\] After subtracting, we have: \[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\] There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{50}$

Answer: C 50
Problem: Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

Solution: Since our list does not end with one, we divide every number by 2 and we end up with \[50-49+48-47+ \ldots +4-3+2-1\] We can group each subtracting pair together: \[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\] There are now $25$ pairs of numbers, and the value of each pair is $1$ .  This sum is $25$ . However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{50}$

Answer: C 50
Problem: What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

Solution: The prime factorization is $2016=2^5\times3^2\times7$ .  Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ .  Their desired sum is then $\boxed{12}$

Answer: B 12
Problem: What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

Solution: We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\boxed{12}$

Answer: B 12
Problem: Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$

Solution: Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \[-9+x=1\] \[x=\boxed{10}\]

Answer: D 10
Problem: Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$

Solution: Let us set a variable $y$ equal to $5 * x$ . Solving for y in the equation $3(2)-y=1$ , we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \[x=\boxed{10}\]

Answer: D 10
Problem: Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \[(10a+b)+(10b+a)=132\] \[11a+11b=132\] \[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\boxed{7}$ ordered pairs.

Answer: B 7
Problem: Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{7}$ total possibilities.

Answer: B 7