https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Logsobolev&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-19T16:31:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_28&diff=155397 1994 AHSME Problems/Problem 28 2021-06-06T17:24:27Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, how many lines whose &lt;math&gt;x&lt;/math&gt;-intercept is a positive prime number and whose &lt;math&gt;y&lt;/math&gt;-intercept is a positive integer pass through the point &lt;math&gt;(4,3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> ==Solution==<br /> <br /> The line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;b&lt;/math&gt; is given by the equation &lt;math&gt;\frac{x}{a} + \frac{y}{b} = 1&lt;/math&gt;. We are told &lt;math&gt;(4,3)&lt;/math&gt; is on the line so <br /> <br /> &lt;cmath&gt;\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, this equation holds only if &lt;math&gt;(a-4)&lt;/math&gt; is a factor of &lt;math&gt;12&lt;/math&gt;. The factors are &lt;math&gt;1, 2, 3, 4, 6, 12&lt;/math&gt; which means &lt;math&gt;a&lt;/math&gt; must be one of &lt;math&gt;5, 6, 7, 8, 10, 16&lt;/math&gt;. The only members of this list which are prime are &lt;math&gt;a=5&lt;/math&gt; and &lt;math&gt;a=7&lt;/math&gt;, so the number of solutions is &lt;math&gt;\boxed{\textbf{(C) } 2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=27|num-a=29}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154397 1994 AHSME Problems/Problem 30 2021-05-28T22:43:12Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's number with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;, so this symmetry is its own inverse. Under this symmetry the sum &lt;math&gt;R=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S = \sum_{i=1}^n 7-d_i = 7n - R&lt;/math&gt;. As a result of this symmetry the sum &lt;math&gt;R&lt;/math&gt; the sum &lt;math&gt;S&lt;/math&gt; have the same probability because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;R&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; which sum to &lt;math&gt;S&lt;/math&gt;, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to &lt;math&gt;R&lt;/math&gt; and the combinations which sum to &lt;math&gt;S&lt;/math&gt;.<br /> <br /> When &lt;math&gt;R=1994&lt;/math&gt; we seek the smallest number &lt;math&gt;S=7n-1994&lt;/math&gt;, which clearly happens when &lt;math&gt;n&lt;/math&gt; is smallest. Therefore we want to find the smallest &lt;math&gt;n&lt;/math&gt; which gives non-zero probability of obtaining &lt;math&gt;R=1994&lt;/math&gt;. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;R=1994&lt;/math&gt; being impossible. Thus &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt; and &lt;math&gt;S = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154396 1994 AHSME Problems/Problem 30 2021-05-28T22:27:50Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's number with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;, so this symmetry is its own inverse. Under this symmetry the sum &lt;math&gt;R=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S = \sum_{i=1}^n 7-d_i = 7n - R&lt;/math&gt;. As a result of this symmetry the probability of obtaining the sum &lt;math&gt;R&lt;/math&gt; and the probability of obtaining the sum &lt;math&gt;S&lt;/math&gt; are equal because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;R&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; to get the sum &lt;math&gt;S&lt;/math&gt;, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to &lt;math&gt;R&lt;/math&gt; and the combinations which sum to &lt;math&gt;S&lt;/math&gt;.<br /> <br /> When &lt;math&gt;R=1994&lt;/math&gt; we seek the smallest number &lt;math&gt;S=7n-1994&lt;/math&gt;, and, examining the formula, this happens when &lt;math&gt;n&lt;/math&gt; is smallest. Therefore we want to find the smallest &lt;math&gt;n&lt;/math&gt; which gives non-zero probability of obtaining &lt;math&gt;R=1994&lt;/math&gt;. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;R=1994&lt;/math&gt; being impossible. Thus &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt; and &lt;math&gt;S = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154393 1994 AHSME Problems/Problem 30 2021-05-28T18:20:48Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's result with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;. Under this symmetry the sum &lt;math&gt;S=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S' = \sum_{i=1}^n 7-d_i = 7n - S&lt;/math&gt;. As a result of this symmetry the probabilities of obtaing the sum &lt;math&gt;S&lt;/math&gt; and the sum &lt;math&gt;S'&lt;/math&gt; are equal because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;S&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; to get the sum &lt;math&gt;S'&lt;/math&gt;, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to &lt;math&gt;S&lt;/math&gt; and the combinations which sum to &lt;math&gt;S'&lt;/math&gt;.<br /> <br /> When &lt;math&gt;S=1994&lt;/math&gt; the smallest number &lt;math&gt;S'=7n-1994&lt;/math&gt; corresponds to the smallest number &lt;math&gt;n&lt;/math&gt;. Thus we want to find the smallest &lt;math&gt;n&lt;/math&gt; which gives non-zero probability of obtaining &lt;math&gt;S=1994&lt;/math&gt;. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;S=1994&lt;/math&gt; being impossible. Clearly &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt; and &lt;math&gt;S' = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154392 1994 AHSME Problems/Problem 30 2021-05-28T18:19:33Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's result with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;. Under this symmetry the sum &lt;math&gt;S=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S' = \sum_{i=1}^n 7-d_i = 7n - S&lt;/math&gt;. As a result of this symmetry the probabilities of obtaing the sum &lt;math&gt;S&lt;/math&gt; and the sum &lt;math&gt;S'&lt;/math&gt; are equal because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;S&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; to get the sum &lt;math&gt;S'&lt;/math&gt;, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to &lt;math&gt;S&lt;/math&gt; and the combinations which sum to &lt;math&gt;S'&lt;/math&gt;.<br /> <br /> When &lt;math&gt;S=1994&lt;/math&gt; the smallest number &lt;math&gt;S'=7n-1994&lt;/math&gt; corresponds to the smallest number &lt;math&gt;n&lt;/math&gt;. Thus we want to find the smallest &lt;math&gt;n&lt;/math&gt; which gives non-zero probability of obtaining &lt;math&gt;S=1994&lt;/math&gt;. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;S=1994&lt;/math&gt; being impossible. Clearly &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt;. By the symmetry mentioned above, the probability of rolling &lt;math&gt;S=1994&lt;/math&gt; is the same as the probability of rolling &lt;math&gt;S' = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154391 1994 AHSME Problems/Problem 30 2021-05-28T18:14:17Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's result with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;. Under this symmetry the sum &lt;math&gt;S=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S' = \sum_{i=1}^n 7-d_i = 7n - S&lt;/math&gt;. As a result of this symmetry the probabilities of obtaing the sum &lt;math&gt;S&lt;/math&gt; and the sum &lt;math&gt;S'&lt;/math&gt; are equal because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;S&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; to get the sum &lt;math&gt;S'&lt;/math&gt;, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to &lt;math&gt;S&lt;/math&gt; and the combinations which sum to &lt;math&gt;S'&lt;/math&gt;.<br /> <br /> <br /> The smallest non-zero probability of obtaining &lt;math&gt;S=1994&lt;/math&gt; occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;S=1994&lt;/math&gt; being impossible. That happens when &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt;. By the symmetry mentioned above, the probability of rolling &lt;math&gt;S=1994&lt;/math&gt; is the same as the probability of rolling &lt;math&gt;S' = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_21&diff=154390 1994 AHSME Problems/Problem 21 2021-05-28T18:11:36Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the number of counter examples to the statement:<br /> &lt;cmath&gt;\text{If N is an odd positive integer the sum of whose digits is 4 and none of whose digits is 0, then N is prime}.&quot;&lt;/cmath&gt;<br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4 &lt;/math&gt;<br /> ==Solution==<br /> Since the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; and none of the digits are &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;'s digits must the elements of one of the sets &lt;math&gt;\{1,1,1,1\},\{1,1,2\}&lt;/math&gt;, &lt;math&gt;\{2,2\}&lt;/math&gt;, &lt;math&gt;\{1,3\}&lt;/math&gt;, or &lt;math&gt;\{4\}&lt;/math&gt;. <br /> <br /> In the first case, &lt;math&gt;N = 1111 = 101 \cdot 11&lt;/math&gt; so this is a counter example. <br /> <br /> In the second case, &lt;math&gt;N=112&lt;/math&gt; is excluded for being even. With &lt;math&gt;N=121=11^2&lt;/math&gt; we have a counterexample. We can check &lt;math&gt;N=211&lt;/math&gt; by trial division, and verify it is indeed prime. <br /> <br /> In the third case, &lt;math&gt;N=22&lt;/math&gt; is excluded for being even.<br /> <br /> In the fourth case, both &lt;math&gt;N=13&lt;/math&gt; and &lt;math&gt;N=31&lt;/math&gt; are prime. <br /> <br /> In the last case &lt;math&gt;N=4&lt;/math&gt; is excluded for being even.<br /> <br /> This gives two counterexamples and the answer is &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=154371 1994 AHSME Problems/Problem 23 2021-05-28T08:22:23Z <p>Logsobolev: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at &lt;math&gt;(0,0), (0,3), (3,3), (3,1), (5,1)&lt;/math&gt; and &lt;math&gt;(5,0)&lt;/math&gt;. The slope of the line through the origin that divides the area of this region exactly in half is<br /> &lt;asy&gt;<br /> Label l;<br /> l.p=fontsize(6);<br /> xaxis(&quot;$x$&quot;,0,6,Ticks(l,1.0,0.5),EndArrow);<br /> yaxis(&quot;$y$&quot;,0,4,Ticks(l,1.0,0.5),EndArrow);<br /> draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));&lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} &lt;/math&gt;<br /> ==Solution==<br /> Let the vertices be &lt;math&gt;A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)&lt;/math&gt;. It is easy to see that the line must pass through &lt;math&gt;CD&lt;/math&gt;. Let the line intersect &lt;math&gt;CD&lt;/math&gt; at the point &lt;math&gt;G=(3,3-x)&lt;/math&gt; (i.e. the point &lt;math&gt;x&lt;/math&gt; units below &lt;math&gt;C&lt;/math&gt;). Since the quadrilateral &lt;math&gt;ABCG&lt;/math&gt; and pentagon &lt;math&gt;GDEFA&lt;/math&gt; must have the same area, we have the equation &lt;math&gt;3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2&lt;/math&gt;. This simplifies into &lt;math&gt;3x=2&lt;/math&gt;, or &lt;math&gt;x=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;G=(3,\frac{7}{3})&lt;/math&gt;. Therefore the slope of the line is &lt;math&gt;\boxed{\textbf{(E)}\ \frac{7}{9}}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> Consider the small rectangle between &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;x=5&lt;/math&gt; with area &lt;math&gt;2&lt;/math&gt;. If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=3&lt;/math&gt; and &lt;math&gt;y=c&lt;/math&gt; for some unknown &lt;math&gt;c&lt;/math&gt;. For the area to be the same we need &lt;math&gt;(3-c) \cdot 3 = 2&lt;/math&gt;, or &lt;math&gt;c=7/3&lt;/math&gt;. After excluding our two offsetting areas, we're left with a rectangle from &lt;math&gt;x=0&lt;/math&gt; to &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; to &lt;math&gt;y=c&lt;/math&gt;. The area of this region is clearly bisected by its diagonal line. The line passes through &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(3,c)&lt;/math&gt; so its slope is &lt;math&gt;c/3 = 7/9&lt;/math&gt; and the answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_30&diff=154370 1994 AHSME Problems/Problem 30 2021-05-28T08:06:09Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> When &lt;math&gt;n&lt;/math&gt; standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of &lt;math&gt;S&lt;/math&gt;. The smallest possible value of &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 &lt;/math&gt;<br /> ==Solution==<br /> <br /> Let &lt;math&gt;d_i&lt;/math&gt; be the number on the &lt;math&gt;i&lt;/math&gt;th die. There is a symmetry where we can replace each die's result with &lt;math&gt;d_i' = 7-d_i&lt;/math&gt;. Note that applying the symmetry twice we get back to where we started since &lt;math&gt;7-(7-d_i)=d_i&lt;/math&gt;. Under this symmetry the sum &lt;math&gt;S=\sum_{i=1}^n d_i&lt;/math&gt; is replaced by &lt;math&gt;S' = \sum_{i=1}^n 7-d_i = 7n - S&lt;/math&gt;. As a result of this symmetry the probabilities of obtaing the sum &lt;math&gt;S&lt;/math&gt; and the sum &lt;math&gt;S'&lt;/math&gt; are equal because any combination of &lt;math&gt;d_i&lt;/math&gt; which sum to &lt;math&gt;S&lt;/math&gt; can be replaced with &lt;math&gt;d_i'&lt;/math&gt; to get the sum &lt;math&gt;S'&lt;/math&gt;, and conversely. In other words, there is a bijection between the combinations of dice which sum to &lt;math&gt;S&lt;/math&gt; and the combinations which sum to &lt;math&gt;S'&lt;/math&gt;.<br /> <br /> <br /> The smallest non-zero probability of obtaining &lt;math&gt;S=1994&lt;/math&gt; occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in &lt;math&gt;S=1994&lt;/math&gt; being impossible. That happens when &lt;math&gt; n = \left\lceil \frac{1994}{6} \right\rceil = 333&lt;/math&gt;. By the symmetry mentioned above, the probability of rolling &lt;math&gt;S=1994&lt;/math&gt; is the same as the probability of rolling &lt;math&gt;S' = 333\cdot 7 - 1994 = 337&lt;/math&gt;. The answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=29|after= Last Problem}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_28&diff=154369 1994 AHSME Problems/Problem 28 2021-05-28T07:49:24Z <p>Logsobolev: /* See Also */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, how many lines whose &lt;math&gt;x&lt;/math&gt;-intercept is a positive prime number and whose &lt;math&gt;y&lt;/math&gt;-intercept is a positive integer pass through the point &lt;math&gt;(4,3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> ==Solution==<br /> <br /> The line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;b&lt;/math&gt; is given by the equation &lt;math&gt;\frac{x}{a} + \frac{y}{b} = 1&lt;/math&gt;. We are told &lt;math&gt;(4,3)&lt;/math&gt; is on the line so <br /> <br /> &lt;cmath&gt;\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, this equation holds only if &lt;math&gt;(a-4)&lt;/math&gt; is a factor of &lt;math&gt;12&lt;/math&gt;. The factors are &lt;math&gt;1, 2, 3, 4, 6, 12&lt;/math&gt; which means &lt;math&gt;a&lt;/math&gt; must be one of &lt;math&gt;5, 6, 7, 8, 10, 16&lt;/math&gt;. The only members of this list which are prime are &lt;math&gt;a=5&lt;/math&gt; and &lt;math&gt;a=6&lt;/math&gt;, so the number of solutions is &lt;math&gt;\boxed{\textbf{(C) } 2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=27|num-a=29}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_28&diff=154368 1994 AHSME Problems/Problem 28 2021-05-28T07:49:12Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, how many lines whose &lt;math&gt;x&lt;/math&gt;-intercept is a positive prime number and whose &lt;math&gt;y&lt;/math&gt;-intercept is a positive integer pass through the point &lt;math&gt;(4,3)&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> ==Solution==<br /> <br /> The line with &lt;math&gt;x&lt;/math&gt;-intercept &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;-intercept &lt;math&gt;b&lt;/math&gt; is given by the equation &lt;math&gt;\frac{x}{a} + \frac{y}{b} = 1&lt;/math&gt;. We are told &lt;math&gt;(4,3)&lt;/math&gt; is on the line so <br /> <br /> &lt;cmath&gt;\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, this equation holds only if &lt;math&gt;(a-4)&lt;/math&gt; is a factor of &lt;math&gt;12&lt;/math&gt;. The factors are &lt;math&gt;1, 2, 3, 4, 6, 12&lt;/math&gt; which means &lt;math&gt;a&lt;/math&gt; must be one of &lt;math&gt;5, 6, 7, 8, 10, 16&lt;/math&gt;. The only members of this list which are prime are &lt;math&gt;a=5&lt;/math&gt; and &lt;math&gt;a=6&lt;/math&gt;, so the number of solutions is &lt;math&gt;\boxed{\textbf{(C) } 2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=27|num-a=29}}<br /> {{MAA Notice}}<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=27|num-a=29}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_25&diff=154367 1994 AHSME Problems/Problem 25 2021-05-28T07:40:01Z <p>Logsobolev: /* See Also */</p> <hr /> <div>==Problem==<br /> If &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are non-zero real numbers such that<br /> &lt;cmath&gt; |x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0, &lt;/cmath&gt;<br /> then the integer nearest to &lt;math&gt;x-y&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> ==Solution==<br /> We have two cases to consider: &lt;math&gt;x&lt;/math&gt; is positive or &lt;math&gt;x&lt;/math&gt; is negative. If &lt;math&gt;x&lt;/math&gt; is positive, we have &lt;math&gt;x+y=3&lt;/math&gt; and &lt;math&gt;xy+x^3=0&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;y&lt;/math&gt; in the top equation gives us &lt;math&gt;3-x&lt;/math&gt;. Plugging this in gives us: &lt;math&gt;x^3-x^2+3x=0&lt;/math&gt;. Since we're told &lt;math&gt;x&lt;/math&gt; is not zero, we can divide by &lt;math&gt;x&lt;/math&gt;, giving us: &lt;math&gt;x^2-x+3=0&lt;/math&gt;<br /> <br /> The discriminant of this is &lt;math&gt;(-1)^2-4(1)(3)=-11&lt;/math&gt;, which means the equation has no real solutions. <br /> <br /> We conclude that &lt;math&gt;x&lt;/math&gt; is negative. In this case &lt;math&gt;-x+y=3&lt;/math&gt; and &lt;math&gt;-xy+x^3=0&lt;/math&gt;. Negating the top equation gives us &lt;math&gt;x-y=-3&lt;/math&gt;. We seek &lt;math&gt;x-y&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{(A) -3}&lt;/math&gt;<br /> <br /> -solution by jmania<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=24|num-a=26}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_25&diff=154366 1994 AHSME Problems/Problem 25 2021-05-28T07:39:43Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> If &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are non-zero real numbers such that<br /> &lt;cmath&gt; |x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0, &lt;/cmath&gt;<br /> then the integer nearest to &lt;math&gt;x-y&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> ==Solution==<br /> We have two cases to consider: &lt;math&gt;x&lt;/math&gt; is positive or &lt;math&gt;x&lt;/math&gt; is negative. If &lt;math&gt;x&lt;/math&gt; is positive, we have &lt;math&gt;x+y=3&lt;/math&gt; and &lt;math&gt;xy+x^3=0&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;y&lt;/math&gt; in the top equation gives us &lt;math&gt;3-x&lt;/math&gt;. Plugging this in gives us: &lt;math&gt;x^3-x^2+3x=0&lt;/math&gt;. Since we're told &lt;math&gt;x&lt;/math&gt; is not zero, we can divide by &lt;math&gt;x&lt;/math&gt;, giving us: &lt;math&gt;x^2-x+3=0&lt;/math&gt;<br /> <br /> The discriminant of this is &lt;math&gt;(-1)^2-4(1)(3)=-11&lt;/math&gt;, which means the equation has no real solutions. <br /> <br /> We conclude that &lt;math&gt;x&lt;/math&gt; is negative. In this case &lt;math&gt;-x+y=3&lt;/math&gt; and &lt;math&gt;-xy+x^3=0&lt;/math&gt;. Negating the top equation gives us &lt;math&gt;x-y=-3&lt;/math&gt;. We seek &lt;math&gt;x-y&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{(A) -3}&lt;/math&gt;<br /> <br /> -solution by jmania<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=24|num-a=25}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_24&diff=154365 1994 AHSME Problems/Problem 24 2021-05-28T07:36:44Z <p>Logsobolev: /* See Also */</p> <hr /> <div>==Problem==<br /> A sample consisting of five observations has an arithmetic mean of &lt;math&gt;10&lt;/math&gt; and a median of &lt;math&gt;12&lt;/math&gt;. The smallest value that the range (largest observation minus smallest) can assume for such a sample is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10 &lt;/math&gt;<br /> ==Solution==<br /> The minimum range occurs in the set &lt;math&gt;\{7,7,12,12,12\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C)}\ 5}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_24&diff=154364 1994 AHSME Problems/Problem 24 2021-05-28T07:36:31Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> A sample consisting of five observations has an arithmetic mean of &lt;math&gt;10&lt;/math&gt; and a median of &lt;math&gt;12&lt;/math&gt;. The smallest value that the range (largest observation minus smallest) can assume for such a sample is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10 &lt;/math&gt;<br /> ==Solution==<br /> The minimum range occurs in the set &lt;math&gt;\{7,7,12,12,12\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C)}\ 5}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_23&diff=154363 1994 AHSME Problems/Problem 23 2021-05-28T07:35:20Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at &lt;math&gt;(0,0), (0,3), (3,3), (3,1), (5,1)&lt;/math&gt; and &lt;math&gt;(5,0)&lt;/math&gt;. The slope of the line through the origin that divides the area of this region exactly in half is<br /> &lt;asy&gt;<br /> Label l;<br /> l.p=fontsize(6);<br /> xaxis(&quot;$x$&quot;,0,6,Ticks(l,1.0,0.5),EndArrow);<br /> yaxis(&quot;$y$&quot;,0,4,Ticks(l,1.0,0.5),EndArrow);<br /> draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));&lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ \frac{2}{7} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{3}{4} \qquad\textbf{(E)}\ \frac{7}{9} &lt;/math&gt;<br /> ==Solution==<br /> Let the vertices be &lt;math&gt;A=(0,0),B=(0,3),C=(3,3),D=(3,1),E=(5,1),F=(5,0)&lt;/math&gt;. It is easy to see that the line must pass through &lt;math&gt;CD&lt;/math&gt;. Let the line intersect &lt;math&gt;CD&lt;/math&gt; at the point &lt;math&gt;G=(3,3-x)&lt;/math&gt; (i.e. the point &lt;math&gt;x&lt;/math&gt; units below &lt;math&gt;C&lt;/math&gt;). Since the quadrilateral &lt;math&gt;ABCG&lt;/math&gt; and pentagon &lt;math&gt;GDEFA&lt;/math&gt; must have the same area, we have the equation &lt;math&gt;3\times\frac{1}{2}\times(x+3)=\frac{1}{2}\times3\times(3-x)+2&lt;/math&gt;. This simplifies into &lt;math&gt;3x=2&lt;/math&gt;, or &lt;math&gt;x=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;G=(3,\frac{7}{3})&lt;/math&gt;. Therefore the slope of the line is &lt;math&gt;\boxed{\textbf{(E)}\ \frac{7}{9}}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> <br /> Consider the small rectangle between &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;x=5&lt;/math&gt; with area &lt;math&gt;2&lt;/math&gt;. If we exclude that area then the shape becomes a much simpler, its just a rectangle. In order to offset the exclusion of that area from the shape below the line, we must also exclude a shape with the same area above the line. We want the remainder after excluding both areas to be simple, so exclude a rectangle between &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=3&lt;/math&gt; and &lt;math&gt;y=c&lt;/math&gt; for some unknown &lt;math&gt;c&lt;/math&gt;. For the area to be the same we need &lt;math&gt;(3-c) \cdot 3 = 2&lt;/math&gt;, or &lt;math&gt;c=7/3&lt;/math&gt;. After excluding our two offsetting areas, we're left with a rectangle from &lt;math&gt;x=0&lt;/math&gt; to &lt;math&gt;x=3&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt; to &lt;math&gt;y=c&lt;/math&gt;. The area of this region is clearly bisected by its diagonal line. The slope of this line is &lt;math&gt;c/3 = 7/9&lt;/math&gt; and the answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_22&diff=154362 1994 AHSME Problems/Problem 22 2021-05-28T07:24:45Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 36 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ 630 &lt;/math&gt;<br /> ==Solution==<br /> Since each professor must sit between two students, they cannot be seated in seats &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; (the seats at either end of the row). Hence, each professor has &lt;math&gt;7&lt;/math&gt; seats they can choose from and must be at least &lt;math&gt;1&lt;/math&gt; seat apart.<br /> <br /> This question is equivalent to choosing &lt;math&gt;3&lt;/math&gt; seats from a row of &lt;math&gt;5&lt;/math&gt; seats with no restrictions because we can simply generate a valid arrangement by inserting a seat right after the first and second seat chosen.<br /> <br /> Hence, the answer is &lt;math&gt;\binom{5}{3} \cdot 3! =&lt;/math&gt; &lt;math&gt;60&lt;/math&gt; &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_21&diff=154361 1994 AHSME Problems/Problem 21 2021-05-28T07:23:24Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Find the number of counter examples to the statement:<br /> &lt;cmath&gt;\text{If N is an odd positive integer the sum of whose digits is 4 and none of whose digits is 0, then N is prime}.&quot;&lt;/cmath&gt;<br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4 &lt;/math&gt;<br /> ==Solution==<br /> Since the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; and none of the digits are &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;'s digits must be the elements of the sets &lt;math&gt;\{1,1,1,1\},\{1,1,2\}&lt;/math&gt;, &lt;math&gt;{2,2}&lt;/math&gt;, &lt;math&gt;\{1,3\}&lt;/math&gt;, or &lt;math&gt;\{4\}&lt;/math&gt;. <br /> <br /> In the first case, &lt;math&gt;N = 1111 = 101 \cdot 11&lt;/math&gt; so this is a counter example. <br /> <br /> In the second case, &lt;math&gt;N=112&lt;/math&gt; is excluded for being even. With &lt;math&gt;N=121=11^2&lt;/math&gt; we have a counterexample. We can check &lt;math&gt;N=211&lt;/math&gt; by trial division, and verify it is indeed prime. <br /> <br /> In the third case, &lt;math&gt;N=22&lt;/math&gt; is excluded for being even.<br /> <br /> In the fourth case, both &lt;math&gt;N=13&lt;/math&gt; and &lt;math&gt;N=31&lt;/math&gt; are prime. <br /> <br /> In the last case &lt;math&gt;N=4&lt;/math&gt; is excluded for being even.<br /> <br /> This gives two counterexamples and the answer is &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_20&diff=154360 1994 AHSME Problems/Problem 20 2021-05-28T07:15:03Z <p>Logsobolev: /* See Also */</p> <hr /> <div>==Problem==<br /> Suppose &lt;math&gt;x,y,z&lt;/math&gt; is a geometric sequence with common ratio &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;x \neq y&lt;/math&gt;. If &lt;math&gt;x, 2y, 3z&lt;/math&gt; is an arithmetic sequence, then &lt;math&gt;r&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> ==Solution==<br /> Let &lt;math&gt;y=xr, z=xr^2&lt;/math&gt;. Since &lt;math&gt;x, 2y, 3z&lt;/math&gt; are an arithmetic sequence, there is a common difference and we have &lt;math&gt;2xr-x=3xr^2-2xr&lt;/math&gt;. Dividing through by &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;2r-1=3r^2-2r&lt;/math&gt; or, rearranging, &lt;math&gt;(r-1)(3r-1)=0&lt;/math&gt;. Since we are given &lt;math&gt;x\neq y\implies r\neq 1&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ \frac{1}{3}}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_20&diff=154359 1994 AHSME Problems/Problem 20 2021-05-28T07:14:38Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose &lt;math&gt;x,y,z&lt;/math&gt; is a geometric sequence with common ratio &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;x \neq y&lt;/math&gt;. If &lt;math&gt;x, 2y, 3z&lt;/math&gt; is an arithmetic sequence, then &lt;math&gt;r&lt;/math&gt; is<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{4} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> ==Solution==<br /> Let &lt;math&gt;y=xr, z=xr^2&lt;/math&gt;. Since &lt;math&gt;x, 2y, 3z&lt;/math&gt; are an arithmetic sequence, there is a common difference and we have &lt;math&gt;2xr-x=3xr^2-2xr&lt;/math&gt;. Dividing through by &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;2r-1=3r^2-2r&lt;/math&gt; or, rearranging, &lt;math&gt;(r-1)(3r-1)=0&lt;/math&gt;. Since we are given &lt;math&gt;x\neq y\implies r\neq 1&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(B)}\ \frac{1}{3}}&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_19&diff=154357 1994 AHSME Problems/Problem 19 2021-05-28T07:10:16Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Label one disk &quot;&lt;math&gt;1&lt;/math&gt;&quot;, two disks &quot;&lt;math&gt;2&lt;/math&gt;&quot;, three disks &quot;&lt;math&gt;3&lt;/math&gt;&quot;&lt;math&gt;, ...,&lt;/math&gt; fifty disks &quot;&lt;math&gt;50&lt;/math&gt;&quot;. Put these &lt;math&gt;1+2+3+ \cdots+50=1275&lt;/math&gt; labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501 &lt;/math&gt;<br /> ==Solution==<br /> We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle. <br /> We can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks. <br /> <br /> From disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply &lt;math&gt;41 \cdot 9 = 369&lt;/math&gt;. In total, the maximum number we can pick up without picking up 10 of the same kind is &lt;math&gt;369+45=414&lt;/math&gt;. We need one more disk to guarantee a complete set of 10. Therefore, the answer is &lt;math&gt;\boxed{415}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_18&diff=154356 1994 AHSME Problems/Problem 18 2021-05-28T07:08:38Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in a circle, and &lt;math&gt;\angle B = \angle C = 4\angle A&lt;/math&gt;. If &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are adjacent vertices of a regular polygon of &lt;math&gt;n&lt;/math&gt; sides inscribed in this circle, then &lt;math&gt;n=&lt;/math&gt;<br /> &lt;asy&gt;<br /> draw(Circle((0,0), 5));<br /> draw((0,5)--(3,-4)--(-3,-4)--cycle);<br /> label(&quot;A&quot;, (0,5), N);<br /> label(&quot;B&quot;, (-3,-4), SW);<br /> label(&quot;C&quot;, (3,-4), SE);<br /> dot((0,5));<br /> dot((3,-4));<br /> dot((-3,-4));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> ==Solution==<br /> We solve for &lt;math&gt;\angle A&lt;/math&gt; as follows: &lt;cmath&gt;4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.&lt;/cmath&gt; That means that minor arc &lt;math&gt;\widehat{BC}&lt;/math&gt; has measure &lt;math&gt;40^\circ&lt;/math&gt;. We can fit a maximum of &lt;math&gt;\frac{360}{40}=\boxed{\textbf{(C) }9}&lt;/math&gt; of these arcs in the circle.<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_17&diff=154355 1994 AHSME Problems/Problem 17 2021-05-28T07:07:30Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> An &lt;math&gt;8&lt;/math&gt; by &lt;math&gt;2\sqrt{2}&lt;/math&gt; rectangle has the same center as a circle of radius &lt;math&gt;2&lt;/math&gt;. The area of the region common to both the rectangle and the circle is<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2 &lt;/math&gt;<br /> ==Solution==<br /> &lt;asy&gt;<br /> import cse5;<br /> import olympiad;<br /> real s=2*sqrt(2);<br /> pair A=(0,0),B=(0,s),C=(8,s),D=(8,0),O=(4,sqrt(2)),X;<br /> D(A--B--C--D--cycle);<br /> D(CR(O,2));<br /> pair[] P;<br /> P=IPs(CR(O,2),box(A,C));<br /> for(int i=0; i&lt;4; i=i+1)<br /> {<br /> D(O--P[i],black);<br /> }<br /> X=foot(O,B,C);<br /> D(O--X);<br /> D(rightanglemark(O,X,C));<br /> D(O);<br /> D(MP(&quot;O&quot;,O,S));&lt;/asy&gt;<br /> <br /> We draw the diagram above. Dropping an altitude from the center of the circle to the top side of the rectangle, yields a segment of length &lt;math&gt;\sqrt{2}&lt;/math&gt;. Since the hypotenuse is the radius of the circle, it has length 2 and by Pythagoras the other leg of the triangle also has length &lt;math&gt;\sqrt{2}&lt;/math&gt;. We deduce that the two triangles formed by two radii of circle &lt;math&gt;O&lt;/math&gt; and the segment of the rectangle are 45-45-90 triangles. <br /> <br /> The overlapping area consists of two sectors with central angle 90 and four 45-45-90 triangles with base &lt;math&gt;\sqrt{2}&lt;/math&gt; and height &lt;math&gt;\sqrt{2}&lt;/math&gt;. The area of the circle is &lt;math&gt;4\pi&lt;/math&gt; so the area of the sectors is &lt;math&gt;2 \cdot 4\pi/4 = 2\pi&lt;/math&gt;. The area of the triangles is &lt;math&gt;4\cdot (\sqrt{2})^2 /2 = 4&lt;/math&gt;. The combined area is &lt;cmath&gt;\boxed{\textbf{(D) }2\pi+4.}&lt;/cmath&gt;<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_16&diff=154354 1994 AHSME Problems/Problem 16 2021-05-28T06:54:48Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71 &lt;/math&gt;<br /> ==Solution==<br /> Let &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be the number of red and blue marbles originally in the bag respectively. After &lt;math&gt;1&lt;/math&gt; red marble is removed, there are &lt;math&gt;r+b-1&lt;/math&gt; marbles left in the bag and &lt;math&gt;r-1&lt;/math&gt; red marbles left. So &lt;cmath&gt;\frac{r-1}{r+b-1}=\frac{1}{7}.&lt;/cmath&gt; When &lt;math&gt;2&lt;/math&gt; blue marbles are removed, there are &lt;math&gt;r&lt;/math&gt; red marbles and &lt;math&gt;r+b-2&lt;/math&gt; total marbles left in the bag. So &lt;cmath&gt;\frac{r}{r+b-2}=\frac{1}{5}.&lt;/cmath&gt; Cross multiplying for each yields &lt;cmath&gt;\begin{align*}7r-7=r+b-1&amp;\implies 7r-6=r+b\\ 5r=r+b-2&amp;\implies 5r+2=r+b.\end{align*}&lt;/cmath&gt; We can equate each of these expressions to yields &lt;cmath&gt;7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.&lt;/cmath&gt; Therefore, the total number of marbles is &lt;cmath&gt;r+b=4+18=\boxed{\textbf{(B) }22.}&lt;/cmath&gt;<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_15&diff=154353 1994 AHSME Problems/Problem 15 2021-05-28T06:53:40Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> For how many &lt;math&gt;n&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, ..., 100 \}&lt;/math&gt; is the tens digit of &lt;math&gt;n^2&lt;/math&gt; odd?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50 &lt;/math&gt;<br /> ==Solution==<br /> Let &lt;math&gt;n=10a+b&lt;/math&gt;. So &lt;math&gt;n^2=(10a+b)^2=100a^2+20ab+b^2&lt;/math&gt;. The term &lt;math&gt;100a^2&lt;/math&gt; only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term &lt;math&gt;20ab&lt;/math&gt; only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of &lt;math&gt;2ab&lt;/math&gt; which is even. So we see this term also does not affect whether the tens digit is odd. This means only &lt;math&gt;b^2&lt;/math&gt; can affect whether the tens digit is odd. We can quickly check &lt;math&gt;1^2=1, \dots, 9^2=81&lt;/math&gt; and discover only for &lt;math&gt;b=4&lt;/math&gt; or &lt;math&gt;b=6&lt;/math&gt; does &lt;math&gt;b^2&lt;/math&gt; have an odd tens digit. The total number of positive integers less than or equal to &lt;math&gt;100&lt;/math&gt; that have &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt; as the units digit is &lt;cmath&gt;10\times 2=\boxed{\textbf{(B) }20.}&lt;/cmath&gt;<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_14&diff=154352 1994 AHSME Problems/Problem 14 2021-05-28T06:45:58Z <p>Logsobolev: /* Brief Introduction */</p> <hr /> <div>==Problem==<br /> Find the sum of the arithmetic series<br /> &lt;cmath&gt; 20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40 &lt;/cmath&gt;<br /> &lt;math&gt; \textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000 &lt;/math&gt;<br /> ==Solution==<br /> <br /> <br /> <br /> ===Brief Introduction===<br /> <br /> For those that do not know the formula, the sum of an arithmetic series with first term &lt;math&gt;a_1&lt;/math&gt;, last term &lt;math&gt;a_n&lt;/math&gt; as &lt;math&gt;n&lt;/math&gt; terms, is &lt;cmath&gt;S = \frac{n(a_1+a_n)}{2}.&lt;/cmath&gt; We can prove this as follows:<br /> <br /> Let &lt;math&gt;d&lt;/math&gt; be the common difference between terms of our series and let &lt;math&gt;n&lt;/math&gt; be the number of terms in our series. Let &lt;math&gt;a_1&lt;/math&gt; be the first term. Our series is &lt;cmath&gt;a_1,~a_1+d,~a_1+2d,\dots,a_1+(n-1)d.&lt;/cmath&gt; Note that we have &lt;math&gt;n-1&lt;/math&gt; in the last term because &lt;math&gt;a_1&lt;/math&gt; is a term. Let &lt;math&gt;S&lt;/math&gt; be our sum such that &lt;cmath&gt;S=a_1+(a_1+d)+\dots+(a_1+(n-1)d).&lt;/cmath&gt; We can rewrite our sums as &lt;cmath&gt;S=(a_1+(n-1)d)+(a_1+(n-2)d)+\dots+(a_1+d)+a_1.&lt;/cmath&gt; Adding these two sums together essentially creates &lt;math&gt;n&lt;/math&gt; pairs of &lt;math&gt;a_1+(a_1+(n-1)d)&lt;/math&gt; as shown below: &lt;cmath&gt;2S=n(a_1+(a_1+(n-1)d))\implies S=\frac{n(a_1+a_n)}{2}&lt;/cmath&gt; We use &lt;math&gt;a_n&lt;/math&gt; in place of &lt;math&gt;a_1+(n-1)d&lt;/math&gt; to represent the last term.<br /> <br /> <br /> ----<br /> <br /> ===Solving===<br /> <br /> Our first term is &lt;math&gt;20&lt;/math&gt; and our last term is &lt;math&gt;40&lt;/math&gt;. To find the number of terms, &lt;math&gt;n&lt;/math&gt;, we note that the common difference between each term is &lt;math&gt;\frac{1}{5}&lt;/math&gt;. So we have &lt;cmath&gt;20+\frac{1}{5}(n-1)=40\implies n-1=100\implies n=101.&lt;/cmath&gt; Using our formula, our sum is &lt;cmath&gt;101\left(\frac{20+40}{2}\right)=101\times 30=\boxed{\textbf{(B) }3030.}&lt;/cmath&gt;<br /> <br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_7&diff=154351 1994 AHSME Problems/Problem 7 2021-05-28T06:36:25Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> Squares &lt;math&gt;ABCD&lt;/math&gt; and &lt;math&gt;EFGH&lt;/math&gt; are congruent, &lt;math&gt;AB=10&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; is the center of square &lt;math&gt;ABCD&lt;/math&gt;. The area of the region in the plane covered by these squares is<br /> &lt;asy&gt;<br /> draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);<br /> draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle);<br /> label(&quot;A&quot;, (0,0), W);<br /> label(&quot;B&quot;, (10,0), E);<br /> label(&quot;C&quot;, (10,10), NE);<br /> label(&quot;D&quot;, (0,10), NW);<br /> label(&quot;G&quot;, (5,5), N);<br /> label(&quot;F&quot;, (12,-2), E);<br /> label(&quot;E&quot;, (5,-9), S);<br /> label(&quot;H&quot;, (-2,-2), W);<br /> dot((-2,-2));<br /> dot((5,-9));<br /> dot((12,-2));<br /> dot((0,0));<br /> dot((10,0));<br /> dot((10,10));<br /> dot((0,10));<br /> dot((5,5));<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A)}\ 75 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 175 &lt;/math&gt;<br /> ==Solution==<br /> The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of &lt;math&gt;\triangle ABG&lt;/math&gt; and subtract this from &lt;math&gt;200&lt;/math&gt;, the total area of the two squares.<br /> <br /> <br /> Since &lt;math&gt;G&lt;/math&gt; is the center of &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;BG&lt;/math&gt; is half of the diagonal of the square. The diagonal of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;10\sqrt{2}&lt;/math&gt; so &lt;math&gt;BG=5\sqrt{2}&lt;/math&gt;. Since &lt;math&gt;EFGH&lt;/math&gt; is a square, &lt;math&gt;\angle G=90^\circ&lt;/math&gt;. So &lt;math&gt;\triangle ABG&lt;/math&gt; is an isosceles right triangle. Its area is &lt;math&gt;\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25&lt;/math&gt;. Therefore, the area of the region is &lt;math&gt;200-25=\boxed{\textbf{(E) }175.}&lt;/math&gt;<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> == Solution 2 ==<br /> <br /> `Note the overlap area &lt;math&gt;\triangle ABG&lt;/math&gt; is exactly &lt;math&gt;1/4&lt;/math&gt; the area of &lt;math&gt;ABCD&lt;/math&gt;. So the total area is &lt;math&gt;100 + 100 - 100/4 = 175&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_4&diff=154350 1994 AHSME Problems/Problem 4 2021-05-28T06:31:21Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> In the &lt;math&gt;xy&lt;/math&gt;-plane, the segment with endpoints &lt;math&gt;(-5,0)&lt;/math&gt; and &lt;math&gt;(25,0)&lt;/math&gt; is the diameter of a circle. If the point &lt;math&gt;(x,15)&lt;/math&gt; is on the circle, then &lt;math&gt;x=&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> ==Solution==<br /> We see that the center of this circle is at &lt;math&gt;\left(\frac{-5+25}{2},0\right)=(10,0)&lt;/math&gt;. The radius is &lt;math&gt;\frac{30}{2}=15&lt;/math&gt;. So the equation of this circle is &lt;cmath&gt;(x-10)^2+y^2=225.&lt;/cmath&gt; Substituting &lt;math&gt;y=15&lt;/math&gt; yields &lt;math&gt;(x-10)^2=0&lt;/math&gt; so &lt;math&gt;x=\boxed{\textbf{(A) }10}&lt;/math&gt;.<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> == Solution 2 ==<br /> <br /> The diameter of the circle is &lt;math&gt;25- (-5)=30&lt;/math&gt; so the radius is &lt;math&gt;15&lt;/math&gt; and the center of the circle is on the &lt;math&gt;x&lt;/math&gt;-axis at &lt;math&gt;x=10&lt;/math&gt;. The only point with &lt;math&gt;y=15&lt;/math&gt; must be the point exactly above the center.<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_2&diff=154349 1994 AHSME Problems/Problem 2 2021-05-28T06:26:32Z <p>Logsobolev: /* Solution */</p> <hr /> <div>==Problem==<br /> A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle?<br /> &lt;asy&gt;<br /> draw((0,0)--(10,0)--(10,7)--(0,7)--cycle);<br /> draw((0,5)--(10,5));<br /> draw((3,0)--(3,7));<br /> label(&quot;6&quot;, (1.5,6));<br /> label(&quot;?&quot;, (1.5,2.5));<br /> label(&quot;14&quot;, (6.5,6));<br /> label(&quot;35&quot;, (6.5,2.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> ==Solution==<br /> &lt;asy&gt;<br /> pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7);<br /> path BG=shift(0,-0.5)*(B--G);<br /> path BF=shift(0.5,0)*(B--F);<br /> path FC=shift(0.5,0)*(F--C);<br /> path DH=shift(0,0.5)*(D--H);<br /> draw(A--B--C--D--cycle);<br /> draw(EE--F);<br /> draw(G--H);<br /> draw(BG,L=Label(&quot;$7$&quot;,position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red);<br /> draw(BF,L=Label(&quot;$5$&quot;,position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red);<br /> draw(FC,L=Label(&quot;$2$&quot;,position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red);<br /> draw(DH,L=Label(&quot;$3$&quot;,position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red);<br /> label(&quot;$6$&quot;, (1.5,6));<br /> label(&quot;$15$&quot;, (1.5,2.5),blue);<br /> label(&quot;$14$&quot;, (6.5,6));<br /> label(&quot;$35$&quot;, (6.5,2.5));<br /> &lt;/asy&gt;<br /> <br /> We can easily see the dimensions of each small rectangle. So the area of the last rectangle is &lt;math&gt;3\times 5=\boxed{\textbf{(B) }15}&lt;/math&gt;.<br /> <br /> --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&amp;u=200685 TheMaskedMagician]<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt;<br /> pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7);<br /> path CH=C--H;<br /> path BF=B--F;<br /> path FC=F--C;<br /> path DH=D--H;<br /> draw(A--B--C--D--cycle);<br /> draw(EE--F);<br /> draw(G--H);<br /> draw(CH,L=Label(&quot;$b$&quot;,position=MidPoint,align=(0,1)));<br /> draw(BF,L=Label(&quot;$y$&quot;,position=MidPoint,align=(1,0)));<br /> draw(FC,L=Label(&quot;$x$&quot;,position=MidPoint,align=(1,0)));<br /> draw(DH,L=Label(&quot;$a$&quot;,position=MidPoint,align=(0,1)));<br /> label(&quot;$6$&quot;, (1.5,6));<br /> label(&quot;$14$&quot;, (6.5,6));<br /> label(&quot;$35$&quot;, (6.5,2.5));<br /> &lt;/asy&gt;<br /> <br /> In case its not immediately obvious from inspection what the dimensions of the small rectangles are, we can work it out. We know &lt;math&gt;ax&lt;/math&gt; and &lt;math&gt;bx&lt;/math&gt; and &lt;math&gt;by&lt;/math&gt; and we want to know &lt;math&gt;ay&lt;/math&gt;. We can compute it as follows &lt;math&gt;ay = \frac{(ax)(by)}{bx} = \frac{ 6\cdot 35 }{14} = 15&lt;/math&gt; and the answer is &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AHSME box|year=1994|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_22&diff=154346 1993 AHSME Problems/Problem 22 2021-05-28T04:28:36Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Twenty cubical blocks are arranged as shown. First, 10 are arranged in a triangular pattern; then a layer of 6, arranged in a triangular pattern, is centered on the 10; then a layer of 3, arranged in a triangular pattern, is centered on the 6; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered 1 through 10 in some order. Each block in layers 2,3 and 4 is assigned the number which is the sum of numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.<br /> <br /> &lt;math&gt;\text{(A) } 55\quad<br /> \text{(B) } 83\quad<br /> \text{(C) } 114\quad<br /> \text{(D) } 137\quad<br /> \text{(E) } 144&lt;/math&gt;<br /> <br /> == Solution ==<br /> The value assigned at the top is the weighted sum of the values in the cubes of a given level. The weight for a given cube is the sum of the weights of the blocks which touch it above, since the number in a cube will get incorporated into the overall sum via each of those blocks, according to the weight of each block. Thus, if we think about how the weights propagate down the block pyramid, its sort of like a three dimensional Pascal's triangle.<br /> <br /> The first layer is &lt;math&gt;1&lt;/math&gt;<br /> <br /> The second layer is <br /> <br /> &lt;math&gt;| \, \, 1 \\ |1 \, 1&lt;/math&gt;<br /> <br /> The third layer is <br /> <br /> &lt;math&gt;| \,\,\,\, 1 \\| \,\, 2 \, 2 \\| 1 \,2 \, 1&lt;/math&gt;<br /> <br /> The fourth layer is <br /> <br /> &lt;math&gt;| \,\,\,\,\,\, 1 \\| \,\,\,\, 3 \, 3 \\| \,\, 3 \,6 \, 3 \\| 1\, 3\, 3\, 1&lt;/math&gt;<br /> <br /> The top level sum is minimized if we associate the block with weight 6 with the smallest number 1, the blocks with weight 1 with the largest numbers 8, 9 and 10, and the rest of the blocks with weight 3 with the rest of the numbers. The sum is &lt;math&gt;6\cdot 1 + 3 \cdot (2+3+4+5+6+7 ) + 1\cdot (8+9+10) = 114&lt;/math&gt; so the answer is &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=21|num-a=23}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_26&diff=154344 1993 AHSME Problems/Problem 26 2021-05-28T03:08:57Z <p>Logsobolev: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Find the largest positive value attained by the function<br /> &lt;math&gt;f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; a real number.<br /> <br /> <br /> &lt;math&gt;\text{(A) } \sqrt{7}-1\quad<br /> \text{(B) } 3\quad<br /> \text{(C) } 2\sqrt{3}\quad<br /> \text{(D) } 4\quad<br /> \text{(E) } \sqrt{55}-\sqrt{5}&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can rewrite the function as &lt;math&gt;f(x) = \sqrt{x (8 - x)} - \sqrt{(x - 6) (8 - x)}&lt;/math&gt; and then factor it to get &lt;math&gt;f(x) = \sqrt{8 - x} \left(\sqrt{x} - \sqrt{x - 6}\right)&lt;/math&gt;. From the expressions under the square roots, it is clear that &lt;math&gt;f(x)&lt;/math&gt; is only defined on the interval &lt;math&gt;[6, 8]&lt;/math&gt;.<br /> <br /> The &lt;math&gt;\sqrt{8 - x}&lt;/math&gt; factor is decreasing on the interval. The behavior of the &lt;math&gt;\sqrt{x} - \sqrt{x - 6}&lt;/math&gt; factor is not immediately clear. But rationalizing the numerator, we find that &lt;math&gt;\sqrt{x} - \sqrt{x - 6} = \frac{6}{\sqrt{x} + \sqrt{x - 6}}&lt;/math&gt;, which is monotonically decreasing. Since both factors are always positive, &lt;math&gt;f(x)&lt;/math&gt; is also positive. Therefore, &lt;math&gt;f(x)&lt;/math&gt; is decreasing on &lt;math&gt;[6, 8]&lt;/math&gt;, and the maximum value occurs at &lt;math&gt;x = 6&lt;/math&gt;. Plugging in, we find that the maximum value is &lt;math&gt;\boxed{\text{(C) } 2\sqrt{3}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Note the form of the function is &lt;math&gt;f(x) = \sqrt{ p(x)} - \sqrt{q(x)}&lt;/math&gt; where &lt;math&gt;p(x)&lt;/math&gt; and &lt;math&gt;q(x)&lt;/math&gt; each describe a parabola. Factoring we have &lt;math&gt;p(x) = x(8-x)&lt;/math&gt; and &lt;math&gt;q(x) = (x-6)(8-x)&lt;/math&gt;. <br /> <br /> The first term &lt;math&gt;\sqrt{p(x)}&lt;/math&gt; is defined only when &lt;math&gt;p(x)\geq 0&lt;/math&gt; which is the interval &lt;math&gt;[0,8]&lt;/math&gt; and the second term &lt;math&gt;\sqrt{q(x)}&lt;/math&gt; is defined only when &lt;math&gt;q(x)\geq 0&lt;/math&gt; which is on the interval &lt;math&gt;[6,8]&lt;/math&gt;, so the domain of &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;[0,8] \cap [6,8] = [6,8]&lt;/math&gt;. <br /> <br /> Now &lt;math&gt;p(x)&lt;/math&gt; peaks at the midpoint of its roots at &lt;math&gt;x=4&lt;/math&gt; and it decreases to 0 at &lt;math&gt;x=8&lt;/math&gt;. Thus, &lt;math&gt;p(x)&lt;/math&gt; is decreasing over the entire domain of &lt;math&gt;f(x)&lt;/math&gt; and it obtains its maximum value at the left boundary &lt;math&gt;x=6&lt;/math&gt; and &lt;math&gt;\sqrt{p(x)}&lt;/math&gt; does as well. On the other hand &lt;math&gt;q(x)&lt;/math&gt; obtains its minimum value of &lt;math&gt;q(x)=0&lt;/math&gt; at the left boundary &lt;math&gt;x=6&lt;/math&gt; and &lt;math&gt;\sqrt{q(x)}&lt;/math&gt; does as well. Therefore &lt;math&gt;\sqrt{p(x)}-\sqrt{q(x)}&lt;/math&gt; is maximized at &lt;math&gt;x=6&lt;/math&gt;. (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest). <br /> <br /> The value at &lt;math&gt;x=6&lt;/math&gt; is &lt;math&gt;\sqrt{ 6\cdot 2 } = 2\sqrt{3}&lt;/math&gt; and the answer is &lt;math&gt;\fbox{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=25|num-a=27}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_6&diff=154343 1993 AHSME Problems/Problem 6 2021-05-28T03:01:44Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;math&gt;\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } \sqrt{2}\quad<br /> \text{(B) } 16\quad<br /> \text{(C) } 32\quad<br /> \text{(D) } (12)^{\tfrac{2}{3}}\quad<br /> \text{(E) } 512.5&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\sqrt{\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(1+2^{10})}} = \sqrt{2^8} = 2^4&lt;/math&gt;<br /> <br /> &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=5|num-a=7}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_26&diff=154342 1993 AHSME Problems/Problem 26 2021-05-28T02:53:44Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find the largest positive value attained by the function<br /> &lt;math&gt;f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; a real number.<br /> <br /> <br /> &lt;math&gt;\text{(A) } \sqrt{7}-1\quad<br /> \text{(B) } 3\quad<br /> \text{(C) } 2\sqrt{3}\quad<br /> \text{(D) } 4\quad<br /> \text{(E) } \sqrt{55}-\sqrt{5}&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can rewrite the function as &lt;math&gt;f(x) = \sqrt{x (8 - x)} - \sqrt{(x - 6) (8 - x)}&lt;/math&gt; and then factor it to get &lt;math&gt;f(x) = \sqrt{8 - x} \left(\sqrt{x} - \sqrt{x - 6}\right)&lt;/math&gt;. From the expressions under the square roots, it is clear that &lt;math&gt;f(x)&lt;/math&gt; is only defined on the interval &lt;math&gt;[6, 8]&lt;/math&gt;.<br /> <br /> The &lt;math&gt;\sqrt{8 - x}&lt;/math&gt; factor is decreasing on the interval. The behavior of the &lt;math&gt;\sqrt{x} - \sqrt{x - 6}&lt;/math&gt; factor is not immediately clear. But rationalizing the numerator, we find that &lt;math&gt;\sqrt{x} - \sqrt{x - 6} = \frac{6}{\sqrt{x} + \sqrt{x - 6}}&lt;/math&gt;, which is monotonically decreasing. Since both factors are always positive, &lt;math&gt;f(x)&lt;/math&gt; is also positive. Therefore, &lt;math&gt;f(x)&lt;/math&gt; is decreasing on &lt;math&gt;[6, 8]&lt;/math&gt;, and the maximum value occurs at &lt;math&gt;x = 6&lt;/math&gt;. Plugging in, we find that the maximum value is &lt;math&gt;\boxed{\text{(C) } 2\sqrt{3}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Note the form of the function is &lt;math&gt;f(x) = \sqrt{ p(x)} - \sqrt{q(x)}&lt;/math&gt; where &lt;math&gt;p(x)&lt;/math&gt; and &lt;math&gt;q(x)&lt;/math&gt; each describe a parabola. Factoring we have &lt;math&gt;p(x) = x(8-x)&lt;/math&gt; and &lt;math&gt;q(x) = (x-6)(8-x)&lt;/math&gt;. <br /> <br /> The first term &lt;math&gt;\sqrt{p(x)}&lt;/math&gt; is defined only when &lt;math&gt;p(x)\geq 0&lt;/math&gt; which is the interval &lt;math&gt;[0,8]&lt;/math&gt; and the second term &lt;math&gt;\sqrt{q(x)}&lt;/math&gt; is defined only when &lt;math&gt;q(x)\geq 0&lt;/math&gt; which is on the interval &lt;math&gt;[6,8]&lt;/math&gt;, so the domain of &lt;math&gt;f(x)&lt;/math&gt; is &lt;math&gt;[0,8] \cap [6,8] = [6,8]&lt;/math&gt;. <br /> <br /> Now &lt;math&gt;p(x)&lt;/math&gt; peaks at the midpoint of its roots at &lt;math&gt;x=4&lt;/math&gt; and it decreases to 0 at &lt;math&gt;x=8&lt;/math&gt;. Thus, &lt;math&gt;p(x)&lt;/math&gt; is decreasing over the entire domain of &lt;math&gt;f(x)&lt;/math&gt; and it obtains its maximum value over the domain of &lt;math&gt;f(x)&lt;/math&gt; at the left boundary &lt;math&gt;x=6&lt;/math&gt;, and &lt;math&gt;\sqrt{p(x)}&lt;/math&gt; does as well. On the other hand &lt;math&gt;q(x)&lt;/math&gt; obtains its minimum value of &lt;math&gt;q(x)=0&lt;/math&gt; at the left boundary &lt;math&gt;x=6&lt;/math&gt;, and &lt;math&gt;\sqrt{q(x)}&lt;/math&gt; does as well. Therefore &lt;math&gt;\sqrt{p(x)}-\sqrt{q(x)}&lt;/math&gt; is maximized at &lt;math&gt;x=6&lt;/math&gt;. (If this seems a little unmotivated, a quick sketch of the two parabolic-like curves makes it clear where the distance between them is greatest). <br /> <br /> The value at &lt;math&gt;x=6&lt;/math&gt; is &lt;math&gt;\sqrt{ 6\cdot 2 } = 2\sqrt{3}&lt;/math&gt; and the answer is &lt;math&gt;\fbox{D}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=25|num-a=27}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_20&diff=154341 1993 AHSME Problems/Problem 20 2021-05-28T02:28:23Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Consider the equation &lt;math&gt;10z^2-3iz-k=0&lt;/math&gt;, where &lt;math&gt;z&lt;/math&gt; is a complex variable and &lt;math&gt;i^2=-1&lt;/math&gt;. Which of the following statements is true?<br /> <br /> &lt;math&gt;\text{(A) For all positive real numbers k, both roots are pure imaginary} \quad\\<br /> \text{(B) For all negative real numbers k, both roots are pure imaginary} \quad\\<br /> \text{(C) For all pure imaginary numbers k, both roots are real and rational} \quad\\<br /> \text{(D) For all pure imaginary numbers k, both roots are real and irrational} \quad\\<br /> \text{(E) For all complex numbers k, neither root is real} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; denote the roots of the polynomial. Then &lt;math&gt;r_1 + r_2 = 3i&lt;/math&gt; is pure imaginary, so &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; have offsetting real parts. Write &lt;math&gt;r_1 = a + bi&lt;/math&gt; and &lt;math&gt;r_2 = -a + ci&lt;/math&gt;. <br /> <br /> Now &lt;math&gt;-k = r_1 r_2 = -a^2 -bc + a(c-b)i&lt;/math&gt;. In the case that &lt;math&gt;k&lt;/math&gt; is real, then &lt;math&gt;a(c-b)=0&lt;/math&gt; so either &lt;math&gt;a=0&lt;/math&gt; or that &lt;math&gt;b=c&lt;/math&gt;. In the first case, the roots are pure imaginary and in the second case we have &lt;math&gt;k = a^2+b^2&lt;/math&gt;, a positive number.<br /> <br /> We can therefore conclude that if &lt;math&gt;k&lt;/math&gt; is real and negative, it must be the first case and the roots are pure imaginary.<br /> <br /> It's possible to rule out the other cases by reasoning through the cases, but this is enough to show that &lt;math&gt;\fbox{B}&lt;/math&gt; is true.<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=19|num-a=21}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_16&diff=154340 1993 AHSME Problems/Problem 16 2021-05-28T02:18:02Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Consider the non-decreasing sequence of positive integers<br /> &lt;cmath&gt;1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots&lt;/cmath&gt;<br /> in which the &lt;math&gt;n^{th}&lt;/math&gt; positive integer appears &lt;math&gt;n&lt;/math&gt; times. The remainder when the &lt;math&gt;1993^{rd} &lt;/math&gt; term is divided by &lt;math&gt;5&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 2\quad<br /> \text{(D) } 3\quad<br /> \text{(E) } 4&lt;/math&gt;<br /> <br /> == Solution ==<br /> The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of &lt;math&gt;n&lt;/math&gt;'s ends at position &lt;math&gt;1+2+\dots+n&lt;/math&gt;.<br /> <br /> Therefore we want to find the smallest integer &lt;math&gt;n&lt;/math&gt; that satisfies &lt;math&gt;\frac{n(n+1)}{2}\geq 1993&lt;/math&gt;.<br /> <br /> By trial and error, the value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;63&lt;/math&gt;, and &lt;math&gt;63 \div 5&lt;/math&gt; has a remainder of &lt;math&gt;3&lt;/math&gt;.<br /> <br /> &lt;math&gt;\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=15|num-a=16}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_16&diff=154339 1993 AHSME Problems/Problem 16 2021-05-28T02:15:37Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Consider the non-decreasing sequence of positive integers<br /> &lt;cmath&gt;1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots&lt;/cmath&gt;<br /> in which the &lt;math&gt;n^{th}&lt;/math&gt; positive integer appears &lt;math&gt;n&lt;/math&gt; times. The remainder when the &lt;math&gt;1993^{rd} &lt;/math&gt; term is divided by &lt;math&gt;5&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 2\quad<br /> \text{(D) } 3\quad<br /> \text{(E) } 4&lt;/math&gt;<br /> <br /> == Solution ==<br /> You want to find the largest integer &lt;math&gt;n&lt;/math&gt; that satisfies &lt;math&gt;\frac{n(n+1)}{2}&lt;1993&lt;/math&gt;.<br /> <br /> By trial and error, the value of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;62&lt;/math&gt;. Therefore, the next value of the sequence is &lt;math&gt;63&lt;/math&gt;, and &lt;math&gt;63 \div 5&lt;/math&gt; has a remainder of &lt;math&gt;3&lt;/math&gt;.<br /> <br /> &lt;math&gt;\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=15|num-a=16}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_11&diff=154338 1993 AHSME Problems/Problem 11 2021-05-28T02:11:54Z <p>Logsobolev: /* Problem */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;\log_2(\log_2(\log_2(x)))=2&lt;/math&gt;, then how many digits are in the base-ten representation for x?<br /> <br /> &lt;math&gt;\text{(A) } 5\quad<br /> \text{(B) } 7\quad<br /> \text{(C) } 9\quad<br /> \text{(D) } 11\quad<br /> \text{(E) } 13&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Taking successive exponentials &lt;math&gt;\log_2(\log_2(x)) = 2^2 = 4&lt;/math&gt; and &lt;math&gt;\log_2(x) = 2^4=16&lt;/math&gt; and &lt;math&gt;x = 2^{16}&lt;/math&gt;. Now &lt;math&gt;2^{10} = 1024 \approx 10^3&lt;/math&gt; and &lt;math&gt;2^6 = 64&lt;/math&gt; so we can approximate &lt;math&gt;2^{16} \approx 64000&lt;/math&gt; which has 5 digits. In general, &lt;math&gt;2^n&lt;/math&gt; has approximately &lt;math&gt;n/3&lt;/math&gt; digits.<br /> <br /> &lt;math&gt;\fbox{A}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=10|num-a=12}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_11&diff=154337 1993 AHSME Problems/Problem 11 2021-05-28T02:10:26Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;log_2(log_2(log_2(x)))=2&lt;/math&gt;, then how many digits are in the base-ten representation for x?<br /> <br /> &lt;math&gt;\text{(A) } 5\quad<br /> \text{(B) } 7\quad<br /> \text{(C) } 9\quad<br /> \text{(D) } 11\quad<br /> \text{(E) } 13&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Taking successive exponentials &lt;math&gt;\log_2(\log_2(x)) = 2^2 = 4&lt;/math&gt; and &lt;math&gt;\log_2(x) = 2^4=16&lt;/math&gt; and &lt;math&gt;x = 2^{16}&lt;/math&gt;. Now &lt;math&gt;2^{10} = 1024 \approx 10^3&lt;/math&gt; and &lt;math&gt;2^6 = 64&lt;/math&gt; so we can approximate &lt;math&gt;2^{16} \approx 64000&lt;/math&gt; which has 5 digits. In general, &lt;math&gt;2^n&lt;/math&gt; has approximately &lt;math&gt;n/3&lt;/math&gt; digits.<br /> <br /> &lt;math&gt;\fbox{A}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=10|num-a=12}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_10&diff=154336 1993 AHSME Problems/Problem 10 2021-05-28T02:00:04Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; be the number that results when both the base and the exponent of &lt;math&gt;a^b&lt;/math&gt; are tripled, where &lt;math&gt;a,b&gt;0&lt;/math&gt;. If &lt;math&gt;r&lt;/math&gt; equals the product of &lt;math&gt;a^b&lt;/math&gt; and &lt;math&gt;x^b&lt;/math&gt; where &lt;math&gt;x&gt;0&lt;/math&gt;, then &lt;math&gt;x=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 3\quad<br /> \text{(B) } 3a^2\quad<br /> \text{(C) } 27a^2\quad<br /> \text{(D) } 2a^{3b}\quad<br /> \text{(E) } 3a^{2b}&lt;/math&gt;<br /> <br /> == Solution ==<br /> We have &lt;math&gt;r=(3a)^{3b}&lt;/math&gt;<br /> <br /> From this we have the equation &lt;math&gt;(3a)^{3b}=a^bx^b&lt;/math&gt;<br /> <br /> Raising both sides to the &lt;math&gt;\frac{1}{b}&lt;/math&gt; power we get that &lt;math&gt;27a^3=ax&lt;/math&gt; or &lt;math&gt;x=27a^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\fbox{C}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=9|num-a=11}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_9&diff=154335 1993 AHSME Problems/Problem 9 2021-05-28T01:59:05Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Country &lt;math&gt;A&lt;/math&gt; has &lt;math&gt;c\%&lt;/math&gt; of the world's population and &lt;math&gt;d\%&lt;/math&gt; of the worlds wealth. Country &lt;math&gt;B&lt;/math&gt; has &lt;math&gt;e\%&lt;/math&gt; of the world's population and &lt;math&gt;f\%&lt;/math&gt; of its wealth. Assume that the citizens of &lt;math&gt;A&lt;/math&gt; share the wealth of &lt;math&gt;A&lt;/math&gt; equally,and assume that those of &lt;math&gt;B&lt;/math&gt; share the wealth of &lt;math&gt;B&lt;/math&gt; equally. Find the ratio of the wealth of a citizen of &lt;math&gt;A&lt;/math&gt; to the wealth of a citizen of &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{(A) } \frac{cd}{ef}\quad<br /> \text{(B) } \frac{ce}{ef}\quad<br /> \text{(C) } \frac{cf}{de}\quad<br /> \text{(D) } \frac{de}{cf}\quad<br /> \text{(E) } \frac{df}{ce}&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;W&lt;/math&gt; be the wealth of the world and &lt;math&gt;P&lt;/math&gt; be the population of the world. Hence the wealth of each citizen of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;w_A = \frac{0.01d W}{0.01cP}=\frac{dW}{cP}&lt;/math&gt;. Similarly the wealth of each citizen of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;w_B =\frac{eW}{fP}&lt;/math&gt;. We divide &lt;math&gt;\frac{w_A}{w_B} = \frac{de}{cf}&lt;/math&gt; and see the answer is &lt;math&gt;\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=8|num-a=10}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_7&diff=154334 1993 AHSME Problems/Problem 7 2021-05-28T01:28:16Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The symbol &lt;math&gt;R_k&lt;/math&gt; stands for an integer whose base-ten representation is a sequence of &lt;math&gt;k&lt;/math&gt; ones. For example, &lt;math&gt;R_3=111,R_5=11111&lt;/math&gt;, etc. When &lt;math&gt;R_{24}&lt;/math&gt; is divided by &lt;math&gt;R_4&lt;/math&gt;, the quotient &lt;math&gt;Q=R_{24}/R_4&lt;/math&gt; is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in &lt;math&gt;Q&lt;/math&gt; is:<br /> <br /> &lt;math&gt;\text{(A) } 10\quad<br /> \text{(B) } 11\quad<br /> \text{(C) } 12\quad<br /> \text{(D) } 13\quad<br /> \text{(E) } 15&lt;/math&gt;<br /> <br /> == Solution ==<br /> Note &lt;math&gt;R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}&lt;/math&gt;. <br /> <br /> Therefore &lt;math&gt;\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}&lt;/math&gt;. <br /> <br /> We can recognize this is also the formula for the sum of a geometric series &lt;math&gt;1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}&lt;/math&gt;<br /> <br /> Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the &lt;math&gt;10^5&lt;/math&gt;, &lt;math&gt;10^6&lt;/math&gt; and &lt;math&gt;10^7&lt;/math&gt; places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives &lt;math&gt;5\times 3=15&lt;/math&gt; zeros altogether.<br /> <br /> The answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=6|num-a=8}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_7&diff=154333 1993 AHSME Problems/Problem 7 2021-05-28T01:27:24Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The symbol &lt;math&gt;R_k&lt;/math&gt; stands for an integer whose base-ten representation is a sequence of &lt;math&gt;k&lt;/math&gt; ones. For example, &lt;math&gt;R_3=111,R_5=11111&lt;/math&gt;, etc. When &lt;math&gt;R_{24}&lt;/math&gt; is divided by &lt;math&gt;R_4&lt;/math&gt;, the quotient &lt;math&gt;Q=R_{24}/R_4&lt;/math&gt; is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in &lt;math&gt;Q&lt;/math&gt; is:<br /> <br /> &lt;math&gt;\text{(A) } 10\quad<br /> \text{(B) } 11\quad<br /> \text{(C) } 12\quad<br /> \text{(D) } 13\quad<br /> \text{(E) } 15&lt;/math&gt;<br /> <br /> == Solution ==<br /> Note &lt;math&gt;R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}&lt;/math&gt;. <br /> <br /> Therefore &lt;math&gt;\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}&lt;/math&gt;. <br /> <br /> But we can recognize this form as the sum of a geometric series &lt;math&gt;1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}&lt;/math&gt;<br /> <br /> Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the &lt;math&gt;10^5&lt;/math&gt;, &lt;math&gt;10^6&lt;/math&gt; and &lt;math&gt;10^7&lt;/math&gt; places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives &lt;math&gt;5\times 3=15&lt;/math&gt; zeros altogether.<br /> <br /> The answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=6|num-a=8}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_6&diff=154332 1993 AHSME Problems/Problem 6 2021-05-28T01:14:41Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> &lt;math&gt;\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } \sqrt{2}\quad<br /> \text{(B) } 16\quad<br /> \text{(C) } 32\quad<br /> \text{(D) } (12)^{\tfrac{2}{3}}\quad<br /> \text{(E) } 512.5&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;\sqrt{\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \frac{ 2^{10} }{2^6} \sqrt{\frac{2^{10}+1}{1+2^{10}}} = 2^4&lt;/math&gt;<br /> <br /> &lt;math&gt;\fbox{B}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=5|num-a=7}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_4&diff=154331 1993 AHSME Problems/Problem 4 2021-05-28T01:07:56Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Define the operation &quot;&lt;math&gt;\circ&lt;/math&gt;&quot; by &lt;math&gt;x\circ y=4x-3y+xy&lt;/math&gt;, for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. For how many real numbers &lt;math&gt;y&lt;/math&gt; does &lt;math&gt;3\circ y=12&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 3\quad<br /> \text{(D) } 4\quad<br /> \text{(E) more than 4} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Note that &lt;math&gt;3 \circ y = 4 \cdot 3 - 3y + 3y = 12&lt;/math&gt;, so &lt;math&gt;3 \circ y = 12&lt;/math&gt; is true for all values of &lt;math&gt;y&lt;/math&gt;. Thus there are more than four solutions, and the answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=3|num-a=5}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_4&diff=154330 1993 AHSME Problems/Problem 4 2021-05-28T01:06:55Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> Define the operation &quot;&lt;math&gt;\circ&lt;/math&gt;&quot; by &lt;math&gt;x\circ y=4x-3y+xy&lt;/math&gt;, for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;. For how many real numbers &lt;math&gt;y&lt;/math&gt; does &lt;math&gt;3\circ y=12&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 3\quad<br /> \text{(D) } 4\quad<br /> \text{(E) more than 4} &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Note that &lt;math&gt;3 \circ y = 4 \cdot 3 - 3y + 3y = 12 - 3y + 3y = 12&lt;/math&gt;, which is satisfied for all values of &lt;math&gt;y&lt;/math&gt;. Thus there are more than four solutions, and the answer is &lt;math&gt;\fbox{E}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=3|num-a=5}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_8&diff=154329 1993 AHSME Problems/Problem 8 2021-05-28T00:59:27Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 2\quad<br /> \text{(B) } 4\quad<br /> \text{(C) } 5\quad<br /> \text{(D) } 6\quad<br /> \text{(E) } 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> There are two radius 3 circles to which &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are both externally tangent. One touches the tops of &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; and extends upward, and the other the other touches the bottoms and extends downward. There are also two radius 3 circles to which &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are both internally tangent, one touching the tops and encircling downward, and the other touching the bottoms and encircling upward. There are two radius 3 circles passing through the point where &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are tangent. For one &lt;math&gt;C_1&lt;/math&gt; is internally tangent and &lt;math&gt;C_2&lt;/math&gt; is externally tangent, and for the other &lt;math&gt;C_1&lt;/math&gt; is externally tangent and &lt;math&gt;C_2&lt;/math&gt; is internally tangent.<br /> <br /> &lt;math&gt;\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=7|num-a=9}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Logsobolev https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_8&diff=154328 1993 AHSME Problems/Problem 8 2021-05-28T00:58:32Z <p>Logsobolev: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 2\quad<br /> \text{(B) } 4\quad<br /> \text{(C) } 5\quad<br /> \text{(D) } 6\quad<br /> \text{(E) } 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> There are two radius 3 circles to which &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are both externally tangent. One touches the tops of &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; and extends upward, and the other the other touches the bottoms and extends downward. There are also two radius 3 circles to which &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are both internally tangent, one touching the tops and encircling downward, and the other touching the bottoms and encircling upward. There are two radius 3 circles passing through the point where &lt;math&gt;C_1&lt;/math&gt; and &lt;math&gt;C_2&lt;/math&gt; are tangent. For one &lt;math&gt;C_1&lt;/math&gt; is internally tangent and &lt;math&gt;C_2&lt;/math&gt; is externally tangent, and for the other &lt;math&gt;C_2&lt;/math&gt; is externally tangent and &lt;math&gt;C_1&lt;/math&gt; is internally tangent.<br /> <br /> &lt;math&gt;\fbox{D}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1993|num-b=7|num-a=9}} <br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Logsobolev