https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Brainpopper&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-14T23:44:13Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_1&diff=119375 2020 AIME II Problems/Problem 1 2020-03-14T15:07:54Z <p>Brainpopper: </p> <hr /> <div>no cheating<br /> <br /> 1. What is &lt;math&gt;17 + 25?&lt;/math&gt;<br /> <br /> A) &lt;math&gt;43&lt;/math&gt;<br /> B) &lt;math&gt;75&lt;/math&gt;<br /> C) &lt;math&gt;102&lt;/math&gt;<br /> D) &lt;math&gt;200&lt;/math&gt;<br /> E) &lt;math&gt;1720 + e^i + 194i&lt;/math&gt;<br /> F) &lt;math&gt;42&lt;/math&gt;<br /> <br /> <br /> Solution<br /> <br /> &lt;math&gt;17 + 25 = \boxed{42}&lt;/math&gt;</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_13&diff=91791 2018 AMC 10B Problems/Problem 13 2018-02-17T13:32:58Z <p>Brainpopper: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> How many of the first &lt;math&gt;2018&lt;/math&gt; numbers in the sequence &lt;math&gt;101, 1001, 10001, 100001, \dots&lt;/math&gt; are divisible by &lt;math&gt;101&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }253 \qquad<br /> \textbf{(B) }504 \qquad<br /> \textbf{(C) }505 \qquad<br /> \textbf{(D) }506 \qquad<br /> \textbf{(E) }1009 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;10^{2k}+1&lt;/math&gt; for some odd &lt;math&gt;k&lt;/math&gt; will suffice &lt;math&gt;\mod {101}&lt;/math&gt;. Each &lt;math&gt;2k \in \{2,4,6,\dots,2018\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> (AOPS12142015)<br /> <br /> ==Solution 2==<br /> If we divide each number by &lt;math&gt;101&lt;/math&gt;, we see a pattern occuring in every 4 numbers. &lt;math&gt;101, 1000001, 10000000001, \dots&lt;/math&gt;. We divide &lt;math&gt;2018&lt;/math&gt; by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;504&lt;/math&gt; with &lt;math&gt;2&lt;/math&gt; left over. One divisible number will be in the &lt;math&gt;2&lt;/math&gt; left over, so out answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_13&diff=91790 2018 AMC 10B Problems/Problem 13 2018-02-17T13:32:45Z <p>Brainpopper: </p> <hr /> <div>==Problem==<br /> How many of the first &lt;math&gt;2018&lt;/math&gt; numbers in the sequence &lt;math&gt;101, 1001, 10001, 100001, \dots&lt;/math&gt; are divisible by &lt;math&gt;101&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }253 \qquad<br /> \textbf{(B) }504 \qquad<br /> \textbf{(C) }505 \qquad<br /> \textbf{(D) }506 \qquad<br /> \textbf{(E) }1009 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that &lt;math&gt;10^{2k}+1&lt;/math&gt; for some odd &lt;math&gt;k&lt;/math&gt; will suffice &lt;math&gt;\mod {101}&lt;/math&gt;. Each &lt;math&gt;2k \in \{2,4,6,\dots,2018\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> (AOPS12142015)<br /> <br /> ==Solution 2==<br /> If we divide each number by &lt;math&gt;101&lt;/math&gt;, we see a pattern occuring in every 4 numbers. &lt;math&gt;101, 1000001, 10000000001, \dots. We divide &lt;/math&gt;2018&lt;math&gt; by &lt;/math&gt;4&lt;math&gt; to get &lt;/math&gt;504&lt;math&gt; with &lt;/math&gt;2&lt;math&gt; left over. One divisible number will be in the &lt;/math&gt;2&lt;math&gt; left over, so out answer is &lt;/math&gt;\boxed{\textbf{(C) } 505}$.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_23&diff=76481 2015 AMC 10A Problems/Problem 23 2016-02-20T03:21:46Z <p>Brainpopper: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> The zeroes of the function &lt;math&gt;f(x)=x^2-ax+2a&lt;/math&gt; are integers .What is the sum of the possible values of a?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> By Vieta's Formula, &lt;math&gt;a&lt;/math&gt; is the sum of the integral zeros of the function, and so &lt;math&gt;a&lt;/math&gt; is integral.<br /> <br /> Because the zeros are integral, the discriminant of the function, &lt;math&gt;a^2 - 8a&lt;/math&gt;, is a perfect square, say &lt;math&gt;k^2&lt;/math&gt;. Then adding 16 to both sides and completing the square yields<br /> &lt;cmath&gt;(a - 4)^2 = k^2 + 16.&lt;/cmath&gt;<br /> Hence &lt;math&gt;(a-4)^2 - k^2 = 16&lt;/math&gt; and<br /> &lt;cmath&gt;((a-4) - k)((a-4) + k) = 16.&lt;/cmath&gt;<br /> Let &lt;math&gt;(a-4) - k = u&lt;/math&gt; and &lt;math&gt;(a-4) + k = v&lt;/math&gt;; then, &lt;math&gt;a-4 = \dfrac{u+v}{2}&lt;/math&gt; and so &lt;math&gt;a = \dfrac{u+v}{2} + 4&lt;/math&gt;. Listing all possible &lt;math&gt;(u, v)&lt;/math&gt; pairs (not counting transpositions because this does not affect (&lt;math&gt;u + v&lt;/math&gt;), &lt;math&gt;(2, 8), (4, 4), (-2, -8), (-4, -4)&lt;/math&gt;, yields &lt;math&gt;a = 9, 8, -1, 0&lt;/math&gt;. These &lt;math&gt;a&lt;/math&gt; sum to &lt;math&gt;16&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(C) }16}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; be the integer zeroes of the quadratic.<br /> <br /> Since the coefficient of the &lt;math&gt;x^2&lt;/math&gt; term is &lt;math&gt;1&lt;/math&gt;, the quadratic can be written as &lt;math&gt;(x - r_1)(x - r_2)&lt;/math&gt; or &lt;math&gt;x^2 - (r_1 + r_2)x + r_1r_2&lt;/math&gt;.<br /> <br /> By comparing this with &lt;math&gt;x^2 - ax + 2a&lt;/math&gt;, &lt;math&gt;r_1 + r_2 = a&lt;/math&gt; and &lt;math&gt;r_1r_2 = 2a&lt;/math&gt;.<br /> <br /> Plugging the first equation in the second, &lt;math&gt;r_1r_2 = 2 (r_1 + r_2)&lt;/math&gt;. Rearranging gives &lt;math&gt;r_1r_2 - 2r_1 - 2r_2 = 0&lt;/math&gt;.<br /> <br /> This can be factored as &lt;math&gt;(r_1 - 2)(r_2 - 2) = 4&lt;/math&gt;.<br /> <br /> These factors can be: &lt;math&gt;(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)&lt;/math&gt;.<br /> <br /> We want the number of distinct &lt;math&gt;a = r_1 + r_2&lt;/math&gt;, and these factors gives &lt;math&gt;a = -1, 0, 8, 9&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt;-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Intermediate Number Theory Problems]]</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_10&diff=76455 2013 AMC 8 Problems/Problem 10 2016-02-19T12:33:05Z <p>Brainpopper: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> To find either the LCM or the GCF of two numbers, always prime factorize first.<br /> <br /> The prime factorization of &lt;math&gt;180 = 3^2 \times 5 \times 2^2&lt;/math&gt;.<br /> <br /> The prime factorization of &lt;math&gt;594 = 3^3 \times 11 \times 2&lt;/math&gt;.<br /> <br /> Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is &lt;math&gt;3^3, 5, 11, 2^2&lt;/math&gt;). Multiply all of these to get 5940. <br /> <br /> For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. &lt;math&gt;3^2 \times 2&lt;/math&gt; = 18.<br /> <br /> Thus the answer = &lt;math&gt;\frac{5940}{18}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==Similar Solution==<br /> We start off with a similar approach as the original solution. From the prime factorizations, the GCF is &lt;math&gt;18&lt;/math&gt;.<br /> <br /> It is a well known fact that &lt;math&gt;\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|&lt;/math&gt;. So we have, &lt;math&gt;18\times \operatorname{lcm} (180,594)=594\times 180&lt;/math&gt;.<br /> <br /> Dividing by &lt;math&gt;18&lt;/math&gt; yields &lt;math&gt;\operatorname{lcm} (180,594)=594\times 10=5940&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\dfrac{\operatorname{lcm} (180,594)}{\gcd(180,594)}=\dfrac{5940}{18}=\boxed{\textbf{(C)}\ 330}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_25&diff=71950 2007 AMC 8 Problems/Problem 25 2015-09-07T03:17:31Z <p>Brainpopper: </p> <hr /> <div>===Problem===<br /> On the dart board shown in the Figure, the outer circle has radius &lt;math&gt;6&lt;/math&gt; and the inner circle has radius &lt;math&gt;3&lt;/math&gt;. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?<br /> <br /> AMC8 2007 25.png<br /> &lt;math&gt;\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}&lt;/math&gt;<br /> <br /> ===Solution===<br /> To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).<br /> <br /> <br /> To find the areas of the sections, notice that the three smaller sections trisect a circle with radius &lt;math&gt;3&lt;/math&gt;. The area of this entire circle is &lt;math&gt;9\pi&lt;/math&gt;. The area of each smaller section then must be &lt;math&gt;\frac{9\pi}{3}&lt;/math&gt; or &lt;math&gt;3\pi&lt;/math&gt;. The larger sections trisect an &quot;ring&quot; which is the difference of two circles, one with radius &lt;math&gt;3&lt;/math&gt;, the other radius &lt;math&gt;6&lt;/math&gt;. So, the area of the ring (''annulus'') is &lt;math&gt;36\pi - 9\pi&lt;/math&gt; or &lt;math&gt;27\pi&lt;/math&gt;. The area of each larger section must be &lt;math&gt;\frac{27\pi}{3}&lt;/math&gt; or &lt;math&gt;9\pi&lt;/math&gt;. Note that the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;.<br /> <br /> <br /> One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is &lt;math&gt;3\pi + (2 \cdot 9\pi) = 21\pi&lt;/math&gt;. Since the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;, the probability of getting an odd is &lt;math&gt;\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}&lt;/math&gt;.<br /> <br /> Since the remaining sections contain even numbers (that is, 2), the probability of throwing an odd is the complement, or &lt;math&gt;1 - \frac{7}{12} = \frac{5}{12}&lt;/math&gt;.<br /> <br /> <br /> Now, the two cases: You could either get an odd then an even, or an even then an odd.<br /> <br /> <br /> CASE 1: Odd then even<br /> Multiply the probabilities to get &lt;math&gt;\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}&lt;/math&gt;.<br /> <br /> <br /> CASE 2: Even then odd<br /> Multiply the probabilities to get &lt;math&gt;\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}&lt;/math&gt;. Notice that this is the same.<br /> <br /> <br /> Thus, the total probability of an odd sum is &lt;math&gt;\frac{35}{144} \cdot \frac{2}{1}&lt;/math&gt; = &lt;math&gt;\boxed{B = \frac{35}{72}}&lt;/math&gt;.<br /> {{MAA Notice}}</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_25&diff=71949 2007 AMC 8 Problems/Problem 25 2015-09-07T03:15:31Z <p>Brainpopper: I put the problem up and showed the letter answer, not just the number answer.</p> <hr /> <div>[b]Problem:[/b]<br /> On the dart board shown in the Figure, the outer circle has radius &lt;math&gt;6&lt;/math&gt; and the inner circle has radius &lt;math&gt;3&lt;/math&gt;. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?<br /> <br /> AMC8 2007 25.png<br /> &lt;math&gt;\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}&lt;/math&gt;<br /> <br /> [b]Solution:[/b]<br /> To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).<br /> <br /> <br /> To find the areas of the sections, notice that the three smaller sections trisect a circle with radius &lt;math&gt;3&lt;/math&gt;. The area of this entire circle is &lt;math&gt;9\pi&lt;/math&gt;. The area of each smaller section then must be &lt;math&gt;\frac{9\pi}{3}&lt;/math&gt; or &lt;math&gt;3\pi&lt;/math&gt;. The larger sections trisect an &quot;ring&quot; which is the difference of two circles, one with radius &lt;math&gt;3&lt;/math&gt;, the other radius &lt;math&gt;6&lt;/math&gt;. So, the area of the ring (''annulus'') is &lt;math&gt;36\pi - 9\pi&lt;/math&gt; or &lt;math&gt;27\pi&lt;/math&gt;. The area of each larger section must be &lt;math&gt;\frac{27\pi}{3}&lt;/math&gt; or &lt;math&gt;9\pi&lt;/math&gt;. Note that the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;.<br /> <br /> <br /> One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is &lt;math&gt;3\pi + (2 \cdot 9\pi) = 21\pi&lt;/math&gt;. Since the area of the whole circle is &lt;math&gt;36\pi&lt;/math&gt;, the probability of getting an odd is &lt;math&gt;\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}&lt;/math&gt;.<br /> <br /> Since the remaining sections contain even numbers (that is, 2), the probability of throwing an odd is the complement, or &lt;math&gt;1 - \frac{7}{12} = \frac{5}{12}&lt;/math&gt;.<br /> <br /> <br /> Now, the two cases: You could either get an odd then an even, or an even then an odd.<br /> <br /> <br /> CASE 1: Odd then even<br /> Multiply the probabilities to get &lt;math&gt;\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}&lt;/math&gt;.<br /> <br /> <br /> CASE 2: Even then odd<br /> Multiply the probabilities to get &lt;math&gt;\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}&lt;/math&gt;. Notice that this is the same.<br /> <br /> <br /> Thus, the total probability of an odd sum is &lt;math&gt;\frac{35}{144} \cdot \frac{2}{1}&lt;/math&gt; = &lt;math&gt;\boxed{B = \frac{35}{72}}&lt;/math&gt;.<br /> {{MAA Notice}}</div> Brainpopper https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_18&diff=71937 2008 AMC 8 Problems/Problem 18 2015-09-06T07:59:53Z <p>Brainpopper: /* Solution */</p> <hr /> <div>==Problem==<br /> Two circles that share the same center have radii &lt;math&gt;10&lt;/math&gt; meters and &lt;math&gt;20&lt;/math&gt; meters. An aardvark runs along the path shown, starting at &lt;math&gt;A&lt;/math&gt; and ending at &lt;math&gt;K&lt;/math&gt;. How many meters does the aardvark run?<br /> &lt;asy&gt;<br /> size((150));<br /> draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle);<br /> draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle);<br /> draw((20,0)--(-20,0));<br /> draw((0,20)--(0,-20));<br /> draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow);<br /> draw((-18,1)--(-12, 1), EndArrow);<br /> draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow);<br /> draw((1,-9)--(1,9), EndArrow);<br /> draw((0,12)..(8.3, 8.3)..(12,0), EndArrow);<br /> draw((12,-1)--(18,-1), EndArrow);<br /> label(&quot;$A$&quot;, (0,20), N);<br /> label(&quot;$K\$&quot;, (20,0), E);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 10\pi+20\qquad\textbf{(B)}\ 10\pi+30\qquad\textbf{(C)}\ 10\pi+40\qquad\textbf{(D)}\ 20\pi+20\qquad \\ \textbf{(E)}\ 20\pi+40&lt;/math&gt;<br /> <br /> ==Solution==<br /> We will deal with this part by part:<br /> Part 1: 1/4 circumference of big circle= &lt;math&gt;\frac{2\pi r}{4}=\frac{\pi r}{2}=\frac{20\pi}{2}=10\pi&lt;/math&gt;<br /> Part 2: Big radius minus small radius= &lt;math&gt;20-10=10&lt;/math&gt;<br /> Part 3: 1/4 circumference of small circle= &lt;math&gt;\frac{\pi r}{2}=\frac{10\pi}{2}=5\pi&lt;/math&gt;<br /> Part 4: Diameter of small circle: &lt;math&gt;2*10=20&lt;/math&gt;<br /> Part 5: Same as part 3: &lt;math&gt;5\pi&lt;/math&gt;<br /> Part 6: Same as part 2: &lt;math&gt;10&lt;/math&gt;<br /> Total: &lt;math&gt;10\pi + 10 + 5\pi + 20 + 5\pi + 10 = \boxed{E = 20\pi + 40}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2008|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Brainpopper