https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Archimedes1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:42:49ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1987_IMO_Problems/Problem_2&diff=297331987 IMO Problems/Problem 22009-01-25T03:05:43Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>==Problem==<br />
In an acute-angled triangle <math>ABC </math> the interior bisector of the angle <math>A </math> intersects <math>BC </math> at <math>L </math> and intersects the [[circumcircle]] of <math>ABC </math> again at <math>N </math>. From point <math>L </math> perpendiculars are drawn to <math>AB </math> and <math>AC </math>, the feet of these perpendiculars being <math>K </math> and <math>M </math> respectively. Prove that the quadrilateral <math>AKNM </math> and the triangle <math>ABC </math> have equal areas.<br />
<br />
==Solution==<br />
<br />
We are to prove that <math>[AKNM]=[ABC]</math> or equivilently, <math>[ABC]+[BNC]-[KNC]-[BMN]=[ABC]</math>. Thus, we are to prove that <math>[BNC]=[KNC]+[BMN]</math>. It is clear that since <math>\angle BAN=\angle NAC</math>, the segments <math>BN</math> and <math>NC</math> are equal. Thus, we have <math>[BNC]=\frac{1}{2}BN^2\sin BNC=\frac{1}{2}BN^2\sin A</math> since cyclic quadrilateral <math>ABNC</math> gives <math>\angle BNC=180-\angle A</math>. Thus, we are to prove that<br />
<br />
<math>\frac{1}{2}BN^2\sin A=[KNC]+[BMN]</math><br />
<br />
<math>\Leftrightarrow \frac{1}{2}BN^2\sin A=\frac{1}{2}CN\cdot CK\sin NCA+\frac{1}{2}BN\cdot BM\sin NBA</math><br />
<br />
<math>\Leftrightarrow BN\sin A=CK\sin NCA+BM\sin NBA</math><br />
<br />
From the fact that <math>\angle BNC=180-\angle A</math> and that <math>BNC</math> is iscoceles, we find that <math>\angle NBC=\angle NCB=\frac{1}{2}A</math>. So, we have <math>BN\cos\frac{1}{2}A=\frac{1}{2}BC\Rightarrow BN=\frac{BC}{2\cos \frac{1}{2}A}</math>. So we are to prove that<br />
<br />
<math>\frac{BC\sin A}{2\cos \frac{1}{2}A}=CK\sin NCA+BM\sin NBA</math><br />
<br />
<math>\Leftrightarrow BC\sin \frac{1}{2}A=CK\sin (C+ \frac{1}{2}A)+BM\sin (C+ \frac{1}{2}A)</math><br />
<br />
<math>\Leftrightarrow BC=CK(\sin C\cot\frac{1}{2}A+\cos C)+BM(\sin B\cot\frac{1}{2}A+\cos B)</math><br />
<br />
We have <math>\sin C=\frac{KL}{CL}</math>,<math>\cos C=\frac{CK}{CL}</math>, <math>\cot\frac{1}{2}A=\frac{AK}{KL}=\frac{AM}{LM}</math>, <math>\sin B=\frac{LM}{BL}</math>,<math>\cos B=\frac{BM}{ML}</math>, and so we are to prove that<br />
<br />
<math>BC=CK(\frac{KL}{CL}\frac{AK}{KL})+\frac{CK}{CL})+BM(\frac{LM}{BL}\frac{AM}{LM}+\frac{BM}{ML})</math><br />
<br />
<math>\Leftrightarrow BC=CK(\frac{AK}{CL}+\frac{CK}{CL})+BM(\frac{AM}{BL}+\frac{BM}{ML})</math><br />
<br />
<math>\Leftrightarrow BC=\frac{CK\cdot AC}{CL}+\frac{BM\cdot AB}{BL}</math><br />
<br />
<math>\Leftrightarrow BC=AC\cos C+AB\cos B</math><br />
<br />
We shall show that this is true: Let the altitude from <math>A</math> touch <math>BC</math> at <math>A^\prime</math>. Then it is obvious that <math>AC\cos C=CA^\prime</math> and <math>AB\cos B=A^\prime B</math> and thus <math>AC\cos C+AB\cos B=BC</math>. <br />
<br />
Thus we have proven that <math>[AKNM]=[ABC]</math>.<br />
<br />
==See also==<br />
{{IMO box|num-b=1|num-a=3|year=1987}}<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Van_Aubel%27s_Theorem&diff=28621Van Aubel's Theorem2008-12-18T00:06:28Z<p>Archimedes1: New page: = Theorem = Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the...</p>
<hr />
<div>= Theorem =<br />
Construct squares <math>ABA'B'</math>, <math>BCB'C'</math>, <math>CDC'D'</math>, and <math>DAD'A'</math> externally on the sides of quadrilateral <math>ABCD</math>, and let the centroids of the four squares be <math>P, Q, R,</math> and <math>S</math>, respectively. Then <math>PR = QS</math> and <math>PR \perp QS</math>.<br />
<geogebra>21cd94f930257bcbd188d1ed7139a9336b3eb9bc</geogebra><br />
<br />
= Proofs =<br />
<br />
== Proof 1: Complex Numbers==<br />
Putting the diagram on the complex plane, let any point <math>X</math> be represented by the complex number <math>x</math>. Note that <math>\angle PAB = \frac{\pi}{4}</math> and that <math>PA = \frac{\sqrt{2}}{2}AB</math>, and similarly for the other sides of the quadrilateral. Then we have<br />
<br />
<br />
<cmath>\begin{eqnarray*} <br />
p &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a \\<br />
q &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+b \\<br />
r &=& \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}+c \\<br />
s &=& \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}+d<br />
\end{eqnarray*}</cmath><br />
<br />
From this, we find that <br />
<cmath>\begin{eqnarray*}<br />
p-r &=& \frac{\sqrt{2}}{2}(b-a)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(d-c)e^{i \frac{\pi}{4}}-c \\<br />
&=& \frac{1+i}{2}(b-d) + \frac{1-i}{2}(a-c).<br />
\end{eqnarray*}</cmath><br />
Similarly,<br />
<cmath>\begin{eqnarray*}<br />
q-s &=& \frac{\sqrt{2}}{2}(c-b)e^{i \frac{\pi}{4}}+a - \frac{\sqrt{2}}{2}(a-d)e^{i \frac{\pi}{4}}-c \\<br />
&=& \frac{1+i}{2}(c-a) + \frac{1-i}{2}(b-d).<br />
\end{eqnarray*}</cmath><br />
<br />
Finally, we have <math>(p-r) = i(q-s) = e^{i \pi/2}(q-r)</math>, which implies <math>PR = QS</math> and <math>PR \perp QS</math>, as desired.<br />
<br />
== Proof 2: Mean Geometry Theorem ==</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Spiral_similarity&diff=28383Spiral similarity2008-11-15T03:39:17Z<p>Archimedes1: New page: A '''spiral similarity''' is the composition of a homothety with a rotation.</p>
<hr />
<div>A '''spiral similarity''' is the [[composition]] of a [[homothety]] with a [[rotation]].</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1977_Canadian_MO_Problems/Problem_3&diff=269041977 Canadian MO Problems/Problem 32008-07-05T18:55:07Z<p>Archimedes1: Moved to Intermediate category</p>
<hr />
<div>== Problem ==<br />
<math>N</math> is an integer whose representation in base <math>b</math> is <math>777.</math> Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer.<br />
<br />
== Solution ==<br />
Rewriting <math>N</math> in base <math>10,</math> <math>N=7(b^2+b+1)=a^4</math> for some integer <math>a.</math> Because <math>7\mid a^4</math> and <math>7</math> is prime, <math>a \ge 7^4.</math> Since we want to minimize <math>b,</math> we check to see if <math>a=7^4</math> works.<br />
<br />
<br />
When <math>a=7^4,</math> <math>b^2+b+1=7^3.</math> Solving this quadratic, <math>b = 18 </math>.<br />
<br />
<br />
{{Old CanadaMO box|num-b=2|num-a=4|year=1977}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1970_IMO_Problems/Problem_4&diff=269031970 IMO Problems/Problem 42008-07-05T18:53:33Z<p>Archimedes1: Solution 2</p>
<hr />
<div>==Problem==<br />
Find the set of all positive integers <math>n</math> with the property that the set <math>\{ n, n+1, n+2, n+3, n+4, n+5 \} </math> can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.<br />
<br />
==Solution==<br />
The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be <math>\{ 1, 2, 3, 4, 5, 6 \}</math>, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.<br />
<br />
==Solution 2==<br />
As in the previous solution, none of the six consecutive numbers can be multiples of <math>7</math>. This means that together, they take on the values <math> \{ 1, 2, 3, 4, 5, 6, \} \mod 7</math>. The product of all the numbers in this set, then, is <math>-1 \mod 7</math>, by [[Wilson's Theorem]]. However, <math>-1</math> is not a [[quadratic residue]] <math>\mod 7</math>, which means that we cannot partition the original set into two sets of equal product. Thus, no such <math>n</math> exist.<br />
{{IMO box|year=1970|num-b=3|num-a=5}}<br />
<br />
[[Category:Olympiad Number Theory Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Roots_of_unity&diff=26771Roots of unity2008-06-25T17:03:09Z<p>Archimedes1: /* Solving the equation */ polar -> polar form</p>
<hr />
<div>{{WotWAnnounce|week=June 20-26}}<br />
<br />
The '''Roots of unity''' are a topic closely related to [[trigonometry]]. Roots of unity come up when we examine the [[complex number|complex]] [[root]]s of the [[polynomial]] <math> x^n=1 </math>.<br />
<br />
== Solving the equation ==<br />
<br />
First, we note that since we have an '''n'''th degree polynomial, there will be '''n''' complex roots.<br />
<br />
Now, we can convert everything to [[polar form]] by letting <math>x = re^{i\theta} </math>, and noting that <math>1 = e^{2\pi ik}</math> for <math> k\in \mathbb{Z}</math>, to get <math>r^ne^{ni\theta} = e^{2\pi ik}</math>. The magnitude of the RHS is 1, making <math>r^n=1\Rightarrow r=1</math> (magnitude is always expressed as a positive real number). This leaves us with <math>e^{ni\theta} = e^{2\pi ik}</math>.<br />
<br />
Taking the [[natural logarithm]] of both sides gives us <math> ni\theta = 2\pi ik</math>. Solving this gives <math> \theta=\frac{2\pi k}n </math>. Additionally, we note that for each of <math> k=0,1,2,\ldots,n-1 </math> we get a distinct value for <math> \theta </math>, but once we get to <math> k > n-1 </math>, we start getting [[coterminal]] angles.<br />
<br />
Thus, the solutions to <math>x^n=1 </math> are given by <math>x = e^{2\pi k i/n} </math> for <math> k=0,1,2,\ldots,n-1</math>. We could also express this in trigonometric form as <math>x=\cos\left(\frac{2\pi k}n\right) + i\sin\left(\frac{2\pi k}n\right) = \mathrm{cis }\left(\frac{2\pi k}n\right). </math><br />
<br />
== Geometry of the roots of unity ==<br />
<br />
All of the roots of unity lie on the [[unit circle]] in the complex plane. This can be seen by considering the magnitudes of both sides of the equation <math> x^n = 1</math>. If we let <math> x = re^{i\theta} </math>, we see that <math> r^n = 1</math>, since the magnitude of the RHS of <math> x^n=1 </math> is 1, and for two complex numbers to be equal, both their magnitudes and arguments must be equivalent.<br />
<br />
Additionally, we can see that when the '''n'''th roots of unity are connected in order (more technically, we would call this their [[convex hull]]), they form a regular '''n'''-sided polygon. This becomes even more evident when we look at the arguments of the roots of unity.<br />
<br />
== Properties of roots of unity ==<br />
Listed below is a quick summary of important properties of roots of unity.<br />
<br />
* They occupy the vertices of a regular ''n''-gon in the [[complex plane]].<br />
* For <math> n>1 </math>, the sum of the ''n''th roots of unity is 0. More generally, if <math>\zeta</math> is a primitive ''n''th root of unity (i.e. <math>\zeta^m\neq 1</math> for <math>1\le m\le n-1</math>), then <math>\sum_{k=0}^{n-1} \zeta^{km}=\begin{cases} n & {n\mid m}, \\ 0 & \mathrm{otherwise.}\end{cases} </math><br />
** This is an immediate result of [[Vieta's formulas]] on the polynomial <math> x^n-1 = 0 </math> and [[Newton sums]].<br />
* If <math>\zeta</math> is a primitive ''n''th root of unity, then the roots of unity can be expressed as <math> 1, \zeta, \zeta^2,\ldots,\zeta^{n-1}</math>.<br />
* Also, don't overlook the most obvious property of all! For each <math>n</math>th root of unity, <math>\zeta</math>, we have that <math> \zeta^n=1 </math><br />
<br />
== Uses of roots of unity ==<br />
Roots of unity show up in many surprising places. Here, we list a few:<br />
<br />
* [[Geometry]]<br />
* [[Factoring]]<br />
* [[Number theory]]<br />
<br />
== See also ==<br />
* [[Complex number]]s</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Roots_of_unity&diff=26770Roots of unity2008-06-25T17:02:30Z<p>Archimedes1: /* Solving the equation */ trig -> trigonometric (nitpick)</p>
<hr />
<div>{{WotWAnnounce|week=June 20-26}}<br />
<br />
The '''Roots of unity''' are a topic closely related to [[trigonometry]]. Roots of unity come up when we examine the [[complex number|complex]] [[root]]s of the [[polynomial]] <math> x^n=1 </math>.<br />
<br />
== Solving the equation ==<br />
<br />
First, we note that since we have an '''n'''th degree polynomial, there will be '''n''' complex roots.<br />
<br />
Now, we can convert everything to [[polar]] by letting <math>x = re^{i\theta} </math>, and noting that <math>1 = e^{2\pi ik}</math> for <math> k\in \mathbb{Z}</math>, to get <math>r^ne^{ni\theta} = e^{2\pi ik}</math>. The magnitude of the RHS is 1, making <math>r^n=1\Rightarrow r=1</math> (magnitude is always expressed as a positive real number). This leaves us with <math>e^{ni\theta} = e^{2\pi ik}</math>.<br />
<br />
Taking the [[natural logarithm]] of both sides gives us <math> ni\theta = 2\pi ik</math>. Solving this gives <math> \theta=\frac{2\pi k}n </math>. Additionally, we note that for each of <math> k=0,1,2,\ldots,n-1 </math> we get a distinct value for <math> \theta </math>, but once we get to <math> k > n-1 </math>, we start getting [[coterminal]] angles.<br />
<br />
Thus, the solutions to <math>x^n=1 </math> are given by <math>x = e^{2\pi k i/n} </math> for <math> k=0,1,2,\ldots,n-1</math>. We could also express this in trigonometric form as <math>x=\cos\left(\frac{2\pi k}n\right) + i\sin\left(\frac{2\pi k}n\right) = \mathrm{cis }\left(\frac{2\pi k}n\right). </math><br />
<br />
== Geometry of the roots of unity ==<br />
<br />
All of the roots of unity lie on the [[unit circle]] in the complex plane. This can be seen by considering the magnitudes of both sides of the equation <math> x^n = 1</math>. If we let <math> x = re^{i\theta} </math>, we see that <math> r^n = 1</math>, since the magnitude of the RHS of <math> x^n=1 </math> is 1, and for two complex numbers to be equal, both their magnitudes and arguments must be equivalent.<br />
<br />
Additionally, we can see that when the '''n'''th roots of unity are connected in order (more technically, we would call this their [[convex hull]]), they form a regular '''n'''-sided polygon. This becomes even more evident when we look at the arguments of the roots of unity.<br />
<br />
== Properties of roots of unity ==<br />
Listed below is a quick summary of important properties of roots of unity.<br />
<br />
* They occupy the vertices of a regular ''n''-gon in the [[complex plane]].<br />
* For <math> n>1 </math>, the sum of the ''n''th roots of unity is 0. More generally, if <math>\zeta</math> is a primitive ''n''th root of unity (i.e. <math>\zeta^m\neq 1</math> for <math>1\le m\le n-1</math>), then <math>\sum_{k=0}^{n-1} \zeta^{km}=\begin{cases} n & {n\mid m}, \\ 0 & \mathrm{otherwise.}\end{cases} </math><br />
** This is an immediate result of [[Vieta's formulas]] on the polynomial <math> x^n-1 = 0 </math> and [[Newton sums]].<br />
* If <math>\zeta</math> is a primitive ''n''th root of unity, then the roots of unity can be expressed as <math> 1, \zeta, \zeta^2,\ldots,\zeta^{n-1}</math>.<br />
* Also, don't overlook the most obvious property of all! For each <math>n</math>th root of unity, <math>\zeta</math>, we have that <math> \zeta^n=1 </math><br />
<br />
== Uses of roots of unity ==<br />
Roots of unity show up in many surprising places. Here, we list a few:<br />
<br />
* [[Geometry]]<br />
* [[Factoring]]<br />
* [[Number theory]]<br />
<br />
== See also ==<br />
* [[Complex number]]s</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1998_PMWC_Problems&diff=267531998 PMWC Problems2008-06-23T20:36:50Z<p>Archimedes1: Added team test--Problem T10 needs diagram!</p>
<hr />
<div>== Problem I1 ==<br />
Calculate: <math>\frac{1*2*3+2*4*6+3*6*9+4*8*12+5*10*15}{1*3*5+2*6*10+3*9*15+4*12*20+5*15*25}</math><br />
<br />
[[1998 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[1998 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[1998 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
Suppose in each day on a certain planet, there are only 10 hours and every hour has 100 minutes. What is the measure, in degrees, of the acute angle formed by the hour hand and the minute hand at 6 o'clock 75 minutes?<br />
<br />
[[1998 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
There were many balls which were distributed into 1998 boxes and all these boxes were arranged in a row. If the second box from the left-hand contained 7 balls and any 4 consecutive boxes always had a total of 30 balls, how many balls were there in the right-hand box? <br />
<br />
[[1998 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
After a mathematics test, each of the 25 students in the class got a quick look at the teacher’s grade sheet. Each student noticed five A’s. No student saw all the grades and no student saw her or his own grade. What is the minimum number of students who scored an A on <br />
this test? <br />
<br />
[[1998 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
<br />
[[1998 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
A boy arranges three kinds of books which are 30 mm, 24 mm, and 18 mm thick, respectively. He places only books of the same thickness into 3 stacks of equal height, and wants to make the height as small as possible. How many books would be used in this arrangement? <br />
<br />
[[1998 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
How many triangles are there with side lengths whole numbers and <br />
with a perimeter of 10 cm ? <br />
<br />
[[1998 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
Find the number of factors of 960.<br />
<br />
[[1998 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
What is the units digit of <math>2^{1998}+3^{1998}</math>?<br />
<br />
[[1998 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[1998 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
Every year there is at least one Friday the thirteenth, but no year has more than three. This year there are exactly three : in February, March and November. When will the next year be that contains exactly three Friday the thirteenths? <br />
<br />
[[1998 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
<br />
[[1998 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
Construct a rectangle by putting together nine squares with sides equal to 1, 4, 7, 8, 9, 10, 14, 15 and 18. What is the sum of the areas of the squares on the 4 corners of the resulting rectangle ? <br />
<br />
[[1998 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
What is the 1998th number in the following sequence ?<br />
1, -2, 2, -3, 3, -3, 4, -4, 4, -4, 5, -5, 5, -5, 5, -6, 6, -6, 6, -6, 6,........<br />
[[1998 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
Tom started work on a job alone for 30 days. Jerry continued the job<br />
alone for 5 days, and finally they worked together for another 10 days<br />
to complete that job. For the same job, if Tom and Jerry work<br />
together, they can complete it in 20 days. Assuming Tom and Jerry<br />
each work at a constant rate throughout, how many days will Tom take<br />
to complete that job alone?<br />
[[1998 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
The set L consists of all positive integers which leave a remainder of 1<br />
when divided by 3. A member of L (other than 1) is called an L-prime<br />
if it is not the product of two members of L, other than itself and 1.<br />
Which is the 8th L-prime?<br />
[[1998 PMWC Problems/Problem T3|Solution]]<br />
<br />
== Problem T4 == <br />
There are many circles on a plane. Each is divided into four parts by<br />
two mutually perpendicular diameters. Each part is painted either red,<br />
yellow or blue. No matter how the circles are rotated in the plane, they<br />
are different from one another. At most how many circles are painted<br />
with all three colors?<br />
[[1998 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
Find the largest positive integer with the following properties :<br />
(a) all the digits are different.<br />
(b) each two consecutive digits form a number divisible by either 17<br />
or 23.<br />
[[1998 PMWC Problems/Problem T5|Solution]]<br />
<br />
== Problem T6 ==<br />
There were 3 students in an athletics competition of at least two<br />
events. Each student participated in all events. In each event, student<br />
who finished second got more points than the student who finished<br />
third but less than the student who finished first. All scores were<br />
positive integers and all the events used the same 3 scores. At the end<br />
of the competition, the total scores of the 3 students were 5, 9 and 16.<br />
Determine the first-place score for each event.<br />
[[1998 PMWC Problems/Problem T6|Solution]]<br />
<br />
== Problem T7 == <br />
A leaf is torn from a book of not more than 500 pages. The sum of the<br />
remaining pages numbers is 19905. What is the sum of the two page<br />
numbers of the leaf torn out ?<br />
[[1998 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
A rectangular lawn is surrounded by a path 1 meter in width and<br />
forming a larger rectangle. The dimensions of the lawn are in whole<br />
number of metres and the area of the path equals the area of the lawn.<br />
Find the smallest possible area of the path in metres .<br />
[[1998 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
A, B, C, D and E play a game in which each is either a lion or a goat.<br />
A lion’s statement is always false and a goat’s statement is always true.<br />
A says B is not a goat.<br />
C says D is a lion.<br />
E says A is not a lion.<br />
B says C is not a goat.<br />
D says that E and A are different kinds of animals.<br />
Who are the lions?<br />
[[1998 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
In the following expression, each letter represents a digit. Same digits<br />
are represented by the same letter, and different letters stand for<br />
different digits. Any digit can replace any square, find the 5-digit<br />
number ABCBA?<br />
<br />
[[1998 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_I11&diff=255861997 PMWC Problems/Problem I112008-04-27T01:32:56Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>==Problem==<br />
A rectangle ABCD is made up of five small congruent rectangles as shown in the given figure. Find the perimeter, in cm, of ABCD if its area is <math>6750 cm^2</math>. <br />
[[Image:ABCD.gif]]<br />
<br />
==Solution==<br />
<br />
Let <math>l</math> and <math>w</math> be the length, and width, respectively, of one of the small rectangles.<br />
<br />
<math>3w=2l</math><br />
<br />
<math>l=\dfrac{3}{2}w</math><br />
<br />
<math>6750= 5lw = \dfrac{15}{2}w^2</math><br />
<br />
<math>w=30</math><br />
<br />
<math>l=45</math><br />
<br />
The perimeter of the big rectangle is<br />
<br />
<math>2(w+l)+6w=330</math><br />
<br />
==See also==<br />
{{PMWC box|year=1997|num-b=I10|num-a=I12}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Primary_Mathematics_World_Contest&diff=25585Primary Mathematics World Contest2008-04-27T01:30:35Z<p>Archimedes1: /* Curriculum */</p>
<hr />
<div>The '''Primary Mathematics World Contest''' ('''PMWC''') is a contest which started in 1997, sponsored by the Po Leung Kuk foundation. It is designed for the middle school level, and consists of an individual round and a team round. Each team consists of four people aged 13 or under. <br />
<br />
The competition itself takes place in Hong Kong.<br />
<br />
*Individual Round: The individual round consists of 15 problems, numbered I1-I15.<br />
*Team Round: The team round consists of 10 problems, numbered T1-T10.<br />
<br />
== Curriculum ==<br />
The problems are similar to [[AMC 10]]/[[AMC 12]] level, with an emphasis on number theory, geometry and logic.<br />
<br />
== See also ==<br />
*[[PMWC Problems and Solutions]]<br />
<br />
[[Category:Mathematics competitions]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_forums&diff=24441AoPS Wiki:AoPS forums2008-04-02T03:51:49Z<p>Archimedes1: /* Post Ranking */</p>
<hr />
<div>== Types of Members ==<br />
There are four types of members on the [[Art of Problem Solving]]-[[MathLinks]] forums.<br />
<br />
# Regular member<br />
# Moderator<br />
# Administrator<br />
# Bot<br />
<br />
=== Regular Member ===<br />
A regular member has the basic abilities on the forum.<br />
<br />
=== Moderator ===<br />
Every forum has at least one moderator (with the exception of the <url>index.php?f=224 Test Forum</url>). The moderator of a certain forum has the ability to edit or delete any post in that forum. Additionally, the moderator can split posts from a topic, merge posts from one topic into another, move an entire thread to a different forum and lock/unlock any topic. Finally, the moderator can make topics become stickies or announcements (and vice versa).<br />
<br />
==== Becoming a Moderator ====<br />
The selection of moderators is done by the administrators. There is no set process. New moderators will be chosen only when there is a need for them such as when a new forum is built, other moderators step down, or a forum begins to require additional supervision.<br />
<br />
The process of choosing moderators is not democratic. There is no election. The administrators choose moderators based on their trust and confidence in a member. Being a good, productive member is the best way for one to improve their likelihood of becoming a moderator.<br />
<br />
=== Administrator ===<br />
The full-time members of the Art of Problem Solving staff are administrators. Administrators basically have unlimited power and complete jurisdiction. The administrators include [[David Patrick]], [[Richard Rusczyk]], [[Vanessa Rusczyk]], [[Naoki Sato]] and [[Valentin Vornicu]]. Mathew Crawford has stepped down as an administrator in fall 2006.<br />
<br />
=== Bot ===<br />
The bots are screen names created by search engines that crawl around the website collecting data so that the [[Art of Problem Solving]]-[[MathLinks]] sites will show up in the search results of their engine.<br />
<br />
Examples of bots are AskJeeves, GigaBlast, Yahoo! Slurp, GoogleBot, and MSNBot<br />
<br />
== Post Ranking ==<br />
On the Art of Problem Solving-MathLinks website, under your username, you will find stars, as well as the name of one of the [[Millenium Problems]]. The number of stars you have, as well as the name of the Millenium Problem, depends on your post count. Here is the table that determines your "rank."<br />
<br />
*0 - 19 New Member (Zero Stars)<br />
<br />
*20 - 49 P versus NP (Half Star)<br />
<br />
*50 - 99 Hodge Conjecture (One Star)<br />
<br />
*100 - 249 Poincare Conjecture (Two Stars)<br />
<br />
*250 - 499 Riemann's Hypothesis (Two and Half stars) <br />
<br />
*500 - 999 Yang Mills Theory (Three Stars)<br />
<br />
*1000 - 2499 Navier-Stokes Equation (Four Stars)<br />
<br />
*2500 - <math>\infty</math> Birch & Swinnerton Dyer. (Five Stars)<br />
<br />
*Administrators have six stars.<br />
<br />
These are the [http://www.claymath.org/millennium/ Clay Mathematics Institute's] unsolved "Millenium Problems."<br />
<br />
See the [[AoPS-Mathlinks Rules and Tips]] page.<br />
<br />
[[Category:AoPSWiki|AoPS forums]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Brute_forcing&diff=24437Brute forcing2008-04-02T00:00:29Z<p>Archimedes1: </p>
<hr />
<div>Brute forcing is generally accepted as the term for solving a problem in a roundabout, time-consuming, uncreative, and inconvenient method.<br />
<br />
<br />
Given the problem "How many outfits can you create with thirteen hats and seven pairs of shoes?", a method involving brute force would be to list all 91 possibilities.<br />
<br />
Another method of brute force is the [[Greedy Algorithm]]. As an example, given two sets <math>\{{a}_1,{a}_2,\ldots,{a}_n\}</math> and <math>\{b_1,b_2,\ldots,b_3\}</math> how can we maximize the sum of <math>\sum_{i,j \in n} a_ib_j</math>? We sort the sets such that they are in increasing or decreasing order; then, the maximal sum is <math>a_1b_1 + a_2b_2 + a_3b_3 + \ldots a_nb_n</math>. The "greedy" part is when we maximize the sum each step by taking the largest possible term to add.<br />
<br />
See the [[Rearrangement Inequality]] for consequences of the example (and a more formal proof).<br />
<br />
== See also ==<br />
* [[Casework]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_9&diff=240192000 AIME II Problems/Problem 92008-03-15T04:09:18Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Given that <math>z</math> is a complex number such that <math>z+\frac 1z=2\cos 3^\circ</math>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>.<br />
<br />
== Solution ==<br />
Note that if z is on the unit circle in the complex plane, then <math>z = e^{i\theta} = \cos \theta + i\sin \theta</math> and <math>\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta</math><br />
<br />
We have <math>z+\frac 1z = 2\cos \theta = 2\cos 3^\circ</math> and <math>\theta = 3^\circ</math><br />
Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math><br />
<br />
Using [[De Moivre's Theorem]] we have <math>z^{2000} = \cos 6000^\circ + i\sin 6000^\circ</math>, <math>6000 = 16(360) + 240</math>, so <br />
<math>z^{2000} = \cos 240^\circ + i\sin 240^\circ</math><br />
<br />
We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math><br />
<br />
Finally, the least integer greater than <math>-1</math> is <math>\boxed{000}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=8|num-a=10}}</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=23308AMC 122008-02-16T00:52:50Z<p>Archimedes1: /* Curriculum */</p>
<hr />
<div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br />
<br />
High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br />
<br />
The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br />
<br />
The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br />
<br />
<br />
== Format ==<br />
<br />
The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. As of 2008, calculators are not permitted.<br />
<br />
The AMC 12 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br />
<br />
== Curriculum ==<br />
The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br />
<br />
== Resources ==<br />
=== Links ===<br />
* [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br />
* The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC12.php AMC 12 guide].<br />
* [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br />
* The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br />
* [[AMC 12 Problems and Solutions]]<br />
<br />
=== Recommended reading ===<br />
* [http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php Problem and solution books for past AMC exams].<br />
* Introduction to Counting & Probability by Dr. [[David Patrick]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3 Information]<br />
* Introduction to Geometry by [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Information]<br />
* The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Information].<br />
* The Art of Problem Solving Volume II by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=2 Information].<br />
<br />
<br />
=== AMC Preparation Classes ===<br />
* [[AoPS]] hosts an [http://www.artofproblemsolving.com/Classes/AoPS_C_About.php online school] teaching introductory classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br />
* [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php Math Jam Schedule]<br />
* [[EPGY]] offers an AMC 12 preparation class.<br />
<br />
<br />
== See also ==<br />
* [[Mathematics competitions]]<br />
* [[ARML]]<br />
* [[Mathematics summer programs]]<br />
<br />
<br />
<br />
[[Category:Mathematics competitions]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_21&diff=232002003 AMC 10A Problems/Problem 212008-02-12T04:32:49Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected? <br />
<br />
<math> \mathrm{(A) \ } 22\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 27\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 729 </math><br />
<br />
== Solution ==<br />
'''Solution 1'''<br />
This solution uses the [[Balls and Urns]] method:<br />
<br />
<br />
'''Solution 2'''<br />
Let the ordered triplet <math>(x,y,z)</math> represent the assortment of <math>x</math> chocolate chip cookies, <math>y</math> oatmeal cookies, and <math>z</math> peanut butter cookies. <br />
<br />
Using [[casework]]: <br />
<br />
Pat selects <math>0</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-0=6</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(0,6,0); (0,5,1); (0,4,2); (0,3,3); (0,2,4); (0,1,5); (0,0,6)\} \rightarrow 7</math> assortments. <br />
<br />
Pat selects <math>1</math> chocolate chip cookie: <br />
<br />
Pat needs to select <math>6-1=5</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(1,5,0); (1,4,1); (1,3,2); (0,2,3); (1,1,4); (1,0,5) \} \rightarrow 6</math> assortments. <br />
<br />
Pat selects <math>2</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-2=4</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(2,4,0); (2,3,1); (2,2,2); (2,1,3); (2,0,4)\} \rightarrow 5</math> assortments. <br />
<br />
Pat selects <math>3</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-3=3</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(3,3,0); (3,2,1); (3,1,2); (3,0,3)\} \rightarrow 4</math> assortments. <br />
<br />
Pat selects <math>4</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-4=2</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(4,2,0); (4,1,1); (4,0,2)\} \rightarrow 3</math> assortments. <br />
<br />
Pat selects <math>5</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-5=1</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(5,1,0); (5,0,1)\} \rightarrow 2</math> assortments. <br />
<br />
Pat selects <math>6</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-6=0</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The only assortment is: <math>\{(6,0,0)\} \rightarrow 1</math> assortment. <br />
<br />
The total number of assortments of cookies that can be collected is <math>7+6+5+4+3+2+1=28 \Rightarrow D</math><br />
<br />
== See also==<br />
{{AMC10 box|year=2003|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_21&diff=231992003 AMC 10A Problems/Problem 212008-02-12T04:32:21Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected? <br />
<br />
<math> \mathrm{(A) \ } 22\qquad \mathrm{(B) \ } 25\qquad \mathrm{(C) \ } 27\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 729 </math><br />
<br />
== Solution ==<br />
'''Solution 1'''<br />
The following solution uses the [[Balls and Urns]] method:<br />
'''Solution 2'''<br />
Let the ordered triplet <math>(x,y,z)</math> represent the assortment of <math>x</math> chocolate chip cookies, <math>y</math> oatmeal cookies, and <math>z</math> peanut butter cookies. <br />
<br />
Using [[casework]]: <br />
<br />
Pat selects <math>0</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-0=6</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(0,6,0); (0,5,1); (0,4,2); (0,3,3); (0,2,4); (0,1,5); (0,0,6)\} \rightarrow 7</math> assortments. <br />
<br />
Pat selects <math>1</math> chocolate chip cookie: <br />
<br />
Pat needs to select <math>6-1=5</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(1,5,0); (1,4,1); (1,3,2); (0,2,3); (1,1,4); (1,0,5) \} \rightarrow 6</math> assortments. <br />
<br />
Pat selects <math>2</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-2=4</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(2,4,0); (2,3,1); (2,2,2); (2,1,3); (2,0,4)\} \rightarrow 5</math> assortments. <br />
<br />
Pat selects <math>3</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-3=3</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(3,3,0); (3,2,1); (3,1,2); (3,0,3)\} \rightarrow 4</math> assortments. <br />
<br />
Pat selects <math>4</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-4=2</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(4,2,0); (4,1,1); (4,0,2)\} \rightarrow 3</math> assortments. <br />
<br />
Pat selects <math>5</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-5=1</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The assortments are: <math>\{(5,1,0); (5,0,1)\} \rightarrow 2</math> assortments. <br />
<br />
Pat selects <math>6</math> chocolate chip cookies: <br />
<br />
Pat needs to select <math>6-6=0</math> more cookies that are either oatmeal or peanut butter. <br />
<br />
The only assortment is: <math>\{(6,0,0)\} \rightarrow 1</math> assortment. <br />
<br />
The total number of assortments of cookies that can be collected is <math>7+6+5+4+3+2+1=28 \Rightarrow D</math><br />
<br />
== See also==<br />
{{AMC10 box|year=2003|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_16&diff=231592004 AMC 12A Problems/Problem 162008-02-11T05:02:43Z<p>Archimedes1: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The [[set]] of all [[real number]]s <math>x</math> for which<br />
<br />
<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath><br />
<br />
is defined is <math>\{x|x > c\}</math>. What is the value of <math>c</math>?<br />
<br />
<math>\text {(A)} 0\qquad \text {(B)}2001^{2002} \qquad \text {(C)}2002^{2003} \qquad \text {(D)}2003^{2004} \qquad \text {(E)}2001^{2002^{2003}}</math><br />
<br />
== Solution ==<br />
We know that the domain of <math>\log_k n</math>, where <math>k</math> is a [[constant]], is <math>n > 0</math>. So <math>\log_{2003}(\log_{2002}(\log_{2001}{x})) > 0</math>. By the definition of [[logarithm]]s, we then have <math>\log_{2002}(\log_{2001}{x})) > 2003^0 = 1</math>. Then <math>\log_{2001}{x} > 2002^1 = 2002</math> and <math>x > 2001^{2002}\ \mathrm{(B)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2004|ab=A|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems&diff=231582004 AMC 12A Problems2008-02-11T05:02:28Z<p>Archimedes1: /* Problem 16 */</p>
<hr />
<div>== Problem 1 ==<br />
Alicia earns <math> 20</math> dollars per hour, of which <math>1.45\%</math> is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?<br />
<br />
<math>\mathrm {(A)} 0.0029 \qquad \mathrm {(B)} 0.029 \qquad \mathrm {(C)} 0.29 \qquad \mathrm {(D)} 2.9 \qquad \mathrm {(E)} 29</math><br />
<br />
[[2004 AMC 12A Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
On the AMC 12, each correct answer is worth <math>6</math> points, each incorrect answer is worth <math>0</math> points, and each problem left unanswered is worth <math>2.5</math> points. If Charlyn leaves <math>8</math> of the <math>25</math> problems unanswered, how many of the remaining problems must she answer correctly in order to score at least <math>100</math>?<br />
<br />
<math>\mathrm {(A)} 11 \qquad \mathrm {(B)} 13 \qquad \mathrm {(C)} 14 \qquad \mathrm {(D)} 16 \qquad \mathrm {(E)} 17</math><br />
<br />
[[2004 AMC 12A Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
For how many ordered pairs of positive integers <math>(x,y)</math> is <math>x+2y=100</math>? <br />
<br />
<math>\mathrm {(A)} 33 \qquad \mathrm {(B)} 49 \qquad \mathrm {(C)} 50 \qquad \mathrm {(D)} 99 \qquad \mathrm {(E)} 100</math><br />
<br />
[[2004 AMC 12A Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Bertha has <math>6</math> daughters and no sons. Some of her daughters have <math>6</math> daughters, and the rest have none. Bertha has a total of <math>30</math> daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no children? <br />
<br />
<math>\mathrm {(A)} 22 \qquad \mathrm {(B)} 23 \qquad \mathrm {(C)} 24 \qquad \mathrm {(D)} 25 \qquad \mathrm {(E)} 26</math><br />
<br />
[[2004 AMC 12A Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The graph of the line <math>y=mx+b</math> is shown. Which of the following is true?<br />
<br />
{{image}}<br />
<br />
<math>\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<0 \qquad \mathrm {(C)} mb=0 \qquad \mathrm {(D)} 0<mb<1 \qquad \mathrm {(E)} mb>1</math><br />
<br />
[[2004 AMC 12A Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Let <math>U=2\cdot 2004^{2005}</math>, <math>V=2004^{2005}</math>, <math>W=2003\cdot 2004^{2004}</math>, <math>X=2\cdot 2004^{2004}</math>, <math>Y=2004^{2004}</math> and <math>Z=2004^{2003}</math>. Which of the following is the largest? <br />
<br />
<math>\mathrm {(A)} U-V \qquad \mathrm {(B)} V-W \qquad \mathrm {(C)} W-X \qquad \mathrm {(D)} X-Y \qquad \mathrm {(E)} Y-Z \qquad</math><br />
<br />
[[2004 AMC 12A Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
A game is played with tokens according to the following rules. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players <math>A</math>, <math>B</math> and <math>C</math> start with <math>15</math>, <math>14</math> and <math>13</math> tokens, respectively. How many rounds will there be in the game? <br />
<br />
<math>\mathrm {(A)} 36 \qquad \mathrm {(B)} 37 \qquad \mathrm {(C)} 38 \qquad \mathrm {(D)} 39 \qquad \mathrm {(E)} 40 \qquad</math><br />
<br />
[[2004 AMC 12A Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
In the overlapping triangles <math>\triangle{ABC}</math> and <math>\triangle{ABE}</math> sharing common side <math>AB</math>, <math>\angle{EAB}</math> and <math>\angle{ABC}</math> are right angles, <math>AB=4</math>, <math>BC=6</math>, <math>AE=8</math>, and <math>\overline{AC}</math> and <math>\overline{BE}</math> intersect at <math>D</math>. What is the difference between the areas of <math>\triangle{ADE}</math> and <math>\triangle{BDC}</math>? <br />
<br />
<math>\mathrm {(A)} 2 \qquad \mathrm {(B)} 4 \qquad \mathrm {(C)} 5 \qquad \mathrm {(D)} 8 \qquad \mathrm {(E)} 9 \qquad</math><br />
<br />
[[2004 AMC 12A Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diameter of the jars is increased by <math>25\%</math> without altering the volume, by what percent must the height be decreased?<br />
<br />
<math>\text {(A)} 10\% \qquad \text {(B)} 25\% \qquad \text {(C)} 36\% \qquad \text {(D)} 50\% \qquad \text {(E)}60\%</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
The sum of <math>49</math> consecutive integers is <math>7^5</math>. What is their median?<br />
<br />
<math>\text {(A)} 7 \qquad \text {(B)} 7^2\qquad \text {(C)} 7^3\qquad \text {(D)} 7^4\qquad \text {(E)}7^5</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is <math>20</math> cents. If she had one more quarter, the average value would be <math>21</math> cents. How many dimes does she have in her purse?<br />
<br />
<math>\text {(A)}0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3\qquad \text {(E)}4</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math>A = (0,9)</math> and <math>B = (0,12)</math>. Points <math>A'</math> and <math>B'</math> are on the line <math>y = x</math>, and <math>\overline{AA'}</math> and <math>\overline{BB'}</math> intersect at <math>C = (2,8)</math>. What is the length of <math>\overline{A'B'}</math>?<br />
<br />
<math>\text {(A)} 2 \qquad \text {(B)} 2\sqrt2 \qquad \text {(C)} 3 \qquad \text {(D)} 2 + \sqrt 2\qquad \text {(E)}3\sqrt 2</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Let <math>S</math> be the set of points <math>(a,b)</math> in the coordinate plane, where each of <math>a</math> and <math>b</math> may be <math>- 1</math>, <math>0</math>, or <math>1</math>. How many distinct lines pass through at least two members of <math>S</math>?<br />
<br />
<math>\text {(A)} 8 \qquad \text {(B)} 20 \qquad \text {(C)} 24 \qquad \text {(D)} 27\qquad \text {(E)}36</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
A sequence of three real numbers forms an arithmetic progression with a first term of <math>9</math>. If <math>2</math> is added to the second term and <math>20</math> is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?<br />
<br />
<math>\text {(A)} 1 \qquad \text {(B)} 4 \qquad \text {(C)} 36 \qquad \text {(D)} 49 \qquad \text {(E)}81</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run <math>100</math> meters. They next meet after Sally has run <math>150</math> meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?<br />
<br />
<math>\text {(A)}250 \qquad \text {(B)}300 \qquad \text {(C)}350 \qquad \text {(D)} 400\qquad \text {(E)}500</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
The set of all real numbers <math>x</math> for which<br />
<br />
<cmath>\log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x})))</cmath><br />
<br />
is defined is <math>\{x|x > c\}</math>. What is the value of <math>c</math>?<br />
<br />
<math>\text {(A)} 0\qquad \text {(B)}2001^{2002} \qquad \text {(C)}2002^{2003} \qquad \text {(D)}2003^{2004} \qquad \text {(E)}2001^{2002^{2003}}</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
Let <math>f</math> be a function with the following properties:<br />
<br />
<math>(i) f(1) = 1</math>, and<br />
<br />
<math>(ii) f(2n) = n\times f(n)</math>, for any positive integer <math>n</math>.<br />
<br />
What is the value of <math>f(2^{100})</math>?<br />
<br />
<math>\text {(A)} 1 \qquad \text {(B)} 2^{99} \qquad \text {(C)} 2^{100} \qquad \text {(D)} 2^{4950} \qquad \text {(E)}2^{9999}</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
Square <math>ABCD</math> has side length <math>2</math>. A semicircle with diameter <math>\overline{AB}</math> is constructed inside the square, and the tangent to the semicricle from <math>C</math> intersects side <math>\overline{AD}</math> at <math>E</math>. What is the length of <math>\overline{CE}</math>?<br />
<br />
<center>[[Image:AMC10_2004A_22.png]]</center><br />
<br />
<math>\text {(A)} \frac {2 + \sqrt5}{2} \qquad \text {(B)} \sqrt 5 \qquad \text {(C)} \sqrt 6 \qquad \text {(D)} \frac52 \qquad \text {(E)}5 - \sqrt5</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Circles <math>A, B</math> and <math>C</math> are externally tangent to each other, and internally tangent to circle <math>D</math>. Circles <math>B</math> and <math>C</math> are congruent. Circle <math>A</math> has radius <math>1</math> and passes through the center of <math>D</math>. What is the radius of circle <math>B</math>?<br />
<br />
<math>\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
Select numbers <math>a</math> and <math>b</math> between <math>0</math> and <math>1</math> independently and at random, and let <math>c</math> be their sum. Let <math>A, B</math> and <math>C</math> be the results when <math>a, b</math> and <math>c</math>, respectively, are rounded to the nearest integer. What is the probability that <math>A + B = C</math>?<br />
<br />
<math>\text {(A)} \frac14 \qquad \text {(B)} \frac13 \qquad \text {(C)} \frac12 \qquad \text {(D)} \frac23 \qquad \text {(E)}\frac34</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
If <math>\sum_{n = 0}^{\infty}{\cos^{2n}\theta = 5</math>, what is the value of <math>\cos{2\theta}</math>?<br />
<br />
<math>\text {(A)} \frac15 \qquad \text {(B)} \frac25 \qquad \text {(C)} \frac {\sqrt5}{5}\qquad \text {(D)} \frac35 \qquad \text {(E)}\frac45</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Three mutually tangent spheres of radius <math>1</math> rest on a horizontal plane. A sphere of radius <math>2</math> rests on them. What is the distance from the plane to the top of the larger sphere?<br />
<br />
<math>\text {(A)} 3 + \frac {\sqrt {30}}{2} \qquad \text {(B)} 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)} 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)} \frac {52}{9}\qquad \text {(E)}3 + 2\sqrt2</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
A polynomial<br />
<br />
<cmath>P(x) = c_{2004}x^{2004} + c_{2003}x^{2003} + ... + c_1x + c_0</cmath><br />
<br />
has real coefficients with <math>c_{2004}\not = 0</math> and <math>2004</math> distinct complex zeroes <math>z_k = a_k + b_ki</math>, <math>1\leq k\leq 2004</math> with <math>a_k</math> and <math>b_k</math> real, <math>a_1 = b_1 = 0</math>, and<br />
<br />
<cmath>\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}.</cmath><br />
<br />
Which of the following quantities can be a nonzero number?<br />
<br />
<math>\text {(A)} c_0 \qquad \text {(B)} c_{2003} \qquad \text {(C)} b_2b_3...b_{2004} \qquad \text {(D)} \sum_{k = 1}^{2004}{a_k} \qquad \text {(E)}\sum_{k = 1}^{2004}{c_k}</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
A plane contains points <math>A</math> and <math>B</math> with <math>AB = 1</math>. Let <math>S</math> be the union of all disks of radius <math>1</math> in the plane that cover <math>\overline{AB}</math>. What is the area of <math>S</math>?<br />
<br />
<math>\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
For each integer <math>n\geq 4</math>, let <math>a_n</math> denote the base-<math>n</math> number <math>0.\overline{133}_n</math>. The product <math>a_4a_5...a_{99}</math> can be expressed as <math>\frac {m}{n!}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is as small as possible. What is the value of <math>m</math>?<br />
<br />
<math>\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962</math><br />
<br />
<br />
[[2004 AMC 12A Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
* [[AMC 12]]<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[2004 AMC 12A]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=25 2004 AMC A Math Jam Transcript]<br />
* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_2&diff=231572004 AMC 12A Problems/Problem 22008-02-11T01:47:13Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>==Problem==<br />
On the AMC 12, each correct answer is worth <math>6</math> points, each incorrect answer is worth <math>0</math> points, and each problem left unanswered is worth <math>2.5</math> points. If Charlyn leaves <math>8</math> of the <math>25</math> problems unanswered, how many of the remaining problems must she answer correctly in order to score at least <math>100</math>?<br />
<br />
<math>\text {(A)} 11 \qquad \text {(B)} 13 \qquad \text {(C)} 14 \qquad \text {(D)} 16\qquad \text {(E)}17</math><br />
<br />
==Solution==<br />
<br />
She gets <math>8*2.5=20</math> points for the problems she didn't answer. She must get <math>\lceil \frac{100-20}{6} \rceil =14 \Rightarrow \text {(C)}</math> problems right to score at least 100.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2004|ab=A|num-b=1|num-a=3}}<br />
[[Category:Introductory Algebra Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_15&diff=227641989 AIME Problems/Problem 152008-01-25T02:23:18Z<p>Archimedes1: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
[[Point]] <math>P^{}_{}</math> is inside <math>\triangle ABC^{}_{}</math>. Line segments <math>APD^{}_{}</math>, <math>BPE^{}_{}</math>, and <math>CPF^{}_{}</math> are drawn with <math>D^{}_{}</math> on <math>BC^{}_{}</math>, <math>E^{}_{}</math> on <math>AC^{}_{}</math>, and <math>F{}{}^{}_{}</math> on <math>AB^{}_{}</math> (see the figure below). Given that <math>AP=6^{}_{}</math>, <math>BP=9^{}_{}</math>, <math>PD=6^{}_{}</math>, <math>PE=3^{}_{}</math>, and <math>CF=20^{}_{}</math>, find the [[area]] of <math>\triangle ABC^{}_{}</math>.<br />
<br />
{|<br />
|-<br />
| __TOC__<br />
| [[Image:AIME_1989_Problem_15.png]]<br />
|}<br />
== Solution ==<br />
=== Solution 1 ===<br />
Because we're given three concurrent [[cevian]]s and their lengths, it seems very tempting to apply [[mass point]]s. We immediately see that <math>w_E = 3</math>, <math>w_B = 1</math>, and <math>w_A = w_D = 2</math>. Now, we recall that the masses on the three sides of the triangle must be balanced out, so <math>w_C = 1</math> and <math>w_E = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>. <br />
<br />
Recalling that <math>w_C = w_B = 1</math>, we see that <math>DC = DB</math> and <math>DP</math> is a [[median]] to <math>BC</math> in <math>\triangle BCP</math>. Applying [[Stewart's Theorem]], <math>BC^2 + 12^2 = 2(15^2 + 9^2)</math>, and <math>BC = 6\sqrt {13}</math>. Now notice that <math>2[BCP] = [ABC]</math>, because both triangle share the same base and the <math>h_{\triangle ABC} = 2h_{\triangle BCP}</math>. Applying [[Heron's formula]] on triangle <math>BCP</math> with sides <math>15</math>, <math>9</math>, and <math>6\sqrt3</math>, <math>[BCP] = 54</math> and <math>[ABC] = \boxed{108}</math>.<br />
<br />
=== Solution 2 ===<br />
Using a different form of [[Ceva's Theorem]], we have <math>\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}</math><br />
<br />
Solving <math>4y = x + y</math> and <math>x + y = 20</math>, we obtain <math>x = BP = 15</math> and <math>y = EP = 5</math>.<br />
<br />
Let <math>Q</math> be the point on <math>AB</math> such that <math>FC\parallel QD</math>.<br />
Since <math>AP = PD</math> and <math>FP\parallel QD</math>, <math>QD = 2FP = 6</math>. (Midline Theorem)<br />
<br />
Also, since <math>FC\parallel QD</math> and <math>\QD = \frac12FC</math>, we see that <math>FQ = QB</math>, <math>BD = DC</math>, etc. ([[Midline Theorem]])<br />
Similarly, we have <math>PR = RB</math> (<math>= \frac12PB = 7.5</math>) and thus <math>RD = \frac12PC = 4.5</math>.<br />
<br />
<math>PDR</math> is a <math>3-4-5</math> [[right triangle]], so <math>\angle PDR</math> (<math>\angle ADQ</math>) is <math>90^\circ</math>.<br />
Therefore, the area of <math>\triangle ADQ = \frac12\cdot 12\cdot 6 = 36</math>.<br />
Using area ratio, <math>\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = 108</math>.<br />
<br />
=== Solution 3 ===<br />
Because the length of cevian <math>BE</math> is unknown, we can examine what happens when we extend it or decrease its length and see that it simply changes the angles between the cevians. Wouldn't it be great if it the length of <math>BE</math> was such that <math>\angle APC = 90^\circ</math>? Let's first assume it's a right angle and hope that everything works out. <br />
<br />
Extend <math>AD</math> to <math>Q</math> so that <math>PD = DQ = 6</math>. The result is that <math>BQ = 9</math>, <math>PQ = 12</math>, and <math>BP = 15</math> because <math>\triangle CDP\cong \triangle BDQ</math>. Now we see that if we are able to show that <math>BE = 20</math>, that is <math>PE = 5</math>, then our right angle assumption will be true.<br />
<br />
Apply the [[Pythagorean Theorem]] on <math>\triangle APC</math> to get <math>AC = 3\sqrt {13}</math>, so <math>AE = \sqrt {13}</math> and <math>CE = 2\sqrt {13}</math>. Now, we apply the [[Law of Cosines]] on triangles <math>CEP</math> and <math>AEP</math>. <br />
<br />
Let <math>PE = x</math>. Notice that <math>\angle CEB = 180^\circ - \angle AEB</math> and <math>\cos CEB = - \cos AEB</math>, so we get two nice equations.<br />
<br />
<math>81 = 52 + y^2 - 2y \sqrt {13}\cos CEF</math><br />
<math>36 = 13 + y^2 + y \sqrt {13} \cos CEF</math><br />
<br />
Solving, <math>y = 5</math> (yay!). <br />
<br />
Now, the area is easy to find. <math>[ABC] = [AQB] + [APC] = \frac12(9)(18) + \frac12(6)(9) = 108</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=14|after=Final Question}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2006_Cyprus_MO/Lyceum/Problem_5&diff=183122006 Cyprus MO/Lyceum/Problem 52007-10-18T02:24:15Z<p>Archimedes1: /* Solution */ added</p>
<hr />
<div>==Problem==<br />
If both integers <math>\alpha,\beta</math> are bigger than 1 and satisfy <math>a^7=b^8</math>, then the minimum value of <math>\alpha+\beta</math> is<br />
<br />
A. <math>384</math><br />
<br />
B. <math>2</math><br />
<br />
C. <math>15</math><br />
<br />
D. <math>56</math><br />
<br />
E. <math>512</math><br />
<br />
==Solution==<br />
Since <math>b</math> is greater than <math>1</math> and therefore not equal to zero, we can divide both sides of the equation by <math>b^7</math> to obtain <math>a^7/b^7=b</math>, or<br />
<cmath>\left( \frac{a}{b} \right) ^7=b</cmath><br />
Since <math>b</math> is an integer, we must have <math>a/b</math> is an integer. So, we can start testing out seventh powers of integers. <br />
<math>a/b=1</math> doesn't work, since <math>a</math> and <math>b</math> are defined to be greater than <math>1</math>. The next smallest thing we try is <math>a/b=2</math>. This gives <math>b=(a/b)^7=2^7=128</math>, so <math>a=2b=2(128)=256</math>. Thus, our sum is <math>128+256=\boxed{384}</math>.<br />
<br />
==See also==<br />
{{CYMO box|year=2006|l=Lyceum|num-b=4|num-a=6}}</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Permutation&diff=17988Permutation2007-10-14T01:06:47Z<p>Archimedes1: /* See also */</p>
<hr />
<div>{{stub}}<br />
<br />
<br />
A '''permutation''' of a [[set]] of <math>r</math> objects is any rearrangement (linear ordering) of the <math>r</math> objects. There are <math>\displaystyle r!</math> (the [[factorial]] of <math>r</math>) permutations of a set with <math>r</math> distinct objects.<br />
<br />
One can also consider permutations of [[infinite]] sets. In this case, a permutation of a set <math>S</math> is simply a [[bijection]] between <math>S</math> and itself.<br />
==Notations==<br />
A given permutation of a [[finite]] set can be denoted in a variety of ways. The most straightforward representation is simply to write down what the permutation looks like. For example, the permutations of the set <math>\{1, 2, 3\}</math> are <math>\{1,2,3\}, \{1, 3,2\}, \{2,1,3\}, \{2,3,1\},\{3,1,2\}</math> and <math>\{3,2,1\}</math>. We often drop the brackets and commas, so the permutation <math>\{2,1,3\}</math> would just be represented by <math>213</math>.<br />
<br />
Another common notation is cycle notation. <br />
<br />
==The Symmetric Group==<br />
The set of all permutations of an <math>n</math>-element set is denoted <math>S_n</math>. In fact, <math>S_n</math> forms a [[group]], known as the [[Symmetric group]], under the operation of permutation composition.<br />
<br />
<br />
<br />
A permutation in which no obect remains in the same place it started is called a [[derangement]].<br />
<br />
==Picking ordered subsets of a set==<br />
An important question is how many ways to pick an <math>r</math>-element [[subset]] of a set with <math>n</math> [[element]]s, where order matters. To find how many ways we can do this, note that for the first of the <math>r</math> elements, we have <math>n</math> different objects we can choose from. For the second element, there are <math>n-1</math> objects we can choose, <math>n-2</math> for the third, and so on. In general, the number of ways to permute <math>r</math> objects from a set of <math>n</math> is given by<br />
<math>P(n,r)=n(n-1)(n-2)\cdots(n-r+1)=\frac{n!}{(n-r)!}</math>.<br />
<br />
== See also ==<br />
*[[Combinations]]<br />
* [[Combinatorics]]<br />
* [[Derangement]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Community_portal&diff=17568AoPS Wiki:Community portal2007-10-08T04:03:14Z<p>Archimedes1: /* Dirty Work */</p>
<hr />
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<br />
[[Category:AoPSWiki]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_28&diff=17515University of South Carolina High School Math Contest/1993 Exam/Problem 282007-10-07T21:29:17Z<p>Archimedes1: /* Solution */ removed "need solution" tag</p>
<hr />
<div>== Problem ==<br />
Suppose <math>\triangle ABC</math> is a triangle with 3 acute angles <math>A, B,</math> and <math>C</math>. Then the point <math>( \cos B - \sin A, \sin B - \cos A)</math><br />
<br />
(A) can be in the 1st quadrant and can be in the 2nd quadrant only<br />
<br />
(B) can be in the 3rd quadrant and can be in the 4th quadrant only<br />
<br />
(C) can be in the 2nd quadrant and can be in the 3rd quadrant only<br />
<br />
(D) can be in the 2nd quadrant only<br />
<br />
(E) can be in any of the 4 quadrants<br />
<br />
== Solution ==<br />
<br />
<br />
First note that <math>\cos{A}, \cos{B}, \sin{A}, \sin{B}>0</math> and <math>A+B>90</math>.<br />
<br />
Since <math>\cos B+\sin A>0</math>, <math>\cos B-\sin A</math> and <math>\cos^2B-\sin^2A</math> have the same sign. Similarly, <math>\sin B-\cos A</math> and <math>\sin^2B-\cos^2A</math> have the same sign.<br />
<br />
Notice that <math>\cos^2B-\sin^2A=-(\sin^2B-\cos^2A)</math>. Then <math>(\cos B-\sin A, \sin B-\cos A)</math> can only lie in the 2nd and 4th quadrants.<br />
<br />
----<br />
<br />
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 27|Previous Problem]]<br />
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Next Problem]]<br />
* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]<br />
<br />
[[Category:Intermediate Trigonometry Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Law_of_Cosines&diff=17468Law of Cosines2007-10-07T02:19:49Z<p>Archimedes1: Added Proofs section</p>
<hr />
<div>{{WotWAnnounce|week=Oct 4-Oct 10}}<br />
<br />
The '''Law of Cosines''' is a theorem which relates the side-[[length]]s and [[angle]]s of a [[triangle]]. For a triangle with [[edge]]s of length <math>a</math>, <math>b</math> and <math>c</math> opposite [[angle]]s of measure <math>A</math>, <math>B</math> and <math>C</math>, respectively, the Law of Cosines states:<br />
<br />
<math>c^2 = a^2 + b^2 - 2ab\cos C</math><br />
<br />
<math>b^2 = a^2 + c^2 - 2ac\cos B</math><br />
<br />
<math>a^2 = b^2 + c^2 - 2bc\cos A</math><br />
<br />
In the case that one of the angles has measure <math>90^\circ</math> (is a [[right angle]]), the corresponding statement reduces to the [[Pythagorean Theorem]].<br />
==Proofs==<br />
==See also==<br />
* [[Law of Sines]]<br />
* [[Trigonometry]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems&diff=174102005 PMWC Problems2007-10-05T19:34:40Z<p>Archimedes1: Team Problems</p>
<hr />
<div>== Problem I1 ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>?<br />
<br />
[[2005 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
Let <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}</math>, where <math>a</math> and <math>b</math> are different four-digit positive integers (natural numbers) and <math>c</math> is a five-digit positive integer (natural number). What is the number <math>c</math>?<br />
<br />
[[2005 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
Let <math>x</math> be a fraction between <math>\frac{35}{36}</math> and <math>\frac{91}{183}</math>. If the denominator of <math>x</math> is <math>455</math> and the numerator and denominator have no common factor except <math>1</math>, how many possible values are there for <math>x</math>?<br />
<br />
[[2005 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2005 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
Consider the following conditions on the positive integer (natural number) <math>a</math>:<br />
<br />
1. <math>3a + 5 > 40</math><br />
<br />
2. <math>49a \ge 301</math><br />
<br />
3. <math>20a \le 999</math><br />
<br />
4. <math>101a + 53 \ge 2332</math><br />
<br />
5. <math>15a – 7 \ge 144</math><br />
<br />
If only three of these conditions are true, what is the value of <math>a</math>?<br />
<br />
[[2005 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
A group of <math>100</math> people consists of men, women and children (at least one of each). Exactly <math>200</math> apples are distributed in such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possible ways can this be done?<br />
<br />
[[2005 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
How many numbers are there in the list <math>1, 2, 3, 4, 5, \dots, 10000</math> which contain exactly two consecutive <math>9</math>'s such as <math>993, 1992</math> and <math>9929</math>, but not <math>9295</math> or <math>1999</math>?<br />
<br />
[[2005 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
Some people in Hong Kong express <math>2/8</math> as 8th Feb and others express <math>2/8</math> as<br />
2nd Aug. This can be confusing as when we see <math>2/8</math>, we don’t know whether it<br />
is 8th Feb or 2nd Aug. However, it is easy to understand <math>9/22</math> or <math>22/9</math> as 22nd Sept, because there are only <math>12</math> months in a year. How many dates in a year can cause this confusion?<br />
<br />
[[2005 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
There are four consecutive positive integers (natural numbers) less than <math>2005</math> such that the first (smallest) number is a multiple of <math>5</math>, the second number is a multiple of <math>7</math>, the third number is a multiple of <math>9</math> and the last number is a multiple of <math>11</math>. What is the first of these four numbers?<br />
<br />
[[2005 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
A long string is folded in half eight times, then cut in the middle. How many<br />
pieces are obtained?<br />
<br />
[[2005 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
There are 4 men: A, B, C and D. Each has a son. The four sons are asked to<br />
enter a dark room. Then A, B, C and D enter the dark room, and each of them<br />
walks out with just one child. If none of them comes out with his own son, in<br />
how many ways can this happen?<br />
<br />
[[2005 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2005 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall?<br />
<br />
[[2005 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
On a balance scale, three green balls balance six blue balls, two yellow balls<br />
balance five blue balls and six blue balls balance four white balls. How many blue balls are needed to balance four green, two yellow and two white balls?<br />
<br />
[[2005 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
The sum of the two three-digit integers, <math>\text{6A2}</math> and <math>\text{B34}</math>, is divisible by <math>18</math>. What is the largest possible product of <math>\text{A}</math> and <math>\text{B}</math>?<br />
<br />
[[2005 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
Call an integer "happy", if the sum of its digits is <math>10</math>. How many<br />
"happy" integers are there between <math>100</math> and <math>1000</math>?<br />
<br />
[[2005 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 ==<br />
Compute the sum of <math>a</math>, <math>b</math> and <math>c</math> given that <math>\frac{a}{2}=\frac{b}{3}=\frac{c}{5}</math> and the product of <math>a</math>, <math>b</math> and <math>c</math> is <math>1920</math>.<br />
<br />
[[2005 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
Replace the letters <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> in the following expression with<br />
the numbers <math>1</math>, <math>2</math>, <math>3</math> and <math>4</math>, without repetition:<br />
<cmath>a+\cfrac{1}{b+\cfrac{1}{c+\cfrac{1}{d}}}</cmath><br />
Find the difference between the maximum value and the minimum<br />
value of the expression.<br />
<br />
[[2005 PMWC Problems/Problem T3|Solution]]<br />
<br />
== Problem T4 ==<br />
Buses from town A to town B leave every hour on the hour<br />
(for example: 6:00, 7:00, …).<br />
Buses from town B to town A leave every hour on the half hour<br />
(for example: 6:30, 7:30, …).<br />
The trip between town A and town B takes 5 hours. Assume the<br />
buses travel on the same road.<br />
If you get on a bus from town A, how many buses from town B<br />
do you pass on the road (not including those at the stations)?<br />
<br />
[[2005 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
Mr. Wong has a <math>7</math>-digit phone number <math>\text{ABCDEFG}</math>. The sum of<br />
the number formed by the first <math>4</math> digits <math>\text{ABCD}</math> and the number<br />
formed by the last <math>3</math> digits <math>\text{EFG}</math> is <math>9063</math>. The sum of the number<br />
formed by the first <math>3</math> digits <math>\text{ABC}</math> and the number formed by the<br />
last <math>4</math> digits <math>\text{DEFG}</math> is <math>2529</math>. What is Mr. Wong’s phone number?<br />
<br />
[[2005 PMWC Problems/Problem T5|Solution]]<br />
<br />
== Problem T6 ==<br />
<cmath><br />
\begin{eqnarray*}<br />
1+2 &=& 3 \\<br />
4+5+6 &=& 7+8 \\<br />
9+10+11+12 &=& 13+14+15</cmath><br />
<cmath>\vdots</cmath><br />
If this pattern is continued, find the last number in the &80&th row<br />
(e.g. the last number of the third row is &15&).<br />
<br />
[[2005 PMWC Problems/Problem T6|Solution]]<br />
<br />
== Problem T7 ==<br />
Skipper’s doghouse has a regular hexagonal base that measures<br />
one metre on each side. Skipper is tethered to a 2-metre rope<br />
which is fixed to a vertex. What is the area of the region outside<br />
the doghouse that Skipper can reach? Calculate an approximate<br />
answer by using <math>\pi=3.14</math> or <math>\pi=22/7</math>.<br />
<br />
[[2005 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
An isosceles right triangle is removed from each corner of a<br />
square piece of paper so that a rectangle of unequal sides remains.<br />
If the sum of the areas of the cut-off pieces is <math>200 \text{cm}^2</math> and the<br />
lengths of the legs of the triangles cut off are integers, find the<br />
area of the rectangle.<br />
<br />
[[2005 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
<br />
[[2005 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
Find the largest 12-digit number for which every two consecutive<br />
digits form a distinct 2-digit prime number.<br />
<br />
[[2005 PMWC Problems/Problem T10|Solution]]<br />
== See Also ==<br />
<br />
* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems/Problem_I1&diff=173482005 PMWC Problems/Problem I12007-10-02T02:24:53Z<p>Archimedes1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112\ldots585960</math>?<br />
<br />
== Solution ==<br />
The length of the resulting number is the same no matter what, so to maximize the number we want to invoke the [[greedy algorithm]] - get as many 9s in the first several digits as possible. There are <math>9 + 2(51) = 111</math> digits in the original number, so after deleting 100 we will be left with <math>11</math> digits. Now the number of nines is <math>6</math>. Applying the greety algorithm, we might expect to get <math>99999\ldots</math>, but we promptly run out of digits. So instead, we start with 5 nines, <math>99999</math>. Naturally we would then try to fit an 8 next, but again it turns out that we run out of digits. So we continue with 7. We still need five more digits, and we have six left to choose from. It quickly becomes apparant that our answer is <math>99999785960</math>.<br />
<br />
== See also ==<br />
{{PMWC box|year=2005|before=First question|num-a=I2}}</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=PMWC_Problems_and_Solutions&diff=17289PMWC Problems and Solutions2007-09-30T21:49:01Z<p>Archimedes1: </p>
<hr />
<div>This is a list of all '''PMWC''' tests in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available at [http://www.txstate.edu/mathworks/camps/pmwc/previoustests.html]. If you find problems there which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome. <br />
<br />
* [[1997 PMWC]]<br />
* [[1998 PMWC]] <br />
* [[1999 PMWC]] <br />
* [[2000 PMWC]] <br />
* [[2001 PMWC]] <br />
* [[2002 PMWC]] <br />
* [[2003 PMWC]] <br />
* [[2004 PMWC]] <br />
* [[2005 PMWC]] <br />
* [[2006 PMWC]] <br />
* [[2007 PMWC]] <br />
[[Category:Math Contest Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2007_PMWC&diff=172882007 PMWC2007-09-30T21:45:55Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[2007 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[2007 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[2007 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2007 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
<br />
[[2007 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
<br />
[[2007 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
<br />
[[2007 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
<br />
[[2007 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
<br />
[[2007 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[2007 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[2007 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2007 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
<br />
[[2007 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
<br />
[[2007 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[2007 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2007 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[2007 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
<br />
[[2007 PMWC Problems/Problem T3|Solution]]<br />
<br />
== Problem T4 == <br />
<br />
[[2007 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
<br />
[[2007 PMWC Problems/Problem T5|Solution]]<br />
<br />
== Problem T6 ==<br />
<br />
[[2007 PMWC Problems/Problem T6|Solution]]<br />
<br />
== Problem T7 == <br />
<br />
[[2007 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
<br />
[[2007 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[2007 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2007 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2006_PMWC&diff=172872006 PMWC2007-09-30T21:45:05Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[2006 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[2006 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[2006 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2006 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
<br />
[[2006 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
<br />
[[2006 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
<br />
[[2006 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
<br />
[[2006 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
<br />
[[2006 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[2006 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[2006 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2006 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
<br />
[[2006 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
<br />
[[2006 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[2006 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2006 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[2006 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
<br />
[[2006 PMWC Problems/Problem T3|Solution]]<br />
<br />
== Problem T4 == <br />
<br />
[[2006 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
<br />
[[2006 PMWC Problems/Problem T5|Solution]]<br />
<br />
== Problem T6 ==<br />
<br />
[[2006 PMWC Problems/Problem T6|Solution]]<br />
<br />
== Problem T7 == <br />
<br />
[[2006 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
<br />
[[2006 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[2006 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2006 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2004_PMWC_Problems&diff=172862004 PMWC Problems2007-09-30T21:43:50Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[2004 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[2004 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[2004 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2004 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
<br />
[[2004 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
<br />
[[2004 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
<br />
[[2004 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
<br />
[[2004 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
<br />
[[2004 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[2004 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[2004 PMWC Problems/Problem I11|Solution]]<br />
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== Problem I12 ==<br />
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[[2004 PMWC Problems/Problem I12|Solution]]<br />
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== Problem I13 ==<br />
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[[2004 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
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[[2004 PMWC Problems/Problem I14|Solution]]<br />
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== Problem I15 ==<br />
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[[2004 PMWC Problems/Problem I15|Solution]]<br />
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== Problem T1 ==<br />
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[[2004 PMWC Problems/Problem T1|Solution]]<br />
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== Problem T2 == <br />
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[[2004 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
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[[2004 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
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[[2004 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[2004 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[2004 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
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[[2004 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
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[[2004 PMWC Problems/Problem T8|Solution]]<br />
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== Problem T9 ==<br />
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[[2004 PMWC Problems/Problem T9|Solution]]<br />
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== Problem T10 ==<br />
<br />
[[2004 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2003_PMWC_Problems&diff=172852003 PMWC Problems2007-09-30T21:43:04Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
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[[2003 PMWC Problems/Problem I1|Solution]]<br />
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== Problem I2 ==<br />
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[[2003 PMWC Problems/Problem I2|Solution]]<br />
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== Problem I3 ==<br />
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[[2003 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
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[[2003 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
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[[2003 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
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[[2003 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
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[[2003 PMWC Problems/Problem I7|Solution]]<br />
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== Problem I8 ==<br />
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[[2003 PMWC Problems/Problem I8|Solution]]<br />
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== Problem I9 ==<br />
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[[2003 PMWC Problems/Problem I9|Solution]]<br />
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== Problem I10 ==<br />
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[[2003 PMWC Problems/Problem I10|Solution]]<br />
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== Problem I11 ==<br />
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[[2003 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
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[[2003 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
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[[2003 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
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[[2003 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
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[[2003 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2003 PMWC Problems/Problem T1|Solution]]<br />
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== Problem T2 == <br />
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[[2003 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
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[[2003 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
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[[2003 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[2003 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[2003 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
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[[2003 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
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[[2003 PMWC Problems/Problem T8|Solution]]<br />
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== Problem T9 ==<br />
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[[2003 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2003 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2002_PMWC_Problems&diff=172842002 PMWC Problems2007-09-30T21:41:45Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
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[[2002 PMWC Problems/Problem I1|Solution]]<br />
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== Problem I2 ==<br />
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[[2002 PMWC Problems/Problem I2|Solution]]<br />
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== Problem I3 ==<br />
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[[2002 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
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[[2002 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
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[[2002 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
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[[2002 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
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[[2002 PMWC Problems/Problem I7|Solution]]<br />
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== Problem I8 ==<br />
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[[2002 PMWC Problems/Problem I8|Solution]]<br />
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== Problem I9 ==<br />
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[[2002 PMWC Problems/Problem I9|Solution]]<br />
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== Problem I10 ==<br />
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[[2002 PMWC Problems/Problem I10|Solution]]<br />
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== Problem I11 ==<br />
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[[2002 PMWC Problems/Problem I11|Solution]]<br />
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== Problem I12 ==<br />
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[[2002 PMWC Problems/Problem I12|Solution]]<br />
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== Problem I13 ==<br />
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[[2002 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
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[[2002 PMWC Problems/Problem I14|Solution]]<br />
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== Problem I15 ==<br />
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[[2002 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
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[[2002 PMWC Problems/Problem T1|Solution]]<br />
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== Problem T2 == <br />
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[[2002 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
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[[2002 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
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[[2002 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[2002 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[2002 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
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[[2002 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
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[[2002 PMWC Problems/Problem T8|Solution]]<br />
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== Problem T9 ==<br />
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[[2002 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2002 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2001_PMWC_Problems&diff=172832001 PMWC Problems2007-09-30T21:39:23Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
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[[2001 PMWC Problems/Problem I1|Solution]]<br />
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== Problem I2 ==<br />
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[[2001 PMWC Problems/Problem I2|Solution]]<br />
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== Problem I3 ==<br />
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[[2001 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
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[[2001 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
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[[2001 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
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[[2001 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
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[[2001 PMWC Problems/Problem I7|Solution]]<br />
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== Problem I8 ==<br />
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[[2001 PMWC Problems/Problem I8|Solution]]<br />
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== Problem I9 ==<br />
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[[2001 PMWC Problems/Problem I9|Solution]]<br />
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== Problem I10 ==<br />
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[[2001 PMWC Problems/Problem I10|Solution]]<br />
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== Problem I11 ==<br />
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[[2001 PMWC Problems/Problem I11|Solution]]<br />
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== Problem I12 ==<br />
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[[2001 PMWC Problems/Problem I12|Solution]]<br />
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== Problem I13 ==<br />
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[[2001 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
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[[2001 PMWC Problems/Problem I14|Solution]]<br />
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== Problem I15 ==<br />
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[[2001 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2001 PMWC Problems/Problem T1|Solution]]<br />
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== Problem T2 == <br />
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[[2001 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
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[[2001 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
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[[2001 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[2001 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[2001 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
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[[2001 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
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[[2001 PMWC Problems/Problem T8|Solution]]<br />
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== Problem T9 ==<br />
<br />
[[2001 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2001 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2000_PMWC_Problems&diff=172822000 PMWC Problems2007-09-30T21:37:47Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[2000 PMWC Problems/Problem I1|Solution]]<br />
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== Problem I2 ==<br />
<br />
[[2000 PMWC Problems/Problem I2|Solution]]<br />
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== Problem I3 ==<br />
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[[2000 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
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[[2000 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
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[[2000 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
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[[2000 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
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[[2000 PMWC Problems/Problem I7|Solution]]<br />
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== Problem I8 ==<br />
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[[2000 PMWC Problems/Problem I8|Solution]]<br />
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== Problem I9 ==<br />
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[[2000 PMWC Problems/Problem I9|Solution]]<br />
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== Problem I10 ==<br />
<br />
[[2000 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[2000 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2000 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
<br />
[[2000 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
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[[2000 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[2000 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2000 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[2000 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
<br />
[[2000 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
<br />
[[2000 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[2000 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[2000 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
<br />
[[2000 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
<br />
[[2000 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[2000 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2000 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1999_PMWC_Problems&diff=172811999 PMWC Problems2007-09-30T21:36:08Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[1999 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[1999 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[1999 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[1999 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
<br />
[[1999 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
<br />
[[1999 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
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[[1999 PMWC Problems/Problem I7|Solution]]<br />
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== Problem I8 ==<br />
<br />
[[1999 PMWC Problems/Problem I8|Solution]]<br />
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== Problem I9 ==<br />
<br />
[[1999 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[1999 PMWC Problems/Problem I10|Solution]]<br />
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== Problem I11 ==<br />
<br />
[[1999 PMWC Problems/Problem I11|Solution]]<br />
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== Problem I12 ==<br />
<br />
[[1999 PMWC Problems/Problem I12|Solution]]<br />
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== Problem I13 ==<br />
<br />
[[1999 PMWC Problems/Problem I13|Solution]]<br />
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== Problem I14 ==<br />
<br />
[[1999 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[1999 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[1999 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[1999 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
<br />
[[1999 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
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[[1999 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
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[[1999 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
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[[1999 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
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[[1999 PMWC Problems/Problem T7|Solution]]<br />
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== Problem T8 ==<br />
<br />
[[1999 PMWC Problems/Problem T8|Solution]]<br />
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== Problem T9 ==<br />
<br />
[[1999 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[1999 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1998_PMWC_Problems&diff=172801998 PMWC Problems2007-09-30T21:35:08Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[1998 PMWC Problems/Problem I1|Solution]]<br />
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== Problem I2 ==<br />
<br />
[[1998 PMWC Problems/Problem I2|Solution]]<br />
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== Problem I3 ==<br />
<br />
[[1998 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
<br />
[[1998 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
<br />
[[1998 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
<br />
[[1998 PMWC Problems/Problem I6|Solution]]<br />
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== Problem I7 ==<br />
<br />
[[1998 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
<br />
[[1998 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
<br />
[[1998 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[1998 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[1998 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[1998 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
<br />
[[1998 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
<br />
[[1998 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[1998 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[1998 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[1998 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
<br />
[[1998 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
<br />
[[1998 PMWC Problems/Problem T4|Solution]]<br />
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== Problem T5 ==<br />
<br />
[[1998 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
<br />
[[1998 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
<br />
[[1998 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
<br />
[[1998 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[1998 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[1998 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=1997_PMWC&diff=172791997 PMWC2007-09-30T21:34:00Z<p>Archimedes1: New page: == Problem I1 == Solution == Problem I2 == Solution == Problem I3 == Solution ==...</p>
<hr />
<div>== Problem I1 ==<br />
<br />
[[1997 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
<br />
[[1997 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
<br />
[[1997 PMWC Problems/Problem I3|Solution]]<br />
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== Problem I4 ==<br />
<br />
[[1997 PMWC Problems/Problem I4|Solution]]<br />
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== Problem I5 ==<br />
<br />
[[1997 PMWC Problems/Problem I5|Solution]]<br />
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== Problem I6 ==<br />
<br />
[[1997 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
<br />
[[1997 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
<br />
[[1997 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
<br />
[[1997 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
<br />
[[1997 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
<br />
[[1997 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[1997 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
<br />
[[1997 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
<br />
[[1997 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
<br />
[[1997 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[1997 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 == <br />
<br />
[[1997 PMWC Problems/Problem T2|Solution]]<br />
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== Problem T3 ==<br />
<br />
[[1997 PMWC Problems/Problem T3|Solution]]<br />
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== Problem T4 == <br />
<br />
[[1997 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
<br />
[[1997 PMWC Problems/Problem T5|Solution]]<br />
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== Problem T6 ==<br />
<br />
[[1997 PMWC Problems/Problem T6|Solution]]<br />
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== Problem T7 == <br />
<br />
[[1997 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
<br />
[[1997 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[1997 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[1997 PMWC Problems/Problem T10|Solution]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems&diff=172402005 PMWC Problems2007-09-30T16:58:01Z<p>Archimedes1: </p>
<hr />
<div>== Problem I1 ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>?<br />
<br />
[[2005 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
Let <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}</math>, where <math>a</math> and <math>b</math> are different four-digit positive integers (natural numbers) and <math>c</math> is a five-digit positive integer (natural number). What is the number <math>c</math>?<br />
<br />
[[2005 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
Let <math>x</math> be a fraction between <math>\frac{35}{36}</math> and <math>\frac{91}{183}</math>. If the denominator of <math>x</math> is <math>455</math> and the numerator and denominator have no common factor except <math>1</math>, how many possible values are there for <math>x</math>?<br />
<br />
[[2005 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2005 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
Consider the following conditions on the positive integer (natural number) <math>a</math>:<br />
<br />
1. <math>3a + 5 > 40</math><br />
<br />
2. <math>49a \ge 301</math><br />
<br />
3. <math>20a \le 999</math><br />
<br />
4. <math>101a + 53 \ge 2332</math><br />
<br />
5. <math>15a – 7 \ge 144</math><br />
<br />
If only three of these conditions are true, what is the value of <math>a</math>?<br />
<br />
[[2005 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
A group of <math>100</math> people consists of men, women and children (at least one of each). Exactly <math>200</math> apples are distributed in such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possible ways can this be done?<br />
<br />
[[2005 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
How many numbers are there in the list <math>1, 2, 3, 4, 5, \dots, 10000</math> which contain exactly two consecutive <math>9</math>'s such as <math>993, 1992</math> and <math>9929</math>, but not <math>9295</math> or <math>1999</math>?<br />
<br />
[[2005 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
Some people in Hong Kong express <math>2/8</math> as 8th Feb and others express <math>2/8</math> as<br />
2nd Aug. This can be confusing as when we see <math>2/8</math>, we don’t know whether it<br />
is 8th Feb or 2nd Aug. However, it is easy to understand <math>9/22</math> or <math>22/9</math> as 22nd Sept, because there are only <math>12</math> months in a year. How many dates in a year can cause this confusion?<br />
<br />
[[2005 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
There are four consecutive positive integers (natural numbers) less than <math>2005</math> such that the first (smallest) number is a multiple of <math>5</math>, the second number is a multiple of <math>7</math>, the third number is a multiple of <math>9</math> and the last number is a multiple of <math>11</math>. What is the first of these four numbers?<br />
<br />
[[2005 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
A long string is folded in half eight times, then cut in the middle. How many<br />
pieces are obtained?<br />
<br />
[[2005 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
There are 4 men: A, B, C and D. Each has a son. The four sons are asked to<br />
enter a dark room. Then A, B, C and D enter the dark room, and each of them<br />
walks out with just one child. If none of them comes out with his own son, in<br />
how many ways can this happen?<br />
<br />
[[2005 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2005 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall?<br />
<br />
[[2005 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
On a balance scale, three green balls balance six blue balls, two yellow balls<br />
balance five blue balls and six blue balls balance four white balls. How many blue balls are needed to balance four green, two yellow and two white balls?<br />
<br />
[[2005 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
The sum of the two three-digit integers, <math>\text{6A2}</math> and <math>\text{B34}</math>, is divisible by <math>18</math>. What is the largest possible product of <math>\text{A}</math> and <math>\text{B}</math>?<br />
<br />
[[2005 PMWC Problems/Problem I15|Solution]]<br />
<br />
== Problem T1 ==<br />
<br />
[[2005 PMWC Problems/Problem T1|Solution]]<br />
<br />
== Problem T2 ==<br />
<br />
[[2005 PMWC Problems/Problem T2|Solution]]<br />
<br />
== Problem T3 ==<br />
<br />
[[2005 PMWC Problems/Problem T3|Solution]]<br />
<br />
== Problem T4 ==<br />
<br />
[[2005 PMWC Problems/Problem T4|Solution]]<br />
<br />
== Problem T5 ==<br />
<br />
[[2005 PMWC Problems/Problem T5|Solution]]<br />
<br />
== Problem T6 ==<br />
<br />
[[2005 PMWC Problems/Problem T6|Solution]]<br />
<br />
== Problem T7 ==<br />
<br />
[[2005 PMWC Problems/Problem T7|Solution]]<br />
<br />
== Problem T8 ==<br />
<br />
[[2005 PMWC Problems/Problem T8|Solution]]<br />
<br />
== Problem T9 ==<br />
<br />
[[2005 PMWC Problems/Problem T9|Solution]]<br />
<br />
== Problem T10 ==<br />
<br />
[[2005 PMWC Problems/Problem T10|Solution]]<br />
== See Also ==<br />
<br />
* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems&diff=172392005 PMWC Problems2007-09-30T16:54:02Z<p>Archimedes1: individual probs I1,I2,I3...</p>
<hr />
<div>== Problem I1 ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>?<br />
<br />
[[2005 PMWC Problems/Problem I1|Solution]]<br />
<br />
== Problem I2 ==<br />
Let <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}</math>, where <math>a</math> and <math>b</math> are different four-digit positive integers (natural numbers) and <math>c</math> is a five-digit positive integer (natural number). What is the number <math>c</math>?<br />
<br />
[[2005 PMWC Problems/Problem I2|Solution]]<br />
<br />
== Problem I3 ==<br />
Let <math>x</math> be a fraction between <math>\frac{35}{36}</math> and <math>\frac{91}{183}</math>. If the denominator of <math>x</math> is <math>455</math> and the numerator and denominator have no common factor except <math>1</math>, how many possible values are there for <math>x</math>?<br />
<br />
[[2005 PMWC Problems/Problem I3|Solution]]<br />
<br />
== Problem I4 ==<br />
<br />
[[2005 PMWC Problems/Problem I4|Solution]]<br />
<br />
== Problem I5 ==<br />
Consider the following conditions on the positive integer (natural number) <math>a</math>:<br />
<br />
1. <math>3a + 5 > 40</math><br />
<br />
2. <math>49a \ge 301</math><br />
<br />
3. <math>20a \le 999</math><br />
<br />
4. <math>101a + 53 \ge 2332</math><br />
<br />
5. <math>15a – 7 \ge 144</math><br />
<br />
If only three of these conditions are true, what is the value of <math>a</math>?<br />
<br />
[[2005 PMWC Problems/Problem I5|Solution]]<br />
<br />
== Problem I6 ==<br />
A group of <math>100</math> people consists of men, women and children (at least one of each). Exactly <math>200</math> apples are distributed in such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possible ways can this be done?<br />
<br />
[[2005 PMWC Problems/Problem I6|Solution]]<br />
<br />
== Problem I7 ==<br />
How many numbers are there in the list <math>1, 2, 3, 4, 5, \dots, 10000</math> which contain exactly two consecutive <math>9</math>'s such as <math>993, 1992</math> and <math>9929</math>, but not <math>9295</math> or <math>1999</math>?<br />
<br />
[[2005 PMWC Problems/Problem I7|Solution]]<br />
<br />
== Problem I8 ==<br />
Some people in Hong Kong express <math>2/8</math> as 8th Feb and others express <math>2/8</math> as<br />
2nd Aug. This can be confusing as when we see <math>2/8</math>, we don’t know whether it<br />
is 8th Feb or 2nd Aug. However, it is easy to understand <math>9/22</math> or <math>22/9</math> as 22nd Sept, because there are only <math>12</math> months in a year. How many dates in a year can cause this confusion?<br />
<br />
[[2005 PMWC Problems/Problem I8|Solution]]<br />
<br />
== Problem I9 ==<br />
There are four consecutive positive integers (natural numbers) less than <math>2005</math> such that the first (smallest) number is a multiple of <math>5</math>, the second number is a multiple of <math>7</math>, the third number is a multiple of <math>9</math> and the last number is a multiple of <math>11</math>. What is the first of these four numbers?<br />
<br />
[[2005 PMWC Problems/Problem I9|Solution]]<br />
<br />
== Problem I10 ==<br />
A long string is folded in half eight times, then cut in the middle. How many<br />
pieces are obtained?<br />
<br />
[[2005 PMWC Problems/Problem I10|Solution]]<br />
<br />
== Problem I11 ==<br />
There are 4 men: A, B, C and D. Each has a son. The four sons are asked to<br />
enter a dark room. Then A, B, C and D enter the dark room, and each of them<br />
walks out with just one child. If none of them comes out with his own son, in<br />
how many ways can this happen?<br />
<br />
[[2005 PMWC Problems/Problem I11|Solution]]<br />
<br />
== Problem I12 ==<br />
<br />
[[2005 PMWC Problems/Problem I12|Solution]]<br />
<br />
== Problem I13 ==<br />
Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall?<br />
<br />
[[2005 PMWC Problems/Problem I13|Solution]]<br />
<br />
== Problem I14 ==<br />
On a balance scale, three green balls balance six blue balls, two yellow balls<br />
balance five blue balls and six blue balls balance four white balls. How many blue balls are needed to balance four green, two yellow and two white balls?<br />
<br />
[[2005 PMWC Problems/Problem I14|Solution]]<br />
<br />
== Problem I15 ==<br />
The sum of the two three-digit integers, <math>\text{6A2}</math> and <math>\text{B34}</math>, is divisible by <math>18</math>. What is the largest possible product of <math>\text{A}</math> and <math>\text{B}</math>?<br />
<br />
[[2005 PMWC Problems/Problem I15|Solution]]<br />
<br />
== See Also ==<br />
<br />
* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems&diff=172372005 PMWC Problems2007-09-30T16:51:03Z<p>Archimedes1: 2005 PMWC Individual Test moved to 2005 PMWC: To put the team contests and individual contests on the ame page</p>
<hr />
<div>== Problem 1 ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Let <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}</math>, where <math>a</math> and <math>b</math> are different four-digit positive integers (natural numbers) and <math>c</math> is a five-digit positive integer (natural number). What is the number <math>c</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>x</math> be a fraction between <math>\frac{35}{36}</math> and <math>\frac{91}{183}</math>. If the denominator of <math>x</math> is <math>455</math> and the numerator and denominator have no common factor except <math>1</math>, how many possible values are there for <math>x</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Consider the following conditions on the positive integer (natural number) <math>a</math>:<br />
<br />
1. <math>3a + 5 > 40</math><br />
<br />
2. <math>49a \ge 301</math><br />
<br />
3. <math>20a \le 999</math><br />
<br />
4. <math>101a + 53 \ge 2332</math><br />
<br />
5. <math>15a – 7 \ge 144</math><br />
<br />
If only three of these conditions are true, what is the value of <math>a</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A group of <math>100</math> people consists of men, women and children (at least one of each). Exactly <math>200</math> apples are distributed in such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possible ways can this be done?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
How many numbers are there in the list <math>1, 2, 3, 4, 5, \dots, 10000</math> which contain exactly two consecutive <math>9</math>'s such as <math>993, 1992</math> and <math>9929</math>, but not <math>9295</math> or <math>1999</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Some people in Hong Kong express <math>2/8</math> as 8th Feb and others express <math>2/8</math> as<br />
2nd Aug. This can be confusing as when we see <math>2/8</math>, we don’t know whether it<br />
is 8th Feb or 2nd Aug. However, it is easy to understand <math>9/22</math> or <math>22/9</math> as 22nd Sept, because there are only <math>12</math> months in a year. How many dates in a year can cause this confusion?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
There are four consecutive positive integers (natural numbers) less than <math>2005</math> such that the first (smallest) number is a multiple of <math>5</math>, the second number is a multiple of <math>7</math>, the third number is a multiple of <math>9</math> and the last number is a multiple of <math>11</math>. What is the first of these four numbers?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A long string is folded in half eight times, then cut in the middle. How many<br />
pieces are obtained?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
There are 4 men: A, B, C and D. Each has a son. The four sons are asked to<br />
enter a dark room. Then A, B, C and D enter the dark room, and each of them<br />
walks out with just one child. If none of them comes out with his own son, in<br />
how many ways can this happen?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
On a balance scale, three green balls balance six blue balls, two yellow balls<br />
balance five blue balls and six blue balls balance four white balls. How many blue balls are needed to balance four green, two yellow and two white balls?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
The sum of the two three-digit integers, <math>\text{6A2}</math> and <math>\text{B34}</math>, is divisible by <math>18</math>. What is the largest possible product of <math>\text{A}</math> and <math>\text{B}</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 15|Solution]]<br />
<br />
== See Also ==<br />
<br />
* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Individual_Test&diff=172382005 PMWC Individual Test2007-09-30T16:51:03Z<p>Archimedes1: 2005 PMWC Individual Test moved to 2005 PMWC: To put the team contests and individual contests on the ame page</p>
<hr />
<div>#REDIRECT [[2005 PMWC]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=PMWC_Problems_and_Solutions&diff=17236PMWC Problems and Solutions2007-09-30T16:50:10Z<p>Archimedes1: UNseparating Idvdl and Team</p>
<hr />
<div>* [[1997 PMWC]]<br />
* [[1998 PMWC]] <br />
* [[1999 PMWC]] <br />
* [[2000 PMWC]] <br />
* [[2001 PMWC]] <br />
* [[2002 PMWC]] <br />
* [[2003 PMWC]] <br />
* [[2004 PMWC]] <br />
* [[2005 PMWC]] <br />
* [[2006 PMWC]] <br />
* [[2007 PMWC]] <br />
[[Category:Math Contest Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Primary_Mathematics_World_Contest&diff=17235Primary Mathematics World Contest2007-09-30T16:48:44Z<p>Archimedes1: /* Format */</p>
<hr />
<div>==Format==<br />
<br />
The '''Primary Mathematics World Contest''' (known to most as the '''PMWC''') is a contest which started in 1997, sponsored by the Po Leung Kuk foundation. It is about middle school level, consisting of an individual round and a team round. Each team consists of four people aged 13 or under. <br />
<br />
The competition itself takes place in Hong Kong.<br />
<br />
Individual Round:<br />
The individual round consists of 15 problems, numbered I1-I15.<br />
<br />
Team Round:<br />
The team round consists of 10 problems, numbered T1-T10.<br />
<br />
==Curriculum==<br />
==Links==<br />
[[PMWC Problems and Solutions]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=PMWC&diff=17234PMWC2007-09-30T16:44:52Z<p>Archimedes1: Redirecting to Primary Mathematics World Contest</p>
<hr />
<div>#REDIRECT[[Primary Mathematics World Contest]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Primary_Mathematics_World_Contest&diff=17233Primary Mathematics World Contest2007-09-30T16:33:48Z<p>Archimedes1: No calculator</p>
<hr />
<div>==Format==<br />
<br />
The Primary Mathematics World Contest (known to most as the PMWC) is a contest which started in 1997. It is about middle school level, consisting of an individual round and a team round. Each team consists of four people aged 13 or under. <br />
<br />
The competition itself takes place in Hong Kong.<br />
<br />
Individual Round:<br />
The individual round consists of 15 problems, numbered I1-I15.<br />
<br />
Team Round:<br />
The team round consists of 10 problems, numbered T1-T10.<br />
<br />
==Curriculum==<br />
==Links==<br />
[[PMWC Problems and Solutions]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=Primary_Mathematics_World_Contest&diff=17232Primary Mathematics World Contest2007-09-30T16:33:18Z<p>Archimedes1: /* Format */</p>
<hr />
<div>==Format==<br />
<br />
The Primary Mathematics World Contest (known to most as the PMWC) is a contest which started in 1997. It is about middle school level, consisting of an individual round and a team round. Each team consists of four people aged 13 or under. <br />
<br />
The competition itself takes place in Hong Kong.<br />
<br />
Individual Round:<br />
The individual round consists of 15 problems, numbered I1-I15.<br />
<br />
Team Round:<br />
The team round consists of 10 problems, numbered T1-T10. These are to be done by a team of four, with the aid of a calculator (confirmation required)<br />
<br />
==Curriculum==<br />
==Links==<br />
[[PMWC Problems and Solutions]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=2005_PMWC_Problems&diff=171942005 PMWC Problems2007-09-29T21:13:10Z<p>Archimedes1: New page: == Problem 1 == What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>? [[2005 PMW...</p>
<hr />
<div>== Problem 1 ==<br />
What is the greatest possible number one can get by discarding <math>100</math> digits, in any order, from the number <math>123456789101112 \dots 585960</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Let <math>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2005}</math>, where <math>a</math> and <math>b</math> are different four-digit positive integers (natural numbers) and <math>c</math> is a five-digit positive integer (natural number). What is the number <math>c</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>x</math> be a fraction between <math>\frac{35}{36}</math> and <math>\frac{91}{183}</math>. If the denominator of <math>x</math> is <math>455</math> and the numerator and denominator have no common factor except <math>1</math>, how many possible values are there for <math>x</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Consider the following conditions on the positive integer (natural number) <math>a</math>:<br />
<br />
1. <math>3a + 5 > 40</math><br />
<br />
2. <math>49a \ge 301</math><br />
<br />
3. <math>20a \le 999</math><br />
<br />
4. <math>101a + 53 \ge 2332</math><br />
<br />
5. <math>15a – 7 \ge 144</math><br />
<br />
If only three of these conditions are true, what is the value of <math>a</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
A group of <math>100</math> people consists of men, women and children (at least one of each). Exactly <math>200</math> apples are distributed in such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possible ways can this be done?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
How many numbers are there in the list <math>1, 2, 3, 4, 5, \dots, 10000</math> which contain exactly two consecutive <math>9</math>'s such as <math>993, 1992</math> and <math>9929</math>, but not <math>9295</math> or <math>1999</math>?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
Some people in Hong Kong express <math>2/8</math> as 8th Feb and others express <math>2/8</math> as<br />
2nd Aug. This can be confusing as when we see <math>2/8</math>, we don’t know whether it<br />
is 8th Feb or 2nd Aug. However, it is easy to understand <math>9/22</math> or <math>22/9</math> as 22nd Sept, because there are only <math>12</math> months in a year. How many dates in a year can cause this confusion?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
There are four consecutive positive integers (natural numbers) less than <math>2005</math> such that the first (smallest) number is a multiple of <math>5</math>, the second number is a multiple of <math>7</math>, the third number is a multiple of <math>9</math> and the last number is a multiple of <math>11</math>. What is the first of these four numbers?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
A long string is folded in half eight times, then cut in the middle. How many<br />
pieces are obtained?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
There are 4 men: A, B, C and D. Each has a son. The four sons are asked to<br />
enter a dark room. Then A, B, C and D enter the dark room, and each of them<br />
walks out with just one child. If none of them comes out with his own son, in<br />
how many ways can this happen?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Sixty meters of rope is used to make three sides of a rectangular camping area with a long wall used as the other side. The length of each side of the rectangle is a natural number. What is the largest area that can be enclosed by the rope and the wall?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
On a balance scale, three green balls balance six blue balls, two yellow balls<br />
balance five blue balls and six blue balls balance four white balls. How many blue balls are needed to balance four green, two yellow and two white balls?<br />
<br />
[[2005 PMWC Individual Test Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
The sum of the two three-digit integers, <math>\text{6A2}</math> and <math>\text{B34}</math>, is divisible by <math>18</math>. What is the largest possible product of <math>\text{A}</math> and <math>\text{B}</math>?<br />
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[[2005 PMWC Individual Test Problems/Problem 15|Solution]]<br />
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== See Also ==<br />
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* [[Mathematics competition resources]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=PMWC_Problems_and_Solutions&diff=17191PMWC Problems and Solutions2007-09-29T20:34:39Z<p>Archimedes1: </p>
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<div>* [[1997 PMWC Individual Test]]<br />
* [[1997 PMWC Team Test]]<br />
* [[1998 PMWC Individual Test]]<br />
* [[1998 PMWC Team Test]]<br />
* [[1999 PMWC Individual Test]]<br />
* [[1999 PMWC Team Test]]<br />
* [[2000 PMWC Individual Test]]<br />
* [[2000 PMWC Team Test]]<br />
* [[2001 PMWC Individual Test]]<br />
* [[2001 PMWC Team Test]]<br />
* [[2002 PMWC Individual Test]]<br />
* [[2002 PMWC Team Test]]<br />
* [[2003 PMWC Individual Test]]<br />
* [[2003 PMWC Team Test]]<br />
* [[2004 PMWC Individual Test]]<br />
* [[2004 PMWC Team Test]]<br />
* [[2005 PMWC Individual Test]]<br />
* [[2005 PMWC Team Test]]<br />
* [[2006 PMWC Individual Test]]<br />
* [[2006 PMWC Team Test]]<br />
* [[2007 PMWC Individual Test]]<br />
* [[2007 PMWC Team Test]]<br />
[[Category:Math Contest Problems]]</div>Archimedes1https://artofproblemsolving.com/wiki/index.php?title=PMWC_Problems_and_Solutions&diff=17190PMWC Problems and Solutions2007-09-29T20:23:50Z<p>Archimedes1: New page: * 1997 PMWC * 1998 PMWC * 1999 PMWC * 2000 PMWC * 2001 PMWC * 2002 PMWC * 2003 PMWC * 2004 PMWC * 2005 PMWC * 2006 PMWC * 2007 PMWC [[Category:...</p>
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<div>* [[1997 PMWC]]<br />
* [[1998 PMWC]]<br />
* [[1999 PMWC]]<br />
* [[2000 PMWC]]<br />
* [[2001 PMWC]]<br />
* [[2002 PMWC]]<br />
* [[2003 PMWC]]<br />
* [[2004 PMWC]]<br />
* [[2005 PMWC]]<br />
* [[2006 PMWC]]<br />
* [[2007 PMWC]]<br />
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[[Category:Math Contest Problems]]</div>Archimedes1