Answer: 148.0

Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.

Solution:
$N$ must be some multiple of $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$
$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ all divide $k$ , so $x, y, z = 9, 11, 13$
We have the following three modulo equations:
$nk\equiv 3\pmod{9}$
$nk\equiv 3\pmod{11}$
$nk\equiv 3\pmod{13}$
To solve the equations, you can notice the answer must be of the form $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer.
This must be divisible by $lcm$ $(14, 15, 16)$ , which is $560\cdot 3$
Therefore, $\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q$ , which is an integer. Factor out $3$ and divide to get $\frac{429m+1}{560} = q$ .
Therefore, $429m+1=560q$ . We can use Bezout's Identity or a Euclidean algorithm bash to solve for the least of $m$ and $q$
We find that the least $m$ is $171$ and the least $q$ is $131$
Since we want to factor $1680\cdot q$ , don't multiply: we already know that the prime factors of $1680$ are $2$ $3$ $5$ , and $7$ , and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$

Solution:
Note that the number of play blocks is a multiple of the LCM of $16$ $15$ , and $14$ . The value of this can be found to be $(16)(15)(7) = 1680$ . This number is also divisible by $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ , thus, the three numbers $x, y, z$ are $9, 11, 13$
Thus, $1680k\equiv 3$ when taken mod $9$ $11$ $13$ . Since $1680$ is congruent to $6$ mod $9$ and $3$ mod $13$ , and congruent to $8$ mod $11$ , the number $k$ must be a number that is congruent to $1$ mod $13$ $2$ mod $3$ (because $6$ is a multiple of $3$ , which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo $11$
Looking at the last condition shows that $k\equiv 10$ mod $11$ (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to $10$ mod $11$ and $1$ mod $13$ yield the following lists:
$10$ mod $11$ $21$ $32$ $43$ $54$ $65$ $76$ $87$ $98$ $109$ $120$ $131$ ...
$1$ mod $13$ $14$ $27$ $40$ $53$ $66$ $79$ $92$ $105$ $118$ $131$ $144$ $157$ $170$ ...
Both lists contain $x$ elements where $x$ is the modulo being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$ . Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = \boxed{148}$

Solution:
It is obvious that $N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$ . Therefore, $N=1287\cdot k+3$ . Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two $N$ equations together, then we get $1680a=1287k+3$ . Divide by $3$ . Note that since the RHS is $1\pmod{3}$ , and since $560$ is $2\pmod{3}$ , then $a=3b+2$ , where $b$ is some nonnegative integer, because $a$ must be $2\pmod{3}$
This reduces to $560 \cdot 3b + 1119 = 429k$ . Now, take out the $11!$ With the same procedure, $b=11c-1$ , where $c$ is some nonnegative integer.
You also get $c=13d+4$ , at which point $k=171+560d$ $d$ cannot be equal to $0$ . Therefore, $c=4, b=43, a=131$ , and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $\boxed{148}$

Solution:
We start by noticing that $N = a\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition.
Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$ , they are not divisors of $1680x$ . Thus, we can eliminate all $y \le 14$ s.t. $\gcd(y, 1680x) \ne 1$ which leaves $(x, y, z) = (9, 11, 13)$ . Thus, $N = 1680a \equiv 3 \pmod 9 \equiv 3 \pmod {11} \equiv 3 \pmod {13}$ . Now, we seek to find the least $a$ which satisfies this set of congruences.
By Chinese Remainder Theorem on the first two congruences, we find that $a \equiv 32 \pmod {33}$ (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on $a \equiv 32 \pmod {33}$ and $1680a \equiv 3 \pmod {13}$ we find that $a \equiv 131 \pmod {429}$
Thus, the minimal value of $N$ is possible at $a = 131$ . The prime factorization of this minimum value is $2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131$ and so the answer is $2 + 3 + 5 + 7 + 131 = \boxed{148}$

Solution:
As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$ , so we assume total number of boxes is $1680k$
Then, according to $(b)$ statement, we get $1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}$ . So we have $lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k$ , we just write it to be $(1+n)lcm(x,y,z)=1680k-3$ Which tells that $x,y,z$ must be all odd number. Moreover, we can see $(1+m)lcm(x,y,z)$ can't be a multiple of $3,5,7$ (as $1680$ is a multiple of $5,7$ ) which means that $lcm(x,y,z)=lcm(9,11,13)=1287$ We let $n=1+m$
Now, we write $1287n+3=1680k, 429n+1=560k$ It is true that $n\equiv 1 \pmod{10}$ , let $n=10p+1$ , it has $429p+43=56k$ Then, $p$ must be odd, let $p=2q+1$ , it indicates $429q+236=28k$ Now, $q$ must be even, $q=2s$ tells $429s+118=14k$ Eventually, $s$ must be even, $s=2y$ $858y+59=7k$ $y=1$ is the smallest. This time, $k=131$
So the number of balls is $1680\cdot 131=2^4\cdot 3\cdot 5\cdot 7 \cdot 131$ , the desired value is $2+3+5+7+131=\boxed{148}$

Solution:
From part (a), we know that $2^4\cdot3\cdot5\cdot7 | N$ . From part (b), we know that $N \equiv 3 \pmod {1287}$ . We can expand on part (a) by saying that $N = 1680k$ for some $k$ . Rather than taking the three modulos together, we take them individually. \[1680k \equiv 6k \equiv 3 \pmod 9\] \[k \equiv 2^{-1} \pmod 9\] The inverse of 2 mod 9 is easily seen to be $5$ \[k \equiv 5 \pmod 9\]
Now moving to the second modulo which we leave as follows, \[1680k \equiv 8k \equiv 3 \pmod {11}\] Now the last modulo, \[1680k \equiv 3k \equiv 3 \pmod {13}\] \[k \equiv 1 \pmod{13}\] CRT on the first and the third one results in $k \equiv 4 \pmod {117}$ . Now doing the second one and the one we just made, $k \equiv 131 \pmod{1287}$ . Thus, the smallest value that works for $k = 131$ . Thus $N = 2^4\cdot3\cdot5\cdot7\cdot131$ $2+3+5+7+131 = \boxed{148}$


Answer: 132.0

Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 128\qquad \textbf{(D)}\ 132\qquad \textbf{(E)}\ 136$

Solution: A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \times 13 = 156$ . Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \times 24 = 24$ . The difference in area is $156-24=\boxed{132}$


Answer: 60000.0

NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292526838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?
$\textbf{(A)}\ 6000 \qquad \textbf{(B)}\ 12000 \qquad \textbf{(C)}\ 60000 \qquad \textbf{(D)}\ 120000 \qquad \textbf{(E)}\ 600000$

Solution: Note that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Therefore, the speed (in miles per hour) is \[\frac{292526838}{6.5\cdot30\cdot24} \approx \frac{300000000}{6.5\cdot30\cdot24} = \frac{10000000}{6.5\cdot24} \approx \frac{10000000}{6.4\cdot25} = \frac{10000000}{160} = 62500 \approx \boxed{60000}.\] As the answer choices are far apart from each other, we can ensure that the approximation is correct.

Solution: Note that $292526838 \approx 300000000$ miles. We also know that $6.5$ months is approximately $6.5\cdot30\cdot24$ hours. Now, we can calculate the speed in miles per hour, which we find is about \[\dfrac{300000000}{6.5\cdot30\cdot24}=\dfrac{10000000}{6.5\cdot24}=\dfrac{10000000}{13\cdot12}=\dfrac{10000000}{156}\approx\dfrac{10000000}{150}\approx\dfrac{200000}{3}\approx\boxed{60000}.\] ~MathFun1000


Answer: 55.0

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$ -cent, $10$ -cent, and $25$ -cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage?
(Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$

Solution:
Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$ -cent (nickels), $10$ -cent (dimes), and $25$ -cent (quarters).
If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ dimes to bring our total cost to $7.10 - 3.00 = 4.10$ . However, when we try to use quarters, the $25$ cents don’t fit evenly, so we have to give back $15$ cents to make the quarter amount $4.25$ . The most efficient way to do this is to give back a $10$ -cent (dime) stamp and a $5$ -cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{55}$

Solution: The value of his entire stamp collection is $8$ dollars. To make $$7.10$ with stamps, he should remove $90$ cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many $25$ cent stamps as possible as they have the greatest denomination. He can remove at most $3$ of these stamps. He still has to remove $90-25\cdot3=15$ cents worth of stamps. This can be done with one $5$ and $10$ cent stamp. In total, he has $20\cdot3=60$ stamps in his entire collection. As a result, the maximum number of stamps he can use is $20\cdot3-5=\boxed{55}$


Answer: 8.0

Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes. If she doesn’t use it any
more but leaves the phone on, how many more hours will the battery last?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution: When not being used, the cell phone uses up $\frac{1}{24}$ of its battery per hour. When being used, the cell phone uses up $\frac{1}{3}$ of its battery per hour. Since Niki's phone has been on for $9$ hours, of those $8$ simply on and $1$ being used to talk, $8(\frac{1}{24}) + 1(\frac{1}{3}) = \frac{2}{3}$ of its battery has been used up. To drain the remaining $\frac{1}{3}$ the phone can last for $\frac{\frac{1}{3}}{\frac{1}{24}}=\boxed{8}$ more hours without being used.


Answer: 97.0

Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
We use complementary counting.
We can order the $9$ people around a circle in $\frac{9!}{9} = 8!$ ways. Now we count when there is at least one delegate surrounded by people from only his/her country.
Let the countries be $A,B,C$ . Suppose that the group $XXX$ (for some country $X$ ) appears. To account for circular over counting we fix this group at the top. There are $6!$ ways to arrange the rest of the delegates and $3!$ ways to arrange inside the group. Since there are three countries this group can belong to, the total is $6!*3!*3$
But notice that when the group $XXX$ AND $YYY$ both appear is over counted. So, fix one group at the top. for the last country, we can insert $0,1,2,3$ people in between the two groups. Since we can choose two countries in $\binom{3}{2} = 3$ ways, the total is $3!*3!*3!*4*3$ ways.
Unfortunately, there is over count in the over count. If each country has their delegates together, this case must be added back. Think of these three groups as a whole, and there are $3!/3 = 2$ ways of arranging. In each group there is $3!$ ways of arrangements. So, $3!*3!*3!*2$ is the total for this case.
We now find the complementary probability total. This is \[\frac{6!*3!*3 - 3!*3!*3!*4*3 + 3!*3!*3!*2}{8!} = \frac{15}{56}\] so the actual probability is $1-\frac{15}{56} = \frac{41}{56}$ for an answer of $\boxed{97}$

Solution:
$1680= \binom{9}{3} \binom{6}{3}$ is the total.
$540 = 3* 9* \binom{6}{3}$ is the case where one country has three people in a row. (Three is for selection of country and nine is for rotation of the seats.)
$108 = 3* 9* 4$ is the case where two countries has three people in a row. (Three and nine are for the same reasons above and four is for putting three people in a row for the remaining six seats.)
$18$ is the case where three countries has three people in a row.
$\frac{1680-540+108-18}{1680}=\frac{41}{56}$ so the answer is $\boxed{97}$


Answer: 216.0

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

Solution:
Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them.
The problem we must solve is to distribute meals $\text{BBCCCFFF}$ to orders $\text{BBCCCFFF}$ with no matches. The two people who ordered $B$ 's can either both get $C$ 's, both get $F$ 's, or get one $C$ and one $F.$ We proceed with casework.
Summing across the cases we see there are $24$ possibilities, so the answer is $9 \cdot 24 = \boxed{216.}$

Solution: We only need to figure out the number of ways to order the string $BBBCCCFFF$ , where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there are $2$ ways to serve their meals. Thus, there are $(3\cdot2)^3=\boxed{216}$ ways to serve their meals.


Answer: 17.0

Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively.  Each of three players randomly selects and keeps three of the tiles, and sums those three values.  The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers .  Find $m+n.$

Solution:
In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile.  Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile.  We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
$\dbinom{5}{3} = 10$ choices for the tiles that he gets.  The other two odd tiles can be distributed to the other two players in $2$ ways, and the even tiles can be distributed between them in $\dbinom{4}{2} \cdot \dbinom{2}{2} = 6$ ways.  This gives us a total of $10 \cdot 2 \cdot 6 = 120$ possibilities in which all three people get odd sums.
In order to calculate the probability, we need to know the total number of possible distributions for the tiles.  The first player needs three tiles which we can give him in $\dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $\dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $\dbinom{9}{3} \cdot \dbinom{6}{3} \cdot 1 = 84 \cdot 20 = 1680$ ways total to distribute the tiles.
We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $\frac{360}{1680} = \frac{3}{14},$ so the answer is $3 + 14 = \boxed{017}$

Solution:
Let $O$ stand for an odd number and $E$ an even.
Therefore, one person must pick $OOO$ , the other person must pick $OEE$ and the last person must pick $OEE$ . Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. $EOE$ instead of $OEE$ ) doesn't change the probability, we just need to multiply the probability of one case by $\binom{3}{2}^3 = 27$ as there are 27 such cases (by cases I mean ordered triples of ordered multisets $A, B, C$ such that one of them has 3 $O$ 's and the other two have two $E$ 's and an $O$ in them, respectively.) . Let's do the case $OOO$ $OEE$ $OEE$ $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = \frac{1}{126}$ . We now multiply by 27 to get $\frac{3}{14} \implies m + n = \boxed{017}$ - whatRthose

Solution:
For this problem, let's think about parity. There are $5$ odd numbers from $1-9$ and there are four even numbers from $1-9$ . Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways.
In this case, there are only two ways to get an odd sum. Either have the sequence $OOO$ or $OEE$ where the letters $O$ and $E$ stand for odd and even respectively.
Since we constrained to only $5$ odds, the only way to do the pairing is \[OOO-OEE-OEE\] There are of couse three ways to choose who gets the three odds.
Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this: \[OEE-OEE-OOO\] Now since we are choosing in groups, we can ignor order of these terms.
For the first $OEE$ , there are $5$ ways to choose which odd to use, and $\dbinom{4}{2}$ or $6$ ways to choose the evens.
Great, let's move on to the second $OEE$ . There are $4$ odds left, so there are $4$ ways choose the odds. Since there are only two even numbers, they have to go here, so there is only $1$ way to choose the evens.
The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have $5*6*4*3$ or $360$
Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players.
Fortunately for us, this is not hard as the first player has $\dbinom{9}{3}$ options and the second player has $\dbinom{6}{3}$ . When we multiply these, we get $1680$ . This is our denominator.
When we make the fraction, we have $\frac{360}{1680}$ . When we simplify it, we have: \[\frac{3}{14}\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \[\boxed{017}\] -Pi_is_3.14


Answer: 465.0

Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall.  Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower.  How many different tower heights can be achieved using all ninety-four of the bricks?

Solution:
We have the smallest stack, which has a height of $94 \times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$ 's, $6$ 's, and $15$ 's. Because $0$ $6$ , and $15$ are all multiples of $3$ , the change will always be a multiple of $3$ , so we just need to find the number of changes we can get from $0$ 's, $2$ 's, and $5$ 's.
From here, we count what we can get:
\[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, \ldots\]
It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem , which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 \times 2 - 5 - 2=3$
But we also have a maximum change ( $94 \times 5$ ), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$ 's, $3$ 's, or $5$ 's. The maximum we can't get is $5 \times 3-5-3=7$ , so the numbers $94 \times 5-8$ and below, except $3$ and $1$ , work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 \times 5-8$ and $94 \times 5$ $94 \times 5-7$ obviously doesn't work, $94 \times 5-6$ does since 6 is a multiple of 3, $94 \times 5-5$ does because it is a multiple of $5$ (and $3$ ), $94 \times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$ $94 \times 5-3$ does since $3=3$ , and $94 \times 5-2$ and $94 \times 5-1$ don't, and $94 \times 5$ does.
Thus the numbers $0$ $2$ $4$ all the way to $94 \times 5-8$ $94 \times 5-6$ $94 \times 5-5$ $94 \times 5-3$ , and $94\times 5$ work. That's $2+(94 \times 5 - 8 - 4 +1)+4=\boxed{465}$ numbers. That's the number of changes you can make to a stack of bricks with dimensions $4 \times 10 \times 19$ , including not changing it at all.

Solution:
Using bricks of dimensions $4''\times10''\times19''$ is comparable to using bricks of dimensions $0''\times6''\times15''$ which is comparable to using  bricks of dimensions $0''\times2''\times5''$ . Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$
It follows that all tower heights can be made by using 4 or fewer bricks of height $2''$ . There are $95+94+93+92+91=465$ ways to build a tower using 4 or fewer bricks of height $2''$ . Taking the heights $\mod 5$ , we see that towers using a different number of bricks of height $2''$ have unequal heights. Thus, all of the $\boxed{465}$ tower heights are different.


Answer: 33.0

Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$

Solution: By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$

Solution: We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$ , respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$ , we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$ . The only answer choice which is a multiple of $11$ is $\boxed{33}$ . ~mathboy282


Answer: 33.0

Solution: By the Angle Bisector Theorem , we know that $\frac{AB}{BC} = \frac{3}{8}$ . If we use the lowest possible integer values for $AB$ and $BC$ (the lengths of $AD$ and $DC$ , respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality . If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = \boxed{33}$

Solution: We find that $\frac{AB}{BC}=\frac{3}{8}$ by the Angle Bisector Theorem so we let the lengths be $3n$ and $8n$ , respectively where $n$ is a positive integer. Also since $AD=3$ and $BC=8$ , we notice that the perimeter of the triangle is the sum of these, namely $3n+8n+3+8=11n+11.$ This can be factored into $11(n+1)$ and so the sum must be a multiple of $11$ . The only answer choice which is a multiple of $11$ is $\boxed{33}$ . ~mathboy282


Answer: 1.0

Nonzero real numbers $x$ $y$ $a$ , and $b$ satisfy $x < a$ and $y < b$ . How many of the following inequalities must be true?
$\textbf{(I)}\ x+y < a+b\qquad$
$\textbf{(II)}\ x-y < a-b\qquad$
$\textbf{(III)}\ xy < ab\qquad$
$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution:
Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$ . We can write that $x + y < x + y + k + l \implies x + y < a + b$
It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$
$\textbf{(II)}$ states that $x-y<a-b \implies -3 - (-4) < 1 - 4  \implies 1 < -3$ .Therefore, $\textbf{(II)}$ is false.
$\textbf{(III)}$ states that $xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4$ . Therefore, $\textbf{(III)}$ is false.
$\textbf{(IV)}$ states that $\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-4} < \frac{1}{4} \implies 0.75 < 0.25$ . Therefore, $\textbf{(IV)}$ is false.
One of our four inequalities is true, hence, our answer is $\boxed{1}$


Answer: 0.2

Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
$\mathrm{(A)} \ 0.10 \qquad \mathrm{(B)} \ 0.15 \qquad \mathrm{(C)} \ 0.20 \qquad \mathrm{(D)} \ 0.25 \qquad \mathrm{(E)} \ 0.30 \qquad$

Solution:
We know that the sum of the vertical steps must be equal to $7$ . We also know that they must take $6$ steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:
\[P(x)=(x+1)^{12}\]
Where we need to extract the $x^5$ coefficient. By the binomial coefficient theorem, that term is $\binom{12}{5}=792$ paths. Since there are also $2^{12}$ paths, we have:
$\frac{792}{2^{12}}=\frac{792}{4096}\approx\frac{800}{4000}\approx\boxed{0.20}$


Answer: 184.0

Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$ , and partition the octagon into 7 triangles by drawing segments $\overline{JB}$ $\overline{JC}$ $\overline{JD}$ $\overline{JE}$ $\overline{JF}$ , and $\overline{JG}$ . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23),  G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]

Solution:
We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$ . Recall that the centroid is $\frac{1}{3}$ way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point $J$ . Furthermore, we can take advantage of the reflective symmetry across the line parallel to $BC$ going through $J$ by dealing with less coordinates and ommiting the $\frac{1}{2}$ in the shoelace formula.
By doing some basic algebra, we find that the coordinates of the centroids of $\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE$ are $\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),$ and $\left(18, -\frac{23}{2}\right)$ , respectively. We'll have to throw in the projection of the centroid of $\bigtriangleup JAB$ to the line of reflection to apply shoelace, and that point is $\left( \frac{8}{3}, -\frac{23}{2}\right)$
Finally, applying Shoelace, we get: $\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|$ $=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|$ $=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|$ $=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}$

Solution:
Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$ .
We have a homothety since:
1. $J$ passes through corresponding vertices of the two heptagons.
2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is \[= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.\] The area of each triangle is \[=  \frac{1}{2}\cdot 17\cdot 4=34.\]
Hence, the area of the large heptagon is \[2\cdot 190+34=414.\] Then, from our homothety, the area of the required heptagon is \[\frac{4}{9}\cdot 414=\boxed{184}.\] ~novus677


Answer: 47.0

Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $250 m/min$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $300 m/min$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$

Solution: Since $d = rt$ , we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{47}$ meeting points.

Solution:
We first find the amount of minutes, $k$ , until Odell and Kershaw's next meeting. Let $a$ be the angle in radians between their starting point and the point where they first meet, measured counterclockwise.
Since Kershaw has traveled $300k$ meters at this point and the circumference of his track is $120\pi$ $a=\frac{300k}{120\pi}\cdot 2\pi$ . Similarly, $2\pi-a=\frac{250k}{100\pi}\cdot{2\pi}$ since Odell has traveled $250k$ meters in the opposite direction and the circumference of his track is $100\pi$
Solving for $a$ in the second equation, we get $a=2\pi-\frac{250k}{100\pi}\cdot 2\pi$ . Then, from the first equation, we have $\frac{300k}{120\pi}\cdot 2\pi=2\pi-\frac{250k}{100\pi}\cdot 2\pi$ . Solving for $k$ , we get $k=\frac{\pi}{5}$ . After $k$ minutes, they are back at the same position, except rotated, so they will meet again in $k$ minutes. So the total amount of meetings is $\lfloor\frac{30}{k}\rfloor=\lfloor\frac{150}{\pi}\rfloor=\boxed{47}$


Answer: 47.0

Solution: Since Odell's rate is $5/6$ that of Kershaw, but Kershaw's lap distance is $6/5$ that of Odell, they each run a lap in the same time. Hence they pass twice each time they circle the track. Odell runs \[(30 \ \text{min})\left(250\frac{\text{m}}{\text{min}}\right)\left(\frac{1}{100\pi}\frac{\text{laps}}{\text{m}}\right)= \frac{75}{\pi}\,\text{laps}\approx 23.87\ \text{laps},\] as does Kershaw. Because $23.5 < 23.87 < 24$ , they pass each other $2(23.5)=\boxed{47}$ times.


Answer: 100.0

Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?
$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$

Solution: Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{100}$

Solution: We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be $x$ , so $\frac{400-x}{500-x}=\frac{3}{4}$ . Cross-multiplying gives us $1600-4x=1500-3x \implies x=100$ , so our answer is $\boxed{100}$

Solution:
Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be $x$ . Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation: $0.75(500–x)=400–x$ .
In words, 75% of the total remaining balls are equal to the remaining red balls.
Solving we get $x=100$ , so our answer is $\boxed{100}$ .
~J.L.L


Answer: 50.0

Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 72$

Solution: There are $36$ students in the class: $12$ prefer chocolate pie, $8$ prefer apple pie, and $6$ prefer blueberry pie. Therefore, $36-12-8-6=10$ students prefer cherry pie or lemon pie. Half of these prefer each, so $5$ students prefer cherry pie. This means that $\frac{5}{36}$ of the students prefer cherry pie, so $\frac{5}{36}$ of the full $360^\circ$ should be used for cherry pie. This is $(\frac{5}{36})(360^\circ)=50^\circ, \boxed{50}$


Answer: 1.0

Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\\textbf{(E)}\ \text{Only 1 is true}$

Solution: The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$ , so $\boxed{1}$


Answer: 3.0

Of the following complex numbers $z$ , which one has the property that $z^5$ has the greatest real part?
$\textbf{(A) }{-}2 \qquad \textbf{(B) }{-}\sqrt3+i \qquad \textbf{(C) }{-}\sqrt2+\sqrt2 i \qquad \textbf{(D) }{-}1+\sqrt3 i\qquad \textbf{(E) }2i$

Solution:
First, $\textbf{(B)}$ is $2\text{cis}(150)$ $\textbf{(C)}$ is $2\text{cis}(135)$ $\textbf{(D)}$ is $2\text{cis}(120)$
Taking the real part of the $5$ th power of each we have:
$\textbf{(A): }(-2)^5=-32$
$\textbf{(B): }32\cos(750)=32\cos(30)=16\sqrt{3}$
$\textbf{(C): }32\cos(675)=32\cos(-45)=16\sqrt{2}$
$\textbf{(D): }32\cos(600)=32\cos(240)<0$
$\textbf{(E): }(2i)^5=32i$ , whose real part is $0$
Thus, the answer is $\boxed{3}$

Solution:
We rewrite each answer choice to the polar form \[z=r\operatorname{cis}\theta=r(\cos\theta+i\sin\theta),\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , the real part of $z^5$ is \[\operatorname{Re}\left(z^5\right)=r^5\cos{(5\theta)}.\] We construct a table as follows: \[\begin{array}{c|ccc|ccc|cclclclcc} & & & & & & & & & & & & & & &  \\ [-2ex] \textbf{Choice} & & \boldsymbol{r} & & & \boldsymbol{\theta} & & & & & & \multicolumn{1}{c}{\boldsymbol{\operatorname{Re}\left(z^5\right)}} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & & & &  \\ [-1ex] \textbf{(A)} & & 2 & & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & 2 & & & \tfrac{5\pi}{6} & & & &32\cos{\tfrac{25\pi}{6}}&=&32\cos{\tfrac{\pi}{6}}&=&32\left(\tfrac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & 2 & & & \tfrac{3\pi}{4} & & & &32\cos{\tfrac{15\pi}{4}}&=&32\cos{\tfrac{7\pi}{4}}&=&32\left(\tfrac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & 2 & & & \tfrac{2\pi}{3} & & & &32\cos{\tfrac{10\pi}{3}}&=&32\cos{\tfrac{4\pi}{3}}&=&32\left(-\tfrac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & 2 & & & \tfrac{\pi}{2} & & & &32\cos{\tfrac{5\pi}{2}}&=&32\cos{\tfrac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\] Clearly, the answer is $\boxed{3}.$

Solution:
To find the real part of $z^5,$ we only need the terms with even powers of $i:$ \begin{align*} \operatorname{Re}\left(z^5\right)&=\operatorname{Re}\left((a+bi)^5\right) \\ &=\sum_{k=0}^{2}\binom{5}{2k}a^{5-2k}(bi)^{2k} \\ &=\binom50 a^{5}(bi)^{0} + \binom52 a^{3}(bi)^{2} + \binom54 a^{1}(bi)^{4} \\ &=a^5 - 10a^3b^2 + 5ab^4. \end{align*} We find the real parts of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)}$ directly:
Therefore, the answer is $\boxed{3}.$

Solution:
The full expansion of $z^5$ is \begin{align*} z^5&=(a+bi)^5 \\ &=\sum_{k=0}^{5}\binom5k a^{5-k}(bi)^k \\ &=\binom50 a^{5}(bi)^0+\binom51 a^{4}(bi)^1+\binom52 a^{3}(bi)^2+\binom53 a^{2}(bi)^3+\binom54 a^{1}(bi)^4+\binom55 a^{0}(bi)^5 \\ &=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i \\ &=\left(a^5-10a^3b^2+5ab^4\right) + \left(5a^4b-10a^2b^3+b^5\right)i. \end{align*} We find the full expansions of $\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)},$ then extract their real parts:
Therefore, the answer is $\boxed{3}.$


Answer: 252.0

Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?

Solution:
Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance.
Thus, $\dfrac{3k}{5k + 20} = \dfrac{29}{50}$ , solving gives $k = 116$ . Thus, the number of people that like to dance is $2k + 20 = \boxed{252}$

Solution:
Let the number of girls be $g$ . Let the number of total people originally be $t$
We know that $\frac{g}{t}=\frac{3}{5}$ from the problem.
We also know that $\frac{g}{t+20}=\frac{29}{50}$ from the problem.
We now have a system and we can solve.
The first equation becomes:
$3t=5g$
The second equation becomes:
$50g=29t+580$
Now we can sub in $30t=50g$ by multiplying the first equation by $10$ . We can plug this into our second equation.
$30t=29t+580$
$t=580$
We know that there were originally $580$ people. Of those, $\frac{2}{5}*580=232$ like to dance.
We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=\boxed{252}$

Solution: Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$ , so the solution is $0.4p+20 = \boxed{252}$


Answer: 7.0

On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$

Solution:
Note that:
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{7}.$

Solution:
Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.
Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that $9$ alternaters truthfully answer here. Because only liars and $9$ alternaters answer yes, we can deduce that there are $15-9=6$ liars.
Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that $9$ alternaters lie here. From the previous part, we know that there are $6$ liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.
The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers answer yes on only the first question, we know that all $7$ of them said yes once, resulting in $\boxed{7}$ pieces of candy.

Solution:
Note that we have $4$ types of people:
Given this information, it is reasonable to ignore the fourth type, because they will never answer yes to any question. Hence, we only consider people of type $1, 2,$ and $3.$
The principal's first question implies that \[T + L + A = 22.\] The second question implies that \[L + A = 15.\] The third question implies that \[A = 9.\]
Solving, we find that $T = 7,$ so $\boxed{7}$ is our answer. We can also note that $T = 7,  L = 6,  A = 9,$ and there are $9$ alternators who answer no to every question.


Answer: 7.0

Solution:
Note that:
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{7}.$

Solution:
Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.
Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that $9$ alternaters truthfully answer here. Because only liars and $9$ alternaters answer yes, we can deduce that there are $15-9=6$ liars.
Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that $9$ alternaters lie here. From the previous part, we know that there are $6$ liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.
The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers answer yes on only the first question, we know that all $7$ of them said yes once, resulting in $\boxed{7}$ pieces of candy.

Solution:
Note that we have $4$ types of people:
Given this information, it is reasonable to ignore the fourth type, because they will never answer yes to any question. Hence, we only consider people of type $1, 2,$ and $3.$
The principal's first question implies that \[T + L + A = 22.\] The second question implies that \[L + A = 15.\] The third question implies that \[A = 9.\]
Solving, we find that $T = 7,$ so $\boxed{7}$ is our answer. We can also note that $T = 7,  L = 6,  A = 9,$ and there are $9$ alternators who answer no to every question.


Answer: 30.0

On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave $2$ candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave $4$ candies to his sister. On the third day he ate his final $8$ candies. How many candies did Casper have at the beginning?
$\textbf{(A)}\ 30 \qquad\textbf{(B)}\ 39 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 66$

Solution:
Let $x$ represent the amount of candies Casper had at the beginning.
\begin{align*} \frac{2}{3} \left(\frac{2}{3} x - 2\right) - 4 - 8 &= 0\\ \frac{2}{3} x - 2 &= 18\\ \frac{2}{3} x &= 20\\ x &= \boxed{30}

Solution: We work backwards. If he had 8 candies at the end, then before he gave candies to his sister he had 12 candies. This means that at the end of Halloween he had 18 candies, so before he gave candies to his brother he had 20 candies. Therefore, at the start he had $\boxed{30}$

Solution: A solve by algebra is more secure and safe (and usually faster), but you can also test the answer choices. We are lucky and option A works $\Longrightarrow \boxed{30}$


Answer: 60.0

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

Solution: The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{60}$

Solution: Since we know the number must be a multiple of $15$ , we can eliminate $A$ . We also know that after $12$ days, the students can't find any more arrangements, meaning the number has $12$ factors. Now, we just list the factors of every number, starting with $30$ \[30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6\] \[60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10\] $60$ has $12$ factors, so the answer is $\boxed{60}$


Answer: 6.0

On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Solution:
How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$ . For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$ , and for $4$ dollars, you have $8$ dollars or $ $7$ . Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$ , for $8$ dollars, you have $16$ dollars or $11$ , for $8$ dollars, you have $16$ dollars or $11$ , and for $7$ dollars, you have $14$ dollars or $10$
On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{6}$ different values.

Solution: Continue as in Solution 1 to get $7$ $8$ , or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$ ) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{6}$


Answer: 2.1

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:
By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$ $B(4,0,0)$ , and $C(0,4,0)$ , we find that the equation of plane $ABC$ is:
\[\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by:
\[\frac{|Aa + Bb + Cc + D|}{\sqrt{A^2 + B^2 + C^2}}.\]
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$ ) to the plane is given by:
\[\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.\]
Since $\sqrt{34} < 6$ , this number should be just a little over $2$ , and the correct answer is $\boxed{2.1}$

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: 29.0

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?
$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 33$

Solution: Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ $4a-c=99$ . Also, $a+c\leq50 \Rightarrow c\leq50-a$ . We can substitute this inequality into the previous equation to obtain another inequality: $4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8$ . Since $a$ is an integer, the maximum value for $a$ is $\boxed{29}$


Answer: 1.0

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points.  What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$

Solution: To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$ . The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$ . The difference between the mean and median, therefore, is $\boxed{1}$

Solution:
The remaining $30\%$ of the students got $95$ points.
The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$ .
The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$ , the median is $85$ . The difference between the mean and the median is $\boxed{1}$


Answer: 1.0

Solution: To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$ . The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$ . The difference between the mean and median, therefore, is $\boxed{1}$

Solution:
The remaining $30\%$ of the students got $95$ points.
The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$ .
The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$ , the median is $85$ . The difference between the mean and the median is $\boxed{1}$


Answer: 375.0

On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by $10$

Solution:
Let $n$ be the number of car lengths that separates each car (it is easy to see that this should be the same between each pair of consecutive cars.) Then their speed is at most $15n$ . Let a unit be the distance between the cars (front to front). Then the length of each unit is $4(n + 1)$ . To maximize, in a unit, the CAR comes first, THEN the empty space. So at time zero, the car is right at the eye.
Hence, we count the number of units that pass the eye in an hour: $\frac {15,000n\frac{\text{meters}}{\text{hour}}}{4(n + 1)\frac{\text{meters}}{\text{unit}}} = \frac {15,000n}{4(n + 1)}\frac{\text{units}}{\text{hour}}$ . We wish to maximize this.
Observe that as $n$ gets larger, the $+ 1$ gets less and less significant, so we take the limit as $n$ approaches infinity
Now, as the speeds are clearly finite, we can never actually reach $3750$ full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the $3750$ th unit has passed, but not all of the space behind it. Hence, $3750$ cars is possible, and the answer is $\boxed{375}$

Solution:
Disclaimer: This is for the people who may not understand calculus, and is also how I did it.
First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we start with one car in front of the photoelectric eye.
We first set the speed of the cars as $15k$ . Then, the distance between them is $\frac{4}{1000} \times k\text{km}$ . Therefore, it takes the car closest to the eye not on the eye $\frac{\frac{k}{250}}{15k}$ hours to get to the eye. There is one hour, so the amount of cars that can pass is $\frac{1}{\frac{\frac{k}{250}}{15k}}$ , or $3750$ cars. When divided by ten, you get the quotient of $\boxed{375}$


Answer: 102.0

On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$

Solution: We set up the proportion $\frac{12  \text{cm}}{72  \text{km}}=\frac{17 \text{cm}}{x \text{km}}$ . Thus $x=102 \Rightarrow \boxed{102}$


Answer: 5.0

On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11$

Solution: Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$


Answer: 5.0

Solution: Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$


Answer: 5.0

On a sheet of paper, Isabella draws a circle of radius $2$ , a circle of radius $3$ , and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge  0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\boxed{5}$ values of $k$


Answer: 5.0

On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }7\qquad\mathrm{(D)\ }8\qquad\mathrm{(E)\ }9$

Solution: Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\boxed{5}$

Solution: Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $\frac{60}{3}=20$ times. Thus $d=7\cdot20=140$ , and the sum of the digits is $\boxed{5}$


Answer: 5.0

Solution: Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $\boxed{5}$

Solution: Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $\frac{60}{3}=20$ times. Thus $d=7\cdot20=140$ , and the sum of the digits is $\boxed{5}$


Answer: 80.0

On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
[mathjax]\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100[/mathjax]

Solution: Since Anh spends half an hour to drive 10 miles on the coastal road, his speed is [mathjax]r=\dfrac dt=\dfrac{10}{0.5}=20[/mathjax] mph. His speed on the highway then is [mathjax]60[/mathjax] mph. He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]50[/mathjax] minutes (Note that [mathjax]60[/mathjax] miles per hour is the same as [mathjax]1[/mathjax] mile per minute). The total amount of minutes spent on his trip is [mathjax]30+50\implies \boxed{80}[/mathjax].

Solution: Since Anh drives [mathjax]3[/mathjax] times as fast on the highway, it takes him [mathjax]\dfrac{1}{3}[/mathjax] of the time to drive [mathjax]10[/mathjax] miles on the highway than on the coastal road. [mathjax]\dfrac{1}{3}[/mathjax] of [mathjax]30[/mathjax] is [mathjax]10[/mathjax], and since he drives [mathjax]50[/mathjax] miles on the highway, we multiply [mathjax]10[/mathjax] by [mathjax]5[/mathjax] to get [mathjax]50[/mathjax]. This means it took him [mathjax]50[/mathjax] minutes to drive on the highway, and if we add the [mathjax]30[/mathjax] minutes it took for him to drive on the coastal road, we would get [mathjax]\boxed{80}[/mathjax].


Answer: 45.0

On a trip, a car traveled $80$ miles in an hour and a half, then was stopped in traffic for $30$ minutes, then traveled $100$ miles during the next $2$ hours.  What was the car's average speed in miles per hour for the $4$ -hour trip?
$\text{(A)}\ 45 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

Solution: The average speed is given by the total distance traveled divided by the total time traveled. \[\frac{80+0+100}{4} = \boxed{45}\]


Answer: 97.0

On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
$\text{(A)}\ 90 \qquad \text{(B)}\ 91 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 95 \qquad \text{(E)}\ 97$

Solution: The highest possible score is if you get every answer right, to get $5(20)=100$ . The second highest possible score is if you get $19$ questions right and leave the remaining one blank, to get a $5(19)+1(1)=96$ . Therefore, no score between $96$ and $100$ , exclusive, is possible, so $97$ is not possible, $\boxed{97}$

Solution: We can equivalently construct the following rules: You have 100 point at first, but if you give the wrong answer, you will lose 5 points, if you don't answer a question you will lose 4 points. Obviously, you can lose 10 points, 9 points, 8 points, 5 points or 4 points, but you cannot lose 3 points. The answer is $\boxed{97}$


Answer: 49.0

On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 112$

Solution: This means the ratio is $4:7$ . If the ratio now is $28:x$ , then that means $x= \boxed{49}$


Answer: 400.0

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$

Solution: First, we have to find how many widgets she makes on Day $20$ . We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$ $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ . The sum of $1,3,5, ... 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{400}$

Solution: The sum is just the first $20$ odd counting/natural numbers, which is $20^2=\boxed{400}$

Solution: We can easily find out she makes $2\cdot20-1 = 39$ widgets on Day $20$ . Then, we make the sum of $1,3, 5, ... ,35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+...+(19+21)$ , which include $10$ pairs of $40$ . So, the sum of $1,3,5, ...~39$ is $(40\cdot10)=\boxed{400}$


Answer: 850.0

On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

Solution:
$269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\cdot 2$ .
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:
\[17=1^2+4^2\] \[34=3^2+5^2\] \[51=\emptyset\] \[68=\emptyset\text{ others}\] \[85=2^2+9^2=6^2+7^2\] \[102=\emptyset\] \[119=\emptyset \dots\]
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$ \[\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}\] \[\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850\] \[\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}\]
Thus, $\boxed{850}$ is the only valid answer.

Solution:
Continue in the same way as solution 1 to get that $POK$ has area $3a$ , and $OK = \frac{d}{10}$ . You can then find $PK$ has length $\frac 32$
Then, if we drop a perpendicular from $H$ to $BC$ at $L$ , We get $\triangle HLF \sim \triangle OPK$
Thus, $LF = \frac{15\cdot 34}{d}$ , and we know $HL = d$ , and $HF = 34$ . Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.
\[\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2\]
\[d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0\]
\[(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0\]
$d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$ . Since the area is $d^2$ , we have it is equal to $34 \cdot 25 = \boxed{850}$


Answer: 13.0

On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 26$

Solution: As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly, $\boxed{13}$


Answer: 28.0

On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution:
If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.
\begin{align*} x^2+(x+2)^2 &= 394\\ x^2+x^2+4x+4 &= 394\\ 2x^2 + 4x &= 390\\ x^2 + 2x &= 195\\ \end{align*}
From here, we can see that $x = 13$ as $13^2 + 26 = 195$ , so there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{28}$ students.


Answer: 181.0

One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$

Solution:
Let the shorter base have length $b$ (so the longer has length $b+100$ ), and let the height be $h$ . The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$ . The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$ . Then,
\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$ . By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$
The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$


Answer: 11.0

One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ( $128$ ounces) of soda?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$

Solution: $10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\mathrm{\boxed{11}$


Answer: 770.0

One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination . Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?
1988-1.png

Solution:
Currently there are ${10 \choose 5}$ possible combinations.
With any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$ .
Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$ .
So the number of additional combinations is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$


Answer: 125.0

One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216$

Solution:
Let the original cube have edge length $a$ . Then its volume is $a^3$ .
The new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ .
The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{125}$


Answer: 125.0

Solution:
Let the original cube have edge length $a$ . Then its volume is $a^3$ .
The new box has dimensions $a-1$ $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ .
The difference between the two volumes is $a$ . As we are given that the difference is $5$ , we have $a=5$ , and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{125}$


Answer: 9.0

One half of the water is poured out of a full container.  Then one third of the remainder is poured out.  Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc.  After how many pourings does exactly one tenth of the original water remain?
$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution:
1)Model the amount left in the container as follows:
After the first pour $\frac12$ remains, after the second $\frac12 \times \frac23$ remains, etc.
This becomes the product $\frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{9}{10}$
Note that the terms cancel out leaving $\frac{1}{10}$
Now all that remains is to count the number of terms or pourings, as the numerators form an arithmetic sequence with a common difference of 1 and endpoints (1,9), the number of pourings is $\boxed{9}$


Answer: 301.0

One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius $100$ can be expressed as $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m + n.$

Solution:
To get the green area, we can color all the circles of radius $100$ or below green, then color all those with radius $99$ or below red, then color all those with radius $98$ or below green, and so forth.  This amounts to adding the area of the circle of radius $100$ , but subtracting the circle of radius $99$ , then adding the circle of radius $98$ , and so forth.
The total green area is thus given by $100^{2} \pi - 99^{2} \pi + 98^{2} \pi - \ldots - 1^{2} \pi$ , while the total area is given by $100^{2} \pi$ , so the ratio is \[\frac{100^{2}\pi - 99^{2}\pi + 98^{2}\pi - \ldots - 1^{2}\pi}{100^{2}\pi}\]
For any $a$ $a^{2}-(a-1)^{2}=a^{2}-(a^{2}-2a+1)=2a-1$ .  We can cancel the factor of pi from the numerator and denominator and simplify the ratio to
\[\frac{(2\cdot100 - 1)+(2\cdot98 - 1) + \ldots + (2\cdot 2 - 1)}{100^{2}} = \frac{2\cdot(100 + 98 + \ldots + 2) - 50}{100^2}.\]
Using the formula for the sum of an arithmetic series , we see that this is equal to
\[\frac{2(50)(51)-50}{100^{2}}=\frac{50(101)}{100^{2}}=\frac{101}{200},\]
so the answer is $101 + 200 =\boxed{301}$

Solution: We want to find $\frac{\sum\limits_{n=1}^{50} (4n-1)\pi}{10000\pi}=\frac{\sum\limits_{n=1}^{50} (4n-1)}{10000}=\frac{(\sum\limits_{n=1}^{50} (4n) )-50}{10000}=\frac{101}{200} \rightarrow 101+200=\boxed{301}$

Solution:
The sum of the areas of the green regions is
\[\left[(2^2-1^2)+(4^2-3^2)+(6^2-5^2)+\cdots+(100^2-99^2)\right]\pi\] \[=\left[(2+1)+(4+3)+(6+5)+\cdots+(100+99)\right]\pi\] \[={1\over2}\cdot100\cdot101\pi.\]
Thus the desired ratio is \[{1\over2}\cdot{{100\cdot101\pi}\over{100^2\pi}}={101\over200},\] and $m+n=\boxed{301}$


Answer: 125.0

One hundred students at Century High School participated in the AHSME last year, and their mean score was 100. The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?
(A) $100$ (B) $112.5$ (C) $120$ (D) $125$ (E) $150$

Solution:
Solution by e_power_pi_times_i
Let $s$ and $\dfrac{3}{2}s$ denote the numbers of seniors and non-seniors, respectively. Then $\dfrac{5}{2}s = 100$ , so $s = 40$ $\dfrac{3}{2}s = 60$ . Let $m$ and $\dfrac{2}{3}m$ denote the mean score of seniors and non-seniors, respectively. Then $40m + 60(\dfrac{2}{3}m) = 40m + 40m = 10000$ . The answer is $\boxed{125}$


Answer: 5.0

One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$

Solution: The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by 8(each person drank 8 ounces) we get 5 members who drank the coffee with milk, or $\boxed{5}$ ~ Mathyguy88


Answer: 5.0

Solution: If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\boxed{5}$


Answer: 5.0

Solution: The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by 8(each person drank 8 ounces) we get 5 members who drank the coffee with milk, or $\boxed{5}$ ~ Mathyguy88


Answer: 5.0

Solution: If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\boxed{5}$


Answer: 144.0

One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$

Solution:
Taking the given equation modulo $2,3,$ and $5,$ respectively, we have \begin{align*} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{align*} By either Fermat's Little Theorem (FLT) or inspection, we get \begin{align*} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{align*} By either the Chinese Remainder Theorem (CRT) or inspection, we get $n\equiv24\pmod{30}.$
It is clear that $n>133,$ so the possible values for $n$ are $144,174,204,\ldots.$ Note that \begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &<133^5+110^5+(84+27)^5 \\ &=133^5+110^5+111^5 \\ &<3\cdot133^5, \end{align*} from which $\left(\frac{n}{133}\right)^5<3.$
If $n\geq174,$ then \begin{align*} \left(\frac{n}{133}\right)^5&>1.3^5 \\ &=1.3^2\cdot1.3^2\cdot1.3 \\ &>1.6\cdot1.6\cdot1.3 \\ &=2.56\cdot1.3 \\ &>2.5\cdot1.2 \\ &=3, \end{align*} which arrives at a contradiction. Therefore, we conclude that $n=\boxed{144}.$

Solution: Note that $n$ is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5\equiv n\pmod{5}.$ Hence, \[n\equiv3+0+4+2\equiv4\pmod{5}.\] Continuing, we examine the equation modulo $3,$ \[n\equiv1-1+0+0\equiv0\pmod{3}.\] Thus, $n$ is divisible by three and leaves a remainder of four when divided by $5.$ It is obvious that $n>133,$ so the only possibilities are $n = 144$ or $n \geq 174.$ It quickly becomes apparent that $174$ is much too large, so $n$ must be $\boxed{144}.$

Solution:
We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, $n^5\equiv n\pmod{5},$ and it is easy to see that $n^5\equiv n\pmod 2.$ Therefore, $133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10},$ so the last digit of $n$ is $4.$
We notice that $133,110,84,$ and $27$ are all very close or equal to multiples of $27.$ We can rewrite $n^5$ as approximately equal to $27^5(5^5+4^5+3^5+1^5) = 27^5(4393).$ This means $\frac{n^5}{27^5}$ must be close to $4393.$
Note that $134$ will obviously be too small, so we try $144$ and get $\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5.$ Bashing through the division, we find that $\frac{1048576}{243}\approx 4315,$ which is very close to $4393.$ It is clear that $154$ will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that $\boxed{144}$ is the answer.

Solution:
In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digits of $133^5, 110^5, 84^5, 27^5$ are $3, 0, 4, 7,$ respectively, so the units digit of $n^5$ is $4.$ This tells us $n$ is even. Since we are dealing with enormous numbers, $n$ should not be that far from $133.$ Note that $n$ 's units digit is $0, 2, 4, 6,$ or $8.$ When to the power of $5,$ they each give $0, 2, 4, 6,$ and $8$ as the units digits. This further clues us that $n$ ends in $4.$
Clearly, $n>133,$ so we start with $134.$ Now we need a way of distinguishing between numbers with units digit $4.$ We can do this by finding the last three digits for each of $133^5, 110^5, 84^5,$ and $27^5,$ which is not that difficult. For $133^5,$ we have \begin{align*} 133^5&=133^2\cdot133^2\cdot133 \\ &\equiv689\cdot689\cdot133 \\ &\equiv721\cdot133 \\ &\equiv893\pmod{1000}. \end{align*} By the same reasoning, we get \begin{align*} n^5&=133^5+110^5+84^5+27^5 \\ &\equiv893+0+424+907 \\ &\equiv224\pmod{1000}. \end{align*} Note that \begin{align*} 134^5&\equiv424\pmod{1000}, \\ 144^5&\equiv224\pmod{1000}, \\ 154^5&\equiv024\pmod{1000}, \\ 164^5&\equiv824\pmod{1000}, \\ 174^5&\equiv624\pmod{1000}, \\ 184^5&\equiv424\pmod{1000}, \\ 194^5&\equiv224\pmod{1000}. \end{align*} By observations, $n=194$ is obviously an overestimate. So, the answer is $n=\boxed{144}.$

Solution: First, we take mod $2$ on both sides to get $n^5\equiv 0\pmod{2}\implies n\equiv 0\pmod{2}$ . Mod $3$ gives $n^5\equiv 0\pmod{3}\implies n\equiv 0\pmod{3}$ . Also, mod $5$ gives $n^5\equiv -1\pmod{5}\implies n\equiv -1\pmod{5}$ (by FLT). Finally, note that mod $7$ gives $n^5\equiv 2\pmod{7}\implies n^{-1}\equiv 2\pmod{7}\implies n\equiv 4\pmod{7}$ . Thus, \begin{align*} n&\equiv 0\pmod{2}, \\ n&\equiv 0\pmod{3}, \\ n&\equiv -1\pmod{5}, \\ n&\equiv 4\pmod{7}. \end{align*} By CRT, $n\equiv 144\pmod{210}$ , so $n$ is one of $144, 354, ...$ . However, $133^5 + 110^5 + 84^5 + 27^5 < 4\cdot 133^5 < (2\cdot 133)^5 < 354^5$ , so $n < 354$ . Thus, $n = \boxed{144}$

Solution: We have \[n^5 = 133^5 + 110^5 + 84^5 +27^5 = 61917364224,\] for which $n = \sqrt [5]{61917364224} = \boxed{144}.$


Answer: 13.0

One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$

Solution: Let the triangle have side lengths $14, 6+x,$ and $8+x$ . The area of this triangle can be computed two ways. We have $A = rs$ , and $A = \sqrt{s(s-a)(s-b)(s-c)}$ , where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\sqrt{(14+x)(x)(8)(6)}$ . Solving gives $x = 7$ as the only valid solution. This triangle has sides $13,14$ and $15$ , so the shortest side is $\boxed{13}$

Solution: Label the tangent points on $BC, CA, AB$ as $D, E, F$ respectively. Let $AF=AE=6$ $BF=BD=8$ , and $CE=CD=x.$ The problem is a matter of solving for $x$ . To this, we use the fact that if $A,B,C$ are the angles of a triangle, then $\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.$ We know that $\tan{\frac{A}{2}} = \frac{2}{3}$ $\tan{\frac{B}{2}} = \frac{1}{2}$ , and $\tan{\frac{C}{2}} = \frac{4}{x},$ so we have the equation $\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.$ Solving this equation yields $x=7$ , so the shortest side has length $\boxed{13}$


Answer: 1.18

One proposal for new postage rates for a letter was $30$ cents for the first ounce and $22$ cents for each additional ounce (or fraction of an ounce).  The postage for a letter weighing $4.5$ ounces was
$\text{(A)}\ \text{96 cents} \qquad \text{(B)}\ \text{1.07 dollars} \qquad \text{(C)}\ \text{1.18 dollars} \qquad \text{(D)}\ \text{1.20 dollars} \qquad \text{(E)}\ \text{1.40 dollars}$

Solution:
After the first ounce, there are $3.5$ ounces left.  Since each additional ounce or fraction of an ounce adds $22$ cents to the total cost, we need to add $4\times 22$ to the cost for the first ounce.
So, the total price is $30+4\times 22=118$ cents. The answer is choice $\boxed{1.18}$


Answer: 488.0

One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$

Solution: The total number of cubes is $10^3$ or $1000$ . Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$ , since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\boxed{488}$ cubes that have at least one face painted.

Solution: Each face has $100$ cubes, so multiply by six to get $600$ .  However, we overcounted each small cube on the edge but not on corner of the big cube once and each small cube on the corner of the big cube twice.  Thus, there are $600 - (12 \cdot 8 + 2 \cdot 8) = \boxed{488}$ cubes that have at least one face painted.


Answer: 15.0

One-inch squares are cut from the corners of this 5 inch square.  What is the area in square inches of the largest square that can be fitted into the remaining space?
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]
$\textbf{(A)  } 9\qquad \textbf{(B)  } 12\frac{1}{2}\qquad \textbf{(C)  } 15\qquad \textbf{(D)  } 15\frac{1}{2}\qquad \textbf{(E)  } 17$

Solution:
Proceed as in solution 1 to get $x^2 - 5x + 5 = 0$
Notice that the side length of the square is $\sqrt{x^2+(5-x)^2}$ by the Pythagorean Theorem. Thus, the area of the square is \[\left(\sqrt{x^2+(5-x)^2}\right)^2\] \[= x^2 + x^2 - 10x + 25\] \[= (2x^2 - 10x + 10) + 15\] \[= 2(x^2 - 5x + 5) + 15\] \[= 2(0) + 15\] \[= \boxed{15}\]


Answer: 12.0

Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$

Solution: First, we multiply $12\cdot9$ . To get that, we need $108$ square feet of carpet to cover the room's floor.  Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{12}$

Solution: Since there are $3$ feet in a yard, we divide $9$ by $3$ to get $3$ , and $12$ by $3$ to get $4$ . To find the area of the carpet, we then multiply these two values together to get $\boxed{12}$


Answer: 36.0

Orvin went to the store with just enough money to buy $30$ balloons. When he arrived he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 39$

Solution: If every balloon costs $n$ dollars, then Orvin has $30n$ dollars.  For every balloon he buys for $n$ dollars, he can buy another for $\frac{2n}{3}$ dollars.  This means it costs him $\frac{5n}{3}$ dollars to buy a bundle of $2$ balloons.  With $30n$ dollars, he can buy $\frac{30n}{\frac{5n}{3}} = 18$ sets of two balloons, so the total number of balloons he can buy is $18\times2  \implies \boxed{36}$


Answer: 13.0

Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$ -popsicle boxes for $$2$ each, and $5$ -popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution: $$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$ , we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We're now out of money. We bought $5+5+3=13$ popsicles, so the answer is $\boxed{13}$


Answer: 13.0

Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, 3-popsicle boxes for $$2$ , and 5-popsicle boxes for $$3$ . What is the greatest number of popsicles that Pablo can buy with $$8$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

Solution: We can take two 5-popsicle boxes and one 3-popsicle box with $$8$ . Note that it is optimal since one popsicle is at the rate of $$1$ per popsicle, three popsicles at $$\frac{2}{3}$ per popsicle and finally, five popsicles at $$\frac{3}{5}$ per popsicle, hence we want as many $$3$ sets as possible. It is clear that the above is the optimal method. $\boxed{13}$


Answer: 784.0

Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$

Solution:
The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$
In our setting where $A=(0,0)$ $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$
In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$ $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$
These can be counted as follows: We have $6$ identical red balls (representing powers of $2$ ), $6$ blue balls (representing powers of $5$ ), and three labeled urns (representing the factors $k-1$ $s$ , and $t$ ). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.)
Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{784}$

Solution:
We know that $A$ is $(0, 0)$ . Since $B$ is on the line $y=x$ , let it be represented by the point $(b, b)$ . Similarly, let $D$ be $(d, kd)$ . Since this is a parallelogram, sides $\overline{AD}$ and $\overline{BC}$ are parallel. Therefore, the distance and relative position of $D$ to $A$ is equivalent to that of $C$ to $B$ (if we take the translation of $A$ to $D$ and apply it to $B$ , we will get the coordinates of $C$ ). This yields $C (b+d, b+kd)$ . Using the Shoelace Theorem we get
$1,000,000 = \frac{1}{2}|\left(b(b+kd) + (b+d)(kd)\right) - \left(b(b+d) + (b+kd)d\right)|$
$\Rightarrow 2,000,000 = |2kbd - 2bd|$
$\Rightarrow 1,000,000 = |kbd - bd|$
Since $k > 1, kbd > bd$ . The equation becomes
$1,000,000 = (k-1)bd$
$\Rightarrow \frac{1,000,000}{k-1} = bd$
Since $k$ must be a positive integer greater than $1$ , we know $k-1$ will be a positive integer. We also know that $bd$ is an integer, so $k-1$ must be a factor of $1,000,000$ . Therefore $bd$ will also be a factor of $1,000,000$
Notice that $1,000,000 = 10^6 = 2^6*5^6$
Let $b$ be $2^x5^y$ such that $x, y$ are integers on the interval $[0, 6]$
Let $d$ be $2^w5^z$ such that $w, z$ are integers, $x+w \le 6$ , and $y+z \le 6$
For a pair $(x, y)$ , there are $7-x$ possibilities for $w$ and $7-y$ possibilites for $z$ $d$ doesn't have to be the co-factor of $1,000,000$ , it just can't be big enough such that $bd > 1,000,000$ ), for a total of $(7-x)(7-y)$ possibilities. So we want
$\sum_{k=0, i=0}^6 (7-k)(7-i)$
$\left(\text{the sum of the number of possible pairs}(w, z) \text{ for all pairs}(k , i)\text{ for } k[0, 6]\text{ and } i[0, 6]\right)$
Notice that if we "fix" the value of $k$ , at, say $6$ , then run through all of the values of $i$ , change the value of $k$ to $5$ , and run through all of the values of $i$ again, and so on until we exhaust all $49$ combinations of $(k, i)$ we get something like this:
$1*1 + 1*2 + ... + 1*6 + 1*7 + 2*1 + 2*2 + ... + 2*6 + 2*7 + ..... + 7*1 + 7*2 + ... + 7*6 + 7*7$
which can be rewritten
$1(1+2+...+7)+2(1+2+...+7)+.....+7(1+2+...+7)$
$\Rightarrow (1+2+...+7)(1+2+...+7)$
$\Rightarrow 28^2$
$\Rightarrow 784$
So there are $784$ possible sets of coordinates $B,$ $C$ , and $D \Rightarrow \boxed{784}$


Answer: 2.0

Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$
2003 12B AMC-20.png
$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$

Solution:
The roots of this equation are $-1, 1, \text{ and } x$ , letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$ . Therefore, $x=\frac{d}{a}$ . Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$ . We know that the y-intercept of the polynomial is $d$ , so $d=2$ . Plugging in for $d$ $\frac{2}{a}=-\frac{b}{a}$
Therefore, $b=-2 \Rightarrow \boxed{2}$


Answer: 119.0

Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's.  How far can he number the pages of his scrapbook with these digits?
$\text{(A)}\ 22 \qquad \text{(B)}\ 99 \qquad \text{(C)}\ 112 \qquad \text{(D)}\ 119 \qquad \text{(E)}\ 199$

Solution:
There is $1$ two in the one-digit numbers.
The number of two-digit numbers with a two in the tens place is $10$ and the number with a two in the ones place is $9$ . Thus the digit two is used $10+9=19$ times for the two digit numbers.
Now, Pat Peano only has $22-1-19=2$ remaining twos. You must subtract 1 because 22 is counted twice. The last numbers with a two that he can write are $102$ and $112$ . He can continue numbering the last couple pages without a two until $120$ , with the last number he writes being $\boxed{119}$


Answer: 1000.0

Pat intended to multiply a number by $6$ but instead divided by $6$ . Pat then meant to add $14$ but instead subtracted $14$ . After these mistakes, the result was $16$ . If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$

Solution:
We reverse the operations that he did and then use the correct operations. His end result is $16$ . Before that, he subtracted $14$ which means that his number after the first operation was $30$ . He divided by $6$ so his number was $180$
Now, we multiply $180$ by $6$ to get $1080$ . Finally, $1080+14=1094$ . Since $1094>1000$ , our answer is $\boxed{1000}$