https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Billyqq&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T06:24:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_20&diff=1140262017 AMC 10B Problems/Problem 202020-01-02T16:25:40Z<p>Billyqq: /* See Also */</p>
<hr />
<div>==Problem==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}</math><br />
==Solution 1==<br />
We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math><br />
<br />
==Solution 2: Constructive counting==<br />
Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= <math>\boxed{1/19, \space \text{B}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_20&diff=1140252017 AMC 10B Problems/Problem 202020-01-02T16:24:57Z<p>Billyqq: /* See Also */</p>
<hr />
<div>==Problem==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}</math><br />
==Solution 1==<br />
We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math><br />
<br />
==Solution 2: Constructive counting==<br />
Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= <math>\boxed{1/19, \space \text{B}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2017|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_20&diff=1140242017 AMC 10B Problems/Problem 202020-01-02T16:24:45Z<p>Billyqq: /* See Also */</p>
<hr />
<div>==Problem==<br />
The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}</math><br />
==Solution 1==<br />
We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math><br />
<br />
==Solution 2: Constructive counting==<br />
Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= <math>\boxed{1/19, \space \text{B}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
{{AMC12 box|year=2017|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_3&diff=1140232017 AMC 10B Problems/Problem 32020-01-02T16:19:46Z<p>Billyqq: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
==Solution==<br />
Notice that <math>y+z</math> must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.<br />
<br />
The other choices:<br />
<br />
<math>\textbf{(A)}</math> As <math>x</math> grows closer to <math>0</math>, <math>x^2</math> decreases and thus becomes less than <math>y</math>.<br />
<br />
<math>\textbf{(B)}</math> <math>x</math> can be as small as possible (<math>x>0</math>), so <math>xz</math> grows close to <math>0</math> as <math>x</math> approaches <math>0</math>.<br />
<br />
<math>\textbf{(C)}</math> For all <math>-1<y<0</math>, <math>|y|>|y^2|</math>, and thus it is always negative.<br />
<br />
<math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=3}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_3&diff=1140222017 AMC 10B Problems/Problem 32020-01-02T16:19:27Z<p>Billyqq: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the inequalities<br />
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.<br />
Which of the following numbers is necessarily positive?<br />
<br />
<math>\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z</math><br />
<br />
==Solution==<br />
Notice that <math>y+z</math> must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.<br />
<br />
The other choices:<br />
<br />
<math>\textbf{(A)}</math> As <math>x</math> grows closer to <math>0</math>, <math>x^2</math> decreases and thus becomes less than <math>y</math>.<br />
<br />
<math>\textbf{(B)}</math> <math>x</math> can be as small as possible (<math>x>0</math>), so <math>xz</math> grows close to <math>0</math> as <math>x</math> approaches <math>0</math>.<br />
<br />
<math>\textbf{(C)}</math> For all <math>-1<y<0</math>, <math>|y|>|y^2|</math>, and thus it is always negative.<br />
|<br />
<math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=3}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_5&diff=1013862019 AMC 10A Problems/Problem 52019-02-09T20:32:15Z<p>Billyqq: Created page with "==Problem 5== What is the greatest number of consecutive integers whose sum is <math>45 ?</math> <math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\te..."</p>
<hr />
<div>==Problem 5==<br />
What is the greatest number of consecutive integers whose sum is <math>45 ?</math><br />
<br />
<math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120</math><br />
<br />
==Solution==<br />
Note that every term in the sequence <math>-44, -43..., 44, 45</math> cancels out except <math>45</math>. This results in <math>\boxed{\textbf{(D) } 90 }</math> integers.</div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems&diff=1013822019 AMC 10A Problems2019-02-09T20:26:45Z<p>Billyqq: /* Problem 21 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
==Problem 2==<br />
What is the hundreds digit of <math>(20!-15!)\ ?</math><br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math><br />
<br />
==Problem 3==<br />
Ana and Bonita were born on the same date in different years, <math>n</math> years apart. Last year Ana was <math>5</math> times as old as Bonita. This year Ana's age is the square of Bonita's age. What is <math>n?</math><br />
<br />
<math>\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15</math><br />
<br />
==Problem 4==<br />
A box contains <math>28</math> red balls, <math>20</math> green balls, <math>19</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least <math>15</math> balls of a single color will be drawn<math>?</math><br />
<br />
<math>\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91</math><br />
<br />
==Problem 5==<br />
What is the greatest number of consecutive integers whose sum is <math>45 ?</math><br />
<br />
<math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120</math><br />
<br />
==Problem 6==<br />
==Problem 7==<br />
==Problem 8==<br />
==Problem 9==<br />
==Problem 10==<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
Let <math>\Delta ABC</math> be an isosceles triangle with <math>BC = AC</math> and <math>\angle ACB = 40^{\circ}</math>. Contruct the circle with diameter <math>\overline{BC}</math>, and let <math>D</math> and <math>E</math> be the other intersection points of the circle with the sides <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively. Let <math>F</math> be the intersection of the diagonals of the quadrilateral <math>BCDE</math>. What is the degree measure of <math>\angle BFC ?</math><br />
<br />
<math>\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120</math><br />
<br />
==Problem 14==<br />
<br />
For a set of four distinct lines in a plane, there are exactly <math>N</math> distinct points that lie on two or more of the lines. What is the sum of all possible values of <math>N</math>?<br />
<br />
<math>\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21</math><br />
<br />
==Problem 15==<br />
<br />
A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and<br />
<cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive inegers. What is <math>p+q ?</math><br />
<br />
<math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math><br />
<br />
==Problem 16==<br />
<br />
The figure below shows <math>13</math> circles of radius <math>1</math> within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius <math>1 ?</math><br />
<br />
<asy>unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white);</asy><br />
<br />
<math>\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi(3\sqrt{3} +2) \qquad\textbf{(D) } 10 \pi (\sqrt{3} - 1) \qquad\textbf{(E) } \pi(\sqrt{3} + 6)</math><br />
<br />
==Problem 17==<br />
<br />
A child builds towers using identically shaped cubes of different color. How many different towers with a height <math>8</math> cubes can the child build with <math>2</math> red cubes, <math>3</math> blue cubes, and <math>4</math> green cubes? (One cube will be left out.)<br />
<br />
<math>\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320</math><br />
<br />
==Problem 18==<br />
<br />
For some positive integer <math>k</math>, the repeating base-<math>k</math> representation of the (base-ten) fraction <math>\frac{7}{51}</math> is <math>0.\overline{23}_k = 0.232323..._k</math>. What is <math>k</math>?<br />
<br />
<br />
<math>\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17</math><br />
<br />
==Problem 19==<br />
<br />
What is the least possible value of<br />
<cmath>(x+1)(x+2)(x+3)(x+4)+2019</cmath>where <math>x</math> is a real number?<br />
<br />
<math>\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021</math><br />
<br />
==Problem 20==<br />
==Problem 21==<br />
A sphere with center <math>O</math> has radius 6. A triangle with sides of length <math>15</math>, <math>15</math>, and <math>24</math> is situated in space so that each of its sides is tangent to the sphere. What is the distance between <math>O</math> and the plane determined by the triangle?<br />
<br />
<math>\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) }4 \qquad \textbf{(C) } 3\sqrt{2} \qquad \textbf{(D) } 2\sqrt{5} \qquad \textbf{(E) } 5</math><br />
<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==<br />
For how many integers <math>n</math> between <math>1</math> and <math>50</math>, inclusive, is <cmath>\frac{(n^2-1)!}{(n!)^{n}}</cmath> an integer? (Recall that <math>0!=1</math>.)<br />
<br />
<math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math></div>Billyqqhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=888932004 AMC 10A Problems/Problem 202017-12-13T22:39:10Z<p>Billyqq: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Points <math>E</math> and <math>F</math> are located on square <math>ABCD</math> so that <math>\triangle BEF</math> is equilateral. What is the ratio of the area of <math>\triangle DEF</math> to that of <math>\triangle ABE</math>?<br />
<br />
<center>[[Image:AMC10_2004A_20.png]]</center><br />
<br />
<math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} </math><br />
<br />
==Solution 1==<br />
Since triangle <math>BEF</math> is equilateral, <math>EA=FC</math>, and <math>EAB</math> and <math>FCB</math> are <math>SAS</math> congruent. Thus, triangle <math>DEF</math> is an isosceles right triangle. So we let <math>DE=x</math>. Thus <math>EF=EB=FB=x\sqrt{2}</math>. If we go angle chasing, we find out that <math>\angle AEB=75^{\circ}</math>, thus <math>\angle ABE=15^{\circ}</math>. <math>\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>. Thus <math>\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}</math>, or <math>AE=\frac{x(\sqrt{3}-1)}{2}</math>. Thus <math>AB=\frac{x(\sqrt{3}+1)}{2}</math>, and <math>[ABE]=\frac{x^2}{4}</math>, and <math>[DEF]=\frac{x^2}{2}</math>. Thus the ratio of the areas is <math>\boxed{\mathrm{(D)}\ 2}</math><br />
<br />
==Solution 2 (Non-trig) ==<br />
Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have<br />
<math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is<br />
<cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.</cmath><br />
<br />
==Solution 3==<br />
<math>\bigtriangleup BEF</math> is equilateral, so <math>\angle EBF = 60^{\circ}</math>, and <math>\angle EBA = \angle FBC</math> so they must each be <math>15^{\circ}</math>. Then let <math>BE=EF=FB=1</math>, which gives <math>EA=\sin{15^{\circ}}</math> and <math>AB=\cos{15^{\circ}}</math>. <br />
The area of <math>\bigtriangleup ABE</math> is then <math>\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}</math>. <br />
<math>\bigtriangleup DEF</math> is an isosceles right triangle with hypotenuse 1, so <math>DE=DF=\frac{1}{\sqrt{2}}</math> and therefore its area is <math>\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}</math>. <br />
The ratio of areas is then <math>\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}</math><br />
<br />
==See also==<br />
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131332 AoPS topic]<br />
{{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Ratio Problems]]<br />
{{MAA Notice}}</div>Billyqq