problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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901ecba54db1e3832a2fae79df10cd17 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_14 | Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$ | 120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
$(5)(4)(3)(2)(1) = 120$ Making the greatest integer,
$(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)$ $= (5)(8)(3)(1)(1) =120$
8 is the largest value and will go in the front.
We can express the number as $85311$
$8+5+3+1+1=\boxed... | D | 18 |
55ccf4688843e354cc85e6dd31f4d979 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15 | In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircl... | Let the radius of the large circle be $R$ . Then, the radius of the smaller circles are $\frac R2$ . The areas of the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is $\frac 14$ . This means the combined area of the 2 smaller circles is half ... | D | 1 |
55ccf4688843e354cc85e6dd31f4d979 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_15 | In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?
[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircl... | Let the radius of the two smaller circles be $r$ . It follows that the area of one of the smaller circles is ${\pi}r^2$ . Thus, the area of the two inner circles combined would evaluate to $2{\pi}r^2$ which is $1$ . Since the radius of the bigger circle is two times that of the smaller circles (the diameter), the radiu... | D | 1 |
dd938fa724a3e05463b56800b7cd7da4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_16 | Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?
$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760... | Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5$ factorial. Now, we multiply this product by $2!$ because there are $... | C | 5760 |
7b71c770e26b596f6166c35d485a1c81 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17 | Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \t... | Every 10 feet Bella goes, Ella goes 50 feet, which means a total of 60 feet. They need to travel that 60 feet $10560\div60=176$ times to travel the entire 2 miles. Since Bella goes 10 feet 176 times, this means that she travels a total of 1760 feet. And since she walks 2.5 feet each step, $1760\div2.5=\boxed{704}$ | A | 704 |
7b71c770e26b596f6166c35d485a1c81 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17 | Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \t... | We know that Bella goes 2.5 feet per step and since Ella rides 5 times faster than Bella she must go 12.5 feet on her bike for every step of Bella's. For Bella, it takes 4,224 steps, and for Ella, it takes 1/5th those steps since Ella goes 5 times faster than Bella, taking her 844.8 steps. The number of steps where the... | A | 704 |
7b71c770e26b596f6166c35d485a1c81 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_17 | Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides $5$ times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \t... | We can turn $2 \tfrac{1}{2}$ into an improper fraction. It will then become 5/2. Since Ella bikes 5 times faster, we multiply 5/2 by 5 to get 25/2. Then we add 5/2 to it in order to find the distance they walk and bike together in total. After adding, you should get 30/2 which is equal to 15. This means that after 15 t... | A | 704 |
cce7bc1f46689306c6f7c0da2b34882f | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_18 | How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | We can first find the prime factorization of $23,232$ , which is $2^6\cdot3^1\cdot11^2$ . Now, we add one to our powers and multiply. Therefore, the answer is $(6+1)\cdot(1+1)\cdot(2+1)=7\cdot2\cdot3=\boxed{42}$ | E | 42 |
cce7bc1f46689306c6f7c0da2b34882f | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_18 | How many positive factors does $23,232$ have?
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | Observe that $69696$ $264^2$ , so this is $\frac{1}{3}$ of $264^2$ which is $88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3$ , which has $3 \cdot 7 \cdot 2 = 42$ factors. The answer is $\boxed{42}$ | E | 42 |
7ebb11e2b7b205ca77bbfd3458785374 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{8}$ | C | 8 |
7ebb11e2b7b205ca77bbfd3458785374 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are $2^3=8$ ways.
Then the left 2 boxes on the row above are determined.
Then the left 1 box on the row above that is determined
Then the right 1 box on that row is determined.
Then the right 1 box on the row below is determined.
Then... | C | 8 |
7ebb11e2b7b205ca77bbfd3458785374 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are $a$ $b$ $c$ , and $d$ , which are either 1 or -1.
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$a$",(0,0)); draw(shift(1,0)*box); label("$b$",(1,0)); draw(shif... | C | 8 |
7ebb11e2b7b205ca77bbfd3458785374 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_19 | In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the ... | The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a $+$ on the above, half of them have a $-$ above. So, For the lowest layer with $4$ blocks, there are $2^4=16$ possible combination and half of them will lead a $+$ (or $-$ ) ... | C | 8 |
75d5a9e42bc7c3b35b54d0e096c0a354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | Looking at the values, we notice that $11-7=4$ $9-5=4$ and $6-2=4$ . This means we are looking for a value that is four less than a multiple of $11$ $9$ , and $6$ . The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$ , so the numbers that fulfill this can be written as $198k-4$ , where $k$ is a posit... | E | 5 |
75d5a9e42bc7c3b35b54d0e096c0a354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | Let us create the equations: $6x+2 = 9y+5 = 11z+7$ , and we know $100 \leq 11z+7 <1000$ , it gives us $9 \leq z \leq 90$ , which is the range of the value of z. Because of $6x+2=11z+7$ , then $6x=11z+5=6z+5(z+1)$ , so $(z+1)$ must be a mutiple of 6. Because of $9y+5=11z+7$ , then $9y=11z+2=9z+2(z+1)$ , so $(z+1)$ must ... | E | 5 |
75d5a9e42bc7c3b35b54d0e096c0a354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | By the Chinese Remainder Theorem , we have that all solutions are in the form $x=198k+194$ where $k\in \mathbb{Z}.$ Counting the number of values, we get $\boxed{5}.$ | E | 5 |
75d5a9e42bc7c3b35b54d0e096c0a354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We can use modular arithmetic. Set up the equations: $x \equiv 2 \mod 6,$ $x \equiv 5 \mod 9,$ and $x \equiv 7 \mod 11.$ These equations can also be written as $x+4 \equiv 0 \mod 6,$ $x+4 \equiv 0 \mod 9,$ and $x+4 \equiv 0 \mod 11.$ Since $x+4$ is congruent to numbers $6, 9,$ and $11,$ then it must also be congruent t... | E | 5 |
75d5a9e42bc7c3b35b54d0e096c0a354 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_21 | How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | Let $N$ be the three digit positive integer. $N = 6a + 2 = 9b + 5 = 11c + 7$ . Then, we add four to all sides and write $N + 4 = 6(a+1) = 9(b+1) = 11(c+1)$ . Now, we know that $N + 4$ is divisible by 6, 9, and 11. The LCM of 6, 9, and 11 is equal to 198, so $N = 198k - 4$ . From this, we can figure out that $N$ can be ... | E | 5 |
20e0908451c3becf4de97446fe4accd4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | Let the area of $\triangle CEF$ be $x$ . Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$
By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$ . Now, consider trapezoid $ABED$ . Its area is $45+4x$ , wh... | B | 108 |
20e0908451c3becf4de97446fe4accd4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | We can use analytic geometry for this problem.
Let us start by giving $D$ the coordinate $(0,0)$ $A$ the coordinate $(0,1)$ , and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving for their intersection gives point $F$ coordinates $\left(\fra... | B | 108 |
20e0908451c3becf4de97446fe4accd4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | $\triangle ABC$ has half the area of the square. $\triangle FEC$ has base equal to half the square side length, and by AA Similarity with $\triangle FBA$ , it has 1/(1+2)= 1/3 the height, so has $\dfrac1{12}$ th area of square. Thus, the area of the quadrilateral is $1-1/2-1/12=5/12$ th the area of the square. The area... | B | 108 |
20e0908451c3becf4de97446fe4accd4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$ . Drop an altitude from $F$ to $\overline{CE}$ and call it $h$ . Also, call $\overline{CE}$ $x$ . As stated before, we have $\triangle ABF \sim \triangle CEF$ , so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2... | B | 108 |
20e0908451c3becf4de97446fe4accd4 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_22 | Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); la... | Solution with Cartesian and Barycentric Coordinates:
We start with the following:
Claim: Given a square $ABCD$ , let $E$ be the midpoint of $\overline{DC}$ and let $BE\cap AC = F$ . Then $\frac {AF}{FC}=2$
Proof: We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$ . We have that $\overline{AC... | null | 108 |
4c757b1760aba59edc146b933c5c0edd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | Choose side "lengths" $a,b,c$ for the triangle, where "length" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\leq b$ and $a < c$ ] or [ $a=b=c$ (but thi... | D | 57 |
4c757b1760aba59edc146b933c5c0edd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent.
If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put th... | D | 57 |
4c757b1760aba59edc146b933c5c0edd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is
$\frac{8\cdot6-8}{{8 \choose 3 }}$ $=\frac{... | D | 57 |
4c757b1760aba59edc146b933c5c0edd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | Let $1$ point of the triangle be fixed at the top. Then, there are ${7 \choose 2} = 21$ ways to choose the other $2$ points. There must be $3$ spaces in the points and $3$ points themselves. This leaves $2$ extra points to be placed anywhere. By stars and bars, there are $3$ triangle points ( $n$ ) and $2$ extra points... | D | 57 |
4c757b1760aba59edc146b933c5c0edd | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_23 | From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[... | We select a vertex of the octagon; this will be the first vertex of our triangle. Define the $distance$ of a vertex from another to be the minimum number of edges that one must travel on to get from one vertex to the other. There are three distinct cases; the second vertex is a distance of 1 away from the selected vert... | D | 57 |
eaf7d3e4623ee9d5f2ae3bc71d450919 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$ | We compute $2^8+1=257$ . We're all familiar with what $6^3$ is, namely $216$ , which is too small. The smallest cube greater than it is $7^3=343$ $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$ , which therefore will clearly be the largest cube less than $2^{18}+1$ . So, the required numb... | E | 58 |
eaf7d3e4623ee9d5f2ae3bc71d450919 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$ | First, $2^8+1=257$ . Then, $2^{18}+1=262145$ . Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ends with $64$ . Now, by counting how many numbers are between these, we find the answer to be $\boxed{58}$ | E | 58 |
eaf7d3e4623ee9d5f2ae3bc71d450919 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$ | First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number $-$ the lowest number $+ 1$ . People will forget the $+1$ so the only possibilities are C and E. We can clearly see... | E | 58 |
eaf7d3e4623ee9d5f2ae3bc71d450919 | https://artofproblemsolving.com/wiki/index.php/2018_AMC_8_Problems/Problem_25 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$ | There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$ , which is over $2^8+1=257$ . We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$ , which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gi... | E | 58 |
f1047259d2cc85e238f4556aeb6d5b90 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2 | Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldr... | Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$ . Solving for $x$ , we get $x=\boxed{120}.$ | E | 120 |
f1047259d2cc85e238f4556aeb6d5b90 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2 | Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldr... | We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes. Brenda received $36$ votes, which is $\frac{30}{100} = \frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\frac{10}{3},$ we get the total number of votes, which is $\boxed{120}.$ | E | 120 |
f1047259d2cc85e238f4556aeb6d5b90 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_2 | Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldr... | If $36$ votes is $\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$ . So, the answer is $\boxed{120}.$ | E | 120 |
8d46178e844feb626548112d4eeb916c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_3 | What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$
$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$ | $\sqrt{16\sqrt{8\sqrt{4}}}$ $\sqrt{16\sqrt{8\cdot 2}}$ $\sqrt{16\sqrt{16}}$ $\sqrt{16\cdot 4}$ $\sqrt{64}$ $\boxed{8}$ | C | 8 |
990e7fe8b23415d94a28c3bb514af8a7 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_4 | When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?
$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$ | We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\boxed{2400}.$ | D | 2400 |
6d130716848b6e6023a3f41e10318bdb | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5 | What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$ | Directly calculating:
We evaluate both the top and bottom: $\frac{40320}{36}$ . This simplifies to $\boxed{1120}$ | B | 1120 |
6d130716848b6e6023a3f41e10318bdb | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5 | What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$ | It is well known that the sum of all numbers from $1$ to $n$ is $\frac{n(n+1)}{2}$ . Therefore, the denominator is equal to $\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6$ . Now, we can cancel the factors of $2$ $3$ , and $6$ from both the numerator and denominator, only leaving $8 \cdot 7 \cdot 5 \cdot 4 \cdot 1... | B | 1120 |
6d130716848b6e6023a3f41e10318bdb | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_5 | What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$ | First, we evaluate $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$ to get 36. We notice that 36 is 6 squared, so we can factor the denominator like $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}$ then cancel the 6s out,\ to get $\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}$ . Now that we have escaped fraction form... | B | 1120 |
df046061239d641158442a2e84eda907 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | The sum of the ratios is $10$ . Since the sum of the angles of a triangle is $180^{\circ}$ , the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle. The question asks for the largest, so the answer is $\boxed{72}$ | D | 72 |
df046061239d641158442a2e84eda907 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | We can denote the angles of the triangle as $3x$ $3x$ $4x$ . Due to the sum of the angles in a triangle, $3x+3x+4x=180^{\circ}\implies x=18^{\circ}$ . The greatest angle is $4x$ and after substitution we get $\boxed{72}$ | D | 72 |
df046061239d641158442a2e84eda907 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | We know the longest side must be denoted by the 4 in the ratio. Since the ratio is 3:3:4, we know that the longest side must be $\frac{4}{3+3+4}$ of the degree total (which for all triangles is 180). Thus, \[\frac{4}{3+3+4} \cdot 180 = \frac{4}{10} \cdot 180 = \boxed{72}\] | D | 72 |
df046061239d641158442a2e84eda907 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_6 | If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$ | Since we see the ratio is $3:3:4$ , we can rule out the answer of ${\textbf{(E) }90}$ because the numbers in the ratio are too big to have $90^\circ$ . Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is $\boxed{72}$ | D | 72 |
6b89b1018464aca0872af1641b5ef845 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19. Or we could just try to divide the exampl... | A | 11 |
6b89b1018464aca0872af1641b5ef845 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | We are given one of the numbers that can represent $Z$ , so we can just try out the options to see which one is a factor of $247247$ . We get $\boxed{11}$ | A | 11 |
6b89b1018464aca0872af1641b5ef845 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, $+2-4+7-2+4-7=0$ . Because the result is 0... | A | 11 |
6b89b1018464aca0872af1641b5ef845 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | Similar to solution 1, let $Z=ABCABC$ . To prove it is divisible by 11, we can compute its alternating sum, which is $A-B+C-A+B-C=0$ , which is divisible by 11. Therefore, the answer is $\boxed{11}$ | A | 11 |
6b89b1018464aca0872af1641b5ef845 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_7 | Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$ | We can find that all numbers like $Z$ are divisible by 1001. 1001 is divisible by 11 because when we divide it, we get a whole number. So, the answer is $\boxed{11}$ | A | 11 |
f22bb94104fb55913f2d1dcb3b03f7c3 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8 | Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."
(1) It is prime.
(2) It is even.
(3) It is divisible by 7.
(4) One of its digi... | Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the $2$ -digit number is even, and t... | D | 8 |
f22bb94104fb55913f2d1dcb3b03f7c3 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8 | Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."
(1) It is prime.
(2) It is even.
(3) It is divisible by 7.
(4) One of its digi... | (Statement 1) Cannot be true, because only one of these four statements is true, and (Statement 1) states that the number is prime, which would make (Statement 2) and (Statement 3) false, which is not possible. And since the number being described is even, it must end with an even number (0,2,4,6,8). And since the numb... | D | 8 |
f22bb94104fb55913f2d1dcb3b03f7c3 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_8 | Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."
(1) It is prime.
(2) It is even.
(3) It is divisible by 7.
(4) One of its digi... | Like solutions 1 and 2, Statement 1 can't be true because it would contradict both Statements 2 and 3. Therefore, the other three must be true. We know the following:
The only multiple of 14 that has a tens digit of 9 is 98. Thus, our answer is $\boxed{8}.$ | D | 8 |
62cb29586bfdbde93bc574b853bc4c35 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9 | All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$ | The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$... | D | 4 |
62cb29586bfdbde93bc574b853bc4c35 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9 | All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$ | Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ $... | D | 4 |
4224582c41851015fef8d4c98d535969 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11 | A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$ | Since the number of tiles lying on both diagonals is $37$ , counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\boxed{361}$ | C | 361 |
4224582c41851015fef8d4c98d535969 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11 | A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$ | Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So, we figure out the length of each separate diagonal, multiply by two, and then add 1. (Realize that we can just join two of the s... | C | 361 |
ccc50eb537d3c969c90f9a187ddafd36 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_13 | Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
$\textbf{(A) }0\quad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | Given $n$ games, there must be a total of $n$ wins and $n$ losses. Hence, $4 + 3 + K = 2 + 3 + 3$ where $K$ is Kyler's wins. $K = 1$ , so our final answer is $\boxed{1}.$ ~CHECKMATE2021 | B | 1 |
f0f0c19317ba24f140b27a1c83900764 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | Let the number of questions that they solved alone be $x$ . Let the percentage of problems they correctly solve together be $a$ %.
As given, \[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\]
Hence, $a = 96$
Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$ . Therefor... | C | 93 |
f0f0c19317ba24f140b27a1c83900764 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | Assume the total amount of problems is $100$ per half homework assignment since we are dealing with percentages, not values. Then, we know that Chloe got $80$ problems correct by herself and got $176$ problems correct overall. We also know that Zoe had $90$ problems she did correctly alone. We can see that the total am... | C | 93 |
f0f0c19317ba24f140b27a1c83900764 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then, Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then, the average of 90 and 96 is $\boxed... | C | 93 |
f0f0c19317ba24f140b27a1c83900764 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | (Slightly different Solution)
Suppose we said that there were $100$ problems in their assignment. Then, Chloe had $40$ correct and $10$ incorrect on her portion, and $48$ correct and $2$ incorrect on the portion she and Zoe solved. Zoe has $45$ correct and $5$ incorrect on her portion, and $48$ correct and $2$ incorrec... | C | 93 |
f0f0c19317ba24f140b27a1c83900764 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had c... | Let the total number of problems be $t$ . Let the percentage of the number of problems that Chloe and Zoe did together and got right be $x$ . As we can see, Chloe got $80$ % of $\frac {1}{2}$ of the total problems right, hence, ${0.80 \cdot \frac{1}{2}t}$ . We also know that Chloe got $88$ % of $t$ problems right altog... | C | 93 |
a96b355871277b3f250047be75c7c1e7 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)-... | Notice that the upper-most section contains a 3 by 3 square that looks like:
[asy]label("$8$", (1, 2)); label("$C$", (2, 2)); label("$8$", (3, 2)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 0)); label("$A$", (2, 0)); label("$M$", (3, 0));[/asy]
It has 6 paths in which you can sp... | D | 24 |
a96b355871277b3f250047be75c7c1e7 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)-... | There are three different kinds of paths that are on this diagram. The first kind is when you directly count $A$ $M$ $C$ in a straight line. The second is when you count $A$ , turn left or right to get $M$ , then go up or down to count $8$ and $C$ . The third is the one where you start with $A$ , move up or down to cou... | D | 24 |
a96b355871277b3f250047be75c7c1e7 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)-... | Notice that the $A$ is adjacent to $4$ $M$ s, each $M$ is adjacent to $3$ $C$ s, and each $C$ is adjacent to $2$ $8$ 's. So for each $A$ , there are $4$ $M$ s, and for each $M$ , there are $3$ $C$ s, and for each $C$ , there are $2$ $8$ s. Thus, the answer is $1\cdot 4\cdot 3\cdot 2 = \boxed{24}.$ | D | 24 |
a96b355871277b3f250047be75c7c1e7 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)-... | We can do this problem by computing how many ways there are to get to each letter (in order). There is $1$ way to get to the $A$ in the center. We can only get to each of the other $M$ s by going there from the $A$ , so there is $1$ way to get to each of the four $M$ s. For the $C$ s, we notice that four $C$ s are surr... | D | 24 |
c2f0eeb731cfbd64e6dc98fc3180100f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17 | Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B)... | We can represent the amount of gold with $g$ and the amount of chests with $c$ . We can use the problem to make the following equations: \[9c-18 = g\] \[6c+3 = g\]
We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.
Therefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ ... | C | 45 |
c2f0eeb731cfbd64e6dc98fc3180100f | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17 | Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B)... | With $9$ coins, there are $\frac{9}{9}+2=1+2=3$ chests, by the first condition. These don't fit in with the second condition, so we move onto $27$ coins. By the same first condition, there are $5$ chests( $\frac{27}{9}+2$ ). This also doesn't fit with the second condition. So, onto $45$ coins. The first condition impli... | C | 45 |
63cd78186fdb28dc1c3a7441e31c14b1 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N)... | We first connect point $B$ with point $D$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle B... | B | 24 |
63cd78186fdb28dc1c3a7441e31c14b1 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N)... | $\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.... | B | 24 |
a28676f2f6063238885b3dabf08d0b43 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Factoring out $98!+99!+100!$ , we have $98! (1+99+99*100)$ , which is $98! (10000)$ . Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$ . The $19$ is because of all the multiples of $5$ .The $3$ is because of all the multiples of $25$ . Now, $1... | D | 26 |
a28676f2f6063238885b3dabf08d0b43 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Also, keep in mind that the number of $5$ ’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$ , which means we need pairs of $5$ ’s and $2$ ’s; we know there will be many more $2$ ’s, so we seek to find the number of $5$ ’s in $98!$ , which the solution tells us. And, that is ... | null | 26 |
a28676f2f6063238885b3dabf08d0b43 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$
Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (sec... | D | 26 |
25df8895c30f4e06855b3631b2f15d6a | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21 | Suppose $a$ $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$
$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }... | There are $2$ cases to consider:
Case $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]
Case $2$ $2... | A | 0 |
e5db1f5688b4ad97f27c76bb4f6f834c | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_23 | Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance trav... | It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , ... | C | 25 |
b1103952e2143c3d216f0d3b13466e2b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }7... | We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor.\]
To this result we add the number... | D | 146 |
b1103952e2143c3d216f0d3b13466e2b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }7... | Note that $\operatorname{lcm}(3,4,5)=60,$ so there is a cycle every $60$ days.
As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green. [asy] /* Made by MRENTHUSIASM */ size(7cm); fill((2,6)--(3,6)--(3,5)--(2,5)--cycle,red); fill((5... | D | 146 |
b1103952e2143c3d216f0d3b13466e2b | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }7... | For any randomly chosen day, there is a $\frac{2}{3}$ chance the first child does not call her, a $\frac{3}{4}$ chance the second child does not call her and a $\frac{4}{5}$ chance the third child does not call her. So, in a randomly chosen day, there is a $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5... | D | 146 |
d91111c6f2e572b284b50dd10fb812d8 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{665}$ | C | 665 |
d91111c6f2e572b284b50dd10fb812d8 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of $\boxed{665}$ | C | 665 |
d91111c6f2e572b284b50dd10fb812d8 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | 11 hours 5 min = $(11 \cdot 60) + 5 \text{min} = 665 \text{min}$ , therefore $\boxed{665}$ | C | 665 |
00fb409c1a9952651965e1a1ed914303 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{12}$ | A | 12 |
00fb409c1a9952651965e1a1ed914303 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | A triangle with the same height and base as a rectangle is half of the rectangle's area. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Therefore, we get $\frac{48}{4} =\boxed{12}$ | A | 12 |
00fb409c1a9952651965e1a1ed914303 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{12}$ | A | 12 |
00fb409c1a9952651965e1a1ed914303 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | A triangle is half of a rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives ... | A | 12 |
00fb409c1a9952651965e1a1ed914303 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); lab... | We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. We add 24 and 12 to get 36. ... | A | 12 |
25a5373d5ded79b44f2954a5bf65ab8d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | Let $r$ be the remaining student's score. We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ . We can use basic algebra to solve for $r$ \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{40}$ | A | 40 |
25a5373d5ded79b44f2954a5bf65ab8d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | Since $90$ is $20$ more than $70$ , and $80$ is $10$ more than $70$ , for $70$ to be the average, the other number must be $30$ less than $70$ , or $\boxed{40}$ | A | 40 |
7d62e9da6097ce726cc22313d2ad9aa1 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_4 | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textb... | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes... | B | 10 |
7ab783235b0faa804f6b17f01f74e44f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\pmod {10}.$ We now look for this one:
$9(1)=9\\ 9(2)=18\\ 9(3... | E | 7 |
7ab783235b0faa804f6b17f01f74e44f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have \[10x+3 \equiv 1 \pmod 9.\] We s... | E | 7 |
7ab783235b0faa804f6b17f01f74e44f | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | We know that the number has to be one more than a multiple of $9$ , because of the remainder of one, and the number has to be $3$ more than a multiple of $10$ , which means that it has to end in a $3$ . Now, if we just list the first few multiples of $9$ adding one to the number we get: $10, 19, 28, 37, 46, 55, 64, 73,... | E | 7 |
9350186cdfe1647070c903ec8c7a8949 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.... | We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{4}$ | B | 4 |
9350186cdfe1647070c903ec8c7a8949 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.... | To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$ , and $3$ names with length $4$ , the $10$ th name would have $4$ letters. Thus our... | B | 4 |
69ce73d9baaee669cab8796b67d3ff0d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | We can group each subtracting pair together: \[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\] After subtracting, we have: \[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\] There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{50}$ | C | 50 |
69ce73d9baaee669cab8796b67d3ff0d | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | Since our list does not end with one, we divide every number by 2 and we end up with \[50-49+48-47+ \ldots +4-3+2-1\] We can group each subtracting pair together: \[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\] There are now $25$ pairs of numbers, and the value of each pair is $1$ . This sum is $25$ . However, we di... | C | 50 |
98e4df9817d0a5ba7265e204c422a47a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9 | What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$ | The prime factorization is $2016=2^5\times3^2\times7$ . Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ . Their desired sum is then $\boxed{12}$ | B | 12 |
98e4df9817d0a5ba7265e204c422a47a | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9 | What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$ | We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both... | B | 12 |
405d510a81d8226f5b9c8fb0b4e6f48b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10 | Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$ | Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \[-9+x=1\] \[x=\boxed{10}\] | D | 10 |
405d510a81d8226f5b9c8fb0b4e6f48b | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10 | Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$ | Let us set a variable $y$ equal to $5 * x$ . Solving for y in the equation $3(2)-y=1$ , we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \[x=\boxed{10}\] | D | 10 |
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