https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ljlbox&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-17T15:58:59Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=131950 2017 AMC 10A Problems/Problem 25 2020-08-16T23:26:47Z <p>Ljlbox: /* Video Solution */</p> <hr /> <div>==Problem==<br /> How many integers between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, inclusive, have the property that some permutation of its digits is a multiple of &lt;math&gt;11&lt;/math&gt; between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999?&lt;/math&gt; For example, both &lt;math&gt;121&lt;/math&gt; and &lt;math&gt;211&lt;/math&gt; have this property.<br /> <br /> &lt;math&gt; \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.<br /> <br /> Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.<br /> <br /> There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each. <br /> <br /> There are 110, 220, 330 ... 990, yielding 9 extra permutations<br /> <br /> Also, there are 209, 308, 407...902, yielding 8 more permutations.<br /> <br /> Now, just subtract these 17 from the total (243), getting 226. &lt;math&gt;\boxed{\textbf{(A) } 226}&lt;/math&gt;<br /> <br /> *Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer. <br /> <br /> ===Solution 2===<br /> <br /> Let the three-digit number be &lt;math&gt;ACB&lt;/math&gt;:<br /> <br /> If a number is divisible by &lt;math&gt;11&lt;/math&gt;, then the difference between the sums of alternating digits is a multiple of &lt;math&gt;11&lt;/math&gt;.<br /> <br /> There are two cases:<br /> &lt;math&gt;A+B=C&lt;/math&gt; and &lt;math&gt;A+B=C+11&lt;/math&gt;<br /> <br /> We now proceed to break down the cases. Note: let &lt;math&gt;A \geq C&lt;/math&gt; so that we avoid counting the same permutations and having to subtract them later.<br /> <br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;: &lt;math&gt;A+B=C&lt;/math&gt;. <br /> <br /> <br /> <br /> &lt;math&gt;\textbf{Part 1}&lt;/math&gt;: &lt;math&gt;B=0, A=C&lt;/math&gt;, this case results in &lt;math&gt;110, 220, 330...990&lt;/math&gt;. There are two ways to arrange the digits in each of those numbers.<br /> &lt;math&gt;2 \cdot 9 = 18&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Part 2}&lt;/math&gt;:<br /> &lt;math&gt;B=1, A+1=C&lt;/math&gt;, this case results in &lt;math&gt;121, 231,... 891&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and &lt;math&gt;3&lt;/math&gt; ways for the first. This leads to &lt;math&gt;45&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 3}&lt;/math&gt;: &lt;math&gt;B=2, A+2=C&lt;/math&gt;, this case results in &lt;math&gt;242, 352,... 792&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to &lt;math&gt;33&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 4}&lt;/math&gt;: &lt;math&gt;B=3, A+3=C&lt;/math&gt;, this case results in &lt;math&gt;363, 473,...693&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to &lt;math&gt;21&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 5}&lt;/math&gt;: &lt;math&gt;B=4, A+4=C&lt;/math&gt;, this case results in &lt;math&gt;484&lt;/math&gt; and &lt;math&gt;594&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in 594 and 3 ways for 484. This leads to &lt;math&gt;9&lt;/math&gt; cases.<br /> <br /> This case has &lt;math&gt;18+45+33+21+9=126&lt;/math&gt; subcases.<br /> <br /> <br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;: &lt;math&gt;A+B=C+11&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\textbf{Part 1}&lt;/math&gt;: &lt;math&gt;C=0, A+B=11&lt;/math&gt;, this cases results in &lt;math&gt;209, 308, 407, 506&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to arrange each of those cases. This leads to &lt;math&gt;16&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 2}&lt;/math&gt;: &lt;math&gt;C=1, A+B=12&lt;/math&gt;, this cases results in &lt;math&gt;319, 418,517,616&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange each of those cases, except the last. This leads to &lt;math&gt;21&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 3}&lt;/math&gt;: &lt;math&gt;C=2, A+B=13&lt;/math&gt;, this cases results in &lt;math&gt;429, 528, 627&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange each of those cases. This leads to &lt;math&gt;18&lt;/math&gt; cases.<br /> <br /> ...<br /> If we continue this counting, we receive &lt;math&gt;16+21+18+15+12+9+6+3=100&lt;/math&gt; subcases.<br /> <br /> &lt;math&gt;100+126=\boxed{\textbf{(A) } 226}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> We note that we only have to consider multiples of &lt;math&gt;11&lt;/math&gt; and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of &lt;math&gt;11&lt;/math&gt; has:<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; All three digits are the same. <br /> By inspection, we find that there are no multiples of &lt;math&gt;11&lt;/math&gt; here.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two of the digits are the same, and the third is different.<br /> <br /> &lt;math&gt;\textbf{Case 2a:}&lt;/math&gt;<br /> There are &lt;math&gt;8&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; without a zero that have this property:<br /> &lt;math&gt;121&lt;/math&gt;, &lt;math&gt;242&lt;/math&gt;, &lt;math&gt;363&lt;/math&gt;, &lt;math&gt;484&lt;/math&gt;, &lt;math&gt;616&lt;/math&gt;, &lt;math&gt;737&lt;/math&gt;, &lt;math&gt;858&lt;/math&gt;, &lt;math&gt;979&lt;/math&gt;.<br /> Each contributes &lt;math&gt;3&lt;/math&gt; valid permutations, so there are &lt;math&gt;8 \cdot 3 = 24&lt;/math&gt; permutations in this subcase.<br /> <br /> &lt;math&gt;\textbf{Case 2b:}&lt;/math&gt;<br /> There are &lt;math&gt;9&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; with a zero that have this property:<br /> &lt;math&gt;110&lt;/math&gt;, &lt;math&gt;220&lt;/math&gt;, &lt;math&gt;330&lt;/math&gt;, &lt;math&gt;440&lt;/math&gt;, &lt;math&gt;550&lt;/math&gt;, &lt;math&gt;660&lt;/math&gt;, &lt;math&gt;770&lt;/math&gt;, &lt;math&gt;880&lt;/math&gt;, &lt;math&gt;990&lt;/math&gt;.<br /> Each one contributes &lt;math&gt;2&lt;/math&gt; valid permutations (the first digit can't be zero), so there are &lt;math&gt;9 \cdot 2 = 18&lt;/math&gt; permutations in this subcase.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; All the digits are different.<br /> Since there are &lt;math&gt;\frac{990-110}{11}+1 = 81&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, there are &lt;math&gt;81-8-9 = 64&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; remaining in this case. However, &lt;math&gt;8&lt;/math&gt; of them contain a zero, namely &lt;math&gt;209&lt;/math&gt;, &lt;math&gt;308&lt;/math&gt;, &lt;math&gt;407&lt;/math&gt;, &lt;math&gt;506&lt;/math&gt;, &lt;math&gt;605&lt;/math&gt;, &lt;math&gt;704&lt;/math&gt;, &lt;math&gt;803&lt;/math&gt;, and &lt;math&gt;902&lt;/math&gt;. Each of those multiples of &lt;math&gt;11&lt;/math&gt; contributes &lt;math&gt;2 \cdot 2=4&lt;/math&gt; valid permutations, but we overcounted by a factor of &lt;math&gt;2&lt;/math&gt;; every permutation of &lt;math&gt;209&lt;/math&gt;, for example, is also a permutation of &lt;math&gt;902&lt;/math&gt;. Therefore, there are &lt;math&gt;8 \cdot 4 / 2 = 16&lt;/math&gt;. Therefore, there are &lt;math&gt;64-8=56&lt;/math&gt; remaining multiples of &lt;math&gt;11&lt;/math&gt; without a &lt;math&gt;0&lt;/math&gt; in this case. Each one contributes &lt;math&gt;3! = 6&lt;/math&gt; valid permutations, but once again, we overcounted by a factor of &lt;math&gt;2&lt;/math&gt; (note that if a number ABC is a multiple of &lt;math&gt;11&lt;/math&gt;, then so is CBA). Therefore, there are &lt;math&gt;56 \cdot 6 / 2 = 168&lt;/math&gt; valid permutations in this subcase.<br /> <br /> Adding up all the permutations from all the cases, we have &lt;math&gt;24+18+16+168 = \boxed{\textbf{(A) } 226}&lt;/math&gt;.<br /> <br /> ==Solution 4 ==<br /> <br /> We can first overcount and then subtract.<br /> We know that there are &lt;math&gt;81&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt;.<br /> <br /> We can then multiply by &lt;math&gt;6&lt;/math&gt; for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)<br /> <br /> Now divide by &lt;math&gt;2&lt;/math&gt;, because if a number &lt;math&gt;abc&lt;/math&gt; with digits &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is a multiple of &lt;math&gt;11&lt;/math&gt;, then &lt;math&gt;cba&lt;/math&gt; is also a multiple of &lt;math&gt;11&lt;/math&gt; so we have counted the same permutations twice. <br /> <br /> Basically, each multiple of &lt;math&gt;11&lt;/math&gt; has its own &lt;math&gt;3&lt;/math&gt; permutations (say &lt;math&gt;abc&lt;/math&gt; has &lt;math&gt;abc&lt;/math&gt; &lt;math&gt;acb&lt;/math&gt; and &lt;math&gt;bac&lt;/math&gt; whereas &lt;math&gt;cba&lt;/math&gt; has &lt;math&gt;cba&lt;/math&gt; &lt;math&gt;cab&lt;/math&gt; and &lt;math&gt;bca&lt;/math&gt;). We know that each multiple of &lt;math&gt;11&lt;/math&gt; has at least &lt;math&gt;3&lt;/math&gt; permutations because it cannot have &lt;math&gt;3&lt;/math&gt; repeating digits.<br /> <br /> Hence we have &lt;math&gt;243&lt;/math&gt; permutations without subtracting for overcounting.<br /> Now note that we overcounted cases in which we have &lt;math&gt;0&lt;/math&gt;'s at the start of each number. So, in theory, we could just answer &lt;math&gt;A&lt;/math&gt; and then move on.<br /> <br /> If we want to solve it, then we continue.<br /> <br /> We overcounted cases where the middle digit of the number is &lt;math&gt;0&lt;/math&gt; and the last digit is &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Note that we assigned each multiple of &lt;math&gt;11&lt;/math&gt; three permutations.<br /> <br /> The last digit is &lt;math&gt;0&lt;/math&gt; gives &lt;math&gt;9&lt;/math&gt; possibilities where we overcounted by &lt;math&gt;1&lt;/math&gt; permutation for each of &lt;math&gt;110, 220, ... , 990&lt;/math&gt;.<br /> <br /> The middle digit is &lt;math&gt;0&lt;/math&gt; gives &lt;math&gt;8&lt;/math&gt; possibilities where we overcount by &lt;math&gt;1&lt;/math&gt;.<br /> &lt;math&gt;605, 704, 803, 902&lt;/math&gt; and &lt;math&gt;506, 407, 308, 209&lt;/math&gt;<br /> <br /> Subtracting &lt;math&gt;17&lt;/math&gt; gives &lt;math&gt;\boxed{\textbf{(A) } 226}&lt;/math&gt;.<br /> <br /> Now, we may ask if there is further overlap (i.e if two of &lt;math&gt;abc&lt;/math&gt; and &lt;math&gt;bac&lt;/math&gt; and &lt;math&gt;acb&lt;/math&gt; were multiples of &lt;math&gt;11&lt;/math&gt;). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod &lt;math&gt;11&lt;/math&gt; and adding, we get that &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;2b&lt;/math&gt;, or &lt;math&gt;2c&lt;/math&gt; is congruent to &lt;math&gt;0\ (mod\ 11)&lt;/math&gt;. Since &lt;math&gt;a, b, c&lt;/math&gt; are digits, this can never happen as none of them can equal &lt;math&gt;11&lt;/math&gt; and they can't equal &lt;math&gt;0&lt;/math&gt; as they are the leading digit of a three-digit number in each of the cases.<br /> <br /> == Solution 5: A Slightly Adjusted Version of Solution 2 ==<br /> <br /> <br /> &lt;math&gt;\textbf{WARNING:}&lt;/math&gt; If you do not feel comfortable looking at a massive amount of casework, please skip the solution.<br /> <br /> <br /> Recalling the divisibility rule for &lt;math&gt;11&lt;/math&gt;, if we have a number &lt;math&gt;ABC&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are digits, then &lt;math&gt;11\mid -A+B-C&lt;/math&gt;.<br /> <br /> Notice that for any three-digit positive integer &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;-A+B-C&lt;11&lt;/math&gt;, thus we have 2 possibilities: &lt;math&gt;-A+B-C=0&lt;/math&gt; and &lt;math&gt;-A+B-C=-11&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; &lt;math&gt;-A+B-C=0\Longrightarrow A+C=B&lt;/math&gt;<br /> <br /> Subcase &lt;math&gt;a&lt;/math&gt;: &lt;math&gt;A\neq B\neq C\neq0&lt;/math&gt;<br /> <br /> We have these values for &lt;math&gt;A+C=B&lt;/math&gt;<br /> &lt;cmath&gt;1+2=3,1+3=4,1+4=5,...,1+8=9&lt;/cmath&gt;<br /> &lt;cmath&gt;2+3=5,2+4=6,...,2+7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;3+4=7,3+5=8,3+6=9&lt;/cmath&gt;<br /> &lt;cmath&gt;4+5=9&lt;/cmath&gt;<br /> From which we get &lt;math&gt;7+5+3+1=16&lt;/math&gt; triples &lt;math&gt;(A,B,C)&lt;/math&gt;. Counting every permutation, we have &lt;math&gt;16\cdot3!=96&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;A=C&lt;/math&gt;, &lt;math&gt;A,B,C\neq0&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;1+1=2,2+2=4,3+3=6,4+4=8&lt;/cmath&gt;<br /> From which we get &lt;math&gt;4\cdot3=12&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;A=B,C=0&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;1+0=1,2+0=2,...,9+0=9&lt;/cmath&gt;<br /> Since &lt;math&gt;0&lt;/math&gt; can't be the hundreds digit, from here we get &lt;math&gt;9\cdot2=18&lt;/math&gt; possibilities. Summing up case &lt;math&gt;1&lt;/math&gt;, we have &lt;math&gt;96+12+18=126&lt;/math&gt; possibilities.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; &lt;math&gt;-A+B-C=-11\Longrightarrow A+C-B=11&lt;/math&gt;<br /> <br /> Subcase &lt;math&gt;a&lt;/math&gt;: &lt;math&gt;A\neq B\neq C\neq0&lt;/math&gt;<br /> <br /> We have these values for &lt;math&gt;A+C-B=11&lt;/math&gt;<br /> &lt;cmath&gt;9+8-6=11,9+7-5=11,...9+3-1=11&lt;/cmath&gt;<br /> &lt;cmath&gt;8+7-4=11,8+6-3=11,8+5-2=11,8+4-1&lt;/cmath&gt;<br /> &lt;cmath&gt;...&lt;/cmath&gt;<br /> From which we get &lt;math&gt;(6+4+2)\cdot3!=72&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;A=C&lt;/math&gt;, &lt;math&gt;A,B,C\neq0&lt;/math&gt;<br /> <br /> We have <br /> &lt;cmath&gt;9+9=7,8+8-5,7+7-3=11,6+6-1=11&lt;/cmath&gt;<br /> From which we get &lt;math&gt;4\cdot3=12&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;B=0\Longrightarrow A+C+11&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;9+2=8+3=7+4=6+5=11&lt;/cmath&gt;<br /> From which we get &lt;math&gt;2\cdot2\cdot4=16&lt;/math&gt; possibilities. Summing up case &lt;math&gt;2&lt;/math&gt;, we have &lt;math&gt;72+12+16=100&lt;/math&gt; possibilities.<br /> <br /> Adding the &lt;math&gt;2&lt;/math&gt; cases, we get a total of &lt;math&gt;126+100=226&lt;/math&gt; possibilities. &lt;math&gt;\boxed{\mathrm{(A)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solution==<br /> Two different variations on solving it.<br /> https://youtu.be/z5KNZEwmrWM<br /> <br /> https://youtu.be/MBcHwu30MX4<br /> -Video Solution by Richard Rusczyk<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_22&diff=131940 2017 AMC 10A Problems/Problem 22 2020-08-16T22:17:28Z <p>Ljlbox: /* Video Solution */</p> <hr /> <div>==Problem==<br /> <br /> Sides &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; of equilateral triangle &lt;math&gt;ABC&lt;/math&gt; are tangent to a circle at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively. What fraction of the area of &lt;math&gt;\triangle ABC&lt;/math&gt; lies outside the circle?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}&lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> real sqrt3 = 1.73205080757;<br /> draw(Circle((4, 4), 4));<br /> draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6));<br /> label(&quot;A&quot;, (4, 12.4));<br /> label(&quot;B&quot;, (-.3, 6.3));<br /> label(&quot;C&quot;, (8.3, 6.3));<br /> label(&quot;O&quot;, (4, 3.4));<br /> &lt;/asy&gt;<br /> Let the radius of the circle be &lt;math&gt;r&lt;/math&gt;, and let its center be &lt;math&gt;O&lt;/math&gt;. Since &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; are tangent to circle &lt;math&gt;O&lt;/math&gt;, then &lt;math&gt;\angle OBA = \angle OCA = 90^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle BOC = 120^{\circ}&lt;/math&gt;. Therefore, since &lt;math&gt;\overline{OB}&lt;/math&gt; and &lt;math&gt;\overline{OC}&lt;/math&gt; are equal to &lt;math&gt;r&lt;/math&gt;, then (pick your favorite method) &lt;math&gt;\overline{BC} = r\sqrt{3}&lt;/math&gt;. The area of the equilateral triangle is &lt;math&gt;\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4&lt;/math&gt;, and the area of the sector we are subtracting from it is &lt;math&gt;\frac 13 \pi r^2 - \frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4&lt;/math&gt;. The area outside of the circle is &lt;math&gt; \frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3&lt;/math&gt;. Therefore, the answer is &lt;cmath&gt;\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}&lt;/cmath&gt;<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=GnJDNtjd57k&amp;feature=youtu.be<br /> <br /> https://youtu.be/ADDAOhNAsjQ<br /> -Video Solution by Richard Rusczyk<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_21&diff=131939 2017 AMC 10A Problems/Problem 21 2020-08-16T22:12:36Z <p>Ljlbox: /* Video Solution */</p> <hr /> <div>==Problem==<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;\tfrac{x}{y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Analyze the first right triangle.<br /> <br /> &lt;asy&gt;<br /> pair A,B,C;<br /> pair D, e, F;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (0,3);<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;$x$&quot;, D/2, W);<br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&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;. This can be written as &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 second 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} = \boxed{\textbf{(D)}\:\frac{37}{35}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/THeq4ZiZxIA<br /> <br /> https://youtu.be/MF2QFOInbYc<br /> -Video Solution by Richard Rusczyk<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}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_25&diff=120494 2013 AMC 10A Problems/Problem 25 2020-04-03T22:43:32Z <p>Ljlbox: /* Solution 3 (Answer choices and reasoning) */</p> <hr /> <div>==Problem==<br /> <br /> All &lt;math&gt;20&lt;/math&gt; diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1 (Drawing)==<br /> <br /> If you draw a clear diagram like the one below, it is easy to see that there are &lt;math&gt;\boxed{\textbf{(A) }49}&lt;/math&gt; points.<br /> <br /> &lt;asy&gt;<br /> size(14cm);<br /> pathpen = brown + 1.337;<br /> // Initialize octagon<br /> pair[] A;<br /> for (int i=0; i&lt;8; ++i) {<br /> A[i] = dir(45*i);<br /> }<br /> D(CR( (0,0), 1));<br /> // Draw diagonals<br /> // choose pen colors<br /> pen[] colors;<br /> colors = orange + 1.337;<br /> colors = purple;<br /> colors = green;<br /> colors = black;<br /> for (int d=1; d&lt;=4; ++d) {<br /> pathpen = colors[d];<br /> for (int j=0; j&lt;8; ++j) {<br /> D(A[j]--A[(j+d) % 8]);<br /> }<br /> }<br /> pathpen = blue + 2;<br /> // Draw all the intersections<br /> pointpen = red + 7;<br /> for (int x1=0; x1&lt;8; ++x1) {<br /> for (int x2=x1+1; x2&lt;8; ++x2) {<br /> for (int x3=x2+1; x3&lt;8; ++x3) {<br /> for (int x4=x3+1; x4&lt;8; ++x4) {<br /> D(IP(A[x1]--A[x2], A[x3]--A[x4]));<br /> D(IP(A[x1]--A[x3], A[x4]--A[x2]));<br /> D(IP(A[x1]--A[x4], A[x2]--A[x3]));<br /> }<br /> }<br /> }<br /> }&lt;/asy&gt;<br /> <br /> ==Solution 2 (Working Backwards)==<br /> <br /> Let the number of intersections be &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;x\le \dbinom{8}{4} = 70&lt;/math&gt;, as every &lt;math&gt;4&lt;/math&gt; vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point. However, four diagonals intersect in the center, so we need to subtract &lt;math&gt;\dbinom{4}{2} -1 = 5&lt;/math&gt; from this count, &lt;math&gt;70-5 = 65&lt;/math&gt;. Note that diagonals like &lt;math&gt;\overline{AD}&lt;/math&gt;, &lt;math&gt;\overline{CG}&lt;/math&gt;, and &lt;math&gt;\overline{BE}&lt;/math&gt; all intersect at the same point. There are &lt;math&gt;8&lt;/math&gt; of this type with three diagonals intersecting at the same point, so we need to subtract &lt;math&gt;2&lt;/math&gt; of the &lt;math&gt;\dbinom{3}{2}&lt;/math&gt; (one is kept as the actual intersection). In the end, we obtain &lt;math&gt;65 - 16 = \boxed{\textbf{(A) }49}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Answer choices and reasoning)==<br /> We know that the amount of intersection points is at most &lt;math&gt;\dbinom{8}{4} = 70&lt;/math&gt;, as in solution &lt;math&gt;2&lt;/math&gt;. There's probably going to be more than &lt;math&gt;5&lt;/math&gt; (to get &lt;math&gt;\boxed{\textbf{(B) }65}&lt;/math&gt;), leading us to the only reasonable answer, &lt;math&gt;\boxed{\textbf{(A) }49}&lt;/math&gt;.<br /> -Lcz<br /> <br /> ==Solution 4 (Drawing but easier)==<br /> Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains &lt;math&gt;8&lt;/math&gt; points (not including the octagon center). However each adjacent region share one side in common and that side contains &lt;math&gt;2&lt;/math&gt; intersection points, so in actuality there are &lt;math&gt;8 - 2 = 6&lt;/math&gt; points per region. We multiply this by &lt;math&gt;8&lt;/math&gt; to get &lt;math&gt;6\cdot 8 = 48&lt;/math&gt; and add the one center point to get &lt;math&gt;48 + 1 = \boxed{\textbf{(A) }49}&lt;/math&gt;.<br /> <br /> ~skyscraper<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://www.youtube.com/watch?v=3ovuMRYs--8<br /> <br /> ~dolphin7<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A&diff=120425 2022 AMC 10A 2020-04-02T18:24:41Z <p>Ljlbox: Created page with &quot;existn't&quot;</p> <hr /> <div>existn't</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems&diff=120374 2022 AMC 10A Problems 2020-04-01T17:31:30Z <p>Ljlbox: </p> <hr /> <div>This shall appear in 2 years time.</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems&diff=120373 2022 AMC 10A Problems 2020-04-01T17:30:20Z <p>Ljlbox: Created page with &quot;&lt;cmath&gt;\lim_(x\to\infty){\frac{x^2}{17e^x-lnx}}&lt;/cmath&gt;&quot;</p> <hr /> <div>&lt;cmath&gt;\lim_(x\to\infty){\frac{x^2}{17e^x-lnx}}&lt;/cmath&gt;</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=114705 2019 AMC 8 Problems 2020-01-14T02:37:47Z <p>Ljlbox: /* Problem 23 */</p> <hr /> <div>== Problem 1 ==<br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?<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 /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is &lt;math&gt;5&lt;/math&gt; feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11}&lt;/math&gt;, &lt;math&gt;\frac{19}{15}&lt;/math&gt;, and &lt;math&gt;\frac{17}{13}&lt;/math&gt;, from least to greatest?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11} &lt; \frac{17}{13} &lt; \frac{19}{15}\qquad\textbf{(B) }\frac{15}{11} &lt; \frac{19}{15} &lt; \frac{17}{13}\qquad\textbf{(C) }\frac{17}{13} &lt; \frac{19}{15} &lt; \frac{15}{11}\qquad\textbf{(D) }\frac{19}{15} &lt; \frac{15}{11} &lt; \frac{17}{13}\qquad\textbf{(E) }\frac{19}{15} &lt; \frac{17}{13} &lt; \frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> <br /> [[File:2019_AMC_8_-4_Image_1.png|900px]]<br /> <br /> [[File:2019_AMC_8_-4_Image_2.png|600px]]<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]]<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> Alice has &lt;math&gt;24&lt;/math&gt; apples. In how many ways can she share them with Becky and Chris so that each of the people has at least &lt;math&gt;2&lt;/math&gt; apples?<br /> <br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> {{MAA Notice}}<br /> <br /> Link back to original page: https://artofproblemsolving.com/wiki/index.php/2019_AMC_8</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_24&diff=111039 1998 AJHSME Problems/Problem 24 2019-11-09T23:46:19Z <p>Ljlbox: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?<br /> <br /> &lt;asy&gt;<br /> unitsize(20);<br /> for(int a = 0; a &lt; 10; ++a)<br /> {<br /> draw((0,a)--(8,a));<br /> }<br /> for (int b = 0; b &lt; 9; ++b)<br /> {<br /> draw((b,0)--(b,9));<br /> }<br /> draw((0,0)--(0,-.5));<br /> draw((1,0)--(1,-1.5));<br /> draw((.5,-1)--(1.5,-1));<br /> draw((2,0)--(2,-.5));<br /> draw((4,0)--(4,-.5));<br /> draw((5,0)--(5,-1.5));<br /> draw((4.5,-1)--(5.5,-1));<br /> draw((6,0)--(6,-.5));<br /> draw((8,0)--(8,-.5));<br /> fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black);<br /> fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black);<br /> fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black);<br /> fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black);<br /> fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black);<br /> label(&quot;$2$&quot;,(1.5,8.2),N);<br /> label(&quot;$4$&quot;,(3.5,8.2),N);<br /> label(&quot;$5$&quot;,(4.5,8.2),N);<br /> label(&quot;$7$&quot;,(6.5,8.2),N);<br /> label(&quot;$8$&quot;,(7.5,8.2),N);<br /> label(&quot;$9$&quot;,(0.5,7.2),N);<br /> label(&quot;$11$&quot;,(2.5,7.2),N);<br /> label(&quot;$12$&quot;,(3.5,7.2),N);<br /> label(&quot;$13$&quot;,(4.5,7.2),N);<br /> label(&quot;$14$&quot;,(5.5,7.2),N);<br /> label(&quot;$16$&quot;,(7.5,7.2),N);&lt;/asy&gt;<br /> &lt;math&gt; \text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> The numbers that are shaded are the triangular numbers, which are numbers in the form &lt;math&gt;\frac{(n)(n+1)}{2}&lt;/math&gt; for positive integers. They can also be generated by starting with &lt;math&gt;1&lt;/math&gt;, and adding &lt;math&gt;1, 2, 3, 4...&lt;/math&gt; as in the description of the problem.<br /> <br /> Squares that have the same remainder after being divided by &lt;math&gt;8&lt;/math&gt; will be in the same column. Thus, we want to find when the last remainder, from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt;, is found.<br /> <br /> So, instead of adding &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10...&lt;/math&gt;, we can effectively either add &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...&lt;/math&gt; or subtract &lt;math&gt;7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...&lt;/math&gt; if we are only concerned about remainders when divided by &lt;math&gt;8&lt;/math&gt;. We will pick the number that keeps the terms on the list between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. We get:<br /> <br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + 2 = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 3 = 6&lt;/math&gt;<br /> <br /> &lt;math&gt;6 - 4 = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 5 = 7&lt;/math&gt;<br /> <br /> &lt;math&gt;7 - 2 = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;5 - 1 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4 + 0 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4 + 1 = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;5 + 2 = 7&lt;/math&gt;<br /> <br /> &lt;math&gt;7 - 5 = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 4 = 6&lt;/math&gt;<br /> <br /> &lt;math&gt;6 - 3 = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;3 - 2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;1 - 1 = 0&lt;/math&gt;<br /> <br /> Finally, a term with &lt;math&gt;0&lt;/math&gt; is found, and checking, all numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;7&lt;/math&gt; are also on the right side of the list. This means the last term in our sequence is the first time that column &lt;math&gt;8&lt;/math&gt; is shaded. There are &lt;math&gt;15&lt;/math&gt; terms in the sequence, leading to an answer of &lt;math&gt;\frac{15\cdot 16}{2} = 120&lt;/math&gt;, which is choice &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Note that the triangular numbers up to &lt;math&gt;120&lt;/math&gt; are &lt;math&gt;1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120&lt;/math&gt;. When you divide each of those numbers by &lt;math&gt;8&lt;/math&gt;, all remainders must be present. We first search for number(s) that are evenly divisible by &lt;math&gt;8&lt;/math&gt;; if two such numbers exist, we search for numbers that leave a remainder of &lt;math&gt;1&lt;/math&gt;, etc.<br /> <br /> Quickly scanning the list, only &lt;math&gt;6, 10, 28, 36, 66, 78&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt; are even. That smaller list doesn't have any multiples of &lt;math&gt;8&lt;/math&gt; until it hits &lt;math&gt;120&lt;/math&gt;. So &lt;math&gt;\boxed{120, E}&lt;/math&gt; must be the answer.<br /> <br /> ===Solution 3===<br /> <br /> The numbers shaded are triangular numbers of the form &lt;math&gt;\frac{(n)(n+1)}{2}&lt;/math&gt;. For this number to be divisible by &lt;math&gt;8&lt;/math&gt;, the numerator must be divisible by &lt;math&gt;16&lt;/math&gt;. Since only one of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; can be even, only one of them can have factors of &lt;math&gt;2&lt;/math&gt;. Therefore, the first time the whole expression is divisible by &lt;math&gt;8&lt;/math&gt; is when either &lt;math&gt;n+1=16&lt;/math&gt; or when &lt;math&gt;n=16&lt;/math&gt;. This gives &lt;math&gt;n=15&lt;/math&gt; as the first time &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt; is divisible by &lt;math&gt;8&lt;/math&gt;, which gives &lt;math&gt;120&lt;/math&gt;. No other triangular number lower than that is divisible by &lt;math&gt;8&lt;/math&gt;, and thus the &lt;math&gt;8^{th}&lt;/math&gt; column on the checkerboard won't be filled until then. That gives &lt;math&gt;\boxed{E}&lt;/math&gt; as the right answer.<br /> <br /> <br /> == See also ==<br /> {{AJHSME box|year=1998|num-b=23|num-a=25}}<br /> * [[AJHSME]]<br /> * [[AJHSME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_24&diff=111038 1998 AJHSME Problems/Problem 24 2019-11-09T23:45:31Z <p>Ljlbox: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> A rectangular board of 8 columns has squared numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that first achieves this result?<br /> <br /> &lt;asy&gt;<br /> unitsize(20);<br /> for(int a = 0; a &lt; 10; ++a)<br /> {<br /> draw((0,a)--(8,a));<br /> }<br /> for (int b = 0; b &lt; 9; ++b)<br /> {<br /> draw((b,0)--(b,9));<br /> }<br /> draw((0,0)--(0,-.5));<br /> draw((1,0)--(1,-1.5));<br /> draw((.5,-1)--(1.5,-1));<br /> draw((2,0)--(2,-.5));<br /> draw((4,0)--(4,-.5));<br /> draw((5,0)--(5,-1.5));<br /> draw((4.5,-1)--(5.5,-1));<br /> draw((6,0)--(6,-.5));<br /> draw((8,0)--(8,-.5));<br /> fill((0,8)--(1,8)--(1,9)--(0,9)--cycle,black);<br /> fill((2,8)--(3,8)--(3,9)--(2,9)--cycle,black);<br /> fill((5,8)--(6,8)--(6,9)--(5,9)--cycle,black);<br /> fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black);<br /> fill((6,7)--(7,7)--(7,8)--(6,8)--cycle,black);<br /> label(&quot;$2$&quot;,(1.5,8.2),N);<br /> label(&quot;$4$&quot;,(3.5,8.2),N);<br /> label(&quot;$5$&quot;,(4.5,8.2),N);<br /> label(&quot;$7$&quot;,(6.5,8.2),N);<br /> label(&quot;$8$&quot;,(7.5,8.2),N);<br /> label(&quot;$9$&quot;,(0.5,7.2),N);<br /> label(&quot;$11$&quot;,(2.5,7.2),N);<br /> label(&quot;$12$&quot;,(3.5,7.2),N);<br /> label(&quot;$13$&quot;,(4.5,7.2),N);<br /> label(&quot;$14$&quot;,(5.5,7.2),N);<br /> label(&quot;$16$&quot;,(7.5,7.2),N);&lt;/asy&gt;<br /> &lt;math&gt; \text{(A)}\ 36\qquad\text{(B)}\ 64\qquad\text{(C)}\ 78\qquad\text{(D)}\ 91\qquad\text{(E)}\ 120 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> <br /> The numbers that are shaded are the triangular numbers, which are numbers in the form &lt;math&gt;\frac{(n)(n+1)}{2}&lt;/math&gt; for positive integers. They can also be generated by starting with &lt;math&gt;1&lt;/math&gt;, and adding &lt;math&gt;1, 2, 3, 4...&lt;/math&gt; as in the description of the problem.<br /> <br /> Squares that have the same remainder after being divided by &lt;math&gt;8&lt;/math&gt; will be in the same column. Thus, we want to find when the last remainder, from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt;, is found.<br /> <br /> So, instead of adding &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9, 10...&lt;/math&gt;, we can effectively either add &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3...&lt;/math&gt; or subtract &lt;math&gt;7, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5...&lt;/math&gt; if we are only concerned about remainders when divided by &lt;math&gt;8&lt;/math&gt;. We will pick the number that keeps the terms on the list between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. We get:<br /> <br /> &lt;math&gt;1&lt;/math&gt;<br /> <br /> &lt;math&gt;1 + 2 = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 3 = 6&lt;/math&gt;<br /> <br /> &lt;math&gt;6 - 4 = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 5 = 7&lt;/math&gt;<br /> <br /> &lt;math&gt;7 - 2 = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;5 - 1 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4 + 0 = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4 + 1 = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;5 + 2 = 7&lt;/math&gt;<br /> <br /> &lt;math&gt;7 - 5 = 2&lt;/math&gt;<br /> <br /> &lt;math&gt;2 + 4 = 6&lt;/math&gt;<br /> <br /> &lt;math&gt;6 - 3 = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;3 - 2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;1 - 1 = 0&lt;/math&gt;<br /> <br /> Finally, a term with &lt;math&gt;0&lt;/math&gt; is found, and checking, all numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;7&lt;/math&gt; are also on the right side of the list. This means the last term in our sequence is the first time that column &lt;math&gt;8&lt;/math&gt; is shaded. There are &lt;math&gt;15&lt;/math&gt; terms in the sequence, leading to an answer of &lt;math&gt;\frac{15\cdot 16}{2} = 120&lt;/math&gt;, which is choice &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> <br /> Note that the triangular numbers up to &lt;math&gt;120&lt;/math&gt; are &lt;math&gt;1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120&lt;/math&gt;. When you divide each of those numbers by &lt;math&gt;8&lt;/math&gt;, all remainders must be present. We first search for number(s) that are evenly divisible by &lt;math&gt;8&lt;/math&gt;; if two such numbers exist, we search for numbers that leave a remainder of &lt;math&gt;1&lt;/math&gt;, etc.<br /> <br /> Quickly scanning the list, only &lt;math&gt;6, 10, 28, 36, 66, 78&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt; are even. That smaller list doesn't have any multiples of &lt;math&gt;8&lt;/math&gt; until it hits &lt;math&gt;120&lt;/math&gt;. So &lt;math&gt;\boxed{120, E}&lt;/math&gt; must be the answer.<br /> <br /> ===Solution 3===<br /> <br /> The numbers shaded are triangular numbers of the form &lt;math&gt;\frac{(n)(n+1)}{2}&lt;/math&gt;. For this number to be divisible by &lt;math&gt;8&lt;/math&gt;, the numerator must be divisible by &lt;math&gt;16&lt;/math&gt;. Since only one of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; can be even, only one of them can have factors of &lt;math&gt;2&lt;/math&gt;. Therefore, the first time the whole expression is divisible by &lt;math&gt;8&lt;/math&gt; is when either &lt;math&gt;n+1=16&lt;/math&gt; or when &lt;math&gt;n=16&lt;/math&gt;. This gives &lt;math&gt;n=15&lt;/math&gt; as the first time &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt; is divisible by &lt;math&gt;8&lt;/math&gt;, which gives &lt;math&gt;120&lt;/math&gt;. No other triangular number lower than that is divisible by &lt;math&gt;8&lt;/math&gt;, and thus the &lt;math&gt;8^{th}&lt;/math&gt; column on the checkerboard won't be filled until then. That gives &lt;math&gt;\boxed{E}&lt;/math&gt; as the right answer.<br /> <br /> <br /> ===Solution 4===<br /> <br /> We can randomly choose an answer. There is a &lt;math&gt;\frac{1}{5}&lt;/math&gt; chance that we get the correct answer, which is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AJHSME box|year=1998|num-b=23|num-a=25}}<br /> * [[AJHSME]]<br /> * [[AJHSME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_19&diff=109462 2013 AMC 10B Problems/Problem 19 2019-08-28T19:14:00Z <p>Ljlbox: /* Problem */</p> <hr /> <div>==Problem==<br /> The real numbers &lt;math&gt;c,b,a&lt;/math&gt; form an arithmetic sequence with &lt;math&gt;a\ge b\ge c&gt; 0&lt;/math&gt;. The quadratic &lt;math&gt;ax^2+bx+c&lt;/math&gt; has exactly one root. What is this root?<br /> <br /> &lt;math&gt; \textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> It is given that &lt;math&gt;ax^2+bx+c=0&lt;/math&gt; has 1 real root, so the discriminant is zero, or &lt;math&gt;b^2=4ac&lt;/math&gt;. Because a, b, c are in arithmetic progression, &lt;math&gt;b-a=c-b&lt;/math&gt;, or &lt;math&gt;b=\frac {a+c} {2} &lt;/math&gt;. We need to find the unique root, or &lt;math&gt;-\frac {b} {2a} &lt;/math&gt; (discriminant is 0). From &lt;math&gt;b^2=4ac&lt;/math&gt;, we can get &lt;math&gt;-\frac {b} {2a} =-\frac {2c} {b}&lt;/math&gt;. Ignoring the negatives(for now), we have <br /> &lt;math&gt;\frac {2c} {b} = \frac {2c} {\frac {a+c} {2}} = \frac {4c} {a+c} = \frac {1} {\frac {1} {\frac {4c} {a+c}}} = \frac {1} {\frac {a+c} {4c}} = \frac {1} {\frac {a} {4c} + \frac {1} {4} }&lt;/math&gt;. Fortunately, finding &lt;math&gt;\frac {a} {c} &lt;/math&gt; is not very hard. Plug in &lt;math&gt;b=\frac {a+c} {2}&lt;/math&gt; to &lt;math&gt;b^2=4ac&lt;/math&gt;, we have &lt;math&gt;a^2+2ac+c^2=16ac&lt;/math&gt;, or &lt;math&gt;a^2-14ac+c^2=0&lt;/math&gt;, and dividing by &lt;math&gt;c^2&lt;/math&gt; gives &lt;math&gt;(\frac {a} {c} ) ^2-14( \frac {a} {c} ) +1 = 0&lt;/math&gt;, so &lt;math&gt;\frac {a} {c} = \frac {14 \pm \sqrt {192} } {2} = 7 \pm 4 \sqrt {3} &lt;/math&gt;. But &lt;math&gt;7-4\sqrt {3} &lt;1&lt;/math&gt;, violating the assumption that &lt;math&gt;a \ge c&lt;/math&gt;. Therefore, &lt;math&gt;\frac {a} {c} = 7 +4\sqrt {3} &lt;/math&gt;. Plugging this in, we have &lt;math&gt;\frac {1} {\frac {a} {4c} + \frac {1} {4} } = \frac {1} {2+ \sqrt {3} } = 2- \sqrt {3} &lt;/math&gt;. But we need the negative of this, so the answer is &lt;math&gt;\boxed {\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Note that we can divide the polynomial by &lt;math&gt;a&lt;/math&gt; to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation must be of the form <br /> &lt;math&gt;(x-r)^2 = x^2 - 2rx + r^2&lt;/math&gt; where &lt;math&gt;1 \ge -2r \ge r^2 \ge 0&lt;/math&gt;.<br /> We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus,<br /> &lt;math&gt;r^2 + 1 = -4r&lt;/math&gt; and &lt;math&gt;r = -2 \pm \sqrt{3}&lt;/math&gt;. Since &lt;math&gt;1 &gt; r^2&lt;/math&gt;, we easily see that &lt;math&gt;|r|&lt;/math&gt; has to be between 1 and 0. Thus, we can eliminate &lt;math&gt;-2 - \sqrt{3}&lt;/math&gt; and are left with &lt;math&gt;\boxed{\textbf{(D)} -2 + \sqrt{3}}&lt;/math&gt; as the answer.<br /> <br /> ===Solution 3===<br /> Given that &lt;math&gt;ax^2+bx+c=0&lt;/math&gt; has only 1 real root, we know that the discriminant must equal 0, or that &lt;math&gt;b^2=4ac&lt;/math&gt;. Because the discriminant equals 0, we have that the root of the quadratic is &lt;math&gt;r=\frac {-b} {2a}&lt;/math&gt;. We are also given that the coefficients of the quadratic are in arithmetic progression, where &lt;math&gt;a \ge b \ge c \ge 0&lt;/math&gt;. Letting the arbitrary difference equal variable &lt;math&gt;d&lt;/math&gt;, we have that &lt;math&gt;a=b+d&lt;/math&gt; and that &lt;math&gt;c=b-d&lt;/math&gt;. Plugging those two equations into &lt;math&gt;b^2=4ac&lt;/math&gt;, we have &lt;math&gt;b^2=4(b^2-d^2)=4b^2-4d^2&lt;/math&gt; which yields &lt;math&gt;3b^2=4d^2&lt;/math&gt;. Isolating &lt;math&gt;d&lt;/math&gt;, we have &lt;math&gt;d=\frac {b \sqrt{3}} {2}&lt;/math&gt;. Substituting that in for &lt;math&gt;d&lt;/math&gt; in &lt;math&gt;a=b+d&lt;/math&gt;, we get &lt;math&gt;a=b+\frac {b \sqrt{3}} {2}=b(1+\frac {\sqrt{3}} {2})&lt;/math&gt;. Once again, substituting that in for &lt;math&gt;a&lt;/math&gt; in &lt;math&gt;r=\frac {-b} {2a}&lt;/math&gt;, we have &lt;math&gt;r=\frac {-b} {2b(1+\frac {\sqrt{3}} {2})}=\frac {-1} {2+\sqrt {3}}=-2+\sqrt {3}&lt;/math&gt;. The answer is: &lt;math&gt;\boxed {\textbf{(D)}}.&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> We have &lt;math&gt;a-b=b-c&lt;/math&gt;<br /> Hence &lt;math&gt;2b=a+c&lt;/math&gt;.<br /> <br /> And we have &lt;math&gt;b^2-4ac=0\implies b^2=4ac&lt;/math&gt;.<br /> Squaring the first expression, and dividing by 4, we get &lt;math&gt;b^2=\frac{a^2+2ac+c^2}{4}&lt;/math&gt;.<br /> <br /> Setting the two equations equal, we have &lt;math&gt;4ac=\frac{a^2+2ac+c^2}{4}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\implies 16ac=a^2+2ac+c^2\implies 14ac=a^2+c^2\implies a^2-14ac+c^2&lt;/math&gt;.<br /> <br /> Dividing by &lt;math&gt;c^2&lt;/math&gt;, we get &lt;math&gt;(\frac{a}{c})^2-14(\frac{a}{c})+1=0&lt;/math&gt;. <br /> <br /> Setting &lt;math&gt;\frac{a}{c}=x,&lt;/math&gt; we have &lt;math&gt;x^2-14x+1=0.&lt;/math&gt; <br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x=\frac{a}{c}=7 \pm 4\sqrt3&lt;/math&gt; We know that &lt;math&gt;a \ge c&lt;/math&gt;, so &lt;math&gt;\frac{a}{c}=7+4\sqrt3&lt;/math&gt;.<br /> <br /> By Vieta's, we know that &lt;math&gt;\frac{c}{a}=r^2&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; equals the double root of the quadratic.<br /> So, we get &lt;math&gt;\frac{c}{a}=\frac{1}{7 + 4\sqrt3}&lt;/math&gt;. After rationalizing the denominator, we get &lt;math&gt;\frac{c}{a} = 7- 4\sqrt 3&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;r^2=7-4\sqrt3&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we have &lt;math&gt;r&lt;/math&gt; equals &lt;math&gt;-2+\sqrt3=\boxed{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=109379 2011 AMC 10A Problems/Problem 13 2019-08-26T02:26:35Z <p>Ljlbox: /* Problem 13 */</p> <hr /> <div>==Problem 13==<br /> How many even integers are there between &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;700&lt;/math&gt; whose digits are all different and come from the set &lt;math&gt;\left\{1,2,5,7,8,9\right\}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We split up into cases of the hundreds digits being &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. If the hundred digits is &lt;math&gt;2&lt;/math&gt;, then the units digits must be &lt;math&gt;8&lt;/math&gt; in order for the number to be even and then there are &lt;math&gt;4&lt;/math&gt; remaining choices (&lt;math&gt;1,5,7,9&lt;/math&gt;) for the tens digit, giving &lt;math&gt;1 \times 4 \times 1=4&lt;/math&gt; possibilities. Similarly, there are &lt;math&gt;1 \times 4 \times 2=8&lt;/math&gt; possibilities for the &lt;math&gt;5&lt;/math&gt; case, giving a total of &lt;math&gt;\boxed{4+8=12 \ \mathbf{(A)}}&lt;/math&gt; possibilities.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=109378 2011 AMC 10A Problems/Problem 13 2019-08-26T02:26:08Z <p>Ljlbox: /* Problem 13 */</p> <hr /> <div>==Problem 13==<br /> How many even integers are there between &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;700&lt;/math&gt; whose digits are all different and come from the set &lt;math&gt;\left\{1,2,5,7,8,9}\right\$?<br /> <br /> &lt;/math&gt;\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$<br /> <br /> == Solution ==<br /> <br /> We split up into cases of the hundreds digits being &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. If the hundred digits is &lt;math&gt;2&lt;/math&gt;, then the units digits must be &lt;math&gt;8&lt;/math&gt; in order for the number to be even and then there are &lt;math&gt;4&lt;/math&gt; remaining choices (&lt;math&gt;1,5,7,9&lt;/math&gt;) for the tens digit, giving &lt;math&gt;1 \times 4 \times 1=4&lt;/math&gt; possibilities. Similarly, there are &lt;math&gt;1 \times 4 \times 2=8&lt;/math&gt; possibilities for the &lt;math&gt;5&lt;/math&gt; case, giving a total of &lt;math&gt;\boxed{4+8=12 \ \mathbf{(A)}}&lt;/math&gt; possibilities.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_13&diff=109377 2011 AMC 10A Problems/Problem 13 2019-08-26T02:25:14Z <p>Ljlbox: /* Problem 13 */</p> <hr /> <div>==Problem 13==<br /> How many even integers are there between &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;700&lt;/math&gt; whose digits are all different and come from the set &lt;math&gt;{1,2,5,7,8,9}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We split up into cases of the hundreds digits being &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. If the hundred digits is &lt;math&gt;2&lt;/math&gt;, then the units digits must be &lt;math&gt;8&lt;/math&gt; in order for the number to be even and then there are &lt;math&gt;4&lt;/math&gt; remaining choices (&lt;math&gt;1,5,7,9&lt;/math&gt;) for the tens digit, giving &lt;math&gt;1 \times 4 \times 1=4&lt;/math&gt; possibilities. Similarly, there are &lt;math&gt;1 \times 4 \times 2=8&lt;/math&gt; possibilities for the &lt;math&gt;5&lt;/math&gt; case, giving a total of &lt;math&gt;\boxed{4+8=12 \ \mathbf{(A)}}&lt;/math&gt; possibilities.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_16&diff=109215 2009 AMC 10A Problems/Problem 16 2019-08-22T19:46:35Z <p>Ljlbox: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; be real numbers with &lt;math&gt;|a-b|=2&lt;/math&gt;, &lt;math&gt;|b-c|=3&lt;/math&gt;, and &lt;math&gt;|c-d|=4&lt;/math&gt;. What is the sum of all possible values of &lt;math&gt;|a-d|&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 12<br /> \qquad<br /> \mathrm{(C)}\ 15<br /> \qquad<br /> \mathrm{(D)}\ 18<br /> \qquad<br /> \mathrm{(E)}\ 24<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> From &lt;math&gt;|a-b|=2&lt;/math&gt; we get that &lt;math&gt;a=b\pm 2&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;b=c\pm3&lt;/math&gt; and &lt;math&gt;c=d\pm4&lt;/math&gt;.<br /> <br /> Substitution gives &lt;math&gt;a=d\pm 4\pm 3\pm 2&lt;/math&gt;. This gives &lt;math&gt;|a-d|=|\pm 4\pm 3\pm 2|&lt;/math&gt;. There are &lt;math&gt;2^3=8&lt;/math&gt; possibilities for the value of &lt;math&gt;\pm 4\pm 3\pm2&lt;/math&gt;:<br /> <br /> &lt;math&gt;4+3+2=\boxed{9}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4+3-2=\boxed{5}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4-3+2=\boxed{3}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4+3+2=\boxed{1}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4-3-2=\boxed{-1}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4+3-2=\boxed{-3}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4-3+2=\boxed{-5}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4-3-2=\boxed{-9}&lt;/math&gt;<br /> <br /> Therefore, the only possible values of &lt;math&gt;|a-d|&lt;/math&gt; are 9, 5, 3, and 1. Their sum is &lt;math&gt;\boxed{18}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> If we add the same constant to all of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;, we will not change any of the differences. Hence we can assume that &lt;math&gt;a=0&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|a-b|=2&lt;/math&gt; we get that &lt;math&gt;|b|=2&lt;/math&gt;, hence &lt;math&gt;b\in\{-2,2\}&lt;/math&gt;.<br /> <br /> If we multiply all four numbers by &lt;math&gt;-1&lt;/math&gt;, we will not change any of the differences. Hence we can [[WLOG]] assume that &lt;math&gt;b=2&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|b-c|=3&lt;/math&gt; we get that &lt;math&gt;c\in\{-1,5\}&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|c-d|=4&lt;/math&gt; we get that &lt;math&gt;d\in\{-5,1,3,9\}&lt;/math&gt;.<br /> <br /> Hence &lt;math&gt;|a-d|=|d|\in\{1,3,5,9\}&lt;/math&gt;, and the sum of possible values is &lt;math&gt;1+3+5+9 = \boxed{18}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_16&diff=109214 2009 AMC 10A Problems/Problem 16 2019-08-22T19:46:24Z <p>Ljlbox: /* == Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; be real numbers with &lt;math&gt;|a-b|=2&lt;/math&gt;, &lt;math&gt;|b-c|=3&lt;/math&gt;, and &lt;math&gt;|c-d|=4&lt;/math&gt;. What is the sum of all possible values of &lt;math&gt;|a-d|&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 12<br /> \qquad<br /> \mathrm{(C)}\ 15<br /> \qquad<br /> \mathrm{(D)}\ 18<br /> \qquad<br /> \mathrm{(E)}\ 24<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> From &lt;math&gt;|a-b|=2&lt;/math&gt; we get that &lt;math&gt;a=b\pm 2&lt;/math&gt;<br /> <br /> Similarly, &lt;math&gt;b=c\pm3&lt;/math&gt; and &lt;math&gt;c=d\pm4&lt;/math&gt;.<br /> <br /> Substitution gives &lt;math&gt;a=d\pm 4\pm 3\pm 2&lt;/math&gt;. This gives &lt;math&gt;|a-d|=|\pm 4\pm 3\pm 2|&lt;/math&gt;. There are &lt;math&gt;2^3=8&lt;/math&gt; possibilities for the value of &lt;math&gt;\pm 4\pm 3\pm2&lt;/math&gt;:<br /> <br /> &lt;math&gt;4+3+2=\boxed{9}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4+3-2=\boxed{5}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4-3+2=\boxed{3}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4+3+2=\boxed{1}&lt;/math&gt;, <br /> <br /> &lt;math&gt;4-3-2=\boxed{-1}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4+3-2=\boxed{-3}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4-3+2=\boxed{-5}&lt;/math&gt;, <br /> <br /> &lt;math&gt;-4-3-2=\boxed{-9}&lt;/math&gt;<br /> <br /> Therefore, the only possible values of &lt;math&gt;|a-d|&lt;/math&gt; are 9, 5, 3, and 1. Their sum is &lt;math&gt;\boxed{18}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> If we add the same constant to all of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;, we will not change any of the differences. Hence we can assume that &lt;math&gt;a=0&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|a-b|=2&lt;/math&gt; we get that &lt;math&gt;|b|=2&lt;/math&gt;, hence &lt;math&gt;b\in\{-2,2\}&lt;/math&gt;.<br /> <br /> If we multiply all four numbers by &lt;math&gt;-1&lt;/math&gt;, we will not change any of the differences. Hence we can [[WLOG]] assume that &lt;math&gt;b=2&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|b-c|=3&lt;/math&gt; we get that &lt;math&gt;c\in\{-1,5\}&lt;/math&gt;.<br /> <br /> From &lt;math&gt;|c-d|=4&lt;/math&gt; we get that &lt;math&gt;d\in\{-5,1,3,9\}&lt;/math&gt;.<br /> <br /> Hence &lt;math&gt;|a-d|=|d|\in\{1,3,5,9\}&lt;/math&gt;, and the sum of possible values is &lt;math&gt;1+3+5+9 = \boxed{18}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2009|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_20&diff=109163 2011 AMC 10A Problems/Problem 20 2019-08-21T20:12:23Z <p>Ljlbox: /* Problem 20 */</p> <hr /> <div>== Problem 20 ==<br /> Two points on the circumference of a circle of radius &lt;math&gt;r&lt;/math&gt; are selected independently and at random. From each point a chord of length &lt;math&gt;r&lt;/math&gt; is drawn in a clockwise direction. What is the probability that the two chords intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution==<br /> <br /> Fix a point &lt;math&gt;A&lt;/math&gt; from which we draw a clockwise chord. In order for the clockwise chord from another point &lt;math&gt;B&lt;/math&gt; to intersect that of point &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; must be no more than &lt;math&gt;r&lt;/math&gt; units apart. By drawing the circle, we quickly see that &lt;math&gt;B&lt;/math&gt; can be on &lt;math&gt;\frac{120}{360}=\boxed{\frac{1}{3} \ \textbf{(D)}}&lt;/math&gt; of the perimeter of the circle. (Imagine a regular hexagon inscribed in the circle)<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=User:Ljlbox&diff=109063 User:Ljlbox 2019-08-20T18:44:04Z <p>Ljlbox: /* practice latex here */</p> <hr /> <div>==practice latex here==<br /> <br /> &lt;math&gt;\frac{1}{2}&lt;/math&gt;<br /> <br /> lol \/\/ho\ t 1s this cl<br /> <br /> &lt;math&gt;encryption key code 993028&lt;/math&gt;<br /> <br /> ==lol what is this==<br /> <br /> brain.exe has failed to respond!<br /> <br /> Running Windows Diagnostics...<br /> <br /> Problem Solved! Solution: git gud</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=User:Ljlbox&diff=109062 User:Ljlbox 2019-08-20T18:43:35Z <p>Ljlbox: /* practice latex here */</p> <hr /> <div>==practice latex here==<br /> <br /> &lt;math&gt;\frac{1}{2}&lt;/math&gt;<br /> <br /> lol \/\/ho\ t 1s this cl<br /> <br /> &lt;math&gt;encryption key code 993028&lt;/math&gt;<br /> <br /> brain.exe has failed to respond!<br /> <br /> Running Windows Diagnostics...<br /> <br /> Problem Solved! Solution: git gud</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=User:Ljlbox&diff=109061 User:Ljlbox 2019-08-20T18:40:32Z <p>Ljlbox: /* practice latex here */</p> <hr /> <div>==practice latex here==<br /> <br /> &lt;math&gt;\frac{1}{2}&lt;/math&gt;<br /> <br /> lol \/\/ho\ t 1s this cl<br /> <br /> &lt;math&gt;encryption key code 993028&lt;/math&gt;</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_7&diff=109060 2001 AMC 10 Problems/Problem 7 2019-08-20T18:24:34Z <p>Ljlbox: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?<br /> <br /> &lt;math&gt; \textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> If &lt;math&gt;x&lt;/math&gt; is the number, then moving the decimal point four places to the right is the same as multiplying &lt;math&gt;x&lt;/math&gt; by &lt;math&gt;10000&lt;/math&gt;. This gives us the equation <br /> &lt;cmath&gt;10000x=4\cdot\frac{1}{x}&lt;/cmath&gt;<br /> This is equivalent to &lt;cmath&gt;x^2=\frac{4}{10000}&lt;/cmath&gt; <br /> Since &lt;math&gt;x&lt;/math&gt; is positive, it follows that &lt;math&gt;x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Alternatively, we could try each solution and see if it fits the problems given statements.<br /> <br /> After testing, we find that &lt;math&gt;\boxed{\textbf{(C)}\ 0.02}&lt;/math&gt; is the correct answer.<br /> <br /> ~ljlbox<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_6&diff=109044 2001 AMC 10 Problems/Problem 6 2019-08-20T13:19:42Z <p>Ljlbox: /* See Also */</p> <hr /> <div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}}<br /> <br /> == Problem ==<br /> Let &lt;math&gt;P(n)&lt;/math&gt; and &lt;math&gt;S(n)&lt;/math&gt; denote the product and the sum, respectively, of the digits<br /> of the integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;P(23) = 6&lt;/math&gt; and &lt;math&gt;S(23) = 5&lt;/math&gt;. Suppose &lt;math&gt;N&lt;/math&gt; is a<br /> two-digit number such that &lt;math&gt;N = P(N)+S(N)&lt;/math&gt;. What is the units digit of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9&lt;/math&gt;<br /> <br /> == Solution ==<br /> Denote &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; as the tens and units digit of &lt;math&gt;N&lt;/math&gt;, respectively. Then &lt;math&gt;N = 10a+b&lt;/math&gt;. It follows that &lt;math&gt;10a+b=ab+a+b&lt;/math&gt;, which implies that &lt;math&gt;9a=ab&lt;/math&gt;. Since &lt;math&gt;a\neq0&lt;/math&gt;, &lt;math&gt;b=9&lt;/math&gt;. So the units digit of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;\boxed{(\text{E})\ 9}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2001|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2001|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=User:Ljlbox&diff=108958 User:Ljlbox 2019-08-18T19:24:05Z <p>Ljlbox: /* practice latex here */</p> <hr /> <div>==practice latex here==<br /> <br /> &lt;math&gt;\frac{1}{2}&lt;/math&gt;<br /> <br /> lol \/\/ho\ t 1s this cl</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=User:Ljlbox&diff=108949 User:Ljlbox 2019-08-17T20:30:14Z <p>Ljlbox: Created page with &quot;==practice latex here== &lt;math&gt;\frac{1}{2}&lt;/math&gt;&quot;</p> <hr /> <div>==practice latex here==<br /> <br /> &lt;math&gt;\frac{1}{2}&lt;/math&gt;</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=108880 2004 AMC 10B Problems/Problem 20 2019-08-14T17:46:35Z <p>Ljlbox: /* Solution (Coordinates) */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. If &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt; so that &lt;math&gt;AT/DT=3&lt;/math&gt; and &lt;math&gt;BT/ET=4&lt;/math&gt;, what is &lt;math&gt;CD/BD&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} &lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,2*WNW);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 (Triangle Areas) ==<br /> <br /> We use the square bracket notation &lt;math&gt;[\cdot]&lt;/math&gt; to denote area.<br /> <br /> WLOG, we can assume &lt;math&gt;[\triangle BTD] = 1&lt;/math&gt;. Then &lt;math&gt;[\triangle BTA] = 3&lt;/math&gt;, and &lt;math&gt;[\triangle ATE] = 3/4&lt;/math&gt;. We have &lt;math&gt;CD/BD = [\triangle ACD]/[\triangle ABD]&lt;/math&gt;, so we need to find the area of quadrilateral &lt;math&gt;TDCE&lt;/math&gt;.<br /> <br /> Draw the line segment &lt;math&gt;TC&lt;/math&gt; to form the two triangles &lt;math&gt;\triangle TDC&lt;/math&gt; and &lt;math&gt;\triangle TEC&lt;/math&gt;. Let &lt;math&gt;x = [\triangle TDC]&lt;/math&gt;, and &lt;math&gt;y = [\triangle TEC]&lt;/math&gt;. By considering triangles &lt;math&gt;\triangle BTC&lt;/math&gt; and &lt;math&gt;\triangle ETC&lt;/math&gt;, we obtain &lt;math&gt;(1+x)/y=4&lt;/math&gt;, and by considering triangles &lt;math&gt;\triangle ATC&lt;/math&gt; and &lt;math&gt;\triangle DTC&lt;/math&gt;, we obtain &lt;math&gt;(3/4+y)/x=3&lt;/math&gt;. Solving, we get &lt;math&gt;x=4/11&lt;/math&gt;, &lt;math&gt;y=15/44&lt;/math&gt;, so the area of quadrilateral &lt;math&gt;TDEC&lt;/math&gt; is &lt;math&gt;x+y=31/44&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}&lt;/math&gt;<br /> <br /> == Solution 2 (Mass points) ==<br /> <br /> The presence of only ratios in the problem essentially cries out for mass points.<br /> <br /> As per the problem, we assign a mass of &lt;math&gt;1&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;, and a mass of &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. Then, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Now, were we to assign a mass of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;, we'd have &lt;math&gt;5T&lt;/math&gt;. Scaling this down by &lt;math&gt;4/5&lt;/math&gt; (to get &lt;math&gt;4T&lt;/math&gt;, which puts &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; in terms of the masses of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;), we assign a mass of &lt;math&gt;\frac{4}{5}&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;\frac{16}{5}&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Now, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;E&lt;/math&gt;, we must give &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;\frac{16}{5}-1=\frac{11}{5}&lt;/math&gt;.<br /> <br /> Finally, the ratio of &lt;math&gt;CD&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; is given by the ratio of the mass of &lt;math&gt;B&lt;/math&gt; to the mass of &lt;math&gt;C&lt;/math&gt;, which is &lt;math&gt;\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}&lt;/math&gt;.<br /> <br /> == Solution 3 (Coordinates) ==<br /> <br /> Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any &lt;math&gt;\triangle ABC&lt;/math&gt;, and we just need to compute it for any single triangle.<br /> <br /> We can choose the points &lt;math&gt;A=(-3,0)&lt;/math&gt;, &lt;math&gt;B=(0,4)&lt;/math&gt;, and &lt;math&gt;D=(1,0)&lt;/math&gt;. This way we will have &lt;math&gt;T=(0,0)&lt;/math&gt;, and &lt;math&gt;E=(0,-1)&lt;/math&gt;. The situation is <br /> shown in the picture below:<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,NW);<br /> label(&quot;$3$&quot;,A--T,N);<br /> label(&quot;$4$&quot;,B--T,W);<br /> label(&quot;$1$&quot;,D--T,N);<br /> label(&quot;$1$&quot;,E--T,W);<br /> <br /> &lt;/asy&gt;<br /> <br /> The point &lt;math&gt;C&lt;/math&gt; is the intersection of the lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. The points on the first line have the form &lt;math&gt;(t,4-4t)&lt;/math&gt;, the points on the second line have the form &lt;math&gt;(t,-1-t/3)&lt;/math&gt;. Solving for &lt;math&gt;t&lt;/math&gt; we get &lt;math&gt;t=15/11&lt;/math&gt;, hence &lt;math&gt;C=(15/11,-16/11)&lt;/math&gt;.<br /> <br /> The ratio &lt;math&gt;CD/BD&lt;/math&gt; can now be computed simply by observing the &lt;math&gt;x&lt;/math&gt; coordinates of &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}<br /> &lt;/cmath&gt;<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_20&diff=108879 2004 AMC 10B Problems/Problem 20 2019-08-14T17:46:24Z <p>Ljlbox: /* Solution (Mass points) */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;, respectively. If &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BE&lt;/math&gt; intersect at &lt;math&gt;T&lt;/math&gt; so that &lt;math&gt;AT/DT=3&lt;/math&gt; and &lt;math&gt;BT/ET=4&lt;/math&gt;, what is &lt;math&gt;CD/BD&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} &lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,2*WNW);<br /> &lt;/asy&gt;<br /> <br /> == Solution 1 (Triangle Areas) ==<br /> <br /> We use the square bracket notation &lt;math&gt;[\cdot]&lt;/math&gt; to denote area.<br /> <br /> WLOG, we can assume &lt;math&gt;[\triangle BTD] = 1&lt;/math&gt;. Then &lt;math&gt;[\triangle BTA] = 3&lt;/math&gt;, and &lt;math&gt;[\triangle ATE] = 3/4&lt;/math&gt;. We have &lt;math&gt;CD/BD = [\triangle ACD]/[\triangle ABD]&lt;/math&gt;, so we need to find the area of quadrilateral &lt;math&gt;TDCE&lt;/math&gt;.<br /> <br /> Draw the line segment &lt;math&gt;TC&lt;/math&gt; to form the two triangles &lt;math&gt;\triangle TDC&lt;/math&gt; and &lt;math&gt;\triangle TEC&lt;/math&gt;. Let &lt;math&gt;x = [\triangle TDC]&lt;/math&gt;, and &lt;math&gt;y = [\triangle TEC]&lt;/math&gt;. By considering triangles &lt;math&gt;\triangle BTC&lt;/math&gt; and &lt;math&gt;\triangle ETC&lt;/math&gt;, we obtain &lt;math&gt;(1+x)/y=4&lt;/math&gt;, and by considering triangles &lt;math&gt;\triangle ATC&lt;/math&gt; and &lt;math&gt;\triangle DTC&lt;/math&gt;, we obtain &lt;math&gt;(3/4+y)/x=3&lt;/math&gt;. Solving, we get &lt;math&gt;x=4/11&lt;/math&gt;, &lt;math&gt;y=15/44&lt;/math&gt;, so the area of quadrilateral &lt;math&gt;TDEC&lt;/math&gt; is &lt;math&gt;x+y=31/44&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}&lt;/math&gt;<br /> <br /> == Solution 2 (Mass points) ==<br /> <br /> The presence of only ratios in the problem essentially cries out for mass points.<br /> <br /> As per the problem, we assign a mass of &lt;math&gt;1&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;, and a mass of &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. Then, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;T&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; has a mass of &lt;math&gt;4&lt;/math&gt;.<br /> <br /> Now, were we to assign a mass of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;, we'd have &lt;math&gt;5T&lt;/math&gt;. Scaling this down by &lt;math&gt;4/5&lt;/math&gt; (to get &lt;math&gt;4T&lt;/math&gt;, which puts &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; in terms of the masses of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;), we assign a mass of &lt;math&gt;\frac{4}{5}&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; and a mass of &lt;math&gt;\frac{16}{5}&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;.<br /> <br /> Now, to balance &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;E&lt;/math&gt;, we must give &lt;math&gt;C&lt;/math&gt; a mass of &lt;math&gt;\frac{16}{5}-1=\frac{11}{5}&lt;/math&gt;.<br /> <br /> Finally, the ratio of &lt;math&gt;CD&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt; is given by the ratio of the mass of &lt;math&gt;B&lt;/math&gt; to the mass of &lt;math&gt;C&lt;/math&gt;, which is &lt;math&gt;\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}&lt;/math&gt;.<br /> <br /> == Solution (Coordinates) ==<br /> <br /> Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any &lt;math&gt;\triangle ABC&lt;/math&gt;, and we just need to compute it for any single triangle.<br /> <br /> We can choose the points &lt;math&gt;A=(-3,0)&lt;/math&gt;, &lt;math&gt;B=(0,4)&lt;/math&gt;, and &lt;math&gt;D=(1,0)&lt;/math&gt;. This way we will have &lt;math&gt;T=(0,0)&lt;/math&gt;, and &lt;math&gt;E=(0,-1)&lt;/math&gt;. The situation is <br /> shown in the picture below:<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);<br /> draw(A--B--C--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> pair T=intersectionpoint(A--D,B--E);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,N);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$T$&quot;,T,NW);<br /> label(&quot;$3$&quot;,A--T,N);<br /> label(&quot;$4$&quot;,B--T,W);<br /> label(&quot;$1$&quot;,D--T,N);<br /> label(&quot;$1$&quot;,E--T,W);<br /> <br /> &lt;/asy&gt;<br /> <br /> The point &lt;math&gt;C&lt;/math&gt; is the intersection of the lines &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt;. The points on the first line have the form &lt;math&gt;(t,4-4t)&lt;/math&gt;, the points on the second line have the form &lt;math&gt;(t,-1-t/3)&lt;/math&gt;. Solving for &lt;math&gt;t&lt;/math&gt; we get &lt;math&gt;t=15/11&lt;/math&gt;, hence &lt;math&gt;C=(15/11,-16/11)&lt;/math&gt;.<br /> <br /> The ratio &lt;math&gt;CD/BD&lt;/math&gt; can now be computed simply by observing the &lt;math&gt;x&lt;/math&gt; coordinates of &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;<br /> \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}<br /> &lt;/cmath&gt;<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_15&diff=108836 2005 AMC 10A Problems/Problem 15 2019-08-13T18:52:00Z <p>Ljlbox: /* See==See also */</p> <hr /> <div>==Problem==<br /> How many positive cubes divide &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; ?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> &lt;math&gt; 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}&lt;/math&gt;<br /> <br /> Therefore, a [[perfect cube]] that divides &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; must be in the form &lt;math&gt;2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are [[nonnegative]] [[multiple]]s of &lt;math&gt;3&lt;/math&gt; that are less than or equal to &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;, respectively. <br /> <br /> So: <br /> <br /> &lt;math&gt;a\in\{0,3,6\}&lt;/math&gt; (&lt;math&gt;3&lt;/math&gt; possibilities)<br /> <br /> &lt;math&gt;b\in\{0,3\}&lt;/math&gt; (&lt;math&gt;2&lt;/math&gt; possibilities)<br /> <br /> &lt;math&gt;c\in\{0\}&lt;/math&gt; (&lt;math&gt;1&lt;/math&gt; possibility)<br /> <br /> &lt;math&gt;d\in\{0\}&lt;/math&gt;(&lt;math&gt;1&lt;/math&gt; possibility)<br /> <br /> <br /> So the number of perfect cubes that divide &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; is &lt;math&gt;3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;math&gt; 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)&lt;/math&gt;<br /> <br /> In the expression, we notice that there are 3 &lt;math&gt;3's&lt;/math&gt;, 3 &lt;math&gt;2's&lt;/math&gt;, and 3 &lt;math&gt;1's&lt;/math&gt;. This gives us our first 3 cubes: &lt;math&gt;3^3&lt;/math&gt;, &lt;math&gt;2^3&lt;/math&gt;, and &lt;math&gt;1^3&lt;/math&gt;. <br /> <br /> However, we can also multiply smaller numbers in the expression to make bigger expressions. For example, &lt;math&gt;(2*2)*4*4=4*4*4=4^3&lt;/math&gt; (one 2 comes from the &lt;math&gt;3!&lt;/math&gt;, and the other from the &lt;math&gt;5!&lt;/math&gt;). Using this method, we find:<br /> <br /> &lt;math&gt;(3*2)*(3*2)*6=6^3&lt;/math&gt;<br /> <br /> and<br /> <br /> &lt;math&gt;(3*4)*(3*4)*(2*6)=12^3&lt;/math&gt;<br /> <br /> So, we have 6 cubes total:&lt;math&gt;1^3 ,2^3, 3^3, 4^3, 6^3,&lt;/math&gt; and &lt;math&gt;12^3&lt;/math&gt; for a total of &lt;math&gt;6&lt;/math&gt; cubes &lt;math&gt;\Rightarrow \mathrm{(E)}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Combinatorics Problems]] <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=108835 2005 AMC 10A Problems/Problem 14 2019-08-13T18:49:42Z <p>Ljlbox: /* See also */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster, slicker and more awesome)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=108834 2005 AMC 10A Problems/Problem 14 2019-08-13T18:47:02Z <p>Ljlbox: /* See also */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster, slicker and more awesome)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=1}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_15&diff=108832 2005 AMC 10A Problems/Problem 15 2019-08-13T18:43:18Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> How many positive cubes divide &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; ?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> &lt;math&gt; 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}&lt;/math&gt;<br /> <br /> Therefore, a [[perfect cube]] that divides &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; must be in the form &lt;math&gt;2^{a}\cdot3^{b}\cdot5^{c}\cdot7^{d}&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are [[nonnegative]] [[multiple]]s of &lt;math&gt;3&lt;/math&gt; that are less than or equal to &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;, respectively. <br /> <br /> So: <br /> <br /> &lt;math&gt;a\in\{0,3,6\}&lt;/math&gt; (&lt;math&gt;3&lt;/math&gt; possibilities)<br /> <br /> &lt;math&gt;b\in\{0,3\}&lt;/math&gt; (&lt;math&gt;2&lt;/math&gt; possibilities)<br /> <br /> &lt;math&gt;c\in\{0\}&lt;/math&gt; (&lt;math&gt;1&lt;/math&gt; possibility)<br /> <br /> &lt;math&gt;d\in\{0\}&lt;/math&gt;(&lt;math&gt;1&lt;/math&gt; possibility)<br /> <br /> <br /> So the number of perfect cubes that divide &lt;math&gt; 3! \cdot 5! \cdot 7! &lt;/math&gt; is &lt;math&gt;3\cdot2\cdot1\cdot1 = 6 \Rightarrow \mathrm{(E)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;math&gt; 3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)&lt;/math&gt;<br /> <br /> In the expression, we notice that there are 3 &lt;math&gt;3's&lt;/math&gt;, 3 &lt;math&gt;2's&lt;/math&gt;, and 3 &lt;math&gt;1's&lt;/math&gt;. This gives us our first 3 cubes: &lt;math&gt;3^3&lt;/math&gt;, &lt;math&gt;2^3&lt;/math&gt;, and &lt;math&gt;1^3&lt;/math&gt;. <br /> <br /> However, we can also multiply smaller numbers in the expression to make bigger expressions. For example, &lt;math&gt;(2*2)*4*4=4*4*4=4^3&lt;/math&gt; (one 2 comes from the &lt;math&gt;3!&lt;/math&gt;, and the other from the &lt;math&gt;5!&lt;/math&gt;). Using this method, we find:<br /> <br /> &lt;math&gt;(3*2)*(3*2)*6=6^3&lt;/math&gt;<br /> <br /> and<br /> <br /> &lt;math&gt;(3*4)*(3*4)*(2*6)=12^3&lt;/math&gt;<br /> <br /> So, we have 6 cubes total:&lt;math&gt;1^3 ,2^3, 3^3, 4^3, 6^3,&lt;/math&gt; and &lt;math&gt;12^3&lt;/math&gt; for a total of &lt;math&gt;6&lt;/math&gt; cubes &lt;math&gt;\Rightarrow \mathrm{(E)}&lt;/math&gt;<br /> <br /> ==See==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> [[Category:Introductory Combinatorics Problems]] <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_14&diff=108827 2005 AMC 10A Problems/Problem 14 2019-08-13T18:37:19Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 41\qquad \mathrm{(B) \ } 42\qquad \mathrm{(C) \ } 43\qquad \mathrm{(D) \ } 44\qquad \mathrm{(E) \ } 45 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits. <br /> <br /> Doing some [[casework]]: <br /> <br /> If the middle digit is &lt;math&gt;1&lt;/math&gt;, possible numbers range from &lt;math&gt;111&lt;/math&gt; to &lt;math&gt;210&lt;/math&gt;. So there are &lt;math&gt;2&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;2&lt;/math&gt;, possible numbers range from &lt;math&gt;123&lt;/math&gt; to &lt;math&gt;420&lt;/math&gt;. So there are &lt;math&gt;4&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;3&lt;/math&gt;, possible numbers range from &lt;math&gt;135&lt;/math&gt; to &lt;math&gt;630&lt;/math&gt;. So there are &lt;math&gt;6&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;4&lt;/math&gt;, possible numbers range from &lt;math&gt;147&lt;/math&gt; to &lt;math&gt;840&lt;/math&gt;. So there are &lt;math&gt;8&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;5&lt;/math&gt;, possible numbers range from &lt;math&gt;159&lt;/math&gt; to &lt;math&gt;951&lt;/math&gt;. So there are &lt;math&gt;9&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;6&lt;/math&gt;, possible numbers range from &lt;math&gt;369&lt;/math&gt; to &lt;math&gt;963&lt;/math&gt;. So there are &lt;math&gt;7&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;7&lt;/math&gt;, possible numbers range from &lt;math&gt;579&lt;/math&gt; to &lt;math&gt;975&lt;/math&gt;. So there are &lt;math&gt;5&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;8&lt;/math&gt;, possible numbers range from &lt;math&gt;789&lt;/math&gt; to &lt;math&gt;987&lt;/math&gt;. So there are &lt;math&gt;3&lt;/math&gt; numbers in this case. <br /> <br /> If the middle digit is &lt;math&gt;9&lt;/math&gt;, the only possible number is &lt;math&gt;999&lt;/math&gt;. So there is &lt;math&gt;1&lt;/math&gt; number in this case. <br /> <br /> So the total number of three-digit numbers that satisfy the property is &lt;math&gt;2+4+6+8+9+7+5+3+1=45\Rightarrow E&lt;/math&gt;<br /> <br /> == Solution 2 (much faster, slicker and more awesome)==<br /> Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:<br /> <br /> If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are &lt;math&gt;5\cdot5=25&lt;/math&gt; numbers in this case.<br /> <br /> If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are &lt;math&gt;4\cdot5=20&lt;/math&gt; numbers here.<br /> <br /> The total number, then, is &lt;math&gt;20+25=45\Rightarrow E&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_13&diff=108826 2005 AMC 10A Problems/Problem 13 2019-08-13T18:35:06Z <p>Ljlbox: /* See also */</p> <hr /> <div>==Problem==<br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy the following condition:<br /> <br /> &lt;math&gt; (130n)^{50} &gt; n^{100} &gt; 2^{200} &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We're given &lt;math&gt; (130n)^{50} &gt; n^{100} &gt; 2^{200} &lt;/math&gt;, so<br /> <br /> &lt;math&gt; \sqrt{(130n)^{50}} &gt; \sqrt{n^{100}} &gt; \sqrt{2^{200}} &lt;/math&gt; (because all terms are positive) and thus<br /> <br /> &lt;math&gt; 130n &gt; n^2 &gt; 2^4 &lt;/math&gt; <br /> <br /> &lt;math&gt; 130n &gt; n^2 &gt; 16 &lt;/math&gt;<br /> <br /> Solving each part separately: <br /> <br /> &lt;math&gt; n^2 &gt; 16 \Longrightarrow n &gt; 4 &lt;/math&gt; <br /> <br /> &lt;math&gt; 130n &gt; n^2 \Longrightarrow 130 &gt; n &lt;/math&gt; <br /> <br /> So &lt;math&gt; 4 &lt; n &lt; 130 &lt;/math&gt;. <br /> <br /> Therefore the answer is the number of [[positive integer]]s over the interval &lt;math&gt; (4,130) &lt;/math&gt; which is &lt;math&gt; 125 \Longrightarrow \mathrm{(E)} &lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_13&diff=108825 2005 AMC 10A Problems/Problem 13 2019-08-13T18:34:56Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy the following condition:<br /> <br /> &lt;math&gt; (130n)^{50} &gt; n^{100} &gt; 2^{200} &lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 7\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 65\qquad \mathrm{(E) \ } 125 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We're given &lt;math&gt; (130n)^{50} &gt; n^{100} &gt; 2^{200} &lt;/math&gt;, so<br /> <br /> &lt;math&gt; \sqrt{(130n)^{50}} &gt; \sqrt{n^{100}} &gt; \sqrt{2^{200}} &lt;/math&gt; (because all terms are positive) and thus<br /> <br /> &lt;math&gt; 130n &gt; n^2 &gt; 2^4 &lt;/math&gt; <br /> <br /> &lt;math&gt; 130n &gt; n^2 &gt; 16 &lt;/math&gt;<br /> <br /> Solving each part separately: <br /> <br /> &lt;math&gt; n^2 &gt; 16 \Longrightarrow n &gt; 4 &lt;/math&gt; <br /> <br /> &lt;math&gt; 130n &gt; n^2 \Longrightarrow 130 &gt; n &lt;/math&gt; <br /> <br /> So &lt;math&gt; 4 &lt; n &lt; 130 &lt;/math&gt;. <br /> <br /> Therefore the answer is the number of [[positive integer]]s over the interval &lt;math&gt; (4,130) &lt;/math&gt; which is &lt;math&gt; 125 \Longrightarrow \mathrm{(E)} &lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=12|num-a=13}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_12&diff=108824 2005 AMC 10A Problems/Problem 12 2019-08-13T18:34:14Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length &lt;math&gt;2&lt;/math&gt;?<br /> <br /> [[Image:2005amc10a12.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} &lt;/math&gt;<br /> <br /> ==Solution==<br /> The area of the ''trefoil'' is equal to the area of a small equilateral triangle plus the area of four &lt;math&gt;60^\circ&lt;/math&gt; sectors with a radius of &lt;math&gt;\frac{2}{2}=1&lt;/math&gt; minus the area of a small equilateral triangle. <br /> <br /> This is equivalent to the area of four &lt;math&gt;60^\circ&lt;/math&gt; sectors with a radius of &lt;math&gt;1&lt;/math&gt;. <br /> <br /> So the answer is: <br /> <br /> &lt;math&gt;4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B &lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_11&diff=108822 2005 AMC 10A Problems/Problem 11 2019-08-13T18:32:37Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> A wooden cube &lt;math&gt;n&lt;/math&gt; units on a side is painted red on all six faces and then cut into &lt;math&gt;n^3&lt;/math&gt; unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since there are &lt;math&gt;n^2&lt;/math&gt; little [[face]]s on each face of the big wooden [[cube (geometry) | cube]], there are &lt;math&gt;6n^2&lt;/math&gt; little faces painted red. <br /> <br /> Since each unit cube has &lt;math&gt;6&lt;/math&gt; faces, there are &lt;math&gt;6n^3&lt;/math&gt; little faces total. <br /> <br /> Since one-fourth of the little faces are painted red, <br /> <br /> &lt;math&gt;\frac{6n^2}{6n^3}=\frac{1}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{1}{n}=\frac{1}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;n=4\Longrightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=10|num-a=12}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=108821 2005 AMC 10A Problems/Problem 10 2019-08-13T18:31:29Z <p>Ljlbox: /* Alternate Solution */</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. &lt;/cmath&gt;<br /> Therefore, we have \implies \boxed{A}.$<br /> <br /> == Solution 3==<br /> There is only one positive value for k such that the quadratic equation would have only one solution.<br /> k-8 and -k-8 are the values of a. I am certain we al know what -8-8 is. That’s correct, &lt;math&gt;\implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=108820 2005 AMC 10A Problems/Problem 10 2019-08-13T18:31:16Z <p>Ljlbox: /* Solution */</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. &lt;/cmath&gt;<br /> Therefore, we have \implies \boxed{A}.$<br /> <br /> == Solution 3==<br /> There is only one positive value for k such that the quadratic equation would have only one solution.<br /> k-8 and -k-8 are the values of a. I am certain we al know what -8-8 is. That’s correct, &lt;math&gt;\implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_10&diff=108818 2005 AMC 10A Problems/Problem 10 2019-08-13T18:29:07Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> There are two values of &lt;math&gt;a&lt;/math&gt; for which the equation &lt;math&gt; 4x^2 + ax + 8x + 9 = 0 &lt;/math&gt; has only one solution for &lt;math&gt;x&lt;/math&gt;. What is the sum of those values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 20 &lt;/math&gt;<br /> <br /> ==Solution==<br /> A [[quadratic equation]] has exactly one [[root]] if and only if it is a [[perfect square]]. So set<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = (mx + n)^2&lt;/math&gt;<br /> <br /> &lt;math&gt;4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2&lt;/math&gt;<br /> <br /> Two [[polynomial]]s are equal only if their [[coefficient]]s are equal, so we must have<br /> <br /> &lt;math&gt;m^2 = 4, n^2 = 9&lt;/math&gt;<br /> <br /> &lt;math&gt;m = \pm 2, n = \pm 3&lt;/math&gt;<br /> <br /> &lt;math&gt;a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12&lt;/math&gt;<br /> <br /> &lt;math&gt;a = 4&lt;/math&gt; or &lt;math&gt;a = -20&lt;/math&gt;.<br /> <br /> So the desired sum is &lt;math&gt; (4)+(-20)=-16 \Longrightarrow \mathrm{(A)} &lt;/math&gt;<br /> <br /> <br /> <br /> Alternatively, note that whatever the two values of &lt;math&gt;a&lt;/math&gt; are, they must lead to equations of the form &lt;math&gt;px^2 + qx + r =0&lt;/math&gt; and &lt;math&gt;px^2 - qx + r = 0&lt;/math&gt;. So the two choices of &lt;math&gt;a&lt;/math&gt; must make &lt;math&gt;a_1 + 8 = q&lt;/math&gt; and &lt;math&gt;a_2 + 8 = -q&lt;/math&gt; so &lt;math&gt;a_1 + a_2 + 16 = 0&lt;/math&gt; and &lt;math&gt;a_1 + a_2 = -16\Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Alternate Solution==<br /> <br /> Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have <br /> &lt;cmath&gt; (a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80. &lt;/cmath&gt; We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the &lt;math&gt;\pm&lt;/math&gt; sign when added). So we must have <br /> &lt;cmath&gt; {-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}. &lt;/cmath&gt;<br /> Therefore, we have \implies \boxed{A}.$<br /> <br /> == Solution 3==<br /> There is only one positive value for k such that the quadratic equation would have only one solution.<br /> k-8 and -k-8 are the values of a. I am certain we al know what -8-8 is. That’s correct, &lt;math&gt;\implies \boxed{A}.&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_9&diff=108817 2005 AMC 10A Problems/Problem 9 2019-08-13T18:28:46Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> Three tiles are marked &lt;math&gt;X&lt;/math&gt; and two other tiles are marked &lt;math&gt;O&lt;/math&gt;. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads &lt;math&gt;XOXOX&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{12}\qquad \mathrm{(B) \ } \frac{1}{10}\qquad \mathrm{(C) \ } \frac{1}{6}\qquad \mathrm{(D) \ } \frac{1}{4}\qquad \mathrm{(E) \ } \frac{1}{3} &lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;\frac{5!}{2!3!}=10&lt;/math&gt; distinct arrangements of three &lt;math&gt;X&lt;/math&gt;'s and two &lt;math&gt;O&lt;/math&gt;'s. <br /> <br /> There is only &lt;math&gt;1&lt;/math&gt; distinct arrangement that reads &lt;math&gt;XOXOX&lt;/math&gt;<br /> <br /> Therefore the desired [[probability]] is &lt;math&gt;\boxed{\frac{1}{10}} \Rightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&diff=108815 2005 AMC 10A Problems/Problem 8 2019-08-13T18:27:27Z <p>Ljlbox: /* See Also */</p> <hr /> <div>== Problem ==<br /> In the figure, the length of side &lt;math&gt;AB&lt;/math&gt; of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\sqrt{50}&lt;/math&gt; and &lt;math&gt;BE=1&lt;/math&gt;. What is the area of the inner square &lt;math&gt;EFGH&lt;/math&gt;?<br /> <br /> [[File:AMC102005Aq.png]]<br /> <br /> &lt;math&gt; \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that side &lt;math&gt;BE&lt;/math&gt;, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, &lt;math&gt;AH = 1&lt;/math&gt;. Then &lt;math&gt;HB = HE + BE = HE + 1&lt;/math&gt;, and &lt;math&gt;HE&lt;/math&gt; is one of the sides of the square whose area we want to find. So:<br /> <br /> &lt;cmath&gt;1^2 + (HE+1)^2=\sqrt{50}^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 + (HE+1)^2=50&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(HE+1)^2=49&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;HE+1=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;HE=6&lt;/cmath&gt; <br /> So, the area of the square is &lt;math&gt;6^2=\boxed{36} \Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=7|num-a=9}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_7&diff=108814 2005 AMC 10A Problems/Problem 7 2019-08-13T18:26:59Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> Josh and Mike live &lt;math&gt;13&lt;/math&gt; miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;m&lt;/math&gt; be the distance in miles that Mike rode. <br /> <br /> Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode &lt;math&gt;2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m&lt;/math&gt; miles. <br /> <br /> Since their combined distance was &lt;math&gt;13&lt;/math&gt; miles, <br /> <br /> &lt;math&gt; \frac{8}{5}m + m = 13 &lt;/math&gt;<br /> <br /> &lt;math&gt; \frac{13}{5}m = 13 &lt;/math&gt; <br /> <br /> &lt;math&gt; m = 5 \Longrightarrow \mathrm{(B)} &lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_6&diff=108813 2005 AMC 10A Problems/Problem 6 2019-08-13T18:26:01Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> The average (mean) of &lt;math&gt;20&lt;/math&gt; numbers is &lt;math&gt;30&lt;/math&gt;, and the average of &lt;math&gt;30&lt;/math&gt; other numbers is &lt;math&gt;20&lt;/math&gt;. What is the average of all &lt;math&gt;50&lt;/math&gt; numbers?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the [[arithmetic mean|average]] of the first &lt;math&gt;20&lt;/math&gt; numbers is &lt;math&gt;30&lt;/math&gt;, their sum is &lt;math&gt;20\cdot30=600&lt;/math&gt;. <br /> <br /> Since the average of &lt;math&gt;30&lt;/math&gt; other numbers is &lt;math&gt;20&lt;/math&gt;, their sum is &lt;math&gt;30\cdot20=600&lt;/math&gt;. <br /> <br /> So the sum of all &lt;math&gt;50&lt;/math&gt; numbers is &lt;math&gt;600+600=1200&lt;/math&gt;<br /> <br /> Therefore, the average of all &lt;math&gt;50&lt;/math&gt; numbers is &lt;math&gt;\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=5|num-a=7}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_2&diff=108812 2005 AMC 10A Problems/Problem 2 2019-08-13T18:25:31Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> For each pair of real numbers &lt;math&gt;a \neq b&lt;/math&gt;, define the [[operation]] &lt;math&gt;\star&lt;/math&gt; as<br /> <br /> &lt;math&gt; (a \star b) = \frac{a+b}{a-b} &lt;/math&gt;.<br /> <br /> What is the value of &lt;math&gt; ((1 \star 2) \star 3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt; ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_4&diff=108811 2005 AMC 10A Problems/Problem 4 2019-08-13T18:23:57Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> A rectangle with a [[diagonal]] of length &lt;math&gt;x&lt;/math&gt; is twice as long as it is wide. What is the area of the rectangle? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let the width of the rectangle be &lt;math&gt;w&lt;/math&gt;. Then the length is &lt;math&gt;2w&lt;/math&gt;.<br /> <br /> Using the [[Pythagorean Theorem]]:<br /> <br /> &lt;math&gt;x^{2}=w^{2}+(2w)^{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2}=5w^{2}&lt;/math&gt;<br /> <br /> The [[area]] of the [[rectangle]] is &lt;math&gt;2w^2=\frac{2}{5}x^2&lt;/math&gt;<br /> <br /> &lt;math&gt;(B)&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_5&diff=108810 2005 AMC 10A Problems/Problem 5 2019-08-13T18:23:39Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> A store normally sells windows at &lt;math&gt;100&lt;/math&gt; each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 &lt;/math&gt;<br /> <br /> ==Solution==<br /> The store's offer means that every &lt;math&gt;5&lt;/math&gt;th window is free. <br /> <br /> Dave would get &lt;math&gt;\lfloor\frac{7}{5}\rfloor=1&lt;/math&gt; free window.<br /> <br /> Doug would get &lt;math&gt;\lfloor\frac{8}{5}\rfloor=1&lt;/math&gt; free window. <br /> <br /> This is a total of &lt;math&gt;2&lt;/math&gt; free windows. <br /> <br /> Together, they would get &lt;math&gt;\lfloor\frac{8+7}{5}\rfloor = \lfloor\frac{15}{5}\rfloor=3&lt;/math&gt; free windows. <br /> <br /> So they get &lt;math&gt;3-2=1&lt;/math&gt; additional window if they purchase the windows together. <br /> <br /> Therefore they save &lt;math&gt;1\cdot100=100\Rightarrow A&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2005|ab=A|num-b=4|num-a=6}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_5&diff=108809 2005 AMC 10A Problems/Problem 5 2019-08-13T18:22:11Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> A store normally sells windows at &lt;math&gt;100&lt;/math&gt; each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 100\qquad \mathrm{(B) \ } 200\qquad \mathrm{(C) \ } 300\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 &lt;/math&gt;<br /> <br /> ==Solution==<br /> The store's offer means that every &lt;math&gt;5&lt;/math&gt;th window is free. <br /> <br /> Dave would get &lt;math&gt;\lfloor\frac{7}{5}\rfloor=1&lt;/math&gt; free window.<br /> <br /> Doug would get &lt;math&gt;\lfloor\frac{8}{5}\rfloor=1&lt;/math&gt; free window. <br /> <br /> This is a total of &lt;math&gt;2&lt;/math&gt; free windows. <br /> <br /> Together, they would get &lt;math&gt;\lfloor\frac{8+7}{5}\rfloor = \lfloor\frac{15}{5}\rfloor=3&lt;/math&gt; free windows. <br /> <br /> So they get &lt;math&gt;3-2=1&lt;/math&gt; additional window if they purchase the windows together. <br /> <br /> Therefore they save &lt;math&gt;1\cdot100=100\Rightarrow A&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2005|ab=A|before=Problem 4|num-a=6}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_4&diff=108808 2005 AMC 10A Problems/Problem 4 2019-08-13T18:21:54Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> A rectangle with a [[diagonal]] of length &lt;math&gt;x&lt;/math&gt; is twice as long as it is wide. What is the area of the rectangle? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let the width of the rectangle be &lt;math&gt;w&lt;/math&gt;. Then the length is &lt;math&gt;2w&lt;/math&gt;.<br /> <br /> Using the [[Pythagorean Theorem]]:<br /> <br /> &lt;math&gt;x^{2}=w^{2}+(2w)^{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2}=5w^{2}&lt;/math&gt;<br /> <br /> The [[area]] of the [[rectangle]] is &lt;math&gt;2w^2=\frac{2}{5}x^2&lt;/math&gt;<br /> <br /> &lt;math&gt;(B)&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2005|ab=A|before=Problem 3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_2&diff=108807 2005 AMC 10A Problems/Problem 2 2019-08-13T18:20:54Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> For each pair of real numbers &lt;math&gt;a \neq b&lt;/math&gt;, define the [[operation]] &lt;math&gt;\star&lt;/math&gt; as<br /> <br /> &lt;math&gt; (a \star b) = \frac{a+b}{a-b} &lt;/math&gt;.<br /> <br /> What is the value of &lt;math&gt; ((1 \star 2) \star 3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt; ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2005|ab=A|before=Problem 1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_2&diff=108806 2005 AMC 10A Problems/Problem 2 2019-08-13T18:20:37Z <p>Ljlbox: /* See Also */</p> <hr /> <div>==Problem==<br /> For each pair of real numbers &lt;math&gt;a \neq b&lt;/math&gt;, define the [[operation]] &lt;math&gt;\star&lt;/math&gt; as<br /> <br /> &lt;math&gt; (a \star b) = \frac{a+b}{a-b} &lt;/math&gt;.<br /> <br /> What is the value of &lt;math&gt; ((1 \star 2) \star 3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } -\frac{2}{3}\qquad \mathrm{(B) \ } -\frac{1}{5}\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \textrm{This\, value\, is\, not\, defined.} &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt; ((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \mathrm{(C)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2005|ab=A|before=Problem 1|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ljlbox https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_8&diff=108651 2000 AMC 12 Problems/Problem 8 2019-08-09T19:28:13Z <p>Ljlbox: /* Solution 5 */</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #8]] and [[2000 AMC 10 Problems|2000 AMC 10 #12]]}}<br /> =Problem=<br /> <br /> Figures &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt; consist of &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;25&lt;/math&gt; nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?<br /> <br /> &lt;asy&gt;<br /> unitsize(8);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);<br /> draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);<br /> draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);<br /> draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);<br /> draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);<br /> draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);<br /> draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);<br /> draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);<br /> draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);<br /> label(&quot;Figure&quot;,(0.5,-1),S);<br /> label(&quot;$0$&quot;,(0.5,-2.5),S);<br /> label(&quot;Figure&quot;,(9.5,-1),S);<br /> label(&quot;$1$&quot;,(9.5,-2.5),S);<br /> label(&quot;Figure&quot;,(19.5,-1),S);<br /> label(&quot;$2$&quot;,(19.5,-2.5),S);<br /> label(&quot;Figure&quot;,(32.5,-1),S);<br /> label(&quot;$3$&quot;,(32.5,-2.5),S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> We can divide up figure &lt;math&gt;n&lt;/math&gt; to get the sum of the sum of the first &lt;math&gt;n+1&lt;/math&gt; odd numbers and the sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers. If you do not see this, here is the example for &lt;math&gt;n=3&lt;/math&gt;:<br /> <br /> &lt;asy&gt;<br /> draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);<br /> draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);<br /> draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);<br /> draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);<br /> draw((3,0)--(3,7));<br /> &lt;/asy&gt;<br /> <br /> The sum of the first &lt;math&gt;n&lt;/math&gt; odd numbers is &lt;math&gt;n^2&lt;/math&gt;, so for figure &lt;math&gt;n&lt;/math&gt;, there are &lt;math&gt;(n+1)^2+n^2&lt;/math&gt; unit squares. We plug in &lt;math&gt;n=100&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Using the recursion from solution 1, we see that the first differences of &lt;math&gt;4, 8, 12, ...&lt;/math&gt; form an arithmetic progression, and consequently that the second differences are constant and all equal to &lt;math&gt;4&lt;/math&gt;. Thus, the original sequence can be generated from a quadratic function.<br /> <br /> If &lt;math&gt;f(n) = an^2 + bn + c&lt;/math&gt;, and &lt;math&gt;f(0) = 1&lt;/math&gt;, &lt;math&gt;f(1) = 5&lt;/math&gt;, and &lt;math&gt;f(2) = 13&lt;/math&gt;, we get a system of three equations in three variables:<br /> <br /> &lt;math&gt;f(0) = 1&lt;/math&gt; gives &lt;math&gt;c = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;f(1) = 5&lt;/math&gt; gives &lt;math&gt;a + b + c = 5&lt;/math&gt;<br /> <br /> &lt;math&gt;f(2) = 13&lt;/math&gt; gives &lt;math&gt;4a + 2b + c = 13&lt;/math&gt;<br /> <br /> Plugging in &lt;math&gt;c=1&lt;/math&gt; into the last two equations gives <br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;4a + 2b = 12&lt;/math&gt;<br /> <br /> Dividing the second equation by 2 gives the system:<br /> <br /> &lt;math&gt;a + b = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;2a + b = 6&lt;/math&gt;<br /> <br /> Subtracting the first equation from the second gives &lt;math&gt;a = 2&lt;/math&gt;, and hence &lt;math&gt;b = 2&lt;/math&gt;. Thus, our quadratic function is:<br /> <br /> &lt;math&gt;f(n) = 2n^2 + 2n + 1&lt;/math&gt;<br /> <br /> Calculating the answer to our problem, &lt;math&gt;f(100) = 20000 + 200 + 1 = 20201&lt;/math&gt;, which is choice &lt;math&gt;\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We can see that each figure &lt;math&gt;n&lt;/math&gt; has a central box and 4 columns of &lt;math&gt;n&lt;/math&gt; boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are &lt;math&gt;\sum_{n=1}^{100} n = 5050&lt;/math&gt; squares. &lt;math&gt;4 \cdot 5050 = 20200&lt;/math&gt;. Adding in the original center box we have &lt;math&gt; 20200 + 1 = \boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a_n&lt;/math&gt; be the number of squares in figure &lt;math&gt;n&lt;/math&gt;. We can easily see that <br /> &lt;cmath&gt;a_0=4\cdot 0+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_1=4\cdot 1+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=4\cdot 3+1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=4\cdot 6+1.&lt;/cmath&gt;<br /> Note that in &lt;math&gt;a_n&lt;/math&gt;, the number multiplied by the 4 is the &lt;math&gt;n&lt;/math&gt;th triangular number. Hence, &lt;math&gt;a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}&lt;/math&gt;.<br /> ~qkddud~<br /> <br /> ==Solution 5==<br /> Let &lt;math&gt;f_n&lt;/math&gt; denote the number of unit cubes in a figure. We have<br /> &lt;cmath&gt;f_0=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f_1=5&lt;/cmath&gt;<br /> &lt;cmath&gt;f_2=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f_3=25&lt;/cmath&gt;<br /> &lt;cmath&gt;f_4=41&lt;/cmath&gt;<br /> &lt;cmath&gt;...&lt;/cmath&gt;<br /> <br /> Computing the difference between the number of cubes in each figure yields<br /> &lt;cmath&gt;4,8,12,16,...&lt;/cmath&gt;<br /> It is easy to notice that this is an arithmetic sequence, with the first term being &lt;math&gt;4&lt;/math&gt; and the difference being &lt;math&gt;4&lt;/math&gt;. Let this sequence be &lt;math&gt;a_n&lt;/math&gt;<br /> <br /> From &lt;math&gt;f_0&lt;/math&gt; to &lt;math&gt;f_{100}&lt;/math&gt;, the sequence will have &lt;math&gt;100&lt;/math&gt; terms. Using the arithmetic sum formula yields<br /> <br /> &lt;cmath&gt;S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;=50(2\cdot 4+99\cdot 4)&lt;/cmath&gt;<br /> &lt;cmath&gt;=50(101\cdot 4)&lt;/cmath&gt;<br /> &lt;cmath&gt;=200\cdot 101&lt;/cmath&gt;<br /> &lt;cmath&gt;=20200&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;f_{100}=1+20200=\boxed{\textbf{(C) }20201}&lt;/math&gt; unit cubes.<br /> <br /> ~ljlbox<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2000|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2000|num-b=11|num-a=13}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ljlbox