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2,301
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_3
| 3
|
In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$
|
We can clearly deduce that $70\%$ of the non-seniors do play an instrument, but, since the total percentage of instrument players is $46.8\%$ , the non-senior population is quite low. By intuition, we can therefore see that the answer is around $\text{B}$ or $\text{C}$ . Testing both of these gives us the answer $\boxed{154}$
| 154
|
2,302
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_3
| 4
|
In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$
|
We know that $40\%$ of the seniors play a musical instrument, and $30\%$ of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is \[46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234\] We can also conclude that $60\%$ of the seniors do not play an instrument, $70\%$ of the non seniors do play an instrument, and $500-234 = 266$ people do play an instrument.
We can now set up the following equations, where $s$ is the number of seniors and $n$ is the number of non-seniors: \[0.3n + 0.6s = 234\] \[0.7n + 0.4s = 266\] By elimination, we get $1.5n$ to be equal to $330$ . This means that $n = \frac{330}{1.5} = 220$ .
The answer is $70$ percent of $220$ . This is equal to \[0.7*220 = 7*22 = 154\] Therefore, the answer is $\boxed{154}$
| 154
|
2,303
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6
| 1
|
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
|
\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$ ). The answer is therefore $1 + 9 = \boxed{10}$
| 10
|
2,304
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6
| 2
|
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
|
Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$
| 10
|
2,305
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6
| 3
|
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
|
Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$
| 10
|
2,306
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6
| 4
|
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
|
Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \boxed{10}$
| 10
|
2,307
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_6
| 5
|
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$
|
Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{10}$
| 10
|
2,308
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_7
| 1
|
Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$
|
If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$
| 21
|
2,309
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_7
| 2
|
Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$
|
We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{21}$
| 21
|
2,310
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_7
| 3
|
Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$
|
We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$
| 21
|
2,311
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_10
| 1
|
In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$
|
Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$ , we have to make sure that its $y$ -coordinate is $\pm20$ , because that's the only way the area of the triangle can be $100$
Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$ . We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$ . The perimeter of this minimal triangle is $2\sqrt{425} + 10$ , which is larger than $50$ . Since the minimum perimeter is greater than $50$ , there is no triangle that satisfies the condition, giving us $\boxed{0}$
| 0
|
2,312
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_10
| 2
|
In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$
|
Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $CD$ , we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ , which is clearly impossible, again giving the answer as $\boxed{0}$
| 0
|
2,313
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_11
| 1
|
Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$ , and the ratio of blue to green marbles in Jar $2$ is $8:1$ . There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$
$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$
|
Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$ , and $\frac{x}{9}$ is the number of green marbles in Jar $2$ . Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$ , we have $\frac{19x}{90}=95$ , so there are $x=450$ marbles in each jar.
Because $\frac{9x}{10}$ is the number of blue marbles in Jar $1$ , and $\frac{8x}{9}$ is the number of blue marbles in Jar $2$ , there are $\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5$ more marbles in Jar $1$ than Jar $2$ . This means the answer is $\boxed{5}$
| 5
|
2,314
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_11
| 3
|
Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$ , and the ratio of blue to green marbles in Jar $2$ is $8:1$ . There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$
$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$
|
Writing out to ratios, we have $9:1$ in jar $1$ and $8:1$ in jar $2$ . Since the jar must have to same amount of marbles, let's make a variable $a$ and $b$ for each of the ratios to be multiplied by. Now we would have $9a + a = 8b + b \rightarrow 10a = 9b$ . We can take the most obvious values of $a$ and $b$ and then scale it from there. We should be able to see that $a$ and $b$ could be $9$ and $10$ respectively. Now remember that there are $95$ green marbles or $x(a + b) = 95$ for some integer $x$ to scale it. Substituting and dividing, we find $x = 5$ . Thus to find the difference of the blue marbles we must do \begin{align*} x(9a - 8b) &= \\ 5(81 - 80) &= \\ 5(1) &= \boxed{5}
| 5
|
2,315
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_12
| 1
|
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$
$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$
|
Observe that $2019_{10} = 5613_7$ . To maximize the sum of the digits, we want as many $6$ s as possible (since $6$ is the highest value in base $7$ ), and this will occur with either of the numbers $4666_7$ or $5566_7$ . Thus, the answer is $4+6+6+6 = 5+5+6+6 = \boxed{22}$
| 22
|
2,316
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_12
| 2
|
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$
$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$
|
Note that all base $7$ numbers with $5$ or more digits are in fact greater than $2019$ . Since the first answer that is possible using a $4$ digit number is $23$ , we start with the smallest base $7$ number that whose digits sum to $23$ , namely $5666_7$ . But this is greater than $2019_{10}$ , so we continue by trying $4666_7$ , which is less than 2019. So the answer is $\boxed{22}$
| 22
|
2,317
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_12
| 3
|
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$
$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$
|
Again note that you want to maximize the number of $6$ s to get the maximum sum. Note that $666_7=342_{10}$ , so you have room to add a thousands digit base $7$ . Fix the $666$ in place and try different thousands digits, to get $4666_7$ as the number with the maximum sum of digits. The answer is $\boxed{22}$
| 22
|
2,318
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_13
| 1
|
What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?
$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$
|
The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$
There are three possibilities for the median: it is either $6$ $8$ , or $x$
Let's start with $6$
$\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.
Now let the median be $8$
$\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.
Finally we let the median be $x$
$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.
Hence the only possible value of $x$ is $\boxed{5}.$
| 5
|
2,319
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_14
| 1
|
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$
|
We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$ . Their divisibility rules (see Solution 2) tell us that $T + M \equiv 3 \;(\bmod\; 9)$ and that $T - M \equiv 7 \;(\bmod\; 11)$ . By guess and checking, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \boxed{12}$
| 12
|
2,320
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_14
| 2
|
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$
|
We know that $H = 0$ , because $19!$ ends in three zeroes (see Solution 1). Furthermore, we know that $9$ and $11$ are both factors of $19!$ . We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$ . For $19!$ to be divisible by $9$ , the sum of digits must simply be divisible by $9$ . Summing the digits, we get that $T + M + 33$ must be divisible by $9$ . This leaves either $\text{A}$ or $\text{C}$ as our answer choice. Now we test for divisibility by $11$ . For a number to be divisible by $11$ , the alternating sum must be divisible by $11$ (for example, with the number $2728$ $2-7+2-8 = -11$ , so $2728$ is divisible by $11$ ). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. By inspection, we can see that this holds if $T=4$ and $M=8$ . The sum is $8 + 4 + 0 = \boxed{12}$
| 12
|
2,321
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_14
| 3
|
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$
|
Multiplying it out, we get $19! = 121,645,100,408,832,000$ . Evidently, $T = 4$ $M = 8$ , and $H = 0$ . The sum is $8 + 4 + 0 = \boxed{12}$
| 12
|
2,322
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_14
| 4
|
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ $M$ , and $H$ denote digits that are not given. What is $T+M+H$
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$
|
7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.
The answer is $4 + 8 + 0 = \boxed{12}$ . ~ AliciaWu
| 12
|
2,323
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_19
| 1
|
Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$
$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$
|
The prime factorization of $100,000$ is $2^5 \cdot 5^5$ . Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$ , whose product is $2^{a+c}5^{b+d}$ , where $0 \le a+c \le 10$ and $0 \le b+d \le 10$
Notice that this is similar to choosing a divisor of $100,000^2 = 2^{10}5^{10}$ , which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two distinct divisors of $2^5 \cdot 5^5$ , namely: $1 = 2^05^0$ $2^{10}5^{10}$ $2^{10}$ , and $5^{10}$ . The last two cannot be written because the maximum factor of $100,000$ containing only $2$ s or $5$ s (and not both) is only $2^5$ or $5^5$ . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$ . The first two would require $1 \cdot 1$ and $2^{5}5^{5} \cdot 2^{5}5^{5}$ , respectively. This gives $121-4 = 117$ candidate numbers. It is not too hard to show that every number of the form $2^p5^q$ , where $0 \le p, q \le 10$ , and $p,q$ are not both $0$ or $10$ , can be written as a product of two distinct elements in $S$ . Hence the answer is $\boxed{117}$
| 117
|
2,324
|
https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_20
| 1
|
As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$
[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$
|
Divide the circle into four parts: the top semicircle by connecting E, F, and G( $A$ ); the bottom sector ( $B$ ), whose arc angle is $120^{\circ}$ because the large circle's radius is $2$ and the short length (the radius of the smaller semicircles) is $1$ , giving a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle; the triangle formed by the radii of $A$ and the chord ( $C$ ); and the four parts which are the corners of a circle inscribed in a square ( $D$ ). Then the area is $A + B - C + D$ (in $B-C$ , we find the area of the bottom shaded region, and in $D$ we find the area of the shaded region above the semicircles but below the diameter).
The area of $A$ is $\frac{1}{2} \pi \cdot 2^2 = 2\pi$
The area of $B$ is $\frac{120^{\circ}}{360^{\circ}} \pi \cdot 2^2 = \frac{4\pi}{3}$
For the area of $C$ , the radius of $2$ , and the distance of $1$ (the smaller semicircles' radius) to $BC$ , creates two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, so $C$ 's area is $2 \cdot \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \sqrt{3}$
The area of $D$ is $4 \cdot 1-\frac{1}{4}\pi \cdot 2^2=4-\pi$
Hence, finding $A+B-C+D$ , the desired area is $\frac{7\pi}{3}-\sqrt{3}+4$ , so the answer is $7+3+3+4=\boxed{17}$
| 17
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2,325
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_20
| 2
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As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $\overarc{AEB},\overarc{BFC},$ and $\overarc{CGD},$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form \[\frac{a}{b}\cdot\pi-\sqrt{c}+d,\] where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d$
[asy] size(6cm); filldraw(circle((0,0),2), grey); filldraw(arc((0,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((-2,-1),1,0,180) -- cycle, gray(1.0)); filldraw(arc((2,-1),1,0,180) -- cycle, gray(1.0)); dot((-3,-1)); label("$A$",(-3,-1),S); dot((-2,0)); label("$E$",(-2,0),NW); dot((-1,-1)); label("$B$",(-1,-1),S); dot((0,0)); label("$F$",(0,0),N); dot((1,-1)); label("$C$",(1,-1), S); dot((2,0)); label("$G$", (2,0),NE); dot((3,-1)); label("$D$", (3,-1), S); [/asy]
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16\qquad\textbf{(E) } 17$
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First we have to solve the area of the non-shaded area(the semicircles) that are in Circle $F$ .The middle semicircle has area $\frac12\pi$ and the other two have about half of their are inside the circle = $\frac14\pi\ + \frac14\pi\ + \frac12\pi\ = \pi$ . Then we subtract the part of the quartercircle that isn't in Circle $F$ . This is an area equal to that of a triangle minus an minor segment. The height of the triangle is the radius of the semicircles, which is $1$ . The length is the radius of Circle $F$ minus the length from the center of the middle semicircle up to until it is on the edge of the circle. Using the Pythagorean Theorem we can figure out that the length is: \[\sqrt{2^2 - 1^2} = \sqrt{3}.\] This means that the length of the triangle is $2 - \sqrt{3}$ and so the area of the triangle is $\frac{2 - \sqrt{3}}{2}$ . For the area of the segment, it's the area of the sector minus the area of the triangle. The triangle's length is the radius of $F$ $2$ , while its height is the radius of the semicircles: $1$ , so the area is 1. The angle is $30^{\circ}$ as the hypotenuse is the radius of $F$ and the opposite side is the radius of the semicircles, which means the area is $\frac{1}{12}$ of the whole area, which is $4\pi$ so the area of the sector is $\frac{\pi}{3}$ and the area of the segment is $\frac{\pi}{3} - 1$ and so the area of the part of the quartercircles that stick out of Circle $F$ is: \[(\frac{2 - \sqrt{3}}{2})-(\frac{\pi}{3} - 1) = \frac{2 - \sqrt{3}}{2} + 1 - \frac{\pi}{3} = \frac{4 - \sqrt{3}}{2} - \frac{\pi}{3}.\]
Since there are two, one for each side, we have to multiply it by 2, so we have ${4 - \sqrt{3}} - \frac{2\pi}{3}$ , which we subtract from $\pi$ which gets us $\frac{5\pi}{3} - 4 + \sqrt{3}$ which we subtract from $4\pi$ $=$ $\frac{12\pi}{3}$ , which is $\frac{7\pi}{3} + 4 - \sqrt{3}$ so we get $7+3+4+3=\boxed{17}$
| 17
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2,326
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25
| 1
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How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
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Let $f(n)$ be the number of valid sequences of length $n$ (satisfying the conditions given in the problem).
We know our valid sequence must end in a $0$ . Then, since we cannot have two consecutive $0$ s, it must end in a $10$ . Now, we only have two cases: it ends with $010$ , or it ends with $110$ which is equivalent to $0110$ . Thus, our sequence must be of the forms $0\ldots010$ or $0\ldots0110$ . In the first case, the first $n-2$ digits are equivalent to a valid sequence of length $n-2$ . In the second, the first $n-3$ digits are equivalent to a valid sequence of length $n-3$ . Therefore, it must be the case that $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the given conditions).
It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ $f(4)=1$ since the only possible valid sequence is $0110$ $f(5)=1$ since the only possible valid sequence is $01010$
The recursive sequence is then as follows:
\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]
So, our answer is $\boxed{65}$
| 65
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2,327
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25
| 2
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How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
|
After any particular $0$ , the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$ , i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$ s and $3$ s to get $18$ . There are a number of ways to do this:
Case 1 : nine $2$ s - there is only $1$ way to arrange them.
Case 2 : two $3$ s and six $2$ s - there are ${8\choose2} = 28$ ways to arrange them.
Case 3 : four $3$ s and three $2$ s - there are ${7\choose4} = 35$ ways to arrange them.
Case 4 : six $3$ s - there is only $1$ way to arrange them.
Summing the four cases gives $1+28+35+1=\boxed{65}$
| 65
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2,328
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25
| 3
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How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
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We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:
Case 1 : Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ .
Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.
Case 2 : There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.
Case 3 : Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.
Case 4 : No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$
Adding these four cases, we have $1+28+35+1=\boxed{65}$ as our final answer.
| 65
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2,329
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25
| 5
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How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
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Suppose the number of $0$ s is $n$ . We can construct the sequence in two steps:
Step 1: put $n-1$ of $1$ s between the $0$ s;
Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$ s in the $n-1$ spots where there is a $1$ . There are $\binom{n-1}{20-2n}$ ways of doing this.
Now we find the possible values of $n$
First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$ s);
And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$ s).
Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{65}.\]
| 65
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2,330
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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25
| 6
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How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0$ , end with a $0$ , contain no two consecutive $0$ s, and contain no three consecutive $1$ s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
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For a valid sequence of length $n$ , the sequence must be in the form of $01xx...xx10$ . By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of:
So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$
We will calculate $f(19)$ by Dynamic Programming
$f(3) = 1$
$f(4) = 1$
$f(5) = 1$
$f(6) = 2$
$f(7) = 2$
$f(8) = 3$
$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$
$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$
$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$
$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$
$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$
$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$
$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$
$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$
$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$
$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$
$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{65}$
| 65
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2,331
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_3
| 1
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A unit of blood expires after $10!=10\cdot 9 \cdot 8 \cdots 1$ seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?
$\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}$
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There are $60 \cdot 60 \cdot 24 = 86400$ seconds in a day, which means that Yasin's blood expires in $10! \div 86400 = 42$ days. Since there are $31$ days in January (consult a calendar), then $42-31+1$ (Jan 1 doesn't count) is $12$ days into February, so $\boxed{12}$
| 12
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2,332
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4
| 1
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How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$
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We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{24}$ ways to choose the classes.
| 24
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2,333
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4
| 2
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How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$
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Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.
Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$ . Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.
Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$ . Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$ . Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{24}$
| 24
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2,334
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4
| 3
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How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$
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We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s: \[10101\] Now we need to place 1 additional 0. There are 4 places to put it: \[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\] It can be placed in any 1 of the underscores.
Since there are $3!=6$ ways to order the math classes, the answer is $6\cdot 4=\boxed{24}$
| 24
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2,335
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6
| 1
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Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$
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If $65\%$ of the votes were likes, then $35\%$ of the votes were dislikes. $65\%-35\%=30\%$ , so $90$ votes is $30\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $\boxed{300}$
| 300
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2,336
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6
| 2
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Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$
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Let's consider that Sangho has received 100 votes. This means he has received 65 upvotes and 35 downvotes. Part of these upvotes and downvotes cancel out, so Sangho is now left with a total of 30 upvotes, or a score increase of 30. In order for his score to be 90, Sangho must receive three sets of 100 votes. Therefore, the answer is $\boxed{300}$
| 300
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2,337
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6
| 3
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Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$
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Let $x$ be the amount of votes cast, $35\%$ of $x$ would be dislikes and $65\%$ of $x$ would be likes. Since a like earns the video 1 point and a dislike takes 1 point away from the video, the amount of points Sangho's video will have in terms of $x$ is $65\%-35\%=30\%$ of $x$ , or $\frac{3}{10}$ of $x$ . Because Songho's video had a score of 90 points, $\frac{3x}{10} = 90$ . After we solve for $x$ , we get $x = \boxed{300}$
| 300
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2,338
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6
| 4
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Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$
|
Let x be the amount of like votes and y be the amount of dislike votes. Then $x-y=90$ . Since $65\%$ of the total votes are like votes, $0.65(x+y)=x$ . then $.65y = .35x$ or $13/7$ y = $x$ . Plugging that in to the original equation, $6/7 y$ $90$ and $y =105$ Then $x=195$ and the total number of votes is $105+195 = 300$ . The answer is $\boxed{300}$
| 300
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2,339
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6
| 5
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Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of $0$ , and the score increases by $1$ for each like vote and decreases by $1$ for each dislike vote. At one point Sangho saw that his video had a score of $90$ , and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
$\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600$
|
Obviously, $90/(65\%-35\%) = 90/30\% = 300$ . The answer is $\boxed{300}$
| 300
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2,340
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7
| 1
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For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$
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Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need:
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$
| 9
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2,341
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7
| 2
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For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$
|
Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$ . We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$ . For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$ . Also keep in mind that the expression will be an integer for $n=0$ , which gives us a total of $\boxed{9}$ for $n.$
| 9
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2,342
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7
| 3
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For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$
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The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$
The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.
In total, there are $\boxed{9}$ values for $n.$
| 9
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2,343
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8
| 1
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Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
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Let $x$ be the number of $5$ -cent coins that Joe has. Therefore, he must have $(x+3) \ 10$ -cent coins and $(23-(x+3)-x) \ 25$ -cent coins. Since the total value of his collection is $320$ cents, we can write \begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. \end{align*} Joe has six $5$ -cent coins, nine $10$ -cent coins, and eight $25$ -cent coins. Thus, our answer is $8-6 = \boxed{2}.$
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2,344
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8
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Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
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Let the number of $5$ -cent coins be $x,$ the number of $10$ -cent coins be $x+3,$ and the number of $25$ -cent coins be $y.$
Set up the following two equations with the information given in the problem: \[5x+10(x+3)+25y=320 \Rightarrow 15x+25y+30=320 \Rightarrow 15x+25y=290\] \[x+x+3+y=23 \Rightarrow 2x+3+y=23 \Rightarrow 2x+y=20\]
From there, multiply the second equation by $25$ to get \[50x+25y=500.\]
Subtract the first equation from the multiplied second equation to get $35x=210,$ or $x=6.$
Substitute $6$ in for $x$ into one of the equations to get $y=8.$
Finally, the answer is $8-6=\boxed{2}.$
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2,345
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8
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Joe has a collection of $23$ coins, consisting of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins. He has $3$ more $10$ -cent coins than $5$ -cent coins, and the total value of his collection is $320$ cents. How many more $25$ -cent coins does Joe have than $5$ -cent coins?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$
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Let $n,d,$ and $q$ be the numbers of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins in Joe's collection, respectively. We are given that \begin{align*} n+d+q&=23, &(1) \\ 5n+10d+25q&=320, &(2) \\ d&=n+3. &(3) \end{align*} Substituting $(3)$ into each of $(1)$ and $(2)$ and then simplifying, we have \begin{align*} 2n+q&=20, \hspace{17.5mm} &(1\star) \\ 3n+5q&=58. &(2\star) \end{align*} Subtracting $(2\star)$ from $5\cdot(1\star)$ gives $7n=42,$ from which $n=6.$ Substituting this into either $(1\star)$ or $(2\star)$ produces $q=8.$
Finally, the answer is $q-n=\boxed{2}.$
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2,346
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \[x=7+\dfrac{9}{16}x.\] This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$
| 24
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2,347
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 2
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \boxed{24}$
| 24
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2,348
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 3
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$ .
Let the base of the small triangles of area 1 be $x$ , then the base length of $\Delta ADE=4x$ . Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$ , then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$
| 24
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2,349
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 4
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\boxed{24}$
| 24
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2,350
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 5
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\boxed{24}$ , like so.
| 24
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2,351
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 6
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\boxed{24}$
| 24
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2,352
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 7
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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You can assume for the base of one of the smaller triangles to be $\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\boxed{24}$
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2,353
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 8
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)$ . Using the infinite geometric series sum formula gives us $7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\boxed{24}$
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2,354
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9
| 9
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All of the triangles in the diagram below are similar to isosceles triangle $ABC$ , in which $AB=AC$ . Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$ . What is the area of trapezoid $DBCE$
[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
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Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$
Let $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\triangle{ADE}$ . The top triangle has a base of $3\cdot{2}=6$ , and $DE=4\cdot{2}=8.$ The height of $\triangle{ADE}$ is $h+1$ , therefore our ratio is $\frac{h}{6}=\frac{h+1}{8}$ , which yields $h=3$ as our answer.
To find the area of the trapezoid, we can take the area of $\triangle{ABC}$ and subtract the area of $\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\boxed{24}$
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2,355
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11
| 1
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When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$
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Add possibilities. There are $3$ ways to sum to $10$ , listed below.
\[4,1,1,1,1,1,1: 7\] \[3,2,1,1,1,1,1: 42\] \[2,2,2,1,1,1,1: 35.\]
Add up the possibilities: $35+42+7=\boxed{84}$
| 84
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2,356
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11
| 2
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When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$
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We can use generating functions, where $(x+x^2+...+x^6)$ is the function for each die. We want to find the coefficient of $x^{10}$ in $(x+x^2+...+x^6)^7$ , which is the coefficient of $x^3$ in $\left(\frac{1-x^7}{1-x}\right)^7$ . This evaluates to $\dbinom{-7}{3} \cdot (-1)^3=\boxed{84}$
| 84
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2,357
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11
| 3
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When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$
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If we let each number take its minimum value of 1, we will get 7 as the minimum sum. So we can do $10$ $7$ $3$ to find the number of balls we need to distribute to get three more added to the minimum to get 10, so the problem is asking how many ways can you put $3$ balls into $7$ boxes. From there we get $\binom{7+3-1}{7-1}=\binom{9}{6}=\boxed{84}$
| 84
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2,358
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11
| 4
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When $7$ fair standard $6$ -sided dice are thrown, the probability that the sum of the numbers on the top faces is $10$ can be written as \[\frac{n}{6^{7}},\] where $n$ is a positive integer. What is $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$
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We can use number separation for this problem. If we set each of the dice value to $D\{a, b, c, d, e, f, g, h\}$ , we can say $D = 10$ and each of $D$ 's elements are larger than $0$ . Using the positive number separation formula, which is $\dbinom{n-1}{r-1}$ , we can make the following equations. \begin{align*} D &= 10 \\ a+b+c+d+e+f+g &= 10 \\ \dbinom{10-1}{7-1} &= \\ \dbinom{9}{6} &= \\ \dbinom{9}{3} &= \\ \dfrac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} &= \\ 12 \cdot 7 &= \boxed{84}
| 84
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2,359
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 1
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
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We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: [asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy] Now, it becomes clear that there are $\boxed{3}$ intersection points.
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2,360
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 2
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
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$x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$ $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
$2y-3$ is negative so $|2y-3| = 3-2y = 1$ $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$
$\textbf{Case 2:}$ $y = 1$ :
It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$ . We already have this solution from Case 2 as $y = 1$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$ $y = \frac{1}{2}$ and $x = \frac{3}{2}$ .
Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$ , and the answer is $\boxed{3}$
| 3
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2,361
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 3
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
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Note that $||x| - |y||$ can take on one of four values: $x + y$ $x - y$ $-x + y$ $-x -y$ . So we have 4 cases:
Case 1: ||x| - |y|| = x+y \[x+3y=3\] \[x+y=1\]
Subtracting:
$2y=2 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$
Case 2: ||x| - |y|| = x-y \[x+3y=3\] \[x-y=1\]
Subtacting:
$4y=2 \Rightarrow y=\dfrac{1}{2}$ an $x=\dfrac{3}{2}$
$\text{Result: } \left(\dfrac{3}{2},\dfrac{1}{2}\right)$
Case 3: ||x| - |y|| = -x+y \[x+3y=3\] \[-x+y=1\]
Adding:
$4y=4 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$ . Since this is the same solution as we got in Case 1, we can not count this (otherwise, we would have overcounted).
Case 4: ||x| - |y|| = -x-y \[x+3y=3\] \[-x-y=1\]
Adding:
$2y=4 \Rightarrow y=2$ and $x=-3$
$\text{Result: } (-3,2)$
Answer
We solved each case by elimination (either adding the two equations or subtracting), to obtain three solutions: $(0, 1)$ $(-3,2)$ $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
Our answer is $\boxed{3}$
| 3
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2,362
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 4
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
|
Just as in solution $2$ , we derive the equation $||3-3y|-|y||=1$ . If we remove the absolute values, the equation collapses into four different possible values. $3-2y$ $3-4y$ $2y-3$ , and $4y-3$ , each equal to either $1$ or $-1$ . Remember that if $P-Q=a$ , then $Q-P=-a$ . Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$ , and $3-4y$ and $4y-3$ . Find the values of $y$ when $3-2y=1$ $3-2y=-1$ $3-4y=1$ and $3-4y=-1$ . We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$ , so the answer is $\boxed{3}$
| 3
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2,363
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 5
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
|
Just as in solution $2$ , we derive the equation $x=3-3y$ . Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$ . Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$ . From here we know that $3y-3y^2$ is either positive or negative.
When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$ .
When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$ . Clearly, there are $3$ different pairs of values and that gives us $\boxed{3}$
| 3
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2,364
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12
| 6
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How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
|
Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$ ) and the other is linear. Subtract to get $\boxed{3}$
| 3
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2,365
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 1
|
What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
We write \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.\] Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\boxed{80}$
| 80
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2,366
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 2
|
What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
\[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}\left(\frac{3^{100}}{2^{96}}\right)+2^{96}\left(2^{4}\right)}{2^{96}\left(\frac{3}{2}\right)^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{\left(\frac{3}{2}\right)^{96}+1}=\frac{\frac{3^{100}}{2^{100}}\cdot2^{4}+2^{4}}{\left(\frac{3}{2}\right)^{96}+1}=\frac{2^{4}\left(\frac{3^{100}}{2^{100}}+1\right)}{\left(\frac{3}{2}\right)^{96}+1}.\]
We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.
\[\frac{2^{4}\left(\frac{3^{100}}{2^{100}}+1\right)}{\left(\frac{3}{2}\right)^{96}+1}<\frac{2^{4}\left(\frac{3^{100}}{2^{100}}\right)}{\left(\frac{3}{2}\right)^{96}},\]
\[\frac{2^{4}\left(\frac{3^{100}}{2^{100}}\right)}{\left(\frac{3}{2}\right)^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81.\]
So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\boxed{80}$
| 80
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2,367
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 3
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What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
Let $x=3^{96}$ and $y=2^{96}$ . Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$ .
Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$ .
So , $16+\frac{65x}{x+y}<16+65=81$ .
And our only answer choice less than 81 is $\boxed{80}$ (RegularHexagon)
| 80
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2,368
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 4
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What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
Dividing by $2^{96}$ in both numerator and denominator, this fraction can be rewritten as \[\frac{81 \times (1.5)^{96} + 16}{(1.5)^{96} + 1}.\] Notice that the $+1$ and the $+16$ will be so insignificant compared to a number such as $(1.5)^{96},$ and that thereby the fraction will be ever so slightly less than $81$ . Thereby, we see that the answer is $\boxed{80}.$
| 80
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2,369
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 5
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What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
If you multiply $(3^{96} + 2^{96})$ by $(3^{4} + 2^{4})$ (to get the exponent up to 100), you'll get $(3^{100} + 2^{100}) + 3^{96} \cdot 2^{4} + 2^{96} \cdot 3^{4}$ . Thus, in the numerator, if you add and subtract by $3^{96} \cdot 2^{4}$ and $2^{96} \cdot 3^{4}$ , you'll get $\frac{(3^{4} + 2^{4})(3^{100} + 2^{100}) - 3^{96} \cdot 2^{4} - 2^{96} \cdot 3^{4}}{3^{96}+2^{96}}$ . You can then take out out the first number to get $3^{4} + 2^{4} - \frac{3^{96} \cdot 2^{4} + 2^{96} \cdot 3^{4}}{3^{96}+2^{96}}$ . This can then be written as $87 - \frac{16 \cdot 3^{96} + 16 \cdot 2^{96} + 75 \cdot 2^{96}}{3^{96}+2^{96}}$ , factoring out the 16 and splitting the fraction will give you $87 - 16 - \frac{65 \cdot 2^{96}}{3^{96}+2^{96}}$ , giving you $81 - \frac{65 \cdot 2^{96}}{3^{96}+2^{96}}$ . While you can roughly say that $\frac{65 \cdot 2^{96}}{3^{96}+2^{96}} < 1$ you can also notice that the only answer choice less than 81 is 80, thus the answer is $\boxed{80}.$
| 80
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2,370
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14
| 6
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What is the greatest integer less than or equal to \[\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?\]
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
|
If you factor out $3^{100}$ from the numerator and $3^{96}$ from the denominator, you will get $\frac{3^{100}\left(1+(\frac{2}{3}\right)^{100})}{3^{96}\left(1+(\frac{2}{3}\right)^{96})}$ . Divide the numerator and denominator by $3^{96}$ to get $\frac{81\left(1+(\frac{2}{3}\right)^{100})}{\left(1+(\frac{2}{3}\right)^{96})}$ . We see that every time we multiply $\frac{2}{3}$ by itself, it slightly decreases, so $1+(\frac{2}{3})^{100}$ will be ever so slightly smaller than $1+(\frac{2}{3})^{96}$ . Thus, the decimal representation of $\frac{\left(1+(\frac{2}{3}\right)^{100})}{\left(1+(\frac{2}{3}\right)^{96})}$ will be extremely close to $1$ , so our solution will be the largest integer that is less than $81$ . Thus, the answer is $\boxed{80}.$
| 80
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2,371
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_15
| 1
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Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$ , as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
[asy] draw(circle((0,0),13)); draw(circle((5,-6.2),5)); draw(circle((-5,-6.2),5)); label("$B$", (9.5,-9.5), S); label("$A$", (-9.5,-9.5), S); [/asy]
$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$
|
[asy] draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$X$", (0,0), N); label("$Y$", (-5,-6.25),NW); label("$Z$", (5,-6.25),NE); [/asy]
Let the center of the surrounding circle be $X$ . The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$ $YZ$ $XA$ , and $XB$ . Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ by SAS.
Writing out the ratios, we get \[\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.\] Therefore, our answer is $65+4= \boxed{69}$
| 69
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2,372
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17
| 1
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Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$
|
We start with $2$ because $1$ is not an answer choice. We would have to include every odd number except $1$ to fill out the set, but then $3$ and $9$ would violate the rule, so that won't work.
Experimentation with $3$ shows it's likewise impossible. You can include $7,11,$ and either $5$ or $10$ (which are always safe). But after adding either $4$ or $8$ we have no more valid numbers.
Finally, starting with $4,$ we find that the sequence $4,5,6,7,9,11$ works, giving us $\boxed{4}.$
| 4
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2,373
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17
| 2
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Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$
|
We know that all odd numbers except $1,$ namely $3, 5, 7, 9, 11,$ can be used.
Now we have $7$ possibilities to choose from for the last number (out of $1, 2, 4, 6, 8, 10, 12$ ). We can eliminate $1, 2, 10,$ and $12,$ and we have $4, 6, 8$ to choose from. However, $9$ is a multiple of $3.$ Now we have to take out either $3$ or $9$ from the list. If we take out $9,$ none of the numbers would work, but if we take out $3,$ we get \[4, 5, 6, 7, 9, 11.\] The least number is $4,$ so the answer is $\boxed{4}.$
| 4
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2,374
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17
| 3
|
Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$
|
We can get the multiples for the numbers in the original set with multiples in the same original set \begin{align*} 1&: \ \text{all elements of }\{1,2,\dots,12\} \\ 2&: \ 4,6,8,10,12 \\ 3&: \ 6,9,12 \\ 4&: \ 8,12 \\ 5&: \ 10 \\ 6&: \ 12 \end{align*} It will be safe to start with $5$ or $6$ since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.
Trying $4,$ we can get $4,5,6,7,9,11.$ So $4$ works.
Trying $3,$ it won't work, so the least is $4.$ This means the answer is $\boxed{4}.$
| 4
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2,375
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17
| 4
|
Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$
|
We partition $\{1,2,\ldots,12\}$ into six nonempty subsets such that for every subset, each element is a multiple of all elements less than or equal to itself: \[\{1,2,4,8\}, \ \{3,6,12\}, \ \{5,10\}, \ \{7\}, \ \{9\}, \ \{11\}.\] Clearly, $S$ must contain exactly one element from each subset:
If $4\in S,$ then the possibilities of $S$ are $\{4,5,6,7,9,11\}$ or $\{4,6,7,9,10,11\}.$ So, the least possible value of an element in $S$ is $\boxed{4}.$
| 4
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2,376
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17
| 5
|
Let $S$ be a set of $6$ integers taken from $\{1,2,\dots,12\}$ with the property that if $a$ and $b$ are elements of $S$ with $a<b$ , then $b$ is not a multiple of $a$ . What is the least possible value of an element in $S$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7$
|
We start with 2 as 1 is not an answer option. Our set would be $\{2,3,5,7,11\}$ . We realize we cannot add 12 to the set because 12 is a multiple of 3. Our set only has 5 elements, so starting with 2 won't work.
We try 3 next. Our set becomes $\{3,4,5,7,11\}$ . We run into the same issue as before so starting with 3 won't work.
We then try 4. Our set becomes $\{4,5,6,7,9,11\}$ . We see we have 6 elements with none being multiples of each other. Therefore our answer is $\boxed{4}.$
| 4
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2,377
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 1
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$
| 281
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2,378
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 2
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$ $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . (RegularHexagon, KLBBC minor changes)
| 281
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2,379
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 3
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic.
Note that if $a_7$ is 1, then it's impossible for \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. (We also must convince ourselves that these $2187$ different sets of coefficients must necessarily yield $2187$ different integer results.)
As A, B, and C are all less than 2187, the answer must be $\boxed{3281}$
| 281
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2,380
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 8
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
To get the number of integers, we can get the highest positive integer that can be represented using \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
Note that the least nonnegative integer that can be represented is $0$ , when all $a_i=0$ . The highest number will be the number when all $a_i=1$ . That will be \[3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}\] \[=3280\]
Therefore, there are $3280$ positive integers and $(3280+1)$ nonnegative integers (while including $0$ ) that can be represented. Our answer is $\boxed{3281}$
| 281
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2,381
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 9
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
Notice that there are $3^8$ options for $a_7, a_6, \cdots a_0$ since each $a_i$ can take on the value $-1$ $0$ , or $1$ . Now we want to find how many of them are positive and then we can add one in the end to account for $0$ (they are asking for non-negative).
By symmetry (look out for these on the contest), we see that exactly half of them are positive. So $\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.$ Now we will add $1$ because of the $0$ to account for the non-negative solutions.
So our final answer is $3280 + 1 = 3281$ which is $\boxed{3281}$
| 281
|
2,382
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https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_18
| 10
|
How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E) } 59,048$
|
Obviously, there are $3^8 = 6561$ possible, and one of them is 0, so other $6560$ are either positive or negative. By the symmetry, we can get the answer is $6560/2 + 1$ which is $\boxed{3281}$
| 281
|
2,383
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_20
| 2
|
A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$
|
\[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$
There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$ . Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{1022}$ -hansenhe
| 22
|
2,384
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 1
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
Substituting $y=x^2-a$ into $x^2+y^2=a^2$ , we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are $4$ total roots (counting multiplicity). We see that $x=0$ always has at least one intersection at $(0,-a)$ (and is in fact a double root).
The other two intersection points have $x$ coordinates $\pm\sqrt{2a-1}$ . We must have $2a-1> 0;$ otherwise we are in the case where the parabola lies entirely above the circle (tangent at the point $(0,a)$ ). This only results in a single intersection point in the real coordinate plane. Thus, we see that $\boxed{12}$
| 12
|
2,385
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 2
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$ , which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$ . This further simplifies to $x^2(1 + x^2 - 2a) = 0$ . Thus, either $x^2 = 0$ , or $x^2 - 2a + 1 = 0$
Since we care about $a$ , we consider the second case. We solve in terms of $a$ , giving $a = \frac{x^2}{2} + \frac{1}{2}$ . We see that in order to find the range in which $a$ lies, we must find the vertex of this equation, which turns out to be $\left(0, \frac{1}{2}\right)$ . Hence, we know that the minimum is $\frac{1}{2}$ , which further implies that $\boxed{12}$
| 12
|
2,386
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 3
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
[asy] Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); [/asy]
Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$ . We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$ . Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$ . Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$ , where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{12}$ as the right answer.
| 12
|
2,387
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 4
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is $2$ , the radius of the circle that matches it has a radius of $\frac{1}{2}$ . This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only $1$ point. When a larger circle is used, it is tangent to $3$ points because the points on either side are now separated from the vertex. Therefore, $\boxed{12}$ is correct.
| 12
|
2,388
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 5
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
Now, let's graph these two equations. We want the blue parabola to be inside this red circle. [asy] import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$a$",(0,5),NE); dot(scale(.7)*"$-a$",(0,-5),N); dot(scale(.7)*"$a$",(5,0),NE); dot(scale(.7)*"$-a$",(-5,0),SE); [/asy] Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$ . Expanding, we get $x^4-2ax^2+x^2=0$ . Factoring out the $x^2$ , we get $x^2(x^2-2a+1)=0$ . Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$ . Therefore, $2a-1>0$ , which means $\boxed{12}$
| 12
|
2,389
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 6
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
In order to solve for the values of $a$ , we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that $\sqrt{a^2 - x^2} = x^2 - a$ . Now, we take square of both sides, and rearrange to obtain $x^4 - (2a - 1)x^2 = 0$ . Now, we may take the second derivative of the equation to obtain $6x^2 - (2a - 1) = 0$ . Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,
\[b^2 - 4ac > 0 \rightarrow 0 - 4(6)(-(2a - 1)) > 0 \rightarrow a > \frac{1}{2}\] The answer is $\boxed{12}$ and we are done.
| 12
|
2,390
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 7
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the $y$ -axis, thus, for them to intersect at exactly $3$ points, it suffices to find the $y$ solution.
First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than $0$ (It is the square root part of the quadratic formula). To find when it is $0$ , we find the roots: \[4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\] Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{12}$
| 12
|
2,391
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 8
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
We can see that if $a = 1$ , we know that the points where the two curves intersect are $(0, -1), (1, 0)$ and $(-1, 0)$ .Because there are only $3$ intersections and $a > 1/2$ , as well as $a > 1/4$ we know that either $\textbf{(D)}$ or $\textbf{(E)}$ is the correct answer. Then we can test a number from $(1/2, 1/4)$ to eliminate the remaining answer. So $\boxed{12}$ is the correct answer.
| 12
|
2,392
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 9
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
Simply plug in $a = \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\boxed{12}$
| 12
|
2,393
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21
| 10
|
Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$
|
An ideal solution come to mind is where they intersect at the $x$ -axis at the same time, which is $(a, 0)$ and $(-a, 0).$ Take the root of our $y=x^2-a$ we get $x=\sqrt{a}$ , set them equal we get $a=\sqrt{a}.$ The only answer is $1$ so it only left us with the answer choice $\boxed{12}$
| 12
|
2,394
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 1
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
The GCD information tells us that $24$ divides $a$ , both $24$ and $36$ divide $b$ , both $36$ and $54$ divide $c$ , and $54$ divides $d$ . Note that we have the prime factorizations: \begin{align*} 24 &= 2^3\cdot 3,\\ 36 &= 2^2\cdot 3^2,\\ 54 &= 2\cdot 3^3. \end{align*}
Hence we have \begin{align*} a &= 2^3\cdot 3\cdot w\\ b &= 2^3\cdot 3^2\cdot x\\ c &= 2^2\cdot 3^3\cdot y\\ d &= 2\cdot 3^3\cdot z \end{align*} for some positive integers $w,x,y,z$ . Now if $3$ divides $w$ , then $\gcd(a,b)$ would be at least $2^3\cdot 3^2$ which is too large, hence $3$ does not divide $w$ . Similarly, if $2$ divides $z$ , then $\gcd(c,d)$ would be at least $2^2\cdot 3^3$ which is too large, so $2$ does not divide $z$ . Therefore, \[\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)\] where neither $2$ nor $3$ divide $\gcd(w,z)$ . In other words, $\gcd(w,z)$ is divisible only by primes that are at least $5$ . The only possible value of $\gcd(a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot3\cdot13$ , so the answer is $\boxed{13}$
| 13
|
2,395
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 2
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
We can say that $a$ and $b$ 'have' $2^3 \cdot 3$ , that $b$ and $c$ have $2^2 \cdot 3^2$ , and that $c$ and $d$ have $3^3 \cdot 2$ . Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 \cdot 3^2$ , and thus $a$ has $2^3 \cdot 3$ (and no more powers of $3$ because otherwise $\gcd(a,b)$ would be different). In addition, $c$ has $3^3 \cdot 2^2$ , and thus $d$ has $3^3 \cdot 2$ (similar to $a$ , we see that $d$ cannot have any other powers of $2$ ). We now assume the simplest scenario, where $a = 2^3 \cdot 3$ and $d = 3^3 \cdot 2$ . According to this base case, we have $\gcd(a, d) = 2 \cdot 3 = 6$ . We want an extra factor between the two such that this number is between $70$ and $100$ , and this new factor cannot be divisible by $2$ or $3$ . Checking through, we see that $6 \cdot 13$ is the only one that works. Therefore the answer is $\boxed{13}$
| 13
|
2,396
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 3
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
First off, note that $24$ $36$ , and $54$ are all of the form $2^x\times3^y$ . The prime factorizations are $2^3\times 3^1$ $2^2\times 3^2$ and $2^1\times 3^3$ , respectively. Now, let $a_2$ and $a_3$ be the number of times $2$ and $3$ go into $a$ , respectively. Define $b_2$ $b_3$ $c_2$ , and $c_3$ similarly. Now, translate the $lcm$ s into the following: \[1) \min(a_2,b_2)=3\] \[2) \min(a_3,b_3)=1\] \[3) \min(b_2,c_2)=2\] \[4) \min(b_3,c_3)=2\] \[5) \min(c_2,d_2)=1\] \[6) \min(c_3,d_3)=3\]
From $4)$ , we see that $b_3 \geq 2$ , thus from $2)$ $a_3 = 1$ . Similarly, from $3)$ $c_2 \geq 2$ , thus from $5)$ $d_2 = 1$
Note also that $d_3 \geq 3$ and $a_2 \geq 3$ . Therefore \[\min(a_2, d_2) = 1\] \[\min(a_3, d_3) = 1\] Thus, $\gcd(a, d) = 2 \times 3 \times k$ for some $k$ having no factors of $2$ or $3$
Since $70 < \gcd(a, d) < 100$ , the only values for k are $12, 13, 14, 15, 16$ , but all have either factors of $2$ or $3$ , except $\boxed{13}$ . And we're done.
| 13
|
2,397
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 4
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
Notice that $\gcd (a,b,c,d)=\gcd(\gcd(a,b),\gcd(b,c),\gcd(c,d))=\gcd(24,36,54)=6$ , so $\gcd(d,a)$ must be a multiple of $6$ . The only answer choice that gives a value between $70$ and $100$ when multiplied by $6$ is $\boxed{13}$ . - mathleticguyyy + einstein
| 13
|
2,398
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 5
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z, W; real R; path quad; //Variable Definitions R = 1; X = R*dir(45); Y = R*dir(135); Z = R*dir(-135); W = R*dir(-45); quad = X--Y--Z--W--cycle; //Diagram draw(quad); label("$b$",X,NE); label("$a=2^3 \cdot 3 \cdot p$",Y,NW); label("$d=2 \cdot 3^3 \cdot q$",Z,SW); label("$c$",W,SE); label("$2^3 \cdot 3$",X--Y); label("$2^2 \cdot 3^2$",X--W); label("$2 \cdot 3^3$",Z--W); label("$2 \cdot 3 \cdot k$",Z--Y); [/asy]
The relationship of $a$ $b$ $c$ , and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$ $70 < 6k < 100$ $12 \le k \le 16$ $k=\boxed{13}$
| 13
|
2,399
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_22
| 6
|
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
|
Just as in solution $1$ , we prime factorize $a, b, c$ and $d$ to observe that
$a=2^3\cdot{3}\cdot{w}$
$b=2^3\cdot{3^2}\cdot{x}$
$c=2^2\cdot{3^3}\cdot{y}$
$d=2\cdot{3^3}\cdot{z}.$
Substituting these expressions for $a$ and $d$ into the final given,
$70<\text{gcd}(2\cdot{3^3}\cdot{z}, 2^3\cdot{3}\cdot{w})<100.$
The greatest common divisor of these two numbers is already $6$ . If $k$ is what we wish to multiply $6$ by to obtain the gcd of these two numbers, then
$70<6k<100$ . Testing the answer choices, only $13$ works for $k$ (in order for the compound inequality to hold). so our gcd is $78$ , which means that $\boxed{13}$ must divide $a$
| 13
|
2,400
|
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_24
| 1
|
Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$
$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$
|
Let $BC = a$ $BG = x$ $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that \[\frac{50}{x} = \frac{10}{y},\] where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{75}$ by substituting.
| 75
|
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