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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2,601 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | 3 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | We can utilize the stars-and-bars distribution technique to solve this problem.
We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have $\binom{2016-1}{2-1}$ "total" answers. We want every third x and second y, so we divide our previous total by 6, which will r... | 337 |
2,602 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_14 | 4 | How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$ | Note that $2016 = 6 \cdot 336$ . In other words, we can write $2016$ as the sum of $336$ sixes.
In turn, we can express each $6$ as either $2 + 2 + 2$ or $3 + 3$
Therefore, we can write $2016$ as $n (2 + 2 + 2) + (336 - n) (3 + 3)$ , where $n$ is an integer between $0$ and $336$ , inclusive. Since each value of $n$ cor... | 337 |
2,603 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_15 | 1 | Seven cookies of radius $1$ inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
[asy] d... | The big cookie has radius $3$ , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is $3^2\pi-7\pi=9\pi-7\pi=2 \pi$ . The scrap cookie has this area, so its radius must be $\boxed{2}$ | 2 |
2,604 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | 1 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the re... | 12 |
2,605 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | 2 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{... | 12 |
2,606 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_17 | 3 | Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is... | We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3... | 12 |
2,607 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | 2 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $... | [asy] size(9cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2, 1), H=(9, 0); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); draw(A--H); draw((6,1)--G); draw(D--H); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,... | 20 |
2,608 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | 4 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $... | Extend $AF$ to meet $CD$ at point $T$ . Since $FC=1$ and $BF=2$ $TC=3$ by similar triangles $\triangle TFC$ and $\triangle AFB$ . It follows that $\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}$ . Now, using similar triangles $\triangle BEP$ and $\triangle DAP$ $\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}$ . WLOG let $BP=1$... | 20 |
2,609 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_19 | 5 | In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $... | Draw line segment $AC$ , and call the intersection between $AC$ and $BD$ point $K$ . In $\delta ABC$ , observe that $BE:EC=1:2$ and $AK:KC=1:1$ . Using mass points, find that $BP:PK=1:1$ . Again utilizing $\delta ABC$ , observe that $BF:FC=2:1$ and $AK:KC=1:1$ . Use mass points to find that $BQ:QK=4:1$ . Now, draw a li... | 20 |
2,610 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_22 | 1 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ . This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ . This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors... | 325 |
2,611 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_22 | 2 | For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$ | $110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a pr... | 325 |
2,612 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | 1 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | We see that $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . So... | 109 |
2,613 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | 2 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | If the given conditions hold for all nonzero numbers $a, b,$ and $c$
Let $a=b=c.$ From the first two givens, this implies that
\[a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.\]
From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$
Let $c=b.$ Substituting this into the fi... | 109 |
2,614 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | 5 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | $2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x = 100$
$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$
$2016 \diamondsuit 2016 = 1$ $2016 \diamondsuit (2016 \diamondsuit 1) = 1$ , so $2016 \diamondsuit 1 = 2016$
$2016 \diamondsuit 1 = (2016 \diamondsu... | 109 |
2,615 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_23 | 6 | A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsui... | Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$ . Hence, $2016 \diamondsuit 6 = 336$ . Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$ . Therefore, the answer is $\boxed{109}$ | 109 |
2,616 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 1 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram dr... | 500 |
2,617 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 2 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(14... | 500 |
2,618 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 3 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
2,619 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 4 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
2,620 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 5 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -... | 500 |
2,621 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 7 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Since all three sides equal $200$ , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100... | 500 |
2,622 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 8 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$ , where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$ . If $M$ and $N$ are the midpoints of $BC$ and $AD$ , respectively, the height of the trapezoid is $OM-ON$ . By the pythagorean theorem, $OM=\sqrt{OB^2-BM... | 500 |
2,623 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 10 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | We first scale down by a factor of $200\sqrt{2}$ . Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ , so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$ , and we let $C$ correspond to the complex number $z$ . Then, $A$ corresp... | 500 |
2,624 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 11 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | Let angle $C$ be $2a$ . This way $BD$ will be $400sin(a)$ . Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$ , we can use the formula \[R=\frac{abc}{4A}\] and \[A=\frac{absin(C)}{2}\] and plug in all the values we got to get \[200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin... | 500 |
2,625 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_24 | 12 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$ . Three of the sides of this quadrilateral have length $200$ . What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | [asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | 500 |
2,626 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | 1 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | We prime factorize $72,600,$ and $900$ . The prime factorizations are $2^3\times 3^2$ $2^3\times 3\times 5^2$ and $2^2\times 3^2\times 5^2$ , respectively. Let $x=2^a\times 3^b\times 5^c$ $y=2^d\times 3^e\times 5^f$ and $z=2^g\times 3^h\times 5^i$ . We know that \[\max(a,d)=3\] \[\max(b,e)=2\] \[\max(a,g)=3\] \[\max(b,... | 15 |
2,627 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | 2 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | It is well known that if the $\text{lcm}(a,b)=c$ and $c$ can be written as $p_1^ap_2^bp_3^c\dots$ , then the highest power of all prime numbers $p_1,p_2,p_3\dots$ must divide into either $a$ and/or $b$ . Or else a lower $c_0=p_1^{a-\epsilon}p_2^{b-\epsilon}p_3^{c-\epsilon}\dots$ is the $\text{lcm}$
Start from $x$ $\tex... | 15 |
2,628 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_25 | 3 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | As said in previous solutions, start by factoring $72, 600,$ and $900$ . The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\] To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows su... | 15 |
2,629 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_1 | 1 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | Factorizing the numerator, $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ then becomes $\frac{\frac{5}{2}}{a^{2}}$ which is equal to $\frac{5}{2}\cdot 2^2$ which is $\boxed{10}$ | 10 |
2,630 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_1 | 2 | What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | Substituting $\frac{1}{2}$ for $a$ in $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ gives us $\boxed{10}$ . ~peelybonehead | 10 |
2,631 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_3 | 1 | Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Substituting carefully, $\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)$
becomes $|4032-2016|+2016=2016+2016=4032$ which is $\boxed{4032}$ | 32 |
2,632 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_3 | 2 | Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$ | Solution by e_power_pi_times_i
Substitute $-y = x = -2016$ into the equation. Now, it is $\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y$ . Since $y = 2016$ , it is a positive number, so $|y| = y$ . Now the equation is $\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y$ . This further simplifies to $2y-y+y = 2y$ , so ... | 32 |
2,633 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_5 | 1 | The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$ | The sum of the ages of the cousins is $4$ times the mean, or $32$ .
There are an even number of cousins, so there is no single median, so $5$ must be the mean of the two in the middle.
Therefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\textbf{(D)}\ \boxed{22}$ | 22 |
2,634 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_6 | 1 | Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$ . The hundreds digits of these numbers must be at least $1$ and $2$ , so $a\ge 100$ and $b\ge 200$
Say $a=100+p$ and $b=200+q$ ; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$ .There are many so... | 4 |
2,635 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_6 | 2 | Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | For this problem, to find the $3$ -digit integer with the smallest sum of digits, one should make the units and tens digit add to $0$ . To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $7$ works best for the top number which makes the bottom d... | 4 |
2,636 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_7 | 1 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into the s... | 135 |
2,637 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_7 | 2 | The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$ . We know $90-4x>90-5x$ since this inequality gives $x>0$ , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so
$90-4x=2(90-5x)$
$90-4x=180-10x$
$6x=90$
$x=15$
Therefore, the angles a... | 135 |
2,638 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_8 | 1 | What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$ | Notice that, for $n\ge 2$ $2015^n\equiv 15^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself). Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$
So the answer is $\textbf{(A)}\ \boxed{0}... | 0 |
2,639 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_9 | 1 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | Let the point where the height of the triangle intersects with the base be $D$ . Now we can guess what $x$ is and find $y$ . If $x$ is $3$ , then $y$ is $9$ . The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$ , respectively. The distance between $B$ and $C$ is $6$ , meaning the area would be $\frac{6 \cdot 9}{2}=2... | 8 |
2,640 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_11 | 1 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | If the dimensions are $4a\times 4b$ , then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).
The total number of posts is $2(a+b)=20$
This diagram represents the number of posts around the garden. [asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<... | 336 |
2,641 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_11 | 2 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the sho... | To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only $16$ posts. Now there are twice as many posts on the longer side then the shorter side. From this, we can see that we can put $8$ posts on the longer side and $4$ posts on th... | 336 |
2,642 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_13 | 1 | At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000... | We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. \[\begin{split} 2a + 3b + 4c & = 1000 \\ b & = 4c \\ a & = 3b \end{split}\]
Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields \[\be... | 100 |
2,643 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_13 | 2 | At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000... | Say there are $12x$ sets of twins, $4x$ sets of triplets, and $x$ sets of quadruplets. That's $12x\cdot2=24x$ twins, $4x\cdot3=12x$ triplets, and $x\cdot4=4x$ quadruplets. A tenth of the babies are quadruplets and that's $\frac{1}{10}(1000)=\boxed{100}$ | 100 |
2,644 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | 1 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | The region is a right triangle which contains the following lattice points: $(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$
[asy]size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0... | 50 |
2,645 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | 2 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ .
This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under t... | 50 |
2,646 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_14 | 3 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}... | The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases.
Case 1: Square has length $\bf1$
The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12... | 50 |
2,647 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | 1 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\te... | Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity: [asy]size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*... | 7 |
2,648 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | 3 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\te... | First let the numbers be \[1 ~8~ 7\] \[2 ~ 9 ~6\] \[3 ~ 4~ 5\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching ea... | 7 |
2,649 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_15 | 4 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\te... | Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is \[1 ~8~ 9\] \[2 ~ 7 ~6\] \[3 ~ 4~ 5\]
$\boxed{7}$ | 7 |
2,650 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | 3 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get \[S=\frac{a}{1-r} \implies S-Sr=a\] Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$ , we would get \[r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0... | 4 |
2,651 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | 4 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Let the first term of the geometric series $x$ . Since it must be decreasing, we have $x>1$ and the third term is $\frac{1}{x}$ . Realize that by AM-GM inequality $x+\frac{1}{x} \ge 2$ with equality if $x = 1$ . However, we established that $x>1$ so that means $x+\frac{1}{x} > 2$ . So the sum of the first three terms o... | 4 |
2,652 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | 5 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Let the first term be $k.$ The sum of the series is $\frac{k}{1- \frac{1}{k}} =\frac{k^2}{k-1}.$ Rewrite this as $\frac{k^2-2k+1}{k-1} +\frac{2k-1}{k-1} = k-1+\frac{2k-2}{k-1} +\frac{1}{k-1} = (k-1) + \left(\frac{1}{k-1}\right) + 2.$ By AM-GM we know that $(k-1) + \left(\frac{1}{k-1}\right) \ge 2$ so the minimum is $2+... | 4 |
2,653 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_16 | 6 | The sum of an infinite geometric series is a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$
$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Set the first term is $a.$ , the common ratio should be $\frac{1}{a}.$
The sum to infinity of the series is $S=\frac{a}{1-\frac{1}{a}}=\frac{a^2}{a-1}.$
Since $S$ is positive, we have $a>1.$ Define the function $f(a)=\frac{a^2}{a-1}$ , the domain of this function is $a>1.$
Let $f^{'}(a)=\frac{2a^2-2a-a^2}{(a-1)^2}=\fra... | 4 |
2,654 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | 1 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | Let us call the six faces of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$ $c$ is opposite $e$ , and $b$ is opposite $f$ .
Thus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$ .
Let us find the sum of these products: \[abc+abe+bcd+bde+acf+cdf+aef+def\] We not... | 729 |
2,655 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | 2 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact t... | 729 |
2,656 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | 3 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$ . To maximize the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{729}$ | 729 |
2,657 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_17 | 4 | All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of th... | First, we intuitivly notice that multiplying large numbers together and smaller numbers together tends to produce larger sums, while multiplying large numbers with smaller numbers tends to produce smaller sums. From this, we guess that it is optimal to have $7, 6,$ and $5$ to be around one vertex to produce at least on... | 729 |
2,658 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | 1 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | Factor $345=3\cdot 5\cdot 23$
Suppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\tfrac12k<m$ .
Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.
Suppose instead we take an even number $2k$ of co... | 7 |
2,659 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | 2 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We need to find consecutive numbers (an arithmetic sequence that increases by $1$ ) that sums to $345$ . This calls for the sum of an arithmetic sequence given that the first term is $k$ , the last term is $g$ and with $n$ elements, which is: $\frac {n \cdot (k+g)}{2}$
We look for sequences of $n$ consecutive numbers s... | 7 |
2,660 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_18 | 4 | In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | The median of the sequence $m$ is either an integer or a half integer. Let $m=\frac{i}{2}, i \in N$ , then $P=i\cdot n=2\cdot 3 \cdot 5 \cdot 23$
On the other hand we have two constraints:
1) $m \geq \frac{n+1}{2} \iff i>n$ because the integers in the sequence are all positive, and $n>1$
2) If $n$ is odd then $m$ is an... | 7 |
2,661 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | 1 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \s... | The center of dilation must lie on the line $A A'$ , which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$ . Note that the center of dilation must have an $x$ -coordinate less than $2$ ; if the $x$ -coordinate were otherwise, then the circle under the transformation would not have an increased $x$ -coordinate in... | 13 |
2,662 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | 2 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \s... | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ /* by adihaya */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmi... | 13 |
2,663 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | 3 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \s... | Using the ratios of radii of the circles, $\frac{3}{2}$ , we find that the scale factor is $1.5$ . If the origin had not moved, this indicates that the center of the
circle would be $(3,3)$ , simply because of $(2 \cdot 1.5, 2 \cdot 1.5)$ . Since the center has moved from $(3,3)$ to $(5,6)$ , we apply the distance for... | 13 |
2,664 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_20 | 5 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \s... | Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\frac{3}{2}$ . The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the cente... | 13 |
2,665 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21 | 1 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | Another way to solve this problem is using cases.
Though this may seem tedious, we only have to do one case since the area enclosed is symmetrical.
The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible,
we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being p... | 2 |
2,666 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_21 | 2 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$ Case 2: $x\geq0, y<0.$ Case 3: $x<0, y\geq0.$ Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can figu... | 2 |
2,667 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | 1 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 6... | There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
[asy]size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0))... | 385 |
2,668 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_22 | 2 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 6... | Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4 \cdot 1330=332.5$ . Looking at the answer choices, we find tha... | 385 |
2,669 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_10B_Problems/Problem_24 | 1 | How many four-digit integers $abcd$ , with $a \neq 0$ , have the property that the three two-digit integers $ab<bc<cd$ form an increasing arithmetic sequence? One such number is $4692$ , where $a=4$ $b=6$ $c=9$ , and $d=2$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(... | The numbers are $10a+b, 10b+c,$ and $10c+d$ . Note that only $d$ can be zero, the numbers $ab$ $bc$ , and $cd$ cannot start with a zero, and $a\le b\le c$
To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$ . This can be rearranged as $10(c-2b+a)=2c-b-d$ . Notice that since the left-hand side is a multiple ... | 17 |
2,670 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | 1 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.
Triangles have $3$ edges and squares have $4$ edges, so we have a system of equations.
We have $a + b$ tiles total, so $a + b = 25$
We have $3a + 4b$ edges total, so $3a + 4b = 84$
Multiplying the first equation by $3$ on both sides gives ... | 9 |
2,671 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | 2 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there needs to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\boxed{9}$ | 9 |
2,672 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_2 | 3 | A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$ | Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$ . There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$ . Together, the total numb... | 9 |
2,673 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_3 | 1 | Ann made a $3$ -step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$ -step staircase?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse... | We can see that a $1$ -step staircase requires $4$ toothpicks and a $2$ -step staircase requires $10$ toothpicks. Thus, to go from a $1$ -step to $2$ -step staircase, $6$ additional toothpicks are needed and to go from a $2$ -step to $3$ -step staircase, $8$ additional toothpicks are needed. Applying this pattern, to g... | 22 |
2,674 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_5 | 1 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textb... | If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ . When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ . So, Payton's score must be $1215-1120 = \boxed{95}$ | 95 |
2,675 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_5 | 2 | Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textb... | The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of... | 95 |
2,676 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_6 | 1 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$ | 32 |
2,677 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_6 | 2 | The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ | Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$ | 32 |
2,678 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | 1 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | $73-13 = 60$ , so the amount of terms in the sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$ is the same as in the sequence $0$ $3$ $6$ $\dotsc$ $57$ $60$
In this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$
However, the number 0 must also be included, adding an... | 21 |
2,679 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | 2 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{21}$ | 21 |
2,680 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | 3 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$ $\frac{61-1}{3}=20, 20 + 1 = 21$
$\boxed{21}$ | 21 |
2,681 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_7 | 4 | How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$ | Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$ . Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$ . Now it is clear to see that there are $21$ terms in the sequence. $\boxed{21}$ | 21 |
2,682 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_8 | 1 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the n... | 4 |
2,683 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_9 | 1 | Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is ... | Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and divid... | 21 |
2,684 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_10 | 1 | How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since:
$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed becau... | 2 |
2,685 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_10 | 2 | How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Case 1: the first letter is A
Subcase 1: the second letter is C
The next letter must either be B or D, both of which do not satisfy the conditions.
Subcase 2: the second letter is D
The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.
Case 2:... | 2 |
2,686 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | 1 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$
$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$
$y^2 + \pi^2 = 2\pi y + 1$
$y^2 - 2\pi y + \pi^2 = 1$
$(y-\pi)^2 = 1$
$y-\pi = \pm 1$
$y = \pi + 1$
$y = \pi - 1$
There are only two solutions to the... | 2 |
2,687 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | 2 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
$y^2 + x^4 = 2x^2 y + 1$ can be written as $x^4-2x^2y+y^2=1$ . Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$ . This gives us two equations:
$x^2-y=1$ and $x^2-y=-1$
One of these $... | 2 |
2,688 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_12 | 3 | Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | This solution is similar to Solution #1 but uses a different way to find $y$ at the end.
Just like Solution #1, we arrive at the conclusion that $y^2 - 2\pi y + \pi^2 = 1$
Simplifying we get:
$y^2 - 2\pi y + \pi^2 -1 = 0$
We now can factor this quadratic. We must find two terms that multiply to $\pi^2 -1$ and add to $2... | 2 |
2,689 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | 1 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not $7,$ because we a... | 5 |
2,690 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | 2 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain... | 5 |
2,691 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_13 | 3 | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$ . Therefore, the answer is $\boxed{5}$ | 5 |
2,692 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 2 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, whic... | 4 |
2,693 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 3 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is $20\pi$ , so that... | 4 |
2,694 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 4 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this: [asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(6... | 4 |
2,695 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 5 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | We can approach this problem with angle measures. As the circumference of the disk is $10\pi,$ and the clock $20 \pi,$ we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks wil... | 4 |
2,696 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 6 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | If the big clock were a flat plane, then the smaller clock could travel $\dfrac{40\pi}{20\pi}=2$ full revolutions.
But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After ... | 4 |
2,697 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 7 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is $60 \pi$ . The other circle is $20 \pi$ and so that is $\frac{1}{3}$ . So $\frac{12}{3} = 4$ which is $\boxed{4}$ | 4 |
2,698 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_14 | 8 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | Let the center of the disk be Planet X with orbit eccentricity $0$ and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the so... | 4 |
2,699 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_15 | 1 | Consider the set of all fractions $\frac{x}{y}$ , where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2... | You can create the equation $\frac{x+1}{y+1}=\frac{11x}{10y}$
Cross multiplying and combining like terms gives $xy + 11x - 10y = 0$
This can be factored into $(x - 10)(y + 11) = -110$
$x$ and $y$ must be positive, so $x > 0$ and $y > 0$ , so $x - 10> -10$ and $y + 11 > 11$
Using the factors of 110, we can get the facto... | 1 |
2,700 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_15 | 2 | Consider the set of all fractions $\frac{x}{y}$ , where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2... | The condition required is $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$
Observe that $x+1 > \frac{11}{10}\cdot x$ so $x$ is at most $9.$
By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11x}{10-x}$ . Since $x$ and $y$ are integer, this only has solutions for $x\in\{5,8,9\}$ ... | 1 |
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