id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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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\tex... | 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 $\boxe... | 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\tex... | 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, ... | 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, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$... | 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=... | 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 s... | 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) } ... | 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) } ... | 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 o... | 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) } ... | 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{infini... | 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 $C$ ,... | 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{infini... | 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 $C... | 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) } ... | 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 $\frac{19x}{90}=95$ ... | 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) } ... | 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 s... | 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... | 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{3... | 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 u... | 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... | 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^{... | 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 $2$... | 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 tri... | 17 |
2,325 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_20 | 2 | 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 $2$... | 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 Ci... | 17 |
2,326 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | 1 | 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?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | 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... | 65 |
2,327 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | 2 | 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 w... | 65 |
2,328 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | 3 | 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?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | 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 ... | 65 |
2,329 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | 5 | 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?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | 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... | 65 |
2,330 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_25 | 6 | 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?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$ | 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 Dy... | 65 |
2,331 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_3 | 1 | 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{... | 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 |
2,332 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4 | 1 | 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\t... | 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 $1, 4, 6$
Peri... | 24 |
2,333 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4 | 2 | 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\t... | 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 ... | 24 |
2,334 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_4 | 3 | 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\t... | 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{\h... | 24 |
2,335 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | 1 | 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 c... | 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 |
2,336 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | 2 | 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 c... | 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,... | 300 |
2,337 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | 3 | 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 c... | 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 Songh... | 300 |
2,338 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | 4 | 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 c... | 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... | 300 |
2,339 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_6 | 5 | 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 c... | Obviously, $90/(65\%-35\%) = 90/30\% = 300$ . The answer is $\boxed{300}$ | 300 |
2,340 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7 | 1 | 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 \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] 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 |
2,341 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7 | 2 | 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... | 9 |
2,342 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_7 | 3 | For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$ | 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 |
2,343 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8 | 1 | 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 \t... | 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 &= ... | 2 |
2,344 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8 | 2 | 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 \t... | 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 \Righta... | 2 |
2,345 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_8 | 3 | 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 \t... | 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{alig... | 2 |
2,346 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 1 | 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... | 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 |
2,347 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 2 | 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... | 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... | 24 |
2,348 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 3 | 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... | 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{... | 24 |
2,349 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 4 | 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... | 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 |
2,350 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 5 | 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... | 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 ar... | 24 |
2,351 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 6 | 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... | 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 proporti... | 24 |
2,352 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 7 | 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... | 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 $40-16=\boxed{24}$ | 24 |
2,353 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 8 | 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... | 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)... | 24 |
2,354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_9 | 9 | 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... | 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 $\triangl... | 24 |
2,355 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | 1 | 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 $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | 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 |
2,356 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | 2 | 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 $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | 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 |
2,357 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | 3 | 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 $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | 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}... | 84 |
2,358 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_11 | 4 | 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 $n$
$\textbf{(A) }42\qquad \textbf{(B) }49\qquad \textbf{(C) }56\qquad \textbf{(D) }63\qquad \textbf{(E) }84\qquad$ | 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 &= 1... | 84 |
2,359 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 1 | 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*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 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 |
2,360 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 2 | 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*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | $x+3y=3$ 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... | 3 |
2,361 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 3 | 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*} $\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 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:
$4y=2 \Rightarrow y=\dfrac{1}{2}$... | 3 |
2,362 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 4 | 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 $-1$ i... | 3 |
2,363 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 5 | 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 the... | 3 |
2,364 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_12 | 6 | 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 |
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... | 80 |
2,366 | 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... | 80 |
2,367 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14 | 3 | 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 |
2,368 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14 | 4 | 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 $81$ . T... | 80 |
2,369 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14 | 5 | 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}) -... | 80 |
2,370 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_14 | 6 | 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)^{9... | 80 |
2,371 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_15 | 1 | 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... | [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,... | 69 |
2,372 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17 | 1 | 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)... | 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... | 4 |
2,373 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_17 | 2 | 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)... | 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 eithe... | 4 |
2,374 | 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)... | 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 ... | 4 |
2,375 | 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)... | 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 pos... | 4 |
2,376 | 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)... | 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 wo... | 4 |
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... | 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... | 281 |
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... | 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 $a_i$ ... | 281 |
2,379 | 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... | 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... | 281 |
2,380 | 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... | 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 $0$ , ... | 281 |
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... | 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 th... | 281 |
2,382 | 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... | 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 m... | \[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $... | 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 ... | 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 inters... | 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 ... | 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}{... | 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 ... | [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 o... | 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 ... | 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 s... | 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 ... | 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) {... | 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 ... | 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 ta... | 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 ... | 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 t... | 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 ... | 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 $(1/2, 1/4)$ t... | 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 ... | 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 ... | 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{... | 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\cdo... | 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{... | 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 addit... | 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{... | 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, tra... | 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{... | 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 $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{... | [asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z, W; real R; path quad; //Variable Definitions R = 1; X = R*dir(45); Y = R*dir(135); Z = R*dir(-135); W = R*dir(-45); quad = X--Y--Z--W--cycle; //Diagram draw(quad); label("$b$",X,NE); label("$a=2^3 \cdot 3 \cdot p$",Y,NW); label... | 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{... | 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 grea... | 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) }6... | 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 o... | 75 |
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