https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Naenaendr&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-13T16:37:39Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_16&diff=145727 2021 AMC 12A Problems/Problem 16 2021-02-12T04:02:39Z <p>Naenaendr: /* Solution 2 */ don't say &quot;obviously&quot; if it isn't obvious</p> <hr /> <div>==Problem==<br /> In the following list of numbers, the integer &lt;math&gt;n&lt;/math&gt; appears &lt;math&gt;n&lt;/math&gt; times in the list for &lt;math&gt;1 \leq n \leq 200&lt;/math&gt;.&lt;cmath&gt;1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200&lt;/cmath&gt;What is the median of the numbers in this list?<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> There are &lt;math&gt;1+2+..+199+200=\frac{(200)(201)}{2}=20100&lt;/math&gt; numbers in total. Let the median be &lt;math&gt;k&lt;/math&gt;. We want to find the median &lt;math&gt;k&lt;/math&gt; such that<br /> &lt;cmath&gt; \frac{k(k+1)}{2}=20100/2,&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt; k(k+1)=20100.&lt;/cmath&gt;<br /> Note that &lt;math&gt;\sqrt{20100} \approx 142&lt;/math&gt;. Plugging this value in as &lt;math&gt;k&lt;/math&gt; gives<br /> &lt;cmath&gt;\frac{1}{2}(142)(143)=10153.&lt;/cmath&gt;<br /> &lt;math&gt;10153-142&lt;10050&lt;/math&gt;, so &lt;math&gt;142&lt;/math&gt; is the &lt;math&gt;152&lt;/math&gt;nd and &lt;math&gt;153&lt;/math&gt;rd numbers, and hence, our desired answer. &lt;math&gt;\fbox{(C) 142}.&lt;/math&gt;.<br /> ===Solution 2===<br /> The &lt;math&gt;x&lt;/math&gt;th number of this sequence is &lt;math&gt;\frac{-1\pm\sqrt{1+8x}}{2}&lt;/math&gt; via the quadratic formula. We can see that if we halve &lt;math&gt;x&lt;/math&gt; we end up getting &lt;math&gt;\frac{-1\pm\sqrt{1+4x}}{2}&lt;/math&gt;. This is approximately the number divided by &lt;math&gt;\sqrt{2}&lt;/math&gt;. &lt;math&gt;\frac{200}{\sqrt{2}} = 141.4&lt;/math&gt; and since &lt;math&gt;142&lt;/math&gt; looks like the only number close to it, it is answer &lt;math&gt;\boxed{(C) 142}&lt;/math&gt; ~Lopkiloinm<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Using Algebra) ==<br /> https://youtu.be/HkwgH9Lc1hE<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_16&diff=145724 2021 AMC 12A Problems/Problem 16 2021-02-12T04:01:25Z <p>Naenaendr: /* Problem */</p> <hr /> <div>==Problem==<br /> In the following list of numbers, the integer &lt;math&gt;n&lt;/math&gt; appears &lt;math&gt;n&lt;/math&gt; times in the list for &lt;math&gt;1 \leq n \leq 200&lt;/math&gt;.&lt;cmath&gt;1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200&lt;/cmath&gt;What is the median of the numbers in this list?<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> There are &lt;math&gt;1+2+..+199+200=\frac{(200)(201)}{2}=20100&lt;/math&gt; numbers in total. Let the median be &lt;math&gt;k&lt;/math&gt;. We want to find the median &lt;math&gt;k&lt;/math&gt; such that<br /> &lt;cmath&gt; \frac{k(k+1)}{2}=20100/2,&lt;/cmath&gt;<br /> or<br /> &lt;cmath&gt; k(k+1)=20100.&lt;/cmath&gt;<br /> Note that &lt;math&gt;\sqrt{20100} \approx 142&lt;/math&gt;. Plugging this value in as &lt;math&gt;k&lt;/math&gt; gives<br /> &lt;cmath&gt;\frac{1}{2}(142)(143)=10153.&lt;/cmath&gt;<br /> &lt;math&gt;10153-142&lt;10050&lt;/math&gt;, so &lt;math&gt;142&lt;/math&gt; is the &lt;math&gt;152&lt;/math&gt;nd and &lt;math&gt;153&lt;/math&gt;rd numbers, and hence, our desired answer. &lt;math&gt;\fbox{(C) 142}.&lt;/math&gt;.<br /> ===Solution 2===<br /> The &lt;math&gt;x&lt;/math&gt;th number of this sequence is obviously &lt;math&gt;\frac{-1\pm\sqrt{1+8x}}{2}&lt;/math&gt; via the quadratic formula. We can see that if we halve &lt;math&gt;x&lt;/math&gt; we end up getting &lt;math&gt;\frac{-1\pm\sqrt{1+4x}}{2}&lt;/math&gt;. This is approximately the number divided by &lt;math&gt;\sqrt{2}&lt;/math&gt;. &lt;math&gt;\frac{200}{\sqrt{2}} = 141.4&lt;/math&gt; and since &lt;math&gt;142&lt;/math&gt; looks like the only number close to it, it is answer &lt;math&gt;\boxed{(C) 142}&lt;/math&gt; ~Lopkiloinm<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Using Algebra) ==<br /> https://youtu.be/HkwgH9Lc1hE<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems&diff=145620 2021 AMC 12A Problems 2021-02-12T02:51:57Z <p>Naenaendr: /* Problem 9 */</p> <hr /> <div>{{AMC12 Problems|year=2021|ab=A}}<br /> ==Problem 1==<br /> What is the value of&lt;cmath&gt;2^{1+2+3}-(2^1+2^2+2^3)?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Under what conditions does &lt;math&gt;\sqrt{a^2+b^2}=a+b&lt;/math&gt; hold, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; It is never true.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; It is true if and only if &lt;math&gt;ab=0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; It is true if and only if &lt;math&gt;a+b\ge 0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; It is true if and only if &lt;math&gt;ab=0&lt;/math&gt; and &lt;math&gt;a+b\ge 0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; It is always true.<br /> <br /> [[2021 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> The sum of two natural numbers is &lt;math&gt;17,402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }10,272 \qquad \textbf{(B) }11,700 \qquad \textbf{(C) }13,362 \qquad \textbf{(D) }14,238 \qquad \textbf{(E) }15,462&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Tom has a collection of &lt;math&gt;13&lt;/math&gt; snakes, &lt;math&gt;4&lt;/math&gt; of which are purple and &lt;math&gt;5&lt;/math&gt; of which are happy. He observes that<br /> <br /> all of his happy snakes can add,<br /> <br /> none of his purple snakes can subtract, and<br /> <br /> all of his snakes that can't subtract also can't add.<br /> <br /> Which of these conclusions can be drawn about Tom's snakes?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; Purple snakes can add.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; Purple snakes are happy.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; Snakes that can add are purple.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; Happy snakes are not purple.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; Happy snakes can't subtract.<br /> <br /> [[2021 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> When a student multiplied the number &lt;math&gt;66&lt;/math&gt; by the repeating decimal&lt;cmath&gt;\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},&lt;/cmath&gt;where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are digits, he did not notice the notation and just multiplied &lt;math&gt;66&lt;/math&gt; times &lt;math&gt;\underline{1}.\underline{a}\underline{b}&lt;/math&gt;. Later he found that his answer is &lt;math&gt;0.5&lt;/math&gt; less than the correct answer. What is the &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\underline{a}\underline{b}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> What is the least possible value of &lt;math&gt;(xy-1)^2+(x+y)^2&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac12 \qquad \textbf{(D) }1 \qquad \textbf{(E) }2&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> A sequence of numbers is defined by &lt;math&gt;D_0=0,D_0=1,D_2=1&lt;/math&gt; and &lt;math&gt;D_n=D_{n-1}+D_{n-3}&lt;/math&gt; for &lt;math&gt;n\ge 3&lt;/math&gt;. What are the parities (evenness or oddness) of the triple of numbers &lt;math&gt;(D_{2021},D_{2022},D_{2023})&lt;/math&gt;, where &lt;math&gt;E&lt;/math&gt; denotes even and &lt;math&gt;O&lt;/math&gt; denotes odd?<br /> <br /> &lt;math&gt;\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Which of the following is equilvalent to&lt;cmath&gt;(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }3^{127}+2^{127} \qquad \textbf{(B) }3^{127}+2^{127}+2\cdot 3^{63}+3\cdot 2^{63} \qquad \textbf{(C) }3^{128}-2^{128} \qquad \textbf{(D) }3^{128}+2^{128} \qquad \textbf{(E) }5^{127}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are &lt;math&gt;3&lt;/math&gt; cm and &lt;math&gt;6&lt;/math&gt; cm. Into each cone is dropped a spherical marble of radius &lt;math&gt;1&lt;/math&gt; cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?<br /> <br /> &lt;asy&gt;<br /> size(350);<br /> defaultpen(linewidth(0.8));<br /> real h1 = 10, r = 3.1, s=0.75;<br /> pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q;<br /> path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9);<br /> draw(ellipse(origin,r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill(origin--Pp--Qp--cycle,gray(0.8));<br /> draw((-r,h1)--(0,0)--(r,h1)^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(0,Qp.y),Arrows(size=8));<br /> draw(origin--(0,12),linetype(&quot;4 4&quot;));<br /> draw(origin--(r*(s-0.1),0));<br /> label(&quot;$3$&quot;,(-0.9,h1*s),N,fontsize(10));<br /> <br /> real h2 = 7.5, r = 6, s=0.6, d = 14;<br /> pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0);<br /> path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1);<br /> draw(ellipse((d,0),r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill((d,0)--Pp--Qp--cycle,gray(0.8));<br /> draw(P--(d,0)--Q^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(d,Qp.y),Arrows(size=8));<br /> draw((d,0)--(d,10),linetype(&quot;4 4&quot;));<br /> draw((d,0)--(d+r*(s-0.1),0));<br /> label(&quot;$6$&quot;,(d-r/4,h2*s-0.06),N,fontsize(10));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> A laser is placed at the point &lt;math&gt;(3,5)&lt;/math&gt;. The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the &lt;math&gt;y&lt;/math&gt;-axis, then hit and bounce off the &lt;math&gt;x&lt;/math&gt;-axis, then hit the point &lt;math&gt;(7,5)&lt;/math&gt;. What is the total distance the beam will travel along this path?<br /> <br /> &lt;math&gt;\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> All the roots of the polynomial &lt;math&gt;z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16&lt;/math&gt; are positive integers, possibly repeated. What is the value of &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Of the following complex numbers &lt;math&gt;z&lt;/math&gt;, which one has the property that &lt;math&gt;z^5&lt;/math&gt; has the greatest real part?<br /> <br /> &lt;math&gt;\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> What is the value of&lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> A choir direction must select a group of singers from among his &lt;math&gt;6&lt;/math&gt; tenors and &lt;math&gt;8&lt;/math&gt; basses. The only<br /> requirements are that the difference between the numbers of tenors and basses must be a multiple<br /> of &lt;math&gt;4&lt;/math&gt;, and the group must have at least one singer. Let &lt;math&gt;N&lt;/math&gt; be the number of different groups that could be<br /> selected. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;100&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> In the following list of numbers, the integer &lt;math&gt;n&lt;/math&gt; appears &lt;math&gt;n&lt;/math&gt; times in the list for &lt;math&gt;1\le n \le 200&lt;/math&gt;.&lt;cmath&gt;1,2,2,3,3,3,4,4,4,...,200,200,...,200&lt;/cmath&gt;What is the median of the numbers in this list?<br /> <br /> &lt;math&gt;\textbf{(A) }100.5 \qquad \textbf{(B) }134 \qquad \textbf{(C) }142 \qquad \textbf{(D) }150.5\qquad \textbf{(E) }167&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Trapezoid &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;\overline{AB}\parallel\overline{CD},BC=CD=43&lt;/math&gt;, and &lt;math&gt;\overline{AD}\perp\overline{BD}&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the intersection of the diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;. Given that &lt;math&gt;OP=11&lt;/math&gt;, the length of &lt;math&gt;AD&lt;/math&gt; can be written in the form &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and &lt;math&gt;n&lt;/math&gt; is not divisible by the quare of any prime. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let &lt;math&gt;f&lt;/math&gt; be a function defined on the set of positive rational numbers with the property that &lt;math&gt;f(a\cdot b)=f(a)+f(b)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Furthermore, suppose that &lt;math&gt;f&lt;/math&gt; also has the property that &lt;math&gt;f(p)=p&lt;/math&gt; for every prime number &lt;math&gt;p&lt;/math&gt;. For which of the following numbers &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;f(x)&lt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> How many solutions does the equation &lt;math&gt;\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)&lt;/math&gt; have in the closed interval &lt;math&gt;[0,\pi]&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Suppose that on a parabola with vertex &lt;math&gt;V&lt;/math&gt; and a focus &lt;math&gt;F&lt;/math&gt; there exists a point &lt;math&gt;A&lt;/math&gt; such that &lt;math&gt;AF=20&lt;/math&gt; and &lt;math&gt;AV=21&lt;/math&gt;. What is the sum of all possible values of the length &lt;math&gt;FV?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> The five solutions to the equation&lt;cmath&gt;(z-1)(z^2+2z+4)(z^2+4z+6)=0&lt;/cmath&gt;may be written in the form &lt;math&gt;x_k+y_ki&lt;/math&gt; for &lt;math&gt;1\le k\le 5,&lt;/math&gt; where &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;y_k&lt;/math&gt; are real. Let &lt;math&gt;\mathcal E&lt;/math&gt; be the unique ellipse that passes through the points &lt;math&gt;(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),&lt;/math&gt; and &lt;math&gt;(x_5,y_5)&lt;/math&gt;. The excentricity of &lt;math&gt;\mathcal E&lt;/math&gt; can be written in the form &lt;math&gt;\sqrt{\frac mn}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;? (Recall that the eccentricity of an ellipse &lt;math&gt;\mathcal E&lt;/math&gt; is the ratio &lt;math&gt;\frac ca&lt;/math&gt;, where &lt;math&gt;2a&lt;/math&gt; is the length of the major axis of &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;2c&lt;/math&gt; is the is the distance between its two foci.)<br /> <br /> &lt;math&gt;\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3\times3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{9}{16} \qquad \textbf{(B) }\frac{5}{8} \qquad \textbf{(C) }\frac34 \qquad \textbf{(D) }\frac{25}{32}\qquad \textbf{(E) }\frac{13}{16}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Semicircle &lt;math&gt;\Gamma&lt;/math&gt; has diameter &lt;math&gt;\overline{AB}&lt;/math&gt; of length &lt;math&gt;14&lt;/math&gt;. Circle &lt;math&gt;\Omega&lt;/math&gt; lies tangent to &lt;math&gt;\overline{AB}&lt;/math&gt; at a point &lt;math&gt;P&lt;/math&gt; and intersects &lt;math&gt;\Gamma&lt;/math&gt; at points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. If &lt;math&gt;QR=3\sqrt3&lt;/math&gt; and &lt;math&gt;\angle QPR=60^\circ&lt;/math&gt;, then the area of &lt;math&gt;\triangle PQR&lt;/math&gt; equals &lt;math&gt;\frac{a\sqrt{b}}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|before=[[2020 AMC 12B Problems]]|after=[[2021 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=112879 AMC 12 2019-12-15T17:03:28Z <p>Naenaendr: /* Format */</p> <hr /> <div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br /> <br /> The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br /> <br /> {{Contest Info|name=AMC 12|region=USA|type=Multiple Choice|difficulty=2-4|breakdown=&lt;u&gt;Problem 1-10&lt;/u&gt;: 2&lt;br&gt;&lt;u&gt;Problem 11-20&lt;/u&gt;: 3&lt;br&gt;&lt;u&gt;Problem 21-25&lt;/u&gt;: 4}}<br /> <br /> == Format ==<br /> <br /> The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Ever since 2008, calculators have been banned from use during the test; however, calculators were never required to solve any problems, and students who did not use calculators were not disadvantaged.<br /> <br /> The AMC 12 is scored in a way that penalizes guessing. Correct answers are worth 6 points, incorrect answers are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].<br /> <br /> == Curriculum ==<br /> The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br /> * [[AMC 12 Problems and Solutions]]<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br /> <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:algebra Intermediate Algebra] by [[Richard Rusczyk]] and [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:counting Intermediate Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc Precalculus] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> <br /> === AMC Preparation Classes ===<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory and intermediate classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 12 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]<br /> [[Category:Intermediate mathematics competitions]]</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_18&diff=105993 2019 AMC 10A Problems/Problem 18 2019-05-28T02:29:31Z <p>Naenaendr: /* Solution 5 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #18]] and [[2019 AMC 12A Problems|2019 AMC 12A #11]]}}<br /> <br /> ==Problem==<br /> <br /> For some positive integer &lt;math&gt;k&lt;/math&gt;, the repeating base-&lt;math&gt;k&lt;/math&gt; representation of the (base-ten) fraction &lt;math&gt;\frac{7}{51}&lt;/math&gt; is &lt;math&gt;0.\overline{23}_k = 0.232323..._k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can expand the fraction &lt;math&gt;0.\overline{23}_k&lt;/math&gt; as follows: &lt;math&gt;0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...&lt;/math&gt; Notice that this is equivalent to <br /> &lt;cmath&gt;2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )&lt;/cmath&gt;<br /> <br /> By summing the geometric series and simplifying, we have &lt;math&gt;\frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;. Solving this quadratic equation (or simply testing the answer choices) yields the answer &lt;math&gt;k = \boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;a = 0.2323\dots_k&lt;/math&gt;. Therefore, &lt;math&gt;k^2a=23.2323\dots_k&lt;/math&gt;.<br /> <br /> From this, we see that &lt;math&gt;k^2a-a=23_k&lt;/math&gt;, so &lt;math&gt;a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;.<br /> <br /> Now, similar to in Solution 1, we can either test if &lt;math&gt;2k+3&lt;/math&gt; is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 3 (bash)==<br /> We can simply plug in all the answer choices as values of &lt;math&gt;k&lt;/math&gt;, and see which one works. After lengthy calculations, this eventually gives us &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt; as the answer.<br /> <br /> ==Solution 4==<br /> Just as in Solution 1, we arrive at the equation &lt;math&gt;\frac{2k+3}{k^2-1}=\frac{7}{51}&lt;/math&gt;.<br /> <br /> We can now rewrite this as &lt;math&gt;\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}&lt;/math&gt;. Notice that &lt;math&gt;2k+3=2(k+1)+1=2(k-1)+5&lt;/math&gt;. As &lt;math&gt;17&lt;/math&gt; is a prime, we therefore must have that one of &lt;math&gt;k-1&lt;/math&gt; and &lt;math&gt;k+1&lt;/math&gt; is divisible by &lt;math&gt;17&lt;/math&gt;. Now, checking each of the answer choices, this gives &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Assuming you are familiar with the rules for basic repeating decimals, &lt;math&gt;0.232323... = \frac{23}{99}&lt;/math&gt;. Now we want our base, &lt;math&gt;k&lt;/math&gt;, to conform to &lt;math&gt;23 = 7\, (mod\, k)&lt;/math&gt; and &lt;math&gt;99 = 51\, (mod\, k)&lt;/math&gt;, the reason being that we wish to convert the number from base &lt;math&gt;10&lt;/math&gt; to base &lt;math&gt;k&lt;/math&gt;. Given the first equation, we know that &lt;math&gt;k&lt;/math&gt; must equal 9, 16, 23, or generally, &lt;math&gt;7n+2&lt;/math&gt;. The only number in this set that is one of the multiple choices is &lt;math&gt;16&lt;/math&gt;. When we test this on the second equation, &lt;math&gt;99 = 51\, (mod\, k)&lt;/math&gt;, it comes to be true. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2019|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_18&diff=105992 2019 AMC 10A Problems/Problem 18 2019-05-28T02:29:06Z <p>Naenaendr: Added my solution</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #18]] and [[2019 AMC 12A Problems|2019 AMC 12A #11]]}}<br /> <br /> ==Problem==<br /> <br /> For some positive integer &lt;math&gt;k&lt;/math&gt;, the repeating base-&lt;math&gt;k&lt;/math&gt; representation of the (base-ten) fraction &lt;math&gt;\frac{7}{51}&lt;/math&gt; is &lt;math&gt;0.\overline{23}_k = 0.232323..._k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can expand the fraction &lt;math&gt;0.\overline{23}_k&lt;/math&gt; as follows: &lt;math&gt;0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + ...&lt;/math&gt; Notice that this is equivalent to <br /> &lt;cmath&gt;2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + ... )&lt;/cmath&gt;<br /> <br /> By summing the geometric series and simplifying, we have &lt;math&gt;\frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;. Solving this quadratic equation (or simply testing the answer choices) yields the answer &lt;math&gt;k = \boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;a = 0.2323\dots_k&lt;/math&gt;. Therefore, &lt;math&gt;k^2a=23.2323\dots_k&lt;/math&gt;.<br /> <br /> From this, we see that &lt;math&gt;k^2a-a=23_k&lt;/math&gt;, so &lt;math&gt;a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}&lt;/math&gt;.<br /> <br /> Now, similar to in Solution 1, we can either test if &lt;math&gt;2k+3&lt;/math&gt; is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 3 (bash)==<br /> We can simply plug in all the answer choices as values of &lt;math&gt;k&lt;/math&gt;, and see which one works. After lengthy calculations, this eventually gives us &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt; as the answer.<br /> <br /> ==Solution 4==<br /> Just as in Solution 1, we arrive at the equation &lt;math&gt;\frac{2k+3}{k^2-1}=\frac{7}{51}&lt;/math&gt;.<br /> <br /> We can now rewrite this as &lt;math&gt;\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}&lt;/math&gt;. Notice that &lt;math&gt;2k+3=2(k+1)+1=2(k-1)+5&lt;/math&gt;. As &lt;math&gt;17&lt;/math&gt; is a prime, we therefore must have that one of &lt;math&gt;k-1&lt;/math&gt; and &lt;math&gt;k+1&lt;/math&gt; is divisible by &lt;math&gt;17&lt;/math&gt;. Now, checking each of the answer choices, this gives &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Assuming you are familiar with the rules for basic repeating decimals, &lt;math&gt;0.232323..._(10) = \frac{23}{99}&lt;/math&gt;. Now we want our base, &lt;math&gt;k&lt;/math&gt;, to conform to &lt;math&gt;23 = 7\, (mod\, k)&lt;/math&gt; and &lt;math&gt;99 = 51\, (mod\, k)&lt;/math&gt;, the reason being that we wish to convert the number from base &lt;math&gt;10&lt;/math&gt; to base &lt;math&gt;k&lt;/math&gt;. Given the first equation, we know that &lt;math&gt;k&lt;/math&gt; must equal 9, 16, 23, or generally, &lt;math&gt;7n+2&lt;/math&gt;. The only number in this set that is one of the multiple choices is &lt;math&gt;16&lt;/math&gt;. When we test this on the second equation, &lt;math&gt;99 = 51\, (mod\, k)&lt;/math&gt;, it comes to be true. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(D) }16}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://www.youtube.com/watch?v=DFfRJolhwN0<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2019|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_12&diff=101584 2019 AMC 10A Problems/Problem 12 2019-02-09T23:19:21Z <p>Naenaendr: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #12]] and [[2019 AMC 12A Problems|2019 AMC 12A #7]]}}<br /> <br /> ==Problem==<br /> <br /> Melanie computes the mean &lt;math&gt;\mu&lt;/math&gt;, the median &lt;math&gt;M&lt;/math&gt;, and the modes of the &lt;math&gt;365&lt;/math&gt; values that are the dates in the months of &lt;math&gt;2019&lt;/math&gt;. Thus her data consist of &lt;math&gt;12&lt;/math&gt; &lt;math&gt;1\text{s}&lt;/math&gt;, &lt;math&gt;12&lt;/math&gt; &lt;math&gt;2\text{s}&lt;/math&gt;, . . . , &lt;math&gt;12&lt;/math&gt; &lt;math&gt;28\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;29\text{s}&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; &lt;math&gt;30\text{s}&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt; &lt;math&gt;31\text{s}&lt;/math&gt;. Let &lt;math&gt;d&lt;/math&gt; be the median of the modes. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A) } \mu &lt; d &lt; M \qquad\textbf{(B) } M &lt; d &lt; \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d &lt; M &lt; \mu \qquad\textbf{(E) } d &lt; \mu &lt; M&lt;/math&gt;<br /> <br /> ==Solution==<br /> First of all, &lt;math&gt;d&lt;/math&gt; obviously has to smaller than &lt;math&gt;M&lt;/math&gt; since when calculating &lt;math&gt;M&lt;/math&gt; you most take into account the &lt;math&gt;29's&lt;/math&gt;, &lt;math&gt;30's&lt;/math&gt;, and &lt;math&gt;31s&lt;/math&gt;. So we can eliminate &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt;. The median, &lt;math&gt;\mu&lt;/math&gt;, is &lt;math&gt;16&lt;/math&gt;, but you realize that the mean (&lt;math&gt;M&lt;/math&gt;) must be smaller than &lt;math&gt;16&lt;/math&gt; since there are much less &lt;math&gt;29's&lt;/math&gt;, &lt;math&gt;30's&lt;/math&gt;, and &lt;math&gt;31s&lt;/math&gt;. Thus the answer is &lt;math&gt;d &lt; \mu &lt; M \implies \boxed{(E)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2019|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> Solution 1:<br /> <br /> Notice that there are 365 total entries, so the median has to be the 183rd one. Then, realize that 12 * 15 is 180, so 16 has to be the median (because 16 is from 181 to 192). Then, look at the modes (1-28) and realize that even if you have 12 of each, the median of those remains the same and you have 14.5. When trying to find the mean, you realize that the mean of the first 28 is simply the same as the median of them, which is 14.5. Then, when you see 29's, 30's, and 31's, you realize that the mean has to be higher. On the other hand, since there are fewer 29's, 30's, and 31's than the rest of the numbers, the mean has to be lower than 16 (the median). Then, you compare those values and you get the answer, which is E.<br /> Edit: Hello can i move this to solution</div> Naenaendr https://artofproblemsolving.com/wiki/index.php?title=2005_Alabama_ARML_TST_Problems/Problem_5&diff=100093 2005 Alabama ARML TST Problems/Problem 5 2019-01-05T00:24:05Z <p>Naenaendr: /* Problem */ Spelling</p> <hr /> <div>==Problem==<br /> A &lt;math&gt;2\times 2&lt;/math&gt; square grid is constructed with four &lt;math&gt;1\times 1&lt;/math&gt; squares. The square on the upper left is labeled &lt;i&gt;A&lt;/i&gt;, the square on the upper right is labeled &lt;i&gt;B&lt;/i&gt;, the square in the lower left is labeled &lt;i&gt;C&lt;/i&gt;, and the square on the lower right is labeled &lt;i&gt;D&lt;/i&gt;. The four squares are to be painted such that 2 are blue, 1 is red, and 1 is green. In how many ways can this be done?<br /> <br /> ==Solution==<br /> We choose two squares to be blue and one to be red; then the green's position is forced. There are &lt;math&gt;{4 \choose 2}{2 \choose 1}=12&lt;/math&gt; ways to do this.<br /> <br /> Equivalently, we could choose one square to be red and one square to be green, then blue is forced: &lt;math&gt;\binom{4}{1} \binom{3}{1}=12&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{ARML box|year=2005|state=Alabama|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Naenaendr