https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jabberwock2&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T05:25:16ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_Pre_2005_Problems/Problem_11&diff=70960Mock AIME 1 Pre 2005 Problems/Problem 112015-07-02T14:51:03Z<p>Jabberwock2: Fixed equation alignment.</p>
<hr />
<div>== Problem ==<br />
Let <math>S</math> denote the value of the sum<br />
<cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath><br />
Determine the remainder obtained when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution ==<br />
Consider the polynomial <cmath>f(x)=(x-1)^{2004}=\sum_{n=0}^{2004}\binom{2004}{n}\cdot(-1)^n x^{2004-n}.</cmath><br />
<br />
Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have <br />
<br />
<cmath> \begin{align*} \frac{f(1)+f(\omega)+f(\omega^2)}{3} &= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \\<br />
&= \frac{1}{3}\sum_{n=0}^{2004}\binom{2004}{n}\cdot(-1)^n\cdot(1^{2004-n}+\omega^{2004-n}+(\omega^2)^{2004-n}) \\<br />
&= \sum_{n=0}^{668}(-1)^n \binom{2004}{3n}.<br />
\end{align*} </cmath><br />
<br />
where the last step follows because <math>1^k+\omega^k+\omega^{2k}</math> is 0 when <math>k</math> is not divisible by 3, and <math>3</math> when <math>k</math> is divisible by 3.<br />
<br />
We now compute <math>\frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3}</math>. WLOG, let <math>\omega = \frac{-1+\sqrt{3}i}{2}, \omega^2=\frac{-1-\sqrt{3}i}{2}</math>. Then <math>\omega-1=\frac{-3+\sqrt{3}i}{2} = \sqrt{3}\cdot \frac{-\sqrt{3}+i}{2}</math>, and <math>\omega^2-1=\sqrt{3}\cdot\frac{-\sqrt{3}-i}{2}</math>. These numbers are both of the form <math>\sqrt{3}\cdot\varphi</math>, where <math>\varphi</math> is a 12th root of unity, so both of these, when raised to the 2004-th power, become <math>3^{1002}</math>. Thus, our desired sum becomes <math>2\cdot3^{1001}</math>.<br />
<br />
To find <math>2\cdot3^{1001} \pmod{1000}</math>, we notice that <math>3^{\phi{500}}\equiv 3^{200}\equiv 1 \pmod{500}</math> so that <math>3^{1001}\equiv 3 \pmod{500}</math>. Then <math>2\cdot3^{1001}=2(500k+3)=1000k+6</math>. Thus, our answer is <math>\boxed{006}</math>.<br />
<br />
== See also ==<br />
{{Mock AIME box|year=Pre 2005|n=1|num-b=10|num-a=12|source=14769}}<br />
<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_Pre_2005_Problems/Problem_11&diff=70959Mock AIME 1 Pre 2005 Problems/Problem 112015-07-02T14:46:41Z<p>Jabberwock2: </p>
<hr />
<div>== Problem ==<br />
Let <math>S</math> denote the value of the sum<br />
<cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath><br />
Determine the remainder obtained when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution ==<br />
Consider the polynomial <cmath>f(x)=(x-1)^{2004}=\sum_{n=0}^{2004}\binom{2004}{n}\cdot(-1)^n x^{2004-n}.</cmath><br />
<br />
Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have <br />
<br />
\begin{align*}<br />
\frac{f(1)+f(\omega)+f(\omega^2)}{3} \\<br />
&= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \\<br />
&= \frac{1}{3}\sum_{n=0}^{2004}\binom{2004}{n}\cdot(-1)^n\cdot(1^{2004-n}+\omega^{2004-n}+(\omega^2)^{2004-n}) \\<br />
&= \sum_{n=0}^{668}(-1)^n \binom{2004}{3n}<br />
\end{align*}<br />
<br />
where the last step follows because <math>1^k+\omega^k+\omega^{2k}</math> is 0 when <math>k</math> is not divisible by 3, and <math>3</math> when <math>k</math> is divisible by 3.<br />
<br />
We now compute <math>\frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3}</math>. WLOG, let <math>\omega = \frac{-1+\sqrt{3}i}{2}, \omega^2=\frac{-1-\sqrt{3}i}{2}</math>. Then <math>\omega-1=\frac{-3+\sqrt{3}i}{2} = \sqrt{3}\cdot \frac{-\sqrt{3}+i}{2}</math>, and <math>\omega^2-1=\sqrt{3}\cdot\frac{-\sqrt{3}-i}{2}</math>. These numbers are both of the form <math>\sqrt{3}\cdot\varphi</math>, where <math>\varphi</math> is a 12th root of unity, so both of these, when raised to the 2004-th power, become <math>3^{1002}</math>. Thus, our desired sum becomes <math>2\cdot3^{1001}</math>.<br />
<br />
To find <math>2\cdot3^{1001} \pmod{1000}</math>, we notice that <math>3^{\phi{500}}\equiv 3^{200}\equiv 1 \pmod{500}</math> so that <math>3^{1001}\equiv 3 \pmod{500}</math>. Then <math>2\cdot3^{1001}=2(500k+3)=1000k+6</math>. Thus, our answer is <math>\boxed{006}</math>.<br />
<br />
== See also ==<br />
{{Mock AIME box|year=Pre 2005|n=1|num-b=10|num-a=12|source=14769}}<br />
<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_Pre_2005_Problems/Problem_3&diff=70957Mock AIME 1 Pre 2005 Problems/Problem 32015-07-02T10:18:04Z<p>Jabberwock2: Corrected absolute values.</p>
<hr />
<div>== Problem ==<br />
<math>A, B, C, D,</math> and <math>E</math> are collinear in that order such that <math>AB = BC = 1, CD = 2,</math> and <math>DE = 9</math>. If <math>P</math> can be any point in space, what is the smallest possible value of <math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2</math>?<br />
<br />
== Solution ==<br />
Let the altitude from <math>P</math> onto <math>AE</math> at <math>Q</math> have lengths <math>PQ = h</math> and <math>AQ = r</math>. It is clear that, for a given <math>r</math> value, <math>AP</math>, <math>BP</math>, <math>CP</math>, <math>DP</math>, and <math>EP</math> are all minimized when <math>h = 0</math>. So <math>P</math> is on <math>AE</math>, and therefore, <math>P = Q</math>. Thus, <math>AP</math>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math> Squaring each of these gives:<br />
<br />
<math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2 = r^2 + (r - 1)^2 + (r - 2)^2 + (r - 4)^2 + (r - 13)^2 = 5r^2 - 40r + 190</math><br />
<br />
This reaches its minimum at <math>r = \frac {40}{2\cdot 5} = 4</math>, at which point the sum of the squares of the distances is <math>\boxed{110}</math>.<br />
<br />
== See also ==<br />
{{Mock AIME box|year=Pre 2005|n=1|num-b=2|num-a=4|source=14769}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=706711990 AIME Problems/Problem 32015-06-07T07:24:10Z<p>Jabberwock2: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>P_1^{}</math> be a [[Regular polygon|regular]] <math>r~\mbox{gon}</math> and <math>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}</math>. What's the largest possible value of <math>s_{}^{}</math>?<br />
<br />
== Solution ==<br />
The formula for the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>. <br />
<br />
Thus, <math>\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}</math>. Cross multiplying and simplifying, we get <math>\frac{58(r-2)}{r} = \frac{59(s-2)}{s}</math>. Cross multiply and combine like terms again to yield <math>58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs</math>. Solving for <math>r</math>, we get <math>r = \frac{116s}{118 - s}</math>.<br />
<br />
<math>r \ge 0</math> and <math>s \ge 0</math>, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], <math>s < 118</math>; the largest possible value of <math>s</math> is <math>117</math>.<br />
<br />
This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>116 \cdot 117</math> and <math>s = \boxed{117}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_8&diff=667222011 AIME I Problems/Problem 82014-12-31T08:49:08Z<p>Jabberwock2: </p>
<hr />
<div>== Problem ==<br />
In triangle <math>ABC</math>, <math>BC = 23</math>, <math>CA = 27</math>, and <math>AB = 30</math>. Points <math>V</math> and <math>W</math> are on <math>\overline{AC}</math> with <math>V</math> on <math> \overline{AW} </math>, points <math>X</math> and <math>Y</math> are on <math> \overline{BC} </math> with <math>X</math> on <math> \overline{CY} </math>, and points <math>Z</math> and <math>U</math> are on <math> \overline{AB} </math> with <math>Z</math> on <math> \overline{BU} </math>. In addition, the points are positioned so that <math> \overline{UV}\parallel\overline{BC} </math>, <math> \overline{WX}\parallel\overline{AB} </math>, and <math> \overline{YZ}\parallel\overline{CA} </math>. Right angle folds are then made along <math> \overline{UV} </math>, <math> \overline{WX} </math>, and <math> \overline{YZ} </math>. The resulting figure is placed on a level floor to make a table with triangular legs. Let <math>h</math> be the maximum possible height of a table constructed from triangle <math>ABC</math> whose top is parallel to the floor. Then <math>h</math> can be written in the form <math> \frac{k\sqrt{m}}{n} </math>, where <math>k</math> and <math>n</math> are relatively prime positive integers and <math>m</math> is a positive integer that is not divisible by the square of any prime. Find <math>k+m+n</math>.<br />
<br />
<br />
<center><asy><br />
unitsize(1 cm);<br />
pair translate;<br />
pair[] A, B, C, U, V, W, X, Y, Z;<br />
A[0] = (1.5,2.8);<br />
B[0] = (3.2,0);<br />
C[0] = (0,0);<br />
U[0] = (0.69*A[0] + 0.31*B[0]);<br />
V[0] = (0.69*A[0] + 0.31*C[0]);<br />
W[0] = (0.69*C[0] + 0.31*A[0]);<br />
X[0] = (0.69*C[0] + 0.31*B[0]);<br />
Y[0] = (0.69*B[0] + 0.31*C[0]);<br />
Z[0] = (0.69*B[0] + 0.31*A[0]);<br />
translate = (7,0);<br />
A[1] = (1.3,1.1) + translate;<br />
B[1] = (2.4,-0.7) + translate;<br />
C[1] = (0.6,-0.7) + translate;<br />
U[1] = U[0] + translate;<br />
V[1] = V[0] + translate;<br />
W[1] = W[0] + translate;<br />
X[1] = X[0] + translate;<br />
Y[1] = Y[0] + translate;<br />
Z[1] = Z[0] + translate;<br />
draw (A[0]--B[0]--C[0]--cycle);<br />
draw (U[0]--V[0],dashed);<br />
draw (W[0]--X[0],dashed);<br />
draw (Y[0]--Z[0],dashed);<br />
draw (U[1]--V[1]--W[1]--X[1]--Y[1]--Z[1]--cycle);<br />
draw (U[1]--A[1]--V[1],dashed);<br />
draw (W[1]--C[1]--X[1]);<br />
draw (Y[1]--B[1]--Z[1]);<br />
dot("$A$",A[0],N);<br />
dot("$B$",B[0],SE);<br />
dot("$C$",C[0],SW);<br />
dot("$U$",U[0],NE);<br />
dot("$V$",V[0],NW);<br />
dot("$W$",W[0],NW);<br />
dot("$X$",X[0],S);<br />
dot("$Y$",Y[0],S);<br />
dot("$Z$",Z[0],NE);<br />
dot(A[1]);<br />
dot(B[1]);<br />
dot(C[1]);<br />
dot("$U$",U[1],NE);<br />
dot("$V$",V[1],NW);<br />
dot("$W$",W[1],NW);<br />
dot("$X$",X[1],dir(-70));<br />
dot("$Y$",Y[1],dir(250));<br />
dot("$Z$",Z[1],NE);</asy></center><br />
<br />
==Solution==<br />
Note that the area is given by Heron's formula and it is <math>20\sqrt{221}</math>. Let <math>h_i</math> denote the length of the altitude dropped from vertex i. It follows that <math>h_b = \frac{40\sqrt{221}}{27}, h_c = \frac{40\sqrt{221}}{30}}, h_a = \frac{40\sqrt{221}}{23}</math>. From similar triangles we can see that <math>\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rightarrow h \le \frac{h_ah_c}{h_a+h_c}</math>. We can see this is true for any combination of a,b,c and thus the minimum of the upper bounds for h yields <math>h = \frac{40\sqrt{221}}{57} \rightarrow \boxed{318}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_12&diff=663902000 AIME I Problems/Problem 122014-11-29T15:43:14Z<p>Jabberwock2: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Given a [[function]] <math>f</math> for which<br />
<cmath><br />
f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath><br />
holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)?</math><br />
<br />
== Solution ==<br />
<cmath>\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\<br />
f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{eqnarray*}</cmath><br />
<br />
Since <math>\mathrm{gcd}(1056, 1760) = 352</math> we can conclude that (by the [[Euclidean algorithm]])<br />
<br />
<cmath>f(x) = f(352 + x)</cmath><br />
<br />
So we need only to consider one period <math>f(0), f(1), ... f(351)</math>, which can have at most <math>352</math> distinct values which determine the value of <math>f(x)</math> at all other integers. <br />
<br />
But we also know that <math>f(x) = f(46 - x) = f(398 - x)</math>, so the values <math>x = 24, 25, ... 46</math> and <math>x = 200, 201, ... 351</math> are repeated. This gives a total of<br />
<br />
<cmath>352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }</cmath><br />
<br />
distinct values.<br />
<br />
To show that it is possible to have <math>f(23), f(24), \ldots, f(199)</math> distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the [[cosine]] function: <math>f(x) = \cos \left(\frac{360}{352}(x-23)\right)</math> (in degrees).<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_20&diff=663861950 AHSME Problems/Problem 202014-11-29T09:49:26Z<p>Jabberwock2: /* Solution 2 */</p>
<hr />
<div>== Problem==<br />
<br />
When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math><br />
<br />
===Solution 2===<br />
<br />
By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(\mathrm{D})\ 2.} </math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1950|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_20&diff=663851950 AHSME Problems/Problem 202014-11-29T09:48:29Z<p>Jabberwock2: /* Solution 2 */</p>
<hr />
<div>== Problem==<br />
<br />
When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math><br />
<br />
===Solution 2===<br />
<br />
By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(D)\ 2.} </math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1950|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jabberwock2https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_20&diff=663841950 AHSME Problems/Problem 202014-11-29T09:47:46Z<p>Jabberwock2: /* Solution 2 */</p>
<hr />
<div>== Problem==<br />
<br />
When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math><br />
<br />
===Solution 2===<br />
<br />
By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{\textbf{(D)}\ 2.} </math><br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1950|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jabberwock2