https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ethanspoon&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T13:01:09ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_1&diff=2016422024 AIME I Problems/Problem 12023-11-10T04:25:36Z<p>Ethanspoon: Created page with "AIME I problem 1: If you tried to cheat by looking at this page, what is the probability you won't succeed? A 0 B 0 C 0 D 0 E 0"</p>
<hr />
<div>AIME I problem 1: If you tried to cheat by looking at this page, what is the probability you won't succeed?<br />
A 0<br />
B 0<br />
C 0<br />
D 0 <br />
E 0</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_13&diff=2016402024 AIME I Problems/Problem 132023-11-10T04:24:44Z<p>Ethanspoon: Created page with "GET OFF MY LAWWWWWWN"</p>
<hr />
<div>GET OFF MY LAWWWWWWN</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=1887631983 AIME Problems/Problem 112023-02-07T01:11:41Z<p>Ethanspoon: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The solid shown has a square base of side length <math>s</math>. The upper edge is parallel to the base and has length <math>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black;<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F);<br />
label("A",A,W);<br />
label("B",B,S);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,N);<br />
label("F",F,N);<br />
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --><br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
First, we find the height of the solid by dropping a perpendicular from the midpoint of <math>AD</math> to <math>EF</math>. The hypotenuse of the triangle formed is the [[median]] of equilateral triangle <math>ADE</math>, and one of the legs is <math>3\sqrt{2}</math>. We apply the Pythagorean Theorem to deduce that the height is <math>6</math>.<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br />
triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F); <br />
draw(B--Ba--Ca--C,dashed+d);<br />
draw(A--Aa--Da--D,dashed+d);<br />
draw(E--(E.x,E.y,0),dashed+l);<br />
draw(F--(F.x,F.y,0),dashed+l);<br />
draw(Aa--E--Da,dashed+d);<br />
draw(Ba--F--Ca,dashed+d);<br />
label("A",A,S);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,NE);<br />
label("E",E,N);<br />
label("F",F,N);<br />
label("$12\sqrt{2}$",(E+F)/2,N);<br />
label("$6\sqrt{2}$",(A+B)/2,S);<br />
label("6",(3*s/2,s/2,3),ENE);<br />
</asy></center><br />
Next, we complete t he figure into a triangular prism, and find its volume, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>.<br />
<br />
Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is <math>2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>.<br />
<br />
Thus, our answer is <math>432-144=\boxed{288}</math>.<br />
<br />
=== Solution 2 ===<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black;<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F);<br />
draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed);<br />
label("A",A,(-1,-1,0));<br />
label("B",B,( 2,-1,0));<br />
label("C",C,( 1, 1,0));<br />
label("D",D,(-1, 1,0));<br />
label("E",E,(0,0,1));<br />
label("F",F,(0,0,1));<br />
label("G",G,(0,0,-1));<br />
label("H",H,(0,0,-1));<br />
</asy></center><br />
Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. Now, we have a regular tetrahedron <math>EFGH</math>, which by symmetry has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is:<br />
<br />
<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is <math>6</math>; thus, we will integrate with respect to height from <math>0</math> to <math>6</math>, noting that each cross section of height <math>dh</math> is a rectangle. The volume is then <math>\int_0^h(wl) \ \text{d}h</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. We can express <math>w</math> in terms of <math>h</math> as <math>w=6\sqrt{2}-\sqrt{2}h</math> since it decreases linearly with respect to <math>h</math>, and <math>l=6\sqrt{2}+\sqrt{2}h</math> since it similarly increases linearly with respect to <math>h</math>. Now we solve:<cmath>\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}</cmath>.<br />
<br />
===Solution 4===<br />
Draw an altitude from a vertex of the square base to the top edge. By using <math>30,60, 90</math> triangle ratios, we obtain that the altitude has a length of <math>3 \sqrt{6}</math>, and that little portion that hangs out has a length of <math>3\sqrt2</math>. This is a triangular pyramid with a base of <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math>, and a height of <math>3\sqrt{2}</math>. Since there are two of these, we can compute the sum of the volumes of these two to be <math>72</math>. Now we are left with a triangular prism with a base of dimensions <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math> and a height of <math>6\sqrt2</math>. We can compute the volume of this to be 216, and thus our answer is <math>\boxed{288}</math>.<br />
<br />
pi_is_3.141<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=1887621983 AIME Problems/Problem 112023-02-07T01:10:34Z<p>Ethanspoon: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The solid shown has a square base of side length <math>s</math>. The upper edge is parallel to the base and has length <math>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black;<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F);<br />
label("A",A,W);<br />
label("B",B,S);<br />
label("C",C,SE);<br />
label("D",D,NE);<br />
label("E",E,N);<br />
label("F",F,N);<br />
<br />
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --><br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 ===<br />
First, we find the height of the solid by dropping a perpendicular from the midpoint of <math>AD</math> to <math>EF</math>. The hypotenuse of the triangle formed is the [[median]] of equilateral triangle <math>ADE</math>, and one of the legs is <math>3\sqrt{2}</math>. We apply the Pythagorean Theorem to deduce that the height is <math>6</math>.<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br />
triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F); <br />
draw(B--Ba--Ca--C,dashed+d);<br />
draw(A--Aa--Da--D,dashed+d);<br />
draw(E--(E.x,E.y,0),dashed+l);<br />
draw(F--(F.x,F.y,0),dashed+l);<br />
draw(Aa--E--Da,dashed+d);<br />
draw(Ba--F--Ca,dashed+d);<br />
label("A",A,S);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,NE);<br />
label("E",E,N);<br />
label("F",F,N);<br />
label("$12\sqrt{2}$",(E+F)/2,N);<br />
label("$6\sqrt{2}$",(A+B)/2,S);<br />
label("6",(3*s/2,s/2,3),ENE);<br />
</asy></center><br />
Next, we complete t he figure into a triangular prism, and find its volume, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>.<br />
<br />
Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is <math>2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>.<br />
<br />
Thus, our answer is <math>432-144=\boxed{288}</math>.<br />
<br />
=== Solution 2 ===<br />
<center><asy><br />
size(180);<br />
import three; pathpen = black+linewidth(0.65); pointpen = black;<br />
currentprojection = perspective(30,-20,10);<br />
real s = 6 * 2^.5;<br />
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6);<br />
draw(A--B--C--D--A--E--D);<br />
draw(B--F--C);<br />
draw(E--F);<br />
draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed);<br />
label("A",A,(-1,-1,0));<br />
label("B",B,( 2,-1,0));<br />
label("C",C,( 1, 1,0));<br />
label("D",D,(-1, 1,0));<br />
label("E",E,(0,0,1));<br />
label("F",F,(0,0,1));<br />
label("G",G,(0,0,-1));<br />
label("H",H,(0,0,-1));<br />
</asy></center><br />
Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. Now, we have a regular tetrahedron <math>EFGH</math>, which by symmetry has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is:<br />
<br />
<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math>.<br />
<br />
=== Solution 3 ===<br />
We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is <math>6</math>; thus, we will integrate with respect to height from <math>0</math> to <math>6</math>, noting that each cross section of height <math>dh</math> is a rectangle. The volume is then <math>\int_0^h(wl) \ \text{d}h</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. We can express <math>w</math> in terms of <math>h</math> as <math>w=6\sqrt{2}-\sqrt{2}h</math> since it decreases linearly with respect to <math>h</math>, and <math>l=6\sqrt{2}+\sqrt{2}h</math> since it similarly increases linearly with respect to <math>h</math>. Now we solve:<cmath>\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}</cmath>.<br />
<br />
===Solution 4===<br />
Draw an altitude from a vertex of the square base to the top edge. By using <math>30,60, 90</math> triangle ratios, we obtain that the altitude has a length of <math>3 \sqrt{6}</math>, and that little portion that hangs out has a length of <math>3\sqrt2</math>. This is a triangular pyramid with a base of <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math>, and a height of <math>3\sqrt{2}</math>. Since there are two of these, we can compute the sum of the volumes of these two to be <math>72</math>. Now we are left with a triangular prism with a base of dimensions <math>3\sqrt6, 3\sqrt6, 3\sqrt2</math> and a height of <math>6\sqrt2</math>. We can compute the volume of this to be 216, and thus our answer is <math>\boxed{288}</math>.<br />
<br />
pi_is_3.141<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_9&diff=1883601996 AIME Problems/Problem 92023-01-30T00:52:15Z<p>Ethanspoon: /* Solution 3 (bash) */</p>
<hr />
<div>== Problem ==<br />
A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?<br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math>. Then he goes ahead and opens all lockers <math>2 \pmod {8}</math>, leaving lockers either <math>6 \pmod {16}</math> or <math>14 \pmod {16}</math>. He then goes ahead and opens all lockers <math>14 \pmod {16}</math>, leaving the lockers either <math>6 \pmod {32}</math> or <math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64}</math>. He then opens <math>54 \pmod {64}</math>, leaving <math>22 \pmod {128}</math> or <math>86 \pmod {128}</math>. He then opens <math>22 \pmod {128}</math> and leaves <math>86 \pmod {256}</math> and <math>214 \pmod {256}</math>. He then opens all <math>214 \pmod {256}</math>, so we have <math>86 \pmod {512}</math> and <math>342 \pmod {512}</math>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</math>, and then goes back and opens locker number <math>854</math>, leaving locker number <math>\boxed{342}</math> untouched. He opens that locker.<br />
<br />
=== Solution 2 ===<br />
We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math><br />
<br />
=== Solution 3 (bash)===<br />
List all the numbers from <math>1</math> through <math>1024</math>, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get <math>\boxed{342}</math>.<br />
<br />
(Note: If you try to do this, first look through all the problems! -Guy)<br />
<br />
--------------------------<br />
Edit from EthanSpoon, Doing this with only even numbers will make it faster.<br />
You'll see.<br />
<br />
== See also ==<br />
{{AIME box|year=1996|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_9&diff=1883591996 AIME Problems/Problem 92023-01-30T00:51:46Z<p>Ethanspoon: /* Solution 3 (bash) */</p>
<hr />
<div>== Problem ==<br />
A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?<br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math>. Then he goes ahead and opens all lockers <math>2 \pmod {8}</math>, leaving lockers either <math>6 \pmod {16}</math> or <math>14 \pmod {16}</math>. He then goes ahead and opens all lockers <math>14 \pmod {16}</math>, leaving the lockers either <math>6 \pmod {32}</math> or <math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64}</math>. He then opens <math>54 \pmod {64}</math>, leaving <math>22 \pmod {128}</math> or <math>86 \pmod {128}</math>. He then opens <math>22 \pmod {128}</math> and leaves <math>86 \pmod {256}</math> and <math>214 \pmod {256}</math>. He then opens all <math>214 \pmod {256}</math>, so we have <math>86 \pmod {512}</math> and <math>342 \pmod {512}</math>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</math>, and then goes back and opens locker number <math>854</math>, leaving locker number <math>\boxed{342}</math> untouched. He opens that locker.<br />
<br />
=== Solution 2 ===<br />
We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math><br />
<br />
=== Solution 3 (bash)===<br />
List all the numbers from <math>1</math> through <math>1024</math>, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get <math>\boxed{342}</math>.<br />
<br />
(Note: If you try to do this, first look through all the problems! -Guy)<br />
<br />
--------------------------<br />
Edit from EthanSpoon, Doing this with only odd numbers will make it faster.<br />
You'll see.<br />
<br />
== See also ==<br />
{{AIME box|year=1996|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_9&diff=1883581996 AIME Problems/Problem 92023-01-30T00:51:14Z<p>Ethanspoon: /* Solution 3 (bash) */</p>
<hr />
<div>== Problem ==<br />
A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?<br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math>. Then he goes ahead and opens all lockers <math>2 \pmod {8}</math>, leaving lockers either <math>6 \pmod {16}</math> or <math>14 \pmod {16}</math>. He then goes ahead and opens all lockers <math>14 \pmod {16}</math>, leaving the lockers either <math>6 \pmod {32}</math> or <math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64}</math>. He then opens <math>54 \pmod {64}</math>, leaving <math>22 \pmod {128}</math> or <math>86 \pmod {128}</math>. He then opens <math>22 \pmod {128}</math> and leaves <math>86 \pmod {256}</math> and <math>214 \pmod {256}</math>. He then opens all <math>214 \pmod {256}</math>, so we have <math>86 \pmod {512}</math> and <math>342 \pmod {512}</math>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</math>, and then goes back and opens locker number <math>854</math>, leaving locker number <math>\boxed{342}</math> untouched. He opens that locker.<br />
<br />
=== Solution 2 ===<br />
We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math><br />
<br />
=== Solution 3 (bash)===<br />
List all the numbers from <math>1</math> through <math>1024</math>, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get <math>\boxed{342}</math>.<br />
<br />
(Note: If you try to do this, first look through all the problems! -Guy)<br />
Edit from EthanSpoon, Doing this with only odd numbers will make it faster.<br />
You'll see.<br />
<br />
== See also ==<br />
{{AIME box|year=1996|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_9&diff=1883571996 AIME Problems/Problem 92023-01-30T00:51:01Z<p>Ethanspoon: /* Solution 3 (bash) */</p>
<hr />
<div>== Problem ==<br />
A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?<br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math>. Then he goes ahead and opens all lockers <math>2 \pmod {8}</math>, leaving lockers either <math>6 \pmod {16}</math> or <math>14 \pmod {16}</math>. He then goes ahead and opens all lockers <math>14 \pmod {16}</math>, leaving the lockers either <math>6 \pmod {32}</math> or <math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64}</math>. He then opens <math>54 \pmod {64}</math>, leaving <math>22 \pmod {128}</math> or <math>86 \pmod {128}</math>. He then opens <math>22 \pmod {128}</math> and leaves <math>86 \pmod {256}</math> and <math>214 \pmod {256}</math>. He then opens all <math>214 \pmod {256}</math>, so we have <math>86 \pmod {512}</math> and <math>342 \pmod {512}</math>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</math>, and then goes back and opens locker number <math>854</math>, leaving locker number <math>\boxed{342}</math> untouched. He opens that locker.<br />
<br />
=== Solution 2 ===<br />
We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math><br />
<br />
=== Solution 3 (bash)===<br />
List all the numbers from <math>1</math> through <math>1024</math>, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get <math>\boxed{342}</math>.<br />
<br />
(Note: If you try to do this, first look through all the problems! -Guy)<br />
Edit from EthanSpoon, Doing this with only odd numbers will make it faster.<br />
You'll see.<br />
<br />
== See also ==<br />
{{AIME box|year=1996|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_1&diff=1866131990 AIME Problems/Problem 12023-01-14T21:24:51Z<p>Ethanspoon: /* Solution 1*/</p>
<hr />
<div>== Problem ==<br />
The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neither the [[perfect square | square]] nor the [[perfect cube | cube]] of a positive integer. Find the 500th term of this sequence.<br />
<br />
== Solution 1==<br />
Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than <math>500</math>. This happens to be <math>23^2=529</math>. Notice that there are <math>23</math> squares and <math>8</math> cubes less than or equal to <math>529</math>, but <math>1</math> and <math>2^6</math> are both squares and cubes. Thus, there are <math>529-23-8+2=500</math> numbers in our sequence less than <math>529</math>. Magically, we want the <math>500th</math> term, so our answer is the biggest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>.<br />
<br />
== Solution 2==<br />
similar as above, but to get the intuition why we chose to consider 23^2 = 529 , consider this<br />
<br />
we need n - T = 500, where n = #integers in the list 1,2,..,n and T is the set of numbers which are either k^2 or k^3 and <=n<br />
firstly, we clearly need n > 500<br />
so we think of taking the smallest square greater than 500 and let that be equal to n( u could try letting n = 512 = 8^3 to with similiar logic, but quickly realise that it fails). This is done so that set T is easy to calculate<br />
so n = 529, set T = 23+8-2 by PIE hence n-T = 500<br />
so our answer is 529-1 = 528<br />
<br />
== See also ==<br />
{{AIME box|year=1990|before=First Question|num-a=2}}<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_13&diff=1860781986 AIME Problems/Problem 132023-01-08T23:44:33Z<p>Ethanspoon: /* Slight Variation */</p>
<hr />
<div>== Problem ==<br />
In a [[sequence]] of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by <tt>TH</tt>, <tt>HH</tt>, and etc. For example, in the sequence <tt>TTTHHTHTTTHHTTH</tt> of 15 coin tosses we observe that there are two <tt>HH</tt>, three <tt>HT</tt>, four <tt>TH</tt>, and five <tt>TT</tt> subsequences. How many different sequences of 15 coin tosses will contain exactly two <tt>HH</tt>, three <tt>HT</tt>, four <tt>TH</tt>, and five <tt>TT</tt> subsequences?<br />
<br />
== Solution ==<br />
Let's consider each of the sequences of two coin tosses as an [[operation]] instead; this operation takes a string and adds the next coin toss on (eg, <tt>THHTH</tt> + <tt>HT</tt> = <tt>THHTHT</tt>). We examine what happens to the last coin toss. Adding <tt>HH</tt> or <tt>TT</tt> is simply an [[identity]] for the last coin toss, so we will ignore them for now. However, adding <tt>HT</tt> or <tt>TH</tt> switches the last coin. <tt>H</tt> switches to <tt>T</tt> three times, but <tt>T</tt> switches to <tt>H</tt> four times; hence it follows that our string will have a structure of <tt>THTHTHTH</tt>. <br />
<br />
Now we have to count all of the different ways we can add the identities back in. There are 5 <tt>TT</tt> subsequences, which means that we have to add 5 <tt>T</tt> into the strings, as long as the new <tt>T</tt>s are adjacent to existing <tt>T</tt>s. There are already 4 <tt>T</tt>s in the sequence, and since order doesn’t matter between different tail flips this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives <math>{{5+3}\choose3} = 56</math> combinations. We do the same with 2 <tt>H</tt>s to get <math>{{2+3}\choose3} = 10</math> combinations; thus there are <math>56 \cdot 10 = \boxed {560}</math> possible sequences.<br />
<br />
=== Slight Variation===<br />
The structure of the final order is <tt>T_H_T_H_T_H_T_H_</tt>, and there are 4 spots to put the 2 heads in, and 4 spots to put the 5 tails in. By using the formula for distributing r identical objects into n distinct boxes <math>\dbinom{r+n-1}{r}</math> and multiplication, the answer is <math>{{2+4-1}\choose2} \cdot{{5+4-1}\choose5}=560</math> ~Slight edits in LaTeX by EthanSpoon<br />
<br />
== See also ==<br />
{{AIME box|year=1986|num-b=12|num-a=14}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185986User:EthanSpoon2023-01-07T05:01:51Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon! If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED AND YOU SAW THIS POST TO HELP, YOU CAN ADD YOUR NAME HERE:<br />
<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
HALL OF FAME(LEVEL 1-4):<br />
<br />
Link: https://artofproblemsolving.com/community/c3101657_the_incredible_forum<br />
<br />
<br />
<br />
<br />
<br />
==Contributes==<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185956User:EthanSpoon2023-01-06T23:29:30Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED AND YOU SAW THIS POST TO HELP, YOU CAN ADD YOUR NAME HERE:<br />
<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
HALL OF FAME(LEVEL 1-4):<br />
<br />
Link: https://artofproblemsolving.com/community/c3101657_the_incredible_forum<br />
<br />
<br />
<br />
<br />
<br />
==Contributes==<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185955User:EthanSpoon2023-01-06T23:28:05Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED AND YOU SAW THIS POST TO HELP, YOU CAN ADD YOUR NAME HERE:<br />
<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
HALL OF FAME(LEVEL 1-4):<br />
<br />
<br />
<br />
<br />
<br />
<br />
==Contributes==<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185954User:EthanSpoon2023-01-06T23:27:31Z<p>Ethanspoon: /* IMPORTANT ANNOUNCMENT */</p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED AND YOU SAW THIS POST TO HELP, YOU CAN ADD YOUR NAME HERE:<br />
<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
HALL OF FAME(LEVEL 1-4):<br />
<br />
<br />
<br />
<br />
<br />
<br />
Contributes:<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185953User:EthanSpoon2023-01-06T23:26:51Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED, YOU CAN ADD YOUR NAME HERE:<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
HALL OF FAME(LEVEL 1-4):<br />
<br />
<br />
<br />
<br />
<br />
<br />
Contributes:<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185952User:EthanSpoon2023-01-06T23:26:21Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center><br />
<br />
==Things I would like to say:==<br />
<br />
Hi again. Well yeah. I joined AOPS a long time ago, I went thought a lot of forums. But now, I am a searching for clues in a mystery in the incredible forum<br />
<br />
==IMPORTANT ANNOUNCMENT==<br />
IF YOU ARE A MEMBER OF AOPS. PLEASE HELP BY SEARCHING FOR CLUES IN THE INCREDIBLE FORUM. DON'T GIVE OUT ANNY CLUES THAT WERE ALREADY FOUND. YOU CAN SEE WHAT IS FOUND ON THE COALITION THEARD OR SOMETHING. IF YOU HELPED, YOU CAN ADD YOUR NAME HERE:<br />
<br />
Level 1: Tried to help(5 upvotes)<br />
Level 2: Found something that was already found(10 upvotes)<br />
Level 3: Found an unfound clue(30 upvotes)<br />
Level 4: Found greater than or equal to 3 unfound clues(100 upvotes)<br />
<br />
[hide=Hall of fame]<br />
[hide=level 3]<br />
[/hide]<br />
[hide=level 4]<br />
[/hide]<br />
<br />
[/hide]<br />
<br />
<br />
<br />
<br />
<br />
Contributes:<br />
1988 AIME problem 9 solution 6</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185951User:EthanSpoon2023-01-06T23:21:10Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">1</font></center></div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185950User:EthanSpoon2023-01-06T23:21:01Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
<br />
<br />
==User Count==<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below by increasing it by 1:</font></div><br />
<center><font size="100px">195</font></center></div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185949User:EthanSpoon2023-01-06T23:19:30Z<p>Ethanspoon: </p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below:<br />
<br />
[size=200]1[\size]</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:EthanSpoon&diff=185948User:EthanSpoon2023-01-06T23:19:22Z<p>Ethanspoon: Created page with "RULES FIRST: If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else. Hi, my usernam..."</p>
<hr />
<div>RULES FIRST:<br />
If you want to edit this(unless it's the user count,) ask EthanSpoon! I warn you, I will know if you are editing the user count or something else.<br />
<br />
Hi, my username is EthanSpoon. If it's your first time here, please edit the number below:<br />
<br />
[size=200]1[/size]</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=1859461988 AIME Problems/Problem 92023-01-06T22:52:48Z<p>Ethanspoon: /* solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO) */</p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>.<br />
<br />
==Solution 1==<br />
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:<br />
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. <br />
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math><br />
<br />
==Solution 2==<br />
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>.<br />
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. <br />
<br />
By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. <br />
<br />
Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>.<br />
<br />
== Solution 3 ==<br />
Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. <br />
<br />
We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam<br />
<br />
== Solution 4 (Bash) ==<br />
<br />
Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature<br />
<br />
<br />
==solution 5==<br />
This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math><br />
<br />
Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math><br />
<br />
~bluesoul<br />
<br />
==solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)==<br />
Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible.<br />
So we do trial and error:<br />
<br />
12(FAIL), 22(FAIL), 32(FAIL), 42(FAIL), 52(FAIL), 62(FAIL), 72(FAIL), 82(FAIL), 92(FAIL)<br />
<br />
Continuing on, we get that 192 is the smallest positive integer.<math>(192^3=7077888)</math><br />
~EthanSpoon<br />
<br />
== See also ==<br />
{{AIME box|year=1988|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=1859451988 AIME Problems/Problem 92023-01-06T22:52:37Z<p>Ethanspoon: /* solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO) */</p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>.<br />
<br />
==Solution 1==<br />
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:<br />
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. <br />
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math><br />
<br />
==Solution 2==<br />
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>.<br />
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. <br />
<br />
By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. <br />
<br />
Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>.<br />
<br />
== Solution 3 ==<br />
Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. <br />
<br />
We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam<br />
<br />
== Solution 4 (Bash) ==<br />
<br />
Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature<br />
<br />
<br />
==solution 5==<br />
This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math><br />
<br />
Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math><br />
<br />
~bluesoul<br />
<br />
==solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)==<br />
Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible.<br />
So we do trial and error:<br />
<br />
12(FAIL), 22(FAIL), 32(FAIL), 42(FAIL), 52(FAIL), 62(FAIL), 72(FAIL), 82(FAIL), 92(FAIL)<br />
Continuing on, we get that 192 is the smallest positive integer.<math>(192^3=7077888)</math><br />
~EthanSpoon<br />
<br />
== See also ==<br />
{{AIME box|year=1988|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=1859441988 AIME Problems/Problem 92023-01-06T22:52:27Z<p>Ethanspoon: /* solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO) */</p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>.<br />
<br />
==Solution 1==<br />
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:<br />
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. <br />
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math><br />
<br />
==Solution 2==<br />
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>.<br />
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. <br />
<br />
By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. <br />
<br />
Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>.<br />
<br />
== Solution 3 ==<br />
Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. <br />
<br />
We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam<br />
<br />
== Solution 4 (Bash) ==<br />
<br />
Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature<br />
<br />
<br />
==solution 5==<br />
This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math><br />
<br />
Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math><br />
<br />
~bluesoul<br />
<br />
==solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)==<br />
Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible.<br />
So we do trial and error:<br />
12(FAIL), 22(FAIL), 32(FAIL), 42(FAIL), 52(FAIL), 62(FAIL), 72(FAIL), 82(FAIL), 92(FAIL)<br />
Continuing on, we get that 192 is the smallest positive integer.<math>(192^3=7077888)</math><br />
~EthanSpoon<br />
<br />
== See also ==<br />
{{AIME box|year=1988|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=1859431988 AIME Problems/Problem 92023-01-06T22:51:05Z<p>Ethanspoon: </p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>.<br />
<br />
==Solution 1==<br />
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:<br />
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. <br />
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math><br />
<br />
==Solution 2==<br />
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>.<br />
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. <br />
<br />
By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. <br />
<br />
Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>.<br />
<br />
== Solution 3 ==<br />
Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. <br />
<br />
We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam<br />
<br />
== Solution 4 (Bash) ==<br />
<br />
Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature<br />
<br />
<br />
==solution 5==<br />
This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math><br />
<br />
Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math><br />
<br />
~bluesoul<br />
<br />
==solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)==<br />
Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible.<br />
So we do trial and error:<br />
<br />
12 no<br />
22 no<br />
32 no<br />
42 yes<br />
52 no<br />
62 no<br />
72 no<br />
82 no<br />
92 no<br />
Continuing on, we get that 192 is the smallest positive integer.<br />
ALSO: <math>192^3=7077888</math><br />
~EthanSpoon<br />
== See also ==<br />
{{AIME box|year=1988|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&diff=1859421988 AIME Problems/Problem 92023-01-06T22:50:40Z<p>Ethanspoon: </p>
<hr />
<div>== Problem ==<br />
Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>.<br />
<br />
==Solution 1==<br />
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:<br />
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. <br />
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math><br />
<br />
==Solution 2==<br />
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>.<br />
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. <br />
<br />
By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. <br />
<br />
Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>.<br />
<br />
== Solution 3 ==<br />
Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>. <br />
<br />
We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam<br />
<br />
== Solution 4 (Bash) ==<br />
<br />
Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature<br />
<br />
<br />
==solution 5==<br />
This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math><br />
<br />
Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math><br />
<br />
~bluesoul<br />
<br />
==solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)==<br />
Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible.<br />
So we do trial and error<br />
12 no<br />
22 no<br />
32 no<br />
42 yes<br />
52 no<br />
62 no<br />
72 no<br />
82 no<br />
92 no<br />
Continuing on, we get that 192 is the smallest positive integer.<br />
ALSO: <math>192^3=7077888</math><br />
~EthanSpoon<br />
== See also ==<br />
{{AIME box|year=1988|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=20&diff=185873202023-01-05T20:55:44Z<p>Ethanspoon: </p>
<hr />
<div>Huh, you found it. That's actually a very good achievement. So, want to go even more in the rabbit hole?? Yes? If so, PM SpeedCuber7 "Please give me another hint."<br />
<br />
interesting<br />
ETHANSPOON WAS HERE<br />
<br />
{{stub}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=7&diff=18586572023-01-05T19:39:28Z<p>Ethanspoon: </p>
<hr />
<div>14639201037200387267 <math>\phantom{erehwemos uoy lliw rebmun emirp tseraen}</math><br />
<br />
ETHANSPOON WAS HERE</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=Talk:Coalition_Under_Lovers_of_Trees&diff=185043Talk:Coalition Under Lovers of Trees2022-12-28T00:48:39Z<p>Ethanspoon: Created page with "Coalition under lovers of trees descussion. Very private."</p>
<hr />
<div>Coalition under lovers of trees descussion. Very private.</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=High-Quality_Games_and_Fun&diff=183168High-Quality Games and Fun2022-11-27T22:29:54Z<p>Ethanspoon: </p>
<hr />
<div>[https://artofproblemsolving.com/community/c3101657_highquality_games_and_fun High-Quality Games and Fun] is an AoPS game forum founded by [https://artofproblemsolving.com/wiki/index.php/User:Megahertz13 megahertz13] on August 12, 2022. It has reached up to 2nd on the Other Forums. It is also administrated by [https://artofproblemsolving.com/wiki/index.php/User:Sotowa Sotowa] and moderated by [https://artofproblemsolving.com/wiki/index.php/User:Bob_Smart Bob_Smart], [https://artofproblemsolving.com/wiki/index.php/User:mynameisJEJ mynameisJEJ], [https://artofproblemsolving.com/wiki/index.php/User:EthanSpoon EthanSpoon], and [https://artofproblemsolving.com/wiki/index.php/User:k1glaucus k1glaucus]. Other notable contributers include former moderators [https://artofproblemsolving.com/wiki/index.php/User:Cothoop cothoop], [https://artofproblemsolving.com/wiki/index.php/User:delicatehunter delicatehunter], and [https://artofproblemsolving.com/wiki/index.php/User:Desertsnow Desertsnow], active users [https://artofproblemsolving.com/wiki/index.php/User:BlueGecko BlueGecko] and [https://artofproblemsolving.com/wiki/index.php/User:Speedcuber7 SpeedCuber7], and game makers [https://artofproblemsolving.com/wiki/index.php/User:PatTheKing806 PatTheKing806] and [https://artofproblemsolving.com/wiki/index.php/User:ryanbear ryanbear]. More data can be found in [https://artofproblemsolving.com/community/c3101657h2928835_data_center the Data Center].<br />
<br />
It was previously known as The Incredible Forum until November 20 and Math Puzzles & Games until around September 1 and later Math, Puzzles, and Games until September 9.</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_25&diff=1815422001 AMC 12 Problems/Problem 252022-11-15T21:09:25Z<p>Ethanspoon: /* See Also */</p>
<hr />
<div>== Problem ==<br />
<br />
Consider sequences of positive real numbers of the form <math>x, 2000, y, \dots</math> in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of <math>x</math> does the term 2001 appear somewhere in the sequence?<br />
<br />
<math><br />
\text{(A) }1<br />
\qquad<br />
\text{(B) }2<br />
\qquad<br />
\text{(C) }3<br />
\qquad<br />
\text{(D) }4<br />
\qquad<br />
\text{(E) more than }4<br />
</math><br />
<br />
== Solution ==<br />
<br />
It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that <math>\forall</math> (for all) <math> n>1:~ a_n = a_{n-1}a_{n+1} - 1</math>. This can be rewritten as <math>a_{n+1} = \frac{a_n +1}{a_{n-1}}</math>. We have <math>a_1=x</math> and <math>a_2=2000</math>, and we compute:<br />
<br />
<cmath><br />
\begin{align*}<br />
a_3 <br />
& = \frac{a_2+1}{a_1} = \frac{2001}x<br />
\\<br />
a_4 <br />
& = \frac{a_3+1}{a_2}<br />
= \frac{ \dfrac{2001}x + 1 }{ 2000 }<br />
= \frac{2001 + x}{2000x}<br />
\\<br />
a_5<br />
& = \frac{a_4+1}{a_3}<br />
= \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x }<br />
= \frac{ \frac{2001 + 2001x}{2000x} }{ \frac{2001}x }<br />
= \frac{1+x}{2000}<br />
\\<br />
a_6<br />
& = \frac{a_5+1}{a_4}<br />
= \frac{ \frac{1+x}{2000} + 1 }{ \frac{2001 + x}{2000x} }<br />
= \frac{ \frac{2001+x}{2000} }{ \frac{2001 + x}{2000x} }<br />
= x<br />
\\<br />
a_7<br />
& = \frac{a_6+1}{a_5}<br />
= \frac{ x+1 }{ \frac{1+x}{2000} }<br />
= 2000<br />
\end{align*}<br />
</cmath><br />
<br />
At this point we see that the sequence will become periodic: we have <math>a_6=a_1</math>, <math>a_7=a_2</math>, and each subsequent term is uniquely determined by the previous two.<br />
<br />
Hence if <math>2001</math> appears, it has to be one of <math>a_1</math> to <math>a_5</math>. As <math>a_2=2000</math>, we only have four possibilities left. Clearly <math>a_1=2001</math> for <math>x=2001</math>, and <math>a_3=2001</math> for <math>x=1</math>. The equation <math>a_4=2001</math> solves to <math>x = \frac{2001}{2000\cdot 2001 - 1}</math>, and the equation <math>a_5=2001</math> to <math>x=2000\cdot 2001 - 1</math>.<br />
<br />
No two values of <math>x</math> we just computed are equal, and therefore there are <math>\boxed{4}</math> different values of <math>x</math> for which the sequence contains the value <math>2001</math>.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2001|num-b=24|after=Last question}}<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:Mrthinker&diff=179297User:Mrthinker2022-10-21T19:14:03Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div>Hello!<br />
==User Count==<br />
If this is your first time seeing this, please increase the user count by 1.<br />
</font></div><center><font size="120px">5</font></center><br />
<!-- righhhtttt here. ^ :D --><br />
<br />
==About==<br />
<br />
Have you ever thought of what happens when you die? Some believe in heaven, some don't, but we don't know. Even if you die and you experience the events that happen, you can't come back to tell the world about your discovery.<br />
<br />
<br />
<br />
<br />
have a good day.<br />
<br />
<br />
[[Special:Contributions/Mrthinker | Contributions]]<br />
<br />
<br />
This article has been proposed for [[promotion]]. The reason given is: '''MrThinker is awesome'''<br />
Sysops: Before promoting this article, please check the article [[discussion]] pages and history.<br />
hehe ^</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=1781792012 AMC 12B Problems/Problem 162022-09-13T17:23:56Z<p>Ethanspoon: /* Solution 5 */</p>
<hr />
<div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br />
<br />
== Problem==<br />
Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br />
<br />
<math> \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 </math><br />
<br />
==Solutions==<br />
<br />
=== Solution 1===<br />
Let the ordered triple <math>(a,b,c)</math> denote that <math>a</math> songs are liked by Amy and Beth, <math>b</math> songs by Beth and Jo, and <math>c</math> songs by Jo and Amy. We claim that the only possible triples are <math>(1,1,1), (2,1,1), (1,2,1)(1,1,2)</math>. <br />
<br />
To show this, observe these are all valid conditions. Second, note that none of <math>a,b,c</math> can be bigger than 3. Suppose otherwise, that <math>a = 3</math>. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either <math>b</math> or <math>c</math> to be at least 1. In fact, we require either <math>b</math> or <math>c</math> to equal 1, otherwise there will be a song liked by all three. Suppose <math>b = 1</math>. Then we must have <math>c=0</math> since no song is liked by all three girls, a contradiction.<br />
<br />
'''Case 1''': How many ways are there for <math>(a,b,c)</math> to equal <math>(1,1,1)</math>? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So <math>(a,b,c)=(1,1,1)</math> in <math>4\cdot3\cdot2\cdot4 = 96</math> ways.<br />
<br />
'''Case 2''': To find the number of ways for <math>(a,b,c) = (2,1,1)</math>, observe there are <math>\binom{4}{2} = 6</math> choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are <math>6\cdot2\cdot3=36</math> ways for the girls to like the songs.<br />
<br />
That gives a total of <math>96 + 36 = 132</math> ways for the girls to like the songs, so the answer is <math>\boxed{(\textrm{\textbf{B}})}</math>.<br />
<br />
=== Solution 2===<br />
<br />
<br />
Let <math>AB, BJ</math>, and <math>AJ</math> denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let <math>A, B, J,</math> and <math>N</math> denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least <math>1\: AB, BJ</math>, and <math>AJ</math>, they must be <math>3</math> songs out of the <math>4</math> that Amy, Beth, and Jo listened to. The fourth song can be of any type <math>N, A, B, J, AB, BJ</math>, and <math>AJ</math> (there is no <math>ABJ</math> because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange <math>AB, BJ, AJ</math>, and a song from the set <math>\{N, A, B, J, AB, BJ, AJ\}</math>.<br />
<br />
Case 1: Fourth song = <math>N, A, B, J </math><br />
<br />
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!)</math> rearrangements for each choice <math>*\: 4</math> choices = <math>96</math>.<br />
<br />
Case 2: Fourth song = <math>AB, BJ, AJ</math><br />
<br />
Note that in Case <math>2</math>, all three of the choices for the fourth song repeat somewhere in the first three songs.<br />
<br />
Number of ways to rearrange = <math>(4!/2!)</math> rearrangements for each choice <math>*\: 3</math> choices = <math>36</math>.<br />
<br />
<math>96 + 36 = \boxed{\textbf{(B)} \: 132}</math>.<br />
<br />
=== Solution 3===<br />
<br />
<br />
There are <math>\binom{4}{3}</math> ways to choose the three songs that are liked by the three pairs of girls.<br />
<br />
There are <math>3!</math> ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br />
<br />
In total, there are <math>\binom{4}{3}\cdot3!</math> possibilities for the first <math>3</math> songs.<br />
<br />
There are <math>3</math> cases for the 4th song, call it song D.<br />
<br />
Case <math>1</math>: D is disliked by all <math>3</math> girls <math>\implies</math> there is only <math>1</math> possibility.<br />
<br />
Case <math>2</math>: D is liked by exactly <math>1</math> girl <math>\implies</math> there are <math>3</math> possibility.<br />
<br />
Case <math>3</math>: D is liked by exactly <math>2</math> girls <math>\implies</math> there are <math>3</math> pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first <math>3</math> songs liked by the same girls.<br />
<br />
Counting the overlaps, there are <math>3</math> ways to choose the pair with overlaps and <math>4\cdot3=12</math> ways to choose what the other <math>2</math> pairs like independently. In total, there are <math>3\cdot12=36</math> overlapped possibilities.<br />
<br />
Finally, there are <math>\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132</math> ways for the songs to be likely by the girls. <math>\boxed{\mathrm{(B)}}</math><br />
<br />
~ Nafer<br />
<br />
===Solution 4===<br />
This is a bipartite graph problem, with the girls as left vertices and songs as right vertices. An edge connecting left vertex and right vertex means that a girl like a song.<br />
<br />
Condition 1: "No song is liked by all three", means that the degree of right vertices is at most 2. <br />
<br />
Condition 2: "for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third", means that for any pair of left vertices, there is at least a right vertex connecting to them.<br />
<br />
To meet condition 2, there are at least 3 right vertices with 2 edges connecting to left vertices. There are 2 cases:<br />
<br />
Case 1: there are only 3 such right vertices. There are <math>\binom{4}{3}</math> such vertices, with <math>3!</math> ways of connections to the left vertices, total arrangements are <math>\binom{4}{3}\cdot3! = 24</math>. The fourth right vertex either has no edge to the 3 left vertices, or 1 edge to 1 of the 3 left vertices. So there are <math>24\cdot(1+3) = 96</math> ways.<br />
<br />
Case 2: there are 4 such right vertices, 2 of them have edges to the same pair of left vertices. There are <math>\binom{4}{2}</math> such vertices, with <math>3!</math> ways of connections. So there are <math>\binom{4}{2}\cdot3! = 36</math> ways.<br />
<br />
Total ways are <math>96+36=132</math>.<br />
<br />
Another way is to overcount then subtract overlap ways. Similar to previous case 1, the fourth right vertex could have all possible connection to the left vertices except connecting to all 3, so it is <math>2^3-1=7</math> ways, so the total ways are <math>\binom{4}{3}\cdot3!\cdot7 = 24\cdot7 = 168</math>. But this overcounts the case 2 with 36 ways. So total ways are <math>168-36=132</math>.<br />
<br />
-[https://artofproblemsolving.com/wiki/index.php/User:Junche junche]<br />
<br />
==Video Solutions:==<br />
==Video Solution by Richard Rusczyk==<br />
https://artofproblemsolving.com/videos/amc/2012amc10b/272<br />
<br />
~dolphin7<br />
<br />
==Solution 5==<br />
<br />
Let's assign to each of the 3 pairs a song that they like, and that song is disliked by the other girl(satisfy the problem's 2nd condition). The first pair has 4 choice, 2nd has 3 choice, and last one has 2 choice. <math>4 \cdot 3 \cdot 2</math>. Then the last song can be liked/disliked freely, so there is <math>2^3 = 8</math> total ways they can like/dislike the songs. Of these 8 cases, one of them is when they all like the song (which isn't allowed by the problem), so we subtract it away to have 7 cases. <math>4 \cdot 3 \cdot 2 \cdot 7 = 168</math>. But, we have overcounts when one pair of girls likes 2 songs, and the other girl dislikes those 2 songs. We can find the amount of cases that this happens by giving 1 pair of girls 2 songs that they like (and the other dislikes), and the other 2 pairs of girls 1 song that they like. There are 3 pairs of girls that we can choose to like 2 songs, so there are <math>3 \cdot \left(\binom {4}{2} \cdot 2 \cdot 1\right) = 36</math> ways. <math>168 - 36 = \boxed {\textbf{(B) 132}}</math>.<br />
<br />
~heheman<br />
~Arcticturn (minor fixes in LaTeX)<br />
~EthanSpoon (minor fixes in LaTeX)<br />
<br />
== See Also ==<br />
<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_12&diff=1769652015 AMC 10B Problems/Problem 122022-08-15T21:28:55Z<p>Ethanspoon: /* Solution= */</p>
<hr />
<div>==Problem==<br />
For how many integers <math>x</math> is the point <math>(x, -x)</math> inside or on the circle of radius <math>10</math> centered at <math>(5, 5)</math>?<br />
<br />
<math>\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15</math><br />
<br />
==Video Solution 1==<br />
https://youtu.be/RZDFs3qrw7Y<br />
<br />
~Education, the Study of Everything=<br />
<br />
=Solution=<br />
The equation of the circle is <math>(x-5)^2+(y-5)^2=100</math>. Plugging in the given conditions we have <math>(x-5)^2+(-x-5)^2 \leq 100</math>. Expanding gives: <math>x^2-10x+25+x^2+10x+25\leq 100</math>, which simplifies to<br />
<math>x^2\leq 25</math> and therefore<br />
<math>x\leq 5</math> and <math>x\geq -5</math>. So <math>x</math> ranges from <math>-5</math> to <math>5</math>, for a total of <math>\boxed{\mathbf{(A)}\ 11}</math> integer values.<br />
<br />
Note by Williamgolly:<br />
Alternatively, draw out the circle and see that these points must be on the line <math>y=-x</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:MRENTHUSIASM&diff=176010User:MRENTHUSIASM2022-07-17T20:36:41Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div>==User Count==<br />
If this is your first time visiting this page, then edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">144</font></center><br />
<br />
==Introduction==<br />
Hi Everyone,<br />
<br />
I am a hedgehog who likes Message Board halping, funny pictures, emoji wars, and AoPS Wiki contributions. Keep these fun things coming!<br />
<br />
This hedgehog represents my attitude towards mathematics--victorious and enthusiastic!<br />
[[File:Commander Hedgehog.png|center]]<br />
And here is my personified self-portrait:<br />
[[File:Enthusiastic.gif|center]]<br />
~MRENTHUSIASM<br />
<br />
==AoPS Bio==<br />
Jerry graduated from Northeastern University in 2018 as a mathematics and computer science combined major. He has eleven years of experience working with students. He has served as an AoPS instructor and a message board moderator since 2019. During his last two years in high school, he served as the math club president and prepared lively lectures on contest math. As a result, he inspired many kids to get interested in math and participate in math competitions (such as AMC and ARML). For the last eight years, he has worked with Math League to write math contests for students from grades 4 through 12 in the USA. In his mind, nothing is more rewarding than educating tomorrow’s bright minds. In 2020 he published the book English-Chinese Mathematics Encyclopedia: Algebra, Geometry, and Pre-Calculus in Tsinghua University Press in China. In his spare time, he loves contributing to the AoPS Wiki and playing chess and all sorts of board games, especially Scotland Yard, and Samurai.<br />
<br />
==AoPS Wiki Contributions==<br />
Below are my contributions to the AMC/AIME Problems' Solutions in the AoPS Wiki. I understand that in this communal Wiki, we all have to collaborate and make compromises. <i><b>In any of my solutions, if you find flaws and/or want major revisions, then please contact me via my [[User_talk:MRENTHUSIASM|user talk page]] or the private messaging system before you take action. I am sure that we can work things out.</b></i><br />
<br />
Thank you for your cooperation, and hope you enjoy reading my solutions.<br />
<br />
Finally, [https://www.youtube.com/channel/UCR0u7fEppRjD-OFOEbtzhKw here] is my YouTube channel (MRENTHUSIASM) for the video solutions. Go ahead and subscribe--you know you want to. <math>\smiley{}</math> <br />
<br />
~MRENTHUSIASM<br />
<br />
===AJHSME / AMC 8===<br />
* [[2007_AMC_8_Problems/Problem_8|2007 AMC 8 Problem 8]] (Solutions 1, 2)<br />
* [[2007_AMC_8_Problems/Problem_10|2007 AMC 8 Problem 10]] (Solution)<br />
* [[2009_AMC_8_Problems/Problem_4|2009 AMC 8 Problem 4]] (Solution)<br />
* [[2017_AMC_8_Problems/Problem_24|2017 AMC 8 Problem 24]] (Solution 2)<br />
* [[2020_AMC_8_Problems/Problem_1|2020 AMC 8 Problem 1]] (Solutions 2, 3)<br />
* [[2020_AMC_8_Problems/Problem_3|2020 AMC 8 Problem 3]] (Solution 1)<br />
* [[2020_AMC_8_Problems/Problem_4|2020 AMC 8 Problem 4]] (Solutions 2, 4)<br />
* [[2022_AMC_8_Problems/Problem_3|2022 AMC 8 Problem 3]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_4|2022 AMC 8 Problem 4]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_5|2022 AMC 8 Problem 5]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_6|2022 AMC 8 Problem 6]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_8|2022 AMC 8 Problem 8]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_9|2022 AMC 8 Problem 9]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_10|2022 AMC 8 Problem 10]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_13|2022 AMC 8 Problem 13]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_14|2022 AMC 8 Problem 14]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_16|2022 AMC 8 Problem 16]] (Solutions 1, 2)<br />
* [[2022_AMC_8_Problems/Problem_18|2022 AMC 8 Problem 18]] (Solution 1)<br />
* [[2022_AMC_8_Problems/Problem_20|2022 AMC 8 Problem 20]] (Solution 2)<br />
* [[2022_AMC_8_Problems/Problem_22|2022 AMC 8 Problem 22]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_24|2022 AMC 8 Problem 24]] (Solution)<br />
* [[2022_AMC_8_Problems/Problem_25|2022 AMC 8 Problem 25]] (Solution 1)<br />
<br />
===AMC 10===<br />
* [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br />
* [[2007_AMC_10A_Problems/Problem_20|2007 AMC 10A Problem 20]] (Solutions 1, 2, 3, 4, 5, 6, 7)<br />
* [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br />
* [[2012_AMC_10B_Problems/Problem_10|2012 AMC 10B Problem 10]] (Solutions 1, 2)<br />
* [[2018_AMC_10A_Problems/Problem_1|2018 AMC 10A Problem 1]] (Solution)<br />
* [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 1)<br />
* [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solutions 1, 2)<br />
* [[2018_AMC_10A_Problems/Problem_8|2018 AMC 10A Problem 8]] (Solution 3)<br />
* [[2018_AMC_10A_Problems/Problem_17|2018 AMC 10A Problem 17]] (Solution 4)<br />
* [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br />
* [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br />
* [[2018_AMC_10A_Problems/Problem_25|2018 AMC 10A Problem 25]] (Solutions 1, 3)<br />
* [[2018_AMC_10B_Problems/Problem_2|2018 AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_15|2018 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2018_AMC_10B_Problems/Problem_19|2018 AMC 10B Problem 19]] (Solution 1)<br />
* [[2018_AMC_10B_Problems/Problem_20|2018 AMC 10B Problem 20]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_21|2018 AMC 10B Problem 21]] (Solutions 1, 2)<br />
* [[2018_AMC_10B_Problems/Problem_24|2018 AMC 10B Problem 24]] (Diagram, Solution 6)<br />
* [[2020_AMC_10A_Problems/Problem_3|2020 AMC 10A Problem 3]] (Solutions 1, 2)<br />
* [[2020_AMC_10A_Problems/Problem_5|2020 AMC 10A Problem 5]] (Solution 3)<br />
* [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br />
* [[2020_AMC_10A_Problems/Problem_17|2020 AMC 10A Problem 17]] (Solution 1)<br />
* [[2020_AMC_10A_Problems/Problem_24|2020 AMC 10A Problem 24]] (Solution 8)<br />
* [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br />
* [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 1)<br />
* [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br />
* [[2020_AMC_10B_Problems/Problem_8|2020 AMC 10B Problem 8]] (Solutions 1, 2)<br />
* [[2020_AMC_10B_Problems/Problem_13|2020 AMC 10B Problem 13]] (Solution 1)<br />
* [[2020_AMC_10B_Problems/Problem_15|2020 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 1)<br />
* [[2021_AMC_10A_Problems/Problem_1|2021 AMC 10A Problem 1]] (Solution 2)<br />
* [[2021_AMC_10A_Problems/Problem_2|2021 AMC 10A Problem 2]] (Solutions 1, 2, 3, 4)<br />
* [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br />
* [[2021_AMC_10A_Problems/Problem_4|2021 AMC 10A Problem 4]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_5|2021 AMC 10A Problem 5]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_6|2021 AMC 10A Problem 6]] (Solutions 1, 2)<br />
* [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 2)<br />
* [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br />
* [[2021_AMC_10A_Problems/Problem_11|2021 AMC 10A Problem 11]] (Solutions 1, 3)<br />
* [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_AMC_10A_Problems/Problem_13|2021 AMC 10A Problem 13]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solutions 1, 2)<br />
* [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br />
* [[2021_AMC_10A_Problems/Problem_20|2021 AMC 10A Problem 20]] (Solutions 2, 3)<br />
* [[2021_AMC_10A_Problems/Problem_21|2021 AMC 10A Problem 21]] (Diagram, Solution, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_22|2021 AMC 10A Problem 22]] (Solution 2, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_24|2021 AMC 10A Problem 24]] (Diagram, Solutions 1, 2, 4, Video Solution)<br />
* [[2021_AMC_10A_Problems/Problem_25|2021 AMC 10A Problem 25]] (Solutions 2, 3, Video Solution)<br />
* [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 2)<br />
* [[2021_AMC_10B_Problems/Problem_7|2021 AMC 10B Problem 7]] (Solution)<br />
* [[2021_AMC_10B_Problems/Problem_8|2021 AMC 10B Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_10B_Problems/Problem_17|2021 AMC 10B Problem 17]] (Solution)<br />
* [[2021_AMC_10B_Problems/Problem_23|2021 AMC 10B Problem 23]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_1|2021 Fall AMC 10A Problem 1]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_2|2021 Fall AMC 10A Problem 2]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_3|2021 Fall AMC 10A Problem 3]] (Solutions 1, 2, 3)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_4|2021 Fall AMC 10A Problem 4]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_6|2021 Fall AMC 10A Problem 6]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_7|2021 Fall AMC 10A Problem 7]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_10|2021 Fall AMC 10A Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_11|2021 Fall AMC 10A Problem 11]] (Solution 3)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_12|2021 Fall AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_15|2021 Fall AMC 10A Problem 15]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_16|2021 Fall AMC 10A Problem 16]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_17|2021 Fall AMC 10A Problem 17]] (Diagrams, Solution 5)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_18|2021 Fall AMC 10A Problem 18]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_19|2021 Fall AMC 10A Problem 19]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_20|2021 Fall AMC 10A Problem 20]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_21|2021 Fall AMC 10A Problem 21]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_22|2021 Fall AMC 10A Problem 22]] (Diagram)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_24|2021 Fall AMC 10A Problem 24]] (Solution 1)<br />
* [[2021_Fall_AMC_10A_Problems/Problem_25|2021 Fall AMC 10A Problem 25]] (Solution 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_2|2021 Fall AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_4|2021 Fall AMC 10B Problem 4]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_10B_Problems/Problem_7|2021 Fall AMC 10B Problem 7]] (Solution 1)<br />
<br />
===AHSME / AMC 12===<br />
* [[1976_AHSME_Problems/Problem_24|1976 AHSME Problem 24]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_25|1976 AHSME Problem 25]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_27|1976 AHSME Problem 27]] (Solutions 1, 2)<br />
* [[1976_AHSME_Problems/Problem_28|1976 AHSME Problem 28]] (Solution)<br />
* [[1976_AHSME_Problems/Problem_30|1976 AHSME Problem 30]] (Solution)<br />
* [[1978_AHSME_Problems/Problem_20|1978 AHSME Problem 20]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_19|1982 AHSME Problem 19]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_21|1982 AHSME Problem 21]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_23|1982 AHSME Problem 23]] (Solutions 1, 2)<br />
* [[1982_AHSME_Problems/Problem_27|1982 AHSME Problem 27]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_28|1982 AHSME Problem 28]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_29|1982 AHSME Problem 29]] (Solution)<br />
* [[1982_AHSME_Problems/Problem_30|1982 AHSME Problem 30]] (Solution)<br />
* [[1990_AHSME_Problems/Problem_26|1990 AHSME Problem 26]] (Solutions 1, 2)<br />
* [[1991_AHSME_Problems/Problem_6|1991 AHSME Problem 6]] (Solution)<br />
* [[1991_AHSME_Problems/Problem_20|1991 AHSME Problem 20]] (Solution)<br />
* [[1991_AHSME_Problems/Problem_27|1991 AHSME Problem 27]] (Solution)<br />
* [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]] (Solution)<br />
* [[2004_AMC_12A_Problems/Problem_17|2004 AMC 12A Problem 17]]: same as [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br />
* [[2009_AMC_12B_Problems/Problem_1|2009 AMC 12B Problem 1]]: same as [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br />
* [[2014_AMC_12A_Problems/Problem_18|2014 AMC 12A Problem 18]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_2|2018 AMC 12A Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_12A_Problems/Problem 4|2018 AMC 12A Problem 4]]: same as [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_7|2018 AMC 12A Problem 7]]: same as [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solutions 1, 2)<br />
* [[2018_AMC_12A_Problems/Problem_12|2018 AMC 12A Problem 12]]: same as [[2018_AMC_10A_Problems/Problem_17|2018 AMC 10A Problem 17]] (Solution 4)<br />
* [[2018_AMC_12A_Problems/Problem_14|2018 AMC 12A Problem 14]] (Solutions 1, 2, 3, 4, 5)<br />
* [[2018_AMC_12A_Problems/Problem_16|2018 AMC 12A Problem 16]]: same as [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br />
* [[2018_AMC_12A_Problems/Problem_17|2018 AMC 12A Problem 17]]: same as [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_19|2018 AMC 12A Problem 19]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_21|2018 AMC 12A Problem 21]] (Solution 1)<br />
* [[2018_AMC_12A_Problems/Problem_22|2018 AMC 12A Problem 22]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12A_Problems/Problem_23|2018 AMC 12A Problem 23]] (Diagram, Solution 2)<br />
* [[2018_AMC_12A_Problems/Problem_24|2018 AMC 12A Problem 24]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12A_Problems/Problem_25|2018 AMC 12A Problem 25]]: same as [[2018_AMC_10A_Problems/Problem_25|2018 AMC 10A Problem 25]] (Solutions 1, 3)<br />
* [[2018_AMC_12B_Problems/Problem_2|2018 AMC 12B Problem 2]]: same as [[2018_AMC_10B_Problems/Problem_2|2018 AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_4|2018 AMC 12B Problem 4]] (Solution)<br />
* [[2018_AMC_12B_Problems/Problem_6|2018 AMC 12B Problem 6]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_8|2018 AMC 12B Problem 8]] (Solution 1)<br />
* [[2018_AMC_12B_Problems/Problem_9|2018 AMC 12B Problem 9]] (Solutions 1, 2, 3, 4, 5, 6)<br />
* [[2018_AMC_12B_Problems/Problem_11|2018 AMC 12B Problem 11]]: same as [[2018_AMC_10B_Problems/Problem_15|2018 AMC 10B Problem 15]] (Solutions 1, 3)<br />
* [[2018_AMC_12B_Problems/Problem_12|2018 AMC 12B Problem 12]] (Solution)<br />
* [[2018_AMC_12B_Problems/Problem_13|2018 AMC 12B Problem 13]] (Solutions 1, 2, 3)<br />
* [[2018_AMC_12B_Problems/Problem_14|2018 AMC 12B Problem 14]]: same as [[2018_AMC_10B_Problems/Problem_19|2018 AMC 10B Problem 19]] (Solution 1)<br />
* [[2018_AMC_12B_Problems/Problem_15|2018 AMC 12B Problem 15]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_16|2018 AMC 12B Problem 16]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_17|2018 AMC 12B Problem 17]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_18|2018 AMC 12B Problem 18]]: same as [[2018_AMC_10B_Problems/Problem_20|2018 AMC 10B Problem 20]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_19|2018 AMC 12B Problem 19]]: same as [[2018_AMC_10B_Problems/Problem_21|2018 AMC 10B Problem 21]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_20|2018 AMC 12B Problem 20]]: same as [[2018_AMC_10B_Problems/Problem_24|2018 AMC 10B Problem 24]] (Diagram, Solution 6)<br />
* [[2018_AMC_12B_Problems/Problem_21|2018 AMC 12B Problem 21]] (Diagram, Solution)<br />
* [[2018_AMC_12B_Problems/Problem_22|2018 AMC 12B Problem 22]] (Solutions 1, 2)<br />
* [[2018_AMC_12B_Problems/Problem_23|2018 AMC 12B Problem 23]] (Diagram, Solutions 1, 2, 3)<br />
* [[2020_AMC_12A_Problems/Problem_1|2020 AMC 12A Problem 1]] (Solution 3)<br />
* [[2020_AMC_12A_Problems/Problem_6|2020 AMC 12A Problem 6]] (Solution 2)<br />
* [[2020_AMC_12A_Problems/Problem_8|2020 AMC 12A Problem 8]]: same as [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br />
* [[2020_AMC_12A_Problems/Problem_9|2020 AMC 12A Problem 9]] (Solution)<br />
* [[2020_AMC_12A_Problems/Problem_10|2020 AMC 12A Problem 10]] (Solutions 1, 2, 5)<br />
* [[2020_AMC_12A_Problems/Problem_15|2020 AMC 12A Problem 15]] (Solutions 1, 2)<br />
* [[2020_AMC_12A_Problems/Problem_23|2020 AMC 12A Problem 23]]: same as [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br />
* [[2020_AMC_12A_Problems/Problem_25|2020 AMC 12A Problem 25]] (Solution 1, Remark)<br />
* [[2020_AMC_12B_Problems/Problem_3|2020 AMC 12B Problem 3]]: same as [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 1)<br />
* [[2020_AMC_12B_Problems/Problem_4|2020 AMC 12B Problem 4]]: same as [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br />
* [[2020_AMC_12B_Problems/Problem_5|2020 AMC 12B Problem 5]] (Solution 1)<br />
* [[2020_AMC_12B_Problems/Problem_6|2020 AMC 12B Problem 6]] (Solution 2)<br />
* [[2020_AMC_12B_Problems/Problem_10|2020 AMC 12B Problem 10]] (Diagram, Solution 2)<br />
* [[2020_AMC_12B_Problems/Problem_12|2020 AMC 12B Problem 12]] (Diagram)<br />
* [[2020_AMC_12B_Problems/Problem_13|2020 AMC 12B Problem 13]] (Solutions 1, 4)<br />
* [[2020_AMC_12B_Problems/Problem_19|2020 AMC 12B Problem 19]]: same as [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_2|2021 AMC 12A Problem 2]] (Solutions 3, 4)<br />
* [[2021_AMC_12A_Problems/Problem_3|2021 AMC 12A Problem 3]]: same as [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br />
* [[2021_AMC_12A_Problems/Problem_4|2021 AMC 12A Problem 4]]: same as [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 1, 3)<br />
* [[2021_AMC_12A_Problems/Problem_5|2021 AMC 12A Problem 5]]: same as [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_6|2021 AMC 12A Problem 6]] (Solutions 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_7|2021 AMC 12A Problem 7]]: same as [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 2)<br />
* [[2021_AMC_12A_Problems/Problem_8|2021 AMC 12A Problem 8]] (Solution)<br />
* [[2021_AMC_12A_Problems/Problem_9|2021 AMC 12A Problem 9]]: same as [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br />
* [[2021_AMC_12A_Problems/Problem_10|2021 AMC 12A Problem 10]]: same as [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_11|2021 AMC 12A Problem 11]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_13|2021 AMC 12A Problem 13]] (Solutions 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_14|2021 AMC 12A Problem 14]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_15|2021 AMC 12A Problem 15]] (Solutions 1, 2, 4)<br />
* [[2021_AMC_12A_Problems/Problem_17|2021 AMC 12A Problem 17]]: same as [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solutions 1, 2)<br />
* [[2021_AMC_12A_Problems/Problem_18|2021 AMC 12A Problem 18]]: same as [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br />
* [[2021_AMC_12A_Problems/Problem_19|2021 AMC 12A Problem 19]] (Solutions 1, 2, 3)<br />
* [[2021_AMC_12A_Problems/Problem_21|2021 AMC 12A Problem 21]] (Solution 2, Video Solution)<br />
* [[2021_AMC_12A_Problems/Problem_22|2021 AMC 12A Problem 22]] (Solutions 1, 2, 3, 4, 5)<br />
* [[2021_AMC_12A_Problems/Problem_23|2021 AMC 12A Problem 23]]: same as [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3, Video Solution)<br />
* [[2021_AMC_12A_Problems/Problem_24|2021 AMC 12A Problem 24]] (Solution 1)<br />
* [[2021_AMC_12A_Problems/Problem_25|2021 AMC 12A Problem 25]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_4|2021 AMC 12B Problem 4]]: same as [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 2)<br />
* [[2021_AMC_12B_Problems/Problem_11|2021 AMC 12B Problem 11]] (Solution 4)<br />
* [[2021_AMC_12B_Problems/Problem_14|2021 AMC 12B Problem 14]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_20|2021 AMC 12B Problem 20]] (Solution 1)<br />
* [[2021_AMC_12B_Problems/Problem_21|2021 AMC 12B Problem 21]] (Solution 2)<br />
* [[2021_AMC_12B_Problems/Problem_23|2021 AMC 12B Problem 23]] (Solution 4)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_1|2021 Fall AMC 12A Problem 1]]: same as [[2021_Fall_AMC_10A_Problems/Problem_1|2021 Fall AMC 10A Problem 1]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_2|2021 Fall AMC 12A Problem 2]]: same as [[2021_Fall_AMC_10A_Problems/Problem_2|2021 Fall AMC 10A Problem 2]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_3|2021 Fall AMC 12A Problem 3]]: same as [[2021_Fall_AMC_10A_Problems/Problem_4|2021 Fall AMC 10A Problem 4]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_5|2021 Fall AMC 12A Problem 5]]: same as [[2021_Fall_AMC_10A_Problems/Problem_6|2021 Fall AMC 10A Problem 6]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_6|2021 Fall AMC 12A Problem 6]]: same as [[2021_Fall_AMC_10A_Problems/Problem_7|2021 Fall AMC 10A Problem 7]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_7|2021 Fall AMC 12A Problem 7]]: same as [[2021_Fall_AMC_10A_Problems/Problem_10|2021 Fall AMC 10A Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_9|2021 Fall AMC 12A Problem 9]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_10|2021 Fall AMC 12A Problem 10]]: same as [[2021_Fall_AMC_10A_Problems/Problem_12|2021 Fall AMC 10A Problem 12]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_12|2021 Fall AMC 12A Problem 12]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_13|2021 Fall AMC 12A Problem 13]] (Diagram, Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_16|2021 Fall AMC 12A Problem 16]] (Solution)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_17|2021 Fall AMC 12A Problem 17]]: same as [[2021_Fall_AMC_10A_Problems/Problem_20|2021 Fall AMC 10A Problem 20]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_18|2021 Fall AMC 12A Problem 18]]: same as [[2021_Fall_AMC_10A_Problems/Problem_21|2021 Fall AMC 10A Problem 21]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_23|2021 Fall AMC 12A Problem 23]]: same as [[2021_Fall_AMC_10A_Problems/Problem_25|2021 Fall AMC 10A Problem 25]] (Solution 2)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_24|2021 Fall AMC 12A Problem 24]] (Solution 1)<br />
* [[2021_Fall_AMC_12A_Problems/Problem_25|2021 Fall AMC 12A Problem 25]] (Solution 1)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_2|2021 Fall AMC 12B Problem 2]]: same as [[2021_Fall_AMC_10B_Problems/Problem_2|2021 Fall AMC 10B Problem 2]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_3|2021 Fall AMC 12B Problem 3]]: same as [[2021_Fall_AMC_10B_Problems/Problem_4|2021 Fall AMC 10B Problem 4]] (Solutions 1, 2)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_5|2021 Fall AMC 12B Problem 5]]: same as [[2021_Fall_AMC_10B_Problems/Problem_7|2021 Fall AMC 10B Problem 7]] (Solution 1)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_10|2021 Fall AMC 12B Problem 10]] (Solution)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_12|2021 Fall AMC 12B Problem 12]] (Solutions 1, 3)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_17|2021 Fall AMC 12B Problem 17]] (Solution 4)<br />
* [[2021_Fall_AMC_12B_Problems/Problem_18|2021 Fall AMC 12B Problem 18]] (Solution 1)<br />
<br />
===AIME===<br />
* [[1984_AIME_Problems/Problem_4|1984 AIME Problem 4]] (Solutions 1, 2)<br />
* [[1984_AIME_Problems/Problem_10|1984 AIME Problem 10]] (Solution 3)<br />
* [[1985_AIME_Problems/Problem_12|1985 AIME Problem 12]] (Solutions 1, 2, 3)<br />
* [[1986_AIME_Problems/Problem_2|1986 AIME Problem 2]] (Solutions 1, 2)<br />
* [[1987_AIME_Problems/Problem_14|1987 AIME Problem 14]] (Solutions 1, 2, 3)<br />
* [[1988_AIME_Problems/Problem_8|1988 AIME Problem 8]] (Solution 1)<br />
* [[1988_AIME_Problems/Problem_13|1988 AIME Problem 13]] (Solution 6)<br />
* [[1989_AIME_Problems/Problem_1|1989 AIME Problem 1]] (Solutions 3, 5)<br />
* [[1989_AIME_Problems/Problem_8|1989 AIME Problem 8]] (Solutions 1, 2, 3)<br />
* [[1989_AIME_Problems/Problem_9|1989 AIME Problem 9]] (Solutions 1, 2, 4)<br />
* [[1990_AIME_Problems/Problem_8|1990 AIME Problem 8]] (Solution, Remark)<br />
* [[1992_AIME_Problems/Problem_6|1992 AIME Problem 6]] (Solution 1)<br />
* [[1993_AIME_Problems/Problem_7|1993 AIME Problem 7]] (Solution 1)<br />
* [[1993_AIME_Problems/Problem_8|1993 AIME Problem 8]] (Solutions 3, 4)<br />
* [[2019_AIME_II_Problems/Problem_7|2019 AIME II Problem 7]] (Diagram)<br />
* [[2020_AIME_I_Problems/Problem_5|2020 AIME I Problem 5]] (Solution 9)<br />
* [[2021_AIME_I_Problems/Problem_1|2021 AIME I Problem 1]] (Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_2|2021 AIME I Problem 2]] (Solution 5)<br />
* [[2021_AIME_I_Problems/Problem_3|2021 AIME I Problem 3]] (Solution 3)<br />
* [[2021_AIME_I_Problems/Problem_7|2021 AIME I Problem 7]] (Remark)<br />
* [[2021_AIME_I_Problems/Problem_8|2021 AIME I Problem 8]] (Solution 1, Remark)<br />
* [[2021_AIME_I_Problems/Problem_9|2021 AIME I Problem 9]] (Diagram, Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_10|2021 AIME I Problem 10]] (Solution 2)<br />
* [[2021_AIME_I_Problems/Problem_11|2021 AIME I Problem 11]] (Diagram, Solution 1)<br />
* [[2021_AIME_I_Problems/Problem_12|2021 AIME I Problem 12]] (Solution)<br />
* [[2021_AIME_I_Problems/Problem_14|2021 AIME I Problem 14]] (Solutions 2, 3)<br />
* [[2021_AIME_I_Problems/Problem_15|2021 AIME I Problem 15]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_1|2021 AIME II Problem 1]] (Solution 3, Remark)<br />
* [[2021_AIME_II_Problems/Problem_2|2021 AIME II Problem 2]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_3|2021 AIME II Problem 3]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_4|2021 AIME II Problem 4]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_5|2021 AIME II Problem 5]] (Solutions 2, 5)<br />
* [[2021_AIME_II_Problems/Problem_6|2021 AIME II Problem 6]] (Solution 3)<br />
* [[2021_AIME_II_Problems/Problem_7|2021 AIME II Problem 7]] (Solution 3)<br />
* [[2021_AIME_II_Problems/Problem_8|2021 AIME II Problem 8]] (Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_9|2021 AIME II Problem 9]] (Solution, Remarks)<br />
* [[2021_AIME_II_Problems/Problem_10|2021 AIME II Problem 10]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_11|2021 AIME II Problem 11]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_12|2021 AIME II Problem 12]] (Solution 2)<br />
* [[2021_AIME_II_Problems/Problem_13|2021 AIME II Problem 13]] (Solutions 1, 3)<br />
* [[2021_AIME_II_Problems/Problem_14|2021 AIME II Problem 14]] (Diagram, Solution 1)<br />
* [[2021_AIME_II_Problems/Problem_15|2021 AIME II Problem 15]] (Solution 2)<br />
* [[2022_AIME_I_Problems/Problem_1|2022 AIME I Problem 1]] (Solutions 1, 2, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_2|2022 AIME I Problem 2]] (Solution 1, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_3|2022 AIME I Problem 3]] (Diagram, Video Solution)<br />
* [[2022_AIME_I_Problems/Problem_4|2022 AIME I Problem 4]] (Solution 1)<br />
* [[2022_AIME_I_Problems/Problem_5|2022 AIME I Problem 5]] (Solution 2)<br />
* [[2022_AIME_I_Problems/Problem_7|2022 AIME I Problem 7]] (Solution)<br />
* [[2022_AIME_I_Problems/Problem_8|2022 AIME I Problem 8]] (Diagram)<br />
<br />
==AoPS Forums==<br />
[https://artofproblemsolving.com/community/c2686156_hedgehog_military HEDGEHOG MILITARY]<br />
<br />
Let's have a good time in this forum. Feel free to discuss math classes, math competitions, AoPS Wiki contributions, funny pictures, emojis, and ... HEDGEHOGS! <math>\smiley{}</math><br />
<br />
[[File:Hedgehog LALALALALA.gif|center]]<br />
<br />
==Publications==<br />
<i><b>English-Chinese Mathematics Encyclopedia: Algebra, Geometry, and Pre-Calculus</b></i> (Tsinghua University Press in Beijing, China)<br />
<br />
[[File:English-Chinese Mathematics Encyclopedia.jpeg|600px|center]]<br />
<br />
==Artworks==<br />
In this section, I will show you my artworks of certificates and posters. Hope you enjoy them! <math>\smiley{}</math><br />
<br />
I especially thank [[User:Flamekhoemberish|Flame Kho]] for teaching me how to use the Chuangkit design platform.<br />
<br />
===Hall of Fame in MATHCOUNTS/AMC 8 Basics (2557)===<br />
[[File:2021 AMC 8 Perfect Scorers.png|center|750px]] <p><br />
<br />
[[File:2557 Gold.png|center|750px]] <p><br />
<br />
[[File:2557 Silver.png|center|750px]] <p><br />
<br />
[[File:2557 Bronze.png|center|750px]]<br />
<br />
===Hall of Fame in MATHCOUNTS/AMC 8 Basics (2786)===<br />
[[File:2846 2022 AMC 8 Perfect Scorers.png|center|500px]] <p><br />
<br />
[[File:2846 Gold.png|center|750px]] <p><br />
<br />
[[File:2846 Silver.png|center|750px]] <p><br />
<br />
[[File:2846 Bronze 1.png|center|750px]] <p><br />
<br />
[[File:2846 Bronze 2.png|center|750px]]<br />
<br />
===Hall of Fame in MATHCOUNTS/AMC 8 Basics (3011)===<br />
We will see about that--designing templates now ...<br />
<br />
===Most Prominent Member of Boston Xiangqi Community===<br />
[[File:Funny Guy.png|center|750px]]<br />
<br />
==Cats==<br />
As you might know, I am particularly fond of cats. My family has raised more than ten cats. These little angels are so adorable! MEOW MEOW MEOW!!! 🐱🐱🐱<br />
<br />
===Luna===<br />
[[File:Luna.jpg|center|300px]]<br />
<br />
===Kidney Bean===<br />
<center>[[File:Kidney Bean 1.jpg|300px]]&nbsp;&nbsp;&nbsp;[[File:Kidney Bean 2.jpg|300px]]</center><br />
<br />
===Jin Bao===<br />
[[File:Jin Bao.jpg|center|300px]]<br />
<br />
===Gray Gray===<br />
[[File:Little Gray Gray 1.jpg|center|300px]] <p><br />
<br />
<center>[[File:Little Gray Gray 2.jpg|450px]]&nbsp;&nbsp;&nbsp;[[File:Little Gray Gray 3.jpg|450px]]</center><br />
<br />
==External Links==<br />
* [[User_talk:MRENTHUSIASM|User Talk Page]] <br> Welcome to my user talk page here. Feel free to leave any message you like.<br />
* [[User:Flamekhoemberish|Flame Kho's User Page]] <br> Recently I am helping my friend Flame Kho to create his user page. Enjoy his AMC Solutions and artworks! <math>\smiley{}</math></div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:Flamekhoemberish&diff=176009User:Flamekhoemberish2022-07-17T20:33:26Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div>==User Count==<br />
If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">15</font></center><br />
<br />
==Introduction==<br />
Hi Everyone,<br />
<br />
I am Flame Kho. I am currently working in England and quite fluent in both English and Mandarin.<br />
<br />
I have a broad range of interests: mathematics, Cover Chess, classic Chinese songs, travels, etc. I also enjoy creating certificates for chess winners. Visit this page for my interests and artworks. :)<br />
<br />
Below is my upcoming avatar, which I will change two weeks from now.<br />
<br />
[[File:FlameKhoAvatar.jpeg|center|300px]]<br />
~FlameKhoEmberish<br />
<br />
==Mathematics==<br />
I am an amateur problem-solver. I am so proud of my contribution to [[2021_AMC_10A_Problems/Problem_25|2021 AMC 10A Problem 25]] (Solution 3).<br />
<br />
==Cover Chess==<br />
Cover Chess is a variant of Xiangqi (Chinese Chess), where all pieces are covered initially. The game has a considerable amount of luck--even amateur players have a chance of beating masters.<br />
<br />
I managed many Cover Chess tournaments and participated in certificate-making. See my artworks below.<br />
<br />
==Guess Songs==<br />
I am a fan of Chinese classic music. I recognize a lot of songs. If you want to challenge me on guessing songs, I will always be here! :)<br />
<br />
==Artworks==<br />
In this section, I will show you my artworks across different areas. I am the original author for all of the artworks below.<br />
<br />
Hope you enjoy them! :)<br />
<br />
===Certificates===<br />
====Cover Chess====<br />
<center>[[File:Cover Chess Certificate 1.jpeg|450px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 2.jpeg|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 3.jpeg|450px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 7.jpeg|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 8.jpeg|center|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 4.jpeg|300px]]&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 5.jpeg|300px]]&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 6.jpeg|300px]]</center><br />
<br />
====Guess Songs====<br />
[[File:GuessSongs.jpeg|center|300px]]<br />
<br />
===Posters===<br />
====Valentine Speed Chess Tournament====<br />
<center>[[File:Season 2.jpeg|300px]][[File:Season 3.jpeg|300px]][[File:Season 4.jpeg|300px]]</center><br />
<center>[[File:Season 5.jpeg|300px]][[File:Season 6.jpeg|300px]][[File:Season 7.jpeg|300px]]</center><br />
<center>[[File:Season 8.jpeg|300px]][[File:Season 9.jpeg|300px]][[File:Season 10.jpeg|300px]]</center><br />
<br />
====Master Challenges====<br />
[[File:Master 1.jpeg|center|600px]]<p><br />
[[File:Master 2.png|center|600px]]<p><br />
[[File:Master 3.jpeg|center|600px]]<br />
<br />
====Other Chess Tournaments====<br />
<center>[[File:Chess Poster 1.jpeg|400px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Chess Poster 2.jpeg|400px]]</center><p><br />
<center>[[File:Chess Poster 3.jpeg|400px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Chess Poster 4.jpeg|400px]]</center><br />
<br />
===Photographs===<br />
[[File:Church.jpeg|center]]<p><br />
[[File:Outdoors.jpeg|center]]<br />
<br />
===Drawings===<br />
[[File:Drawing.jpeg|center|300px]]<br />
<br />
==Cartoons==<br />
[[File:Cartoon.jpeg|center]]<br />
<br />
==Stickers==<br />
Finally, I am a huge fan of stickers. Below are my favorite ones. As you can tell, I love cats in particular.<br />
<center>[[File:Coming.png|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Hahahahaha.png|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Bad Words.jpg|250px]]</center><p><br />
<center>[[File:Yes.gif|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Fighting.gif|250px]]</center><br />
<br />
==Acknowledgements==<br />
I especially thank [[User:MRENTHUSIASM|MRENTHUSIASM]] for creating and editing my user page.</div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:Aops-g5-gethsemanea2&diff=163698User:Aops-g5-gethsemanea22021-10-18T04:39:45Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div><br />
{{Template:Aops-g5-gethsemanea2/Header}}<br />
<br />
{{User:CreativeHedgehog/Templates/SocialMedia}}<br />
<br />
{{shortcut|[[AGG]]}}<br />
<br />
Welcome to my user page! <br />
<br />
<br />
==About Me==<br />
<br />
<div style="padding-left:16px; padding-right:16px; border:15px #009fad; background-color:#eaeaea; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;"><br />
<br />
Aops-g5-gethsemanea2 is just an <b>Algebra A</b> user <s>who is not as legendary as Piphi</s>. <br />
<br />
Like CreativeHedgehog, Aops-g5-gethsemanea2 is making a lot of projects on this wiki including the Wiki Math Games and his very own main page.<br />
<br />
Aops-g5-gethsemanea2 got an 18 in the AMC 8 and a three-time national qualifier in mathleague.org.<br />
<br />
Aops-g5-gethsemanea2 is an International Countdown Champion in mathleague.org elementary 2021.<br />
<br />
==Acknowledgements==<br />
Thanks to the following users!<br />
<br />
1. Piphi for the inspiration<br />
<br />
2. CreativeHedgehog for the social media buttons, the box code (the one used in his Viewer Count and in my User Count) and supporting this User Page<br />
</div><br />
<br />
==User Count==<br />
<div style="padding-left:16px; padding-right:16px; border:15px #1b365d dashed; background-color:white; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;"><br />
<br />
If this is the first time you visited this page, increase the following number by 1 and change the username written to yours:<br />
<br />
<b><i>Latest user: kante314</i></b><br />
<br />
<div style="font-size:10em; text-align:center;"><b>25</b></div><br />
<br />
[https://artofproblemsolving.com/wiki/index.php?title=User:Aops-g5-gethsemanea2&action=edit&section=3 (Edit this User Count)]<br />
<br />
[https://artofproblemsolving.com/online/?login=1 (Log in Here)]<br />
[https://artofproblemsolving.com/community/logout (Log out Here)]<br />
<br />
</div><br />
<br />
<br />
<br />
<div style="padding-left:16px; padding-right:16px; border:15px #009fad; background-color:#ffe4e1; padding-top:10px; padding-bottom:10px; max-height:200px; overflow:auto; display:block; width:100%;"><br />
<br />
=WARNINGS!=<br />
<br />
==WARNING 1==<br />
This wiki page will NOT work if you use a mobile device with the social media buttons.<br />
<br />
If you want it to work on a mobile device, you may put everything in a div and then align it to the center.<br />
<br />
Then make the width 50%.<br />
<br />
Example: Put the following text before every page:<br />
<br />
<nowiki><br />
<br />
<div style="position:fixed; left:50%; transform:translateX(-50%); width:50%;"><br />
<br />
</nowiki><br />
<br />
And this after it:<br />
<br />
<nowiki><br />
</div><br />
</nowiki><br />
<br />
I just didn't do that because I had 40 pages to edit that out with. [https://artofproblemsolving.com/wiki/index.php/User:CreativeHedgehog CreativeHedgehog] 22:32, 4 August 2020 (EDT)<br />
<br />
==WARNING 2==<br />
If you followed the instruction above, then it will not work this time on a computer. Remove the div at the start and the end to make it back to normal.<br />
<br />
==WARNING 3==<br />
This will only work on the Aops2 skin style. [https://artofproblemsolving.com/wiki/index.php/User:CreativeHedgehog CreativeHedgehog] 21:56, 8 August 2020 (EDT)<br />
<br />
=REMINDER=<br />
Most of these pages require you to sign in to edit the pages, especially the Wiki Math Games.<br />
</div></div>Ethanspoonhttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=163696User:Piphi2021-10-18T04:35:38Z<p>Ethanspoon: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">533<br />
</font></center><br />
</div><br />
<div style="border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black"><font color="black"><br />
Piphi created the [[User:Piphi/Games|AoPS Wiki Games]].<br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18, 2nd in game #19, 10th in game #20, and 2nd in game #21. Piphi made a the Greed Control Hall of Fame [https://artofproblemsolving.com/community/h2445343 here].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br />
<br />
Piphi is part of the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] staff.<br />
</font></div><br />
</div><br />
<div style="border:2px solid black; background:#333333;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="#f0f2f3" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px"><font color="#f0f2f3"><br />
You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br />
<br />
A User Count of 1000<br />
{{User:Piphi/Template:Progress_Bar|53.2|width=100%}}<br />
<br />
500 subpages of [[User:Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|41.6|width=100%}}<br />
<br />
200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|54|width=100%}}<br />
<br />
Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|23.49|width=100%}}</font></div><br />
</div></div>Ethanspoon