https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Qntty&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T22:10:38ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Problem_solving&diff=31427Problem solving2009-04-23T23:05:29Z<p>Qntty: Undo revision 28037 by FameofLight (Talk)</p>
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<div>At its roots, '''problem solving''' is exactly what it sounds like, the process of solving problems. However, problem solving methods permeate the studies of [[mathematics]], [[science]], and [[technology]]. The human processes involved in problem solving are often studied by [[cognitive science | cognitive scientists]].<br />
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{{stub}}</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=Category:Olympiad_Combinatorics_Problems&diff=31225Category:Olympiad Combinatorics Problems2009-04-11T06:16:15Z<p>Qntty: </p>
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<div>This page lists all of the olympiad [[combinatorics]] problems in the [[AoPSWiki]].<br />
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[[Category:Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=Category:Intermediate_Combinatorics_Problems&diff=31224Category:Intermediate Combinatorics Problems2009-04-11T06:16:01Z<p>Qntty: </p>
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<div>This page lists all of the intermediate [[combinatorics]] problems in the [[AoPSWiki]].<br />
[[Category:Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=Category:Introductory_Combinatorics_Problems&diff=31223Category:Introductory Combinatorics Problems2009-04-11T06:15:31Z<p>Qntty: </p>
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<div>This page lists all the introductory [[combinatorics]] problems in the [[AoPSWiki]].<br />
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[[Category:Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_20&diff=312222002 AMC 12B Problems/Problem 202009-04-11T06:13:20Z<p>Qntty: </p>
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<div>== Problem ==<br />
Let <math>\triangle XOY</math> be a [[right triangle|right-angled triangle]] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, respectively. Given that <math>XN = 19</math> and <math>YM = 22</math>, find <math>XY</math>.<br />
<br />
<math>\mathrm{(A)}\ 24<br />
\qquad\mathrm{(B)}\ 26<br />
\qquad\mathrm{(C)}\ 28<br />
\qquad\mathrm{(D)}\ 30<br />
\qquad\mathrm{(E)}\ 32</math><br />
== Solution ==<br />
[[Image:2002_12B_AMC-20.png]]<br />
<br />
Let <math>OM = x</math>, <math>ON = y</math>. By the [[Pythagorean Theorem]] on <math>\triangle XON, MOY</math> respectively,<br />
<cmath>\begin{align*}<br />
(2x)^2 + y^2 &= 19^2\\<br />
x^2 + (2y)^2 &= 22^2\end{align*}<br />
</cmath><br />
<br />
Summing these gives <math>5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169</math>. <br />
<br />
By the Pythagorean Theorem again, we have <br />
<br />
<cmath>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = 26 \Rightarrow \mathrm{(B)}</cmath><br />
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== See also ==<br />
{{AMC12 box|year=2002|ab=B|num-b=19|num-a=21}}</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_21&diff=312212008 AMC 12A Problems/Problem 212009-04-11T05:35:42Z<p>Qntty: </p>
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<div>==Problem==<br />
A permutation <math>(a_1,a_2,a_3,a_4,a_5)</math> of <math>(1,2,3,4,5)</math> is <u>heavy-tailed</u> if <math>a_1 + a_2 < a_4 + a_5</math>. What is the number of heavy-tailed permutations?<br />
<br />
<math>\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52</math><br />
<br />
==Solution==<br />
There are <math>5!=120</math> total permutations. <br />
<br />
For every permutation <math>(a_1,a_2,a_3,a_4,a_5)</math> such that <math>a_1 + a_2 < a_4 + a_5</math>, there is exactly one permutation such that <math>a_1 + a_2 > a_4 + a_5</math>. Thus it suffices to count the permutations such that <math>a_1 + a_2 = a_4 + a_5</math>. <br />
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<math>1+4=2+3</math>, <math>1+5=2+4</math>, and <math>2+5=3+4</math> are the only combinations of numbers that can satisfy <math>a_1 + a_2 = a_4 + a_5</math>. <br />
<br />
There are <math>3</math> combinations of numbers, <math>2</math> possibilities of which side of the equation is <math>a_1+a_2</math> and which side is <math>a_4+a_5</math>, and <math>2^2=4</math> possibilities for rearranging <math>a_1,a_2</math> and <math>a_4,a_5</math>. Thus, there are <math>3\cdot2\cdot4=24</math> permutations such that <math>a_1 + a_2 = a_4 + a_5</math>. <br />
<br />
Thus, the number of <u>heavy-tailed</u> permutations is <math>\frac{120-24}{2}=48 \Rightarrow D</math>. <br />
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== See also ==<br />
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_20&diff=312202009 AMC 12B Problems/Problem 202009-04-11T05:31:35Z<p>Qntty: </p>
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<div>== Problem ==<br />
A convex polyhedron <math>Q</math> has vertices <math>V_1,V_2,\ldots,V_n</math>, and <math>100</math> edges. The polyhedron is cut by planes <math>P_1,P_2,\ldots,P_n</math> in such a way that plane <math>P_k</math> cuts only those edges that meet at vertex <math>V_k</math>. In addition, no two planes intersect inside or on <math>Q</math>. The cuts produce <math>n</math> pyramids and a new polyhedron <math>R</math>. How many edges does <math>R</math> have?<br />
<br />
<math>\mathrm{(A)}\ 200\qquad<br />
\mathrm{(B)}\ 2n\qquad<br />
\mathrm{(C)}\ 300\qquad<br />
\mathrm{(D)}\ 400\qquad<br />
\mathrm{(E)}\ 4n</math><br />
<br />
== Solution ==<br />
<br />
=== Solution 1 ===<br />
Each edge of <math>Q</math> is cut by two planes, so <math>R</math> has <math>200</math> vertices. Three edges of <math>R</math> meet at each vertex, so <math>R</math> has <math>\frac 12 \cdot 3 \cdot 200 = \boxed {300}</math> edges.<br />
<br />
=== Solution 2 ===<br />
At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is <math>200</math>. A middle portion of each original edge is also present in <math>R</math>, so <math>R</math> has <math>100 + 200 = \boxed {300}</math> edges.<br />
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=== Solution 3 ===<br />
Euler's Polyhedron Formula applied to <math>Q</math> gives <math>n - 100 + F = 2</math>, where F is the number of faces of <math>Q</math>. Each edge of <math>Q</math> is cut by two planes, so <math>R</math> has <math>200</math> vertices. Each cut by a plane <math>P_k</math> creates an additional face on <math>R</math>, so Euler's Polyhedron Formula applied to <math>R</math> gives <math>200 - E + (F+n) = 2</math>, where <math>E</math> is the number of edges of <math>R</math>. Subtracting the first equation from the second gives <math>300 - E = 0</math>, whence <math>E = \boxed {300}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
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== See also ==<br />
{{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_23&diff=312192009 AMC 10B Problems/Problem 232009-04-11T05:31:25Z<p>Qntty: </p>
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<div>{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #23]] and [[2009 AMC 12B Problems|2009 AMC 12B #18]]}}<br />
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== Problem ==<br />
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?<br />
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<math>\mathrm{(A)}\frac {1}{16}\qquad<br />
\mathrm{(B)}\frac 18\qquad<br />
\mathrm{(C)}\frac {3}{16}\qquad<br />
\mathrm{(D)}\frac 14\qquad<br />
\mathrm{(E)}\frac {5}{16}</math><br />
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== Solution ==<br />
After 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So the probability that both runners are in the picture is <math>\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}</math>. The answer is <math>\mathrm{(C)}</math>.<br />
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== See also ==<br />
{{AMC10 box|year=2009|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2009|ab=B|num-b=17|num-a=19}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_22&diff=312182008 AMC 12B Problems/Problem 222009-04-11T05:30:10Z<p>Qntty: </p>
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<div>==Problem 22==<br />
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?<br />
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<math>\textbf{(A)} \; \frac {11}{20} \qquad \textbf{(B)} \; \frac {4}{7} \qquad \textbf{(C)} \; \frac {81}{140} \qquad \textbf{(D)} \; \frac {3}{5} \qquad \textbf{(E)} \; \frac {17}{28}</math><br />
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==Solution==<br />
Auntie Em won't be able to park only when none of the four available spots touch. We can form a [[bijection]] between all such cases and the number of ways to pick four spots out of 13: since none of the spots touch, remove a spot from between each of the cars. From the other direction, given four spots out of 13, simply add a spot between each.<br />
So the probability she can park is <cmath>1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}=\frac{17}{28},\qquad\boxed{E}</cmath><br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=21|num-a=23}}<br />
[[Category:Introductory Combinatorics Problems]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=Inequality&diff=31117Inequality2009-04-05T21:33:14Z<p>Qntty: </p>
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<div>The subject of mathematical '''inequalities''' is tied closely with [[optimization]] methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in [[mathematics Olympiads]].<br />
<br />
<br />
==Overview==<br />
Inequalities are arguably a branch of [[elementary algebra]], and relate slightly to [[number theory]]. They deal with [[relations]] of [[variable]]s denoted by four signs: <math>>,<,\ge,\le</math>.<br />
<br />
For two [[number]]s <math>a</math> and <math>b</math>:<br />
*<math>a>b</math> if <math>a</math> is greater than <math>b</math>, that is, <math>a-b</math> is positive.<br />
*<math>a<b</math> if <math>a</math> is smaller than <math>b</math>, that is, <math>a-b</math> is negative.<br />
*<math>a\ge b</math> if <math>a</math> is greater than or equal to <math>b</math>, that is, <math>a-b</math> is either positive or <math>0</math>.<br />
*<math>a\le b</math> if <math>a</math> is less than or equal to <math>b</math>, that is, <math>a-b</math> is either negative or <math>0</math>.<br />
<br />
Note that if and only if <math>a>b</math>, <math>b<a</math>, and vice versa. The same applies to the latter two signs: if and only if <math>a\ge b</math>, <math>b\le a</math>, and vice versa.<br />
<br />
Some properties of inequalities are:<br />
*If <math>a>b</math>, then <math>a+c>b</math>, where <math>c\ge 0</math>.<br />
*If <math>a \ge b</math>, then <math>a+c\ge b</math>, where <math>c\ge 0</math>.<br />
*If <math>a \ge b</math>, then <math>a+c>b</math>, where <math>c>0</math>.<br />
<br />
==Solving Inequalities==<br />
A common application of inequalities is solving them for a variable. For example, consider the inequality <math>5x+7>3x+8</math>. We can solve for the variable <math>x</math> here and get <math>2x-1>0\Leftrightarrow x>\frac{1}{2}</math>, thus placing implicit restrictions upon the variable <math>x</math>. A more complex example is <math>\frac{x-8}{x+5}+4\ge 3</math>. Here is a common mistake: <math>\frac{x-8}{x+5}+4\ge 3 \Rightarrow \frac{x+5-13}{x+5}+4\ge 3 \Rightarrow 1-\frac{13}{x+5}+4\ge 3 \Rightarrow x+5-13+4x+20\ge 3x+15\Rightarrow x\ge \frac{3}{2}</math>.<br />
<br />
The problem here is that we multiplied by <math>x+5</math> as one of the last steps. We also kept the inequality sign in the same direction. However, we don't ''know'' if the quantity <math>x+5</math> is negative or not; we can't assume that it is positive for all real <math>x</math>. Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either.<br />
A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as <math>x</math> varies:<br />
<br />
<math>\frac{x-8}{x+5}+4\ge 3 \Rightarrow \frac{x-8}{x+5}+1\ge 0 \Rightarrow \frac{2x-3}{x+5}\ge 0</math>. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. <math>0</math> makes a good starting point, but does not solve the inequality. Nor does <math>1</math>. Therefore, these two aren't solutions. Then we begin to test numbers such as <math>2</math>, <math>3</math>, and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as <math>x</math> gets larger and larger. But just where does <math>x</math>, which causes a negative fraction at <math>0</math> and <math>1</math>, begin to cause a positive fraction? We can't just assume that <math>2</math> is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when <math>2x-3=0</math> (the numerator), then the fraction is <math>0</math>, and begins to be positive for all higher values of <math>x</math>. Solving the equation reveals that <math>x=\frac{3}{2}</math> is the turning point. After more of this type of work, we realize that <math>x=-5</math> brings about division by <math>0</math>, so it certainly isn't a solution. However, it also tells us that any value of <math>x</math> that is less than <math>-5</math> brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between <math>x=-5</math> and <math>x=\frac{3}{2}</math> (except <math>\frac{3}{2}</math> itself) seems to be a solution.<br />
Therefore, we conclude that the solutions are the intervals <math>(-\infty,-5)\cup[\frac{3}{2},+\infty)</math>.<br />
<br />
For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of <math>x</math> that cause the numerator and/or the denominator to be <math>0</math>.<br />
To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as <math>0</math> in the region <math>(-5,\frac{3}{2})</math>, as well as one value in the region <math>(-\infty,-5]</math> and <math>[\frac{3}{2},+\infty)</math>; then we see which regions are part of the solution set. This does indeed give the complete solution set.<br />
<br />
One must be careful about the boundaries of the solutions. In the example problem, the value <math>x=\frac{3}{2}</math> was a solution only because the inequality was nonstrict. Also, the value <math>x=-5</math> was not a solution because it would bring about division by <math>0</math>. Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by <math>0</math>.<br />
<br />
==Complete Inequalities==<br />
A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called [[Trivial Inequality]], which states that for any real <math>x</math>, <math>x^2\ge 0</math>. Most inequalities of this type hold only for positive numbers, and this type of inequality often has very clever problems and applications.<br />
<br />
== List of Theorems ==<br />
Here are some of the more useful inequality theorems, as well as general inequality topics.<br />
* [[Arithmetic Mean-Geometric Mean | Arithmetic Mean-Geometric Mean Inequality]]<br />
* [[Cauchy-Schwarz Inequality]]<br />
* [[Carleman's Inequality]]<br />
* [[Chebyshevs inequality | Chebyshev's Inequality]]<br />
* [[Geometric inequalities]]<br />
* [[Hölder's inequality]]<br />
* [[Isoperimetric inequalities]]<br />
* [[Jensen's Inequality]]<br />
* [[Maclaurin's Inequality]]<br />
* [[Minkowski Inequality]]<br />
* [[Muirhead's Inequality]]<br />
* [[Nesbitt's Inequality]]<br />
* [[Newton's Inequality]]<br />
* [[Power mean inequality]]<br />
* [[Ptolemy's Inequality]]<br />
* [[Rearrangement Inequality]]<br />
* [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]]<br />
* [[Schur's Inequality]]<br />
* [[Triangle Inequality]]<br />
* [[Trivial inequality]]<br />
<br />
==Problems==<br />
===Introductory===<br />
*A tennis player computes her win [[ratio]] by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly <math>.500</math>. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>.503</math>. What's the largest number of matches she could've won before the weekend began? ([[1992 AIME Problems/Problem 3|Source]])<br />
===Intermediate===<br />
*Given that <math>(a+1)(b+1)(c+1) = 8</math>, and <math>a, b, c \ge 0</math> show that <math>abc \le 1</math>. (<url>weblog_entry.php?t=172070 Source</url>)<br />
===Olympiad===<br />
*Let <math>a,b,c</math> be positive real numbers. Prove that<br />
<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math> ([[2001 IMO Problems/Problem 2|Source]])<br />
<br />
== Resources ==<br />
=== Books ===<br />
==== Intermediate ====<br />
* [http://www.amazon.com/exec/obidos/ASIN/0883856034/artofproblems-20 Introduction to Inequalities]<br />
* [http://www.amazon.com/exec/obidos/ASIN/0883856042/artofproblems-20 Geometric Inequalities]<br />
<br />
==== Olympiad ====<br />
* [http://www.amazon.com/exec/obidos/ASIN/052154677X/artofproblems-20 The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities] by J. Michael Steele.<br />
* [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequalities.<br />
* [http://www.amazon.com/exec/obidos/ASIN/0521358809/artofproblems-20 Inequalities] by [[G. H. Hardy]], [[J. E. Littlewood]], [[G. Pólya]].<br />
<br />
=== Articles ===<br />
==== Olympiad ====<br />
* [http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf Inequalities] by [[MIT]] Professor Kiran Kedlaya.<br />
* [http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf Inequalities] by [[IMO]] gold medalist [[Thomas Mildorf]].<br />
<br />
<br />
=== Classes ===<br />
==== Olympiad ====<br />
* The [http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php Worldwide Online Olympiad Training Program] is designed to help students learn to tackle [[mathematical Olympiad]] problems in topics such as inequalities.<br />
<br />
<br />
== See also ==<br />
* [[Mathematics competitions]]<br />
* [[Math books]]<br />
<br />
[[Category:Inequality]]</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_6&diff=281632002 AMC 12A Problems/Problem 62008-10-06T22:35:20Z<p>Qntty: New page: ==Problem== For how many positive integers <math>m</math> does there exist at least one positive integer n such that <math>m \cdot n \le m + n</math>? <math> \mathrm{(A) \ } 4\qquad \math...</p>
<hr />
<div>==Problem==<br />
For how many positive integers <math>m</math> does there exist at least one positive integer n such that <math>m \cdot n \le m + n</math>?<br />
<br />
<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ }</math> infinitely many<br />
<br />
<br />
==Solution==<br />
<math>m \cdot 1 \le m + 1</math> for <math>n=1</math><br />
<br />
<math>m \le m + 1</math><br />
<br />
<math> \Rightarrow \mathrm{(E) \ }</math> infinitely many</div>Qnttyhttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_3&diff=281622002 AMC 12A Problems/Problem 32008-10-06T22:26:06Z<p>Qntty: New page: ==Problem== According to the standard convention for exponentiation, <cmath> 2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536. </cmath> If the order in which the exponentiations are perf...</p>
<hr />
<div>==Problem==<br />
According to the standard convention for exponentiation, <br />
<cmath> 2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536. </cmath><br />
<br />
If the order in which the exponentiations are performed is changed, how many other values are possible?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4 </math></div>Qntty