https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Eatingstuff&feedformat=atom AoPS Wiki - User contributions [en] 2022-10-05T19:01:24Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&diff=100636 2016 AMC 12A Problems/Problem 17 2019-01-19T22:19:08Z <p>Eatingstuff: /* Solution 2 */</p> <hr /> <div>==Problem 17==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a square. Let &lt;math&gt;E, F, G&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; be the centers, respectively, of equilateral triangles with bases &lt;math&gt;\overline{AB}, \overline{BC}, \overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; each exterior to the square. What is the ratio of the area of square &lt;math&gt;EFGH&lt;/math&gt; to the area of square &lt;math&gt;ABCD&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The center of an equilateral triangle is its centroid, where the three medians meet.<br /> <br /> The distance along the median from the centroid to the base is one third the length of the median.<br /> <br /> Let the side length of the square be &lt;math&gt;1&lt;/math&gt;. The height of &lt;math&gt;\triangle E&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2},&lt;/math&gt; so the distance from &lt;math&gt;E&lt;/math&gt; to the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}&lt;/math&gt;<br /> <br /> &lt;math&gt;EG = 2 \cdot \frac{\sqrt{3}}{6}&lt;/math&gt; (from above) &lt;math&gt; + 1&lt;/math&gt; (side length of the square).<br /> <br /> Since &lt;math&gt;EG&lt;/math&gt; is the diagonal of square &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> The Original Solution 2 is bs. Someone find an alternate solution for this problem.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_19&diff=99795 2017 AMC 12A Problems/Problem 19 2018-12-27T16:05:44Z <p>Eatingstuff: /* Solution 18 */</p> <hr /> <div>==Problem==<br /> <br /> A square with side length &lt;math&gt;x&lt;/math&gt; is inscribed in a right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length &lt;math&gt;y&lt;/math&gt; is inscribed in another right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one side of the square lies on the hypotenuse of the triangle. What is &lt;math&gt;\frac{x}{y}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{12}{13}<br /> \qquad \textbf{(B)}\ \frac{35}{37}<br /> \qquad\textbf{(C)}\ 1<br /> \qquad\textbf{(D)}\ \frac{37}{35}<br /> \qquad\textbf{(E)}\ \frac{13}{12}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Analyze the first right triangle.<br /> <br /> &lt;asy&gt;<br /> pair A,B,C;<br /> pair D, e, F;<br /> A = (0,1);<br /> B = (4,0);<br /> C = (2,3);<br /> <br /> D = (0, 12/7);<br /> e = (12/7 , 12/7);<br /> F = (12/7, 0);<br /> <br /> draw(A--B--C--cycle);<br /> draw(D--e--F);<br /> <br /> label(&quot;$q$&quot;, D/2, W);<br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$3$&quot;, B, SE);<br /> label(&quot;$e$&quot;, C, N);<br /> label(&quot;$D$&quot;, D, W);<br /> label(&quot;$E$&quot;, e, NE);<br /> label(&quot;$F$&quot;, F, S);<br /> &lt;/asy&gt;<br /> <br /> Note that &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle FBE&lt;/math&gt; are similar, so &lt;math&gt;\frac{BF}{FE} = \frac{AB}{AC}&lt;/math&gt;. Th1s can be wr1tten 4s &lt;math&gt;\frac{4-x}{x}=\frac{4}{3}&lt;/math&gt;. Solving, &lt;math&gt;x = \frac{12}{7}&lt;/math&gt;.<br /> <br /> Now we analyze the third triangle.<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C;<br /> pair q, R, S, T;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (0,3);<br /> <br /> q = (1.297, 0);<br /> R = (2.27 , 1.297);<br /> S = (0.973, 2.27);<br /> T = (0, 0.973);<br /> <br /> draw(A--B--C--cycle);<br /> draw(q--R--S--T--cycle);<br /> <br /> label(&quot;$y$&quot;, (q+R)/2, NW);<br /> label(&quot;$A'$&quot;, A, SW);<br /> label(&quot;$B'$&quot;, B, SE);<br /> label(&quot;$C'$&quot;, C, N);<br /> label(&quot;$Q$&quot;, (q-(0,0.3)));<br /> label(&quot;$R$&quot;, R, NE);<br /> label(&quot;$S$&quot;, S, NE);<br /> label(&quot;$T$&quot;, T, W);<br /> &lt;/asy&gt;<br /> <br /> Similarly, &lt;math&gt;\triangle A'B'C'&lt;/math&gt; and &lt;math&gt;\triangle RB'Q&lt;/math&gt; are similar, so &lt;math&gt;RB' = \frac{4}{3}y&lt;/math&gt;, and &lt;math&gt;C'S = \frac{3}{4}y&lt;/math&gt;. Thus, &lt;math&gt;C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y = \frac{60}{37}&lt;/math&gt;. Thus, &lt;math&gt;\frac{x}{y} = \frac{37}{35}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2017|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_20&diff=93525 2018 AMC 10A Problems/Problem 20 2018-03-27T00:46:30Z <p>Eatingstuff: /* Solution 2 */</p> <hr /> <div>A scanning code consists of a &lt;math&gt;7 \times 7&lt;/math&gt; grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of &lt;math&gt;49&lt;/math&gt; squares. A scanning code is called &lt;math&gt;\textit{symmetric}&lt;/math&gt; if its look does not change when the entire square is rotated by a multiple of &lt;math&gt;90 ^{\circ}&lt;/math&gt; counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw a &lt;math&gt;7 \times 7&lt;/math&gt; square.<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|c|c|c|c|}<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> G &amp; D &amp; B &amp; A &amp; B &amp; D &amp; G \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> \end{tabular} &lt;/math&gt;<br /> <br /> Start from the center and label all protruding cells symmetrically.<br /> <br /> More specifically, since there are &lt;math&gt;4&lt;/math&gt; given lines of symmetry (&lt;math&gt;2&lt;/math&gt; diagonals, &lt;math&gt;1&lt;/math&gt; vertical, &lt;math&gt;1&lt;/math&gt; horizontal) and they split the plot into &lt;math&gt;8&lt;/math&gt; equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has &lt;math&gt;10&lt;/math&gt; distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose &lt;math&gt;2^{10}=1024&lt;/math&gt; but then subtract the &lt;math&gt;2&lt;/math&gt; cases where all are white or all are black. That leaves us with &lt;math&gt;\boxed{(B)}, 1022&lt;/math&gt;. ∎<br /> <br /> There are only ten squares we get to actually choose, and two independent choices for each, for a total of &lt;math&gt;2^{10} = 1024&lt;/math&gt; codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of &lt;math&gt;\fbox{\textbf{(B)} \text{ 1022}}&lt;/math&gt;.<br /> ~Nosysnow<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt;<br /> size(100pt);<br /> draw((1,0)--(8,0),linewidth(0.5));<br /> draw((1,2)--(6,2),linewidth(0.5));<br /> draw((1,4)--(4,4),linewidth(0.5));<br /> draw((1,6)--(2,6),linewidth(0.5));<br /> draw((2,6)--(2,0),linewidth(0.5));<br /> draw((4,4)--(4,0),linewidth(0.5));<br /> draw((6,2)--(6,0),linewidth(0.5));<br /> draw((1,0)--(1,7),dashed+linewidth(0.5));<br /> draw((1,7)--(8,0),dashed+linewidth(0.5));<br /> &lt;/asy&gt;<br /> <br /> Imagine folding the scanning code along its lines of symmetry. There will be &lt;math&gt;10&lt;/math&gt; regions which you have control over coloring. Since we must subtract off &lt;math&gt;2&lt;/math&gt; cases for the all-black and all-white cases, the answer is &lt;math&gt;2^{10}-2=\boxed{\textbf{(B) } 1022.}&lt;/math&gt;<br /> <br /> -EatingStuff<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_4&diff=93523 2018 AMC 10A Problems/Problem 4 2018-03-26T23:16:24Z <p>Eatingstuff: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.<br /> <br /> Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:<br /> <br /> Periods &lt;math&gt;1, 3, 5&lt;/math&gt;<br /> <br /> Periods &lt;math&gt;1, 3, 6&lt;/math&gt;<br /> <br /> Periods &lt;math&gt;1, 4, 6&lt;/math&gt;<br /> <br /> Periods &lt;math&gt;2, 4, 6&lt;/math&gt;<br /> <br /> There are &lt;math&gt;4&lt;/math&gt; ways to place &lt;math&gt;3&lt;/math&gt; nondistinguishable classes into &lt;math&gt;6&lt;/math&gt; periods such that no two classes are in consecutive periods. For each of these ways, there are &lt;math&gt;3! = 6&lt;/math&gt; orderings of the classes among themselves.<br /> <br /> Therefore, there are &lt;math&gt;4 \cdot 6 = \boxed{\textbf{(E) } 24}&lt;/math&gt; ways to choose the classes.<br /> <br /> -Versailles15625<br /> <br /> ==Solution 2==<br /> <br /> <br /> Sup people<br /> <br /> ==Solution 3==<br /> Realize that the number of ways of placing, regardless of order, the 3 mathematics courses in a 6-period day so that no two are consecutive is the same as the number of ways of placing 3 mathematics courses in a sequence of 4 periods regardless of order and whether or not they are consecutive.<br /> <br /> To see that there is a one to one correlation, note that for every way of placing 3 mathematics courses in 4 total periods (as above) one can add a non-mathematics course between each pair (2 total) of consecutively occurring mathematics courses (not necessarily back to back) to ensure there will be no two consecutive mathematics courses in the resulting 6-period day.<br /> For example, where &lt;math&gt;M&lt;/math&gt; denotes a math course and &lt;math&gt;O&lt;/math&gt; denotes a non-math course:<br /> &lt;math&gt;M O M M \rightarrow M O O M O M&lt;/math&gt;<br /> <br /> For each 6-period sequence consisting of &lt;math&gt;M&lt;/math&gt;s and &lt;math&gt;O&lt;/math&gt;s, we have &lt;math&gt;3!&lt;/math&gt; orderings of the 3 distinct mathematics courses.<br /> <br /> So, our answer is &lt;math&gt;\dbinom{4}{3}(3!)= \boxed{\textbf{(E) } 24}&lt;/math&gt;<br /> <br /> - Gregwwl<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=3|num-a=5}}<br /> {{AMC12 box|year=2018|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_12&diff=91854 2017 AMC 10A Problems/Problem 12 2018-02-17T21:55:34Z <p>Eatingstuff: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be a set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3,~x+2,&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\\qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\\qquad\textbf{(D)}\ \text{a triangle} \qquad\textbf{(E)}\ \text{three rays with a common endpoint}&lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y \le 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y \le 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x\le1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x\le 1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\le y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\ge 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y\ge 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(E) }\text{three rays with a common endpoint}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_12&diff=91853 2017 AMC 10A Problems/Problem 12 2018-02-17T21:54:14Z <p>Eatingstuff: /* See Also */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be a set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3,~x+2,&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\\qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\\qquad\textbf{(D)}\ \text{a triangle} \qquad\textbf{(E)}\ \text{three rays with a common endpoint}&lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y \le 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y \le 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\le 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x\le1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x\le 1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\le y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\ge 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y\ge 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(E) }\text{three rays with a common endpoint}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13..................................}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_13&diff=91852 2017 AMC 10A Problems/Problem 13 2018-02-17T21:49:42Z <p>Eatingstuff: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Define a sequence recursively by &lt;math&gt;F_{0}=0,~F_{1}=1,&lt;/math&gt; and &lt;math&gt;F_{n}=&lt;/math&gt; the remainder when &lt;math&gt;F_{n-1}+F_{n-2}&lt;/math&gt; is divided by &lt;math&gt;3,&lt;/math&gt; for all &lt;math&gt;n\geq 2.&lt;/math&gt; Thus the sequence starts &lt;math&gt;0,1,1,2,0,2,\ldots&lt;/math&gt; What is &lt;math&gt;F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> A pattern starts to emerge as the function is continued. The repeating pattern is &lt;math&gt;0,1,1,2,0,2,2,1\ldots&lt;/math&gt; The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is &lt;math&gt;\boxed{\textbf{(D)}\ 9}&lt;/math&gt;.....................<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Eatingstuff https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_13&diff=91851 2017 AMC 10A Problems/Problem 13 2018-02-17T21:49:21Z <p>Eatingstuff: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Define a sequence recursively by &lt;math&gt;F_{0}=0,~F_{1}=1,&lt;/math&gt; and &lt;math&gt;F_{n}=&lt;/math&gt; the remainder when &lt;math&gt;F_{n-1}+F_{n-2}&lt;/math&gt; is divided by &lt;math&gt;3,&lt;/math&gt; for all &lt;math&gt;n\geq 2.&lt;/math&gt; Thus the sequence starts &lt;math&gt;0,1,1,2,0,2,\ldots&lt;/math&gt; What is &lt;math&gt;F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> A pattern starts to emerge as the function is continued. The repeating pattern is &lt;math&gt;0,1,1,2,0,2,2,1\ldots&lt;/math&gt; The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is &lt;math&gt;\boxed{\textbf{(D)}\ 9}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Eatingstuff