https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hdang5104&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-18T09:53:17Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_11&diff=49991 2012 AMC 8 Problems/Problem 11 2012-12-22T22:47:12Z <p>Hdang5104: /* Problem */</p> <hr /> <div>==Problem==<br /> The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and &lt;math&gt;x&lt;/math&gt; are all equal. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}5\qquad\textbf{(B)}\hspace{.05in}6\qquad\textbf{(C)}\hspace{.05in}7\qquad\textbf{(D)}\hspace{.05in}11\qquad\textbf{(E)}\hspace{.05in}12 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since there must be an unique mode, we can eliminate answer choices &lt;math&gt; {\textbf{(A)}\ 5} &lt;/math&gt; and &lt;math&gt; {\textbf{(C)}\ 7} &lt;/math&gt;. Now we need to test the remaining answer choices.<br /> <br /> Case 1: &lt;math&gt; x = 6 &lt;/math&gt;<br /> <br /> Mode: &lt;math&gt; 6 &lt;/math&gt;<br /> <br /> Median: &lt;math&gt; 6 &lt;/math&gt;<br /> <br /> Mean: &lt;math&gt; \frac{37}{7} &lt;/math&gt;<br /> <br /> Since the mean does not equal the median or mode, &lt;math&gt; {\textbf{(B)}\ 6} &lt;/math&gt; can also be eliminated.<br /> <br /> Case 2: &lt;math&gt; x = 11 &lt;/math&gt;<br /> <br /> Mode: &lt;math&gt; 6 &lt;/math&gt;<br /> <br /> Median: &lt;math&gt; 6 &lt;/math&gt;<br /> <br /> Mean: &lt;math&gt; 6 &lt;/math&gt;<br /> <br /> We are done with this problem, because we have found when &lt;math&gt; x = 11 &lt;/math&gt;, the condition is satisfied. Therefore, the answer is &lt;math&gt; \boxed{{\textbf{(D)}\ 11}} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=10|num-a=12}}</div> Hdang5104 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_12&diff=49990 2012 AMC 8 Problems/Problem 12 2012-12-22T22:43:33Z <p>Hdang5104: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the units digit of &lt;math&gt;13^{2012}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}3\qquad\textbf{(C)}\hspace{.05in}5\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}9 &lt;/math&gt;<br /> <br /> ==Solution==<br /> The problem wants us to find the units digit of &lt;math&gt; 13^{2012} &lt;/math&gt;, therefore, we can eliminate the tens digit of &lt;math&gt; 13 &lt;/math&gt;, because the tens digit will not affect the final result. So our new expression is &lt;math&gt; 3^{2012} &lt;/math&gt;. Now we need to look for a pattern in the units digit.<br /> <br /> &lt;math&gt; 3^1 \implies 3 &lt;/math&gt;<br /> <br /> &lt;math&gt; 3^2 \implies 9 &lt;/math&gt;<br /> <br /> &lt;math&gt; 3^3 \implies 7 &lt;/math&gt;<br /> <br /> &lt;math&gt; 3^4 \implies 1 &lt;/math&gt;<br /> <br /> &lt;math&gt; 3^5 \implies 3 &lt;/math&gt;<br /> <br /> We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we divide can subtract 1 from 2012 and divide by 4. The remainder is the power of three that we are looking for, minus 1. &lt;math&gt;2011&lt;/math&gt; divided by &lt;math&gt;4&lt;/math&gt; leaves a remainder of &lt;math&gt;3&lt;/math&gt;, so the answer is the units digit of &lt;math&gt;3^{3+1}&lt;/math&gt;, or &lt;math&gt;3^4&lt;/math&gt;. Thus, we find that the units digit of &lt;math&gt; 13^{2012} &lt;/math&gt; is <br /> &lt;math&gt; \boxed{{\textbf{(A)}\ 1}} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=11|num-a=13}}</div> Hdang5104 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_15&diff=49989 2012 AMC 8 Problems/Problem 15 2012-12-22T22:32:37Z <p>Hdang5104: /* Solution */</p> <hr /> <div>==Problem==<br /> The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?<br /> <br /> &lt;math&gt; \textbf{(A)}\hspace{.05in}40\text{ and }50\qquad\textbf{(B)}\hspace{.05in}51\text{ and }55\qquad\textbf{(C)}\hspace{.05in}56\text{ and }60\qquad\textbf{(D)}\hspace{.05in}61\text{ and }65\qquad\textbf{(E)}\hspace{.05in}66\text{ and }99 &lt;/math&gt;<br /> <br /> ==Solution==<br /> To find the answer to this problem, we need to find the least common multiple of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt; and add &lt;math&gt;2&lt;/math&gt; to the result. The least common multiple of the four numbers is &lt;math&gt;60&lt;/math&gt;, and by adding &lt;math&gt;2&lt;/math&gt;, we find that that such number is &lt;math&gt;62&lt;/math&gt;. Now we need to find the only given range that contains &lt;math&gt;62&lt;/math&gt;. The only such range is answer &lt;math&gt;{\textbf{(D)}&lt;/math&gt;, and so our final answer is &lt;math&gt; \boxed{\textbf{(D)}\ 61\text{ and }65} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2012|num-b=14|num-a=16}}</div> Hdang5104