https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Prestonh&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T01:10:45ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:MRENTHUSIASM&diff=177261User:MRENTHUSIASM2022-08-21T16:06:53Z<p>Prestonh: /* 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">156</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, 3)<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, 3)<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>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=User:Prestonh&diff=143452User:Prestonh2021-01-27T20:15:38Z<p>Prestonh: </p>
<hr />
<div>It is rumored that he doesn't exist</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=1995_AJHSME_Problems&diff=1355521995 AJHSME Problems2020-10-22T03:09:49Z<p>Prestonh: /* Problem 10 */</p>
<hr />
<div>{{AJHSME Problems<br />
|year = 1995<br />
}}<br />
<br />
==Problem 1==<br />
<br />
Walter has exactly one penny, one nickel, one dime and one quarter in his pocket. What percent of one dollar is in his pocket?<br />
<br />
<math>\text{(A)}\ 4\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 41\% \qquad \text{(E)}\ 59\%</math><br />
<br />
[[1995 AJHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
Jose is <math>4</math> years younger than Zack. Zack is <math>3</math> years older than Inez. Inez is <math>15</math> years old. How old is Jose?<br />
<br />
<math>\text{(A)}\ 8 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22</math><br />
<br />
[[1995 AJHSME Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
Which of the following operations has the same effect on a number as multiplying by <math>\dfrac{3}{4}</math> and then dividing by <math>\dfrac{3}{5}</math>?<br />
<br />
<math>\text{(A)}\ \text{dividing by }\dfrac{4}{3} \qquad \text{(B)}\ \text{dividing by }\dfrac{9}{20} \qquad \text{(C)}\ \text{multiplying by }\dfrac{9}{20} \qquad \text{(D)}\ \text{dividing by }\dfrac{5}{4} \qquad \text{(E)}\ \text{multiplying by }\dfrac{5}{4}</math><br />
<br />
[[1995 AJHSME Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
A teacher tells the class,<br />
<br />
<center><br />
''"Think of a number, add 1 to it, and double the result. Give the answer to your partner. Partner, subtract 1 from the number you are given and double the result to get your answer."''<br />
</center><br />
<br />
Ben thinks of <math>6</math>, and gives his answer to Sue. What should Sue's answer be?<br />
<br />
<math>\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 30</math><br />
<br />
[[1995 AJHSME Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
Find the smallest whole number that is larger than the sum<br />
<br />
<cmath>2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.</cmath><br />
<br />
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18</math><br />
<br />
[[1995 AJHSME Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
Figures <math>I</math>, <math>II</math>, and <math>III</math> are squares. The perimeter of <math>I</math> is <math>12</math> and the perimeter of <math>II</math> is <math>24</math>. The perimeter of <math>III</math> is<br />
<br />
<asy><br />
draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle);<br />
draw((9,0)--(9,9));<br />
draw((9,6)--(12,6));<br />
label("$III$",(4.5,4),N);<br />
label("$II$",(12,2.5),N);<br />
label("$I$",(10.5,6.75),N);<br />
</asy><br />
<br />
<math>\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81</math><br />
<br />
[[1995 AJHSME Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
At Clover View Junior High, one half of the students go home on the school bus. One fourth go home by automobile. One tenth go home on their bicycles. The rest walk home. What fractional part of the students walk home?<br />
<br />
<math>\text{(A)}\ \dfrac{1}{16} \qquad \text{(B)}\ \dfrac{3}{20} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{17}{20} \qquad \text{(E)}\ \dfrac{9}{10}</math><br />
<br />
[[1995 AJHSME Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
An American traveling in Italy wishes to exchange American money (dollars) for Italian money (lire). If 3000 lire = \$1.60, how much lire will the traveler receive in exchange for \$1.00?<br />
<br />
<math>\text{(A)}\ 180 \qquad \text{(B)}\ 480 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 1875 \qquad \text{(E)}\ 4875</math><br />
<br />
[[1995 AJHSME Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
Three congruent circles with centers <math>P</math>, <math>Q</math>, and <math>R</math> are tangent to the sides of rectangle <math>ABCD</math> as shown. The circle centered at <math>Q</math> has diameter <math>4</math> and passes through points <math>P</math> and <math>R</math>. The area of the rectangle is<br />
<br />
<asy><br />
pair A,B,C,D,P,Q,R;<br />
A = (0,4); B = (8,4); C = (8,0); D = (0,0);<br />
P = (2,2); Q = (4,2); R = (6,2);<br />
dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R);<br />
draw(A--B--C--D--cycle);<br />
draw(circle(P,2));<br />
draw(circle(Q,2));<br />
draw(circle(R,2));<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$P$",P,W);<br />
label("$Q$",Q,W);<br />
label("$R$",R,W);<br />
</asy><br />
<br />
<math>\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128</math><br />
<br />
[[1995 AJHSME Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
A jacket and a shirt originally sold for 80 dollars and 40 dollars, respectively. During a sale Chris bought the 80 dollar jacket at a <math>40\%</math> discount and the 40 dollar shirt at a <math>55\%</math> discount. The total amount saved was what percent of the total of the original prices?<br />
<br />
<math>\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 79\dfrac{1}{6}\% \qquad \text{(E)}\ 95\%</math>.<br />
<br />
[[1995 AJHSME Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be closest to<br />
<br />
<asy><br />
for(int i = -2; i <= 2; ++i)<br />
{<br />
draw((i,0)--(i,3),dashed);<br />
}<br />
draw((-3,1)--(3,1),dashed);<br />
draw((-3,2)--(3,2),dashed);<br />
draw((-3,0)--(-3,3)--(3,3)--(3,0)--cycle);<br />
dot((-3,0)); label("$A$",(-3,0),SW);<br />
dot((-3,3)); label("$B$",(-3,3),NW);<br />
dot((0,3)); label("$C$",(0,3),N);<br />
dot((3,3)); label("$D$",(3,3),NE);<br />
dot((3,0)); label("$E$",(3,0),SE);<br />
dot((0,0)); label("start",(0,0),S);<br />
label("$\longrightarrow$",(0,-0.75),E);<br />
label("$\longleftarrow$",(0,-0.75),W);<br />
label("$\textbf{Jane}$",(0,-1.25),W);<br />
label("$\textbf{Hector}$",(0,-1.25),E);<br />
</asy><br />
<br />
<math>\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E</math><br />
<br />
[[1995 AJHSME Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
A ''lucky'' year is one in which at least one date, when written in the form month/day/year, has the following property: ''The product of the month times the day equals the last two digits of the year''. For example, 1956 is a lucky year because it has the date 7/8/56 and <math>7\times 8 = 56</math>. Which of the following is NOT a lucky year?<br />
<br />
<math>\text{(A)}\ 1990 \qquad \text{(B)}\ 1991 \qquad \text{(C)}\ 1992 \qquad \text{(D)}\ 1993 \qquad \text{(E)}\ 1994</math><br />
<br />
[[1995 AJHSME Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
In the figure, <math>\angle A</math>, <math>\angle B</math>, and <math>\angle C</math> are right angles. If <math>\angle AEB = 40^\circ </math> and <math>\angle BED = \angle BDE</math>, then <math>\angle CDE = </math><br />
<br />
<asy><br />
dot((0,0)); label("$E$",(0,0),SW);<br />
dot(dir(85)); label("$A$",dir(85),NW);<br />
dot((4,0)); label("$D$",(4,0),SE);<br />
dot((4.05677,0.648898)); label("$C$",(4.05677,0.648898),NE);<br />
draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle);<br />
<br />
dot((2,2)); label("$B$",(2,2),N);<br />
draw((0,0)--(2,2)--(4,0));<br />
pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898));<br />
dot(x[0]); dot(x[1]);<br />
label("$F$",x[0],SE);<br />
label("$G$",x[1],SW);<br />
</asy><br />
<br />
<math>\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ</math><br />
<br />
[[1995 AJHSME Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
A team won 40 of its first 50 games. How many of the remaining 40 games must this team win so it will have won exactly 70% of its games for the season?<br />
<br />
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35</math><br />
<br />
[[1995 AJHSME Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
What is the <math>100^\text{th}</math> digit to the right of the decimal point in the decimal form of <math>4/37</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math><br />
<br />
[[1995 AJHSME Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Students from three middle schools worked on a summer project.<br />
<br />
*Seven students from Allen school worked for 3 days.<br />
*Four students from Balboa school worked for 5 days.<br />
*Five students from Carver school worked for 9 days.<br />
<br />
The total amount paid for the students' work was <math>774</math>. Assuming each student received the same amount for a day's work, how much did the students from Balboa school earn altogether?<br />
<br />
<math>\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \qquad \text{(C)}\ 180.00\text{ dollars} \qquad \text{(D)}\ 193.50\text{ dollars} \qquad \text{(E)}\ 258.00\text{ dollars}</math><br />
<br />
[[1995 AJHSME Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
The table below gives the percent of students in each grade at Annville and Cleona elementary schools:<br />
<br />
<cmath>\begin{tabular}{rccccccc}<br />
& \textbf{\underline{K}} & \textbf{\underline{1}} & \textbf{\underline{2}} & \textbf{\underline{3}} & \textbf{\underline{4}} & \textbf{\underline{5}} & \textbf{\underline{6}} \\<br />
\textbf{Annville:} & 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\% \\<br />
\textbf{Cleona:} & 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\% <br />
\end{tabular}</cmath><br />
<br />
Annville has 100 students and Cleona has 200 students. In the two schools combined, what percent of the students are in grade 6?<br />
<br />
<math>\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%</math><br />
<br />
[[1995 AJHSME Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
The area of each of the four congruent L-shaped regions of this 100-inch by 100-inch square is 3/16 of the total area. How many inches long is the side of the center square?<br />
<br />
<asy><br />
draw((2,2)--(2,-2)--(-2,-2)--(-2,2)--cycle);<br />
draw((1,1)--(1,-1)--(-1,-1)--(-1,1)--cycle);<br />
draw((0,1)--(0,2));<br />
draw((1,0)--(2,0));<br />
draw((0,-1)--(0,-2));<br />
draw((-1,0)--(-2,0));<br />
</asy><br />
<br />
<math>\text{(A)}\ 25 \qquad \text{(B)}\ 44 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 75</math><br />
<br />
[[1995 AJHSME Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
The graph shows the distribution of the number of children in the families of the students in Ms. Jordan's English class. The median number of children in the family for this distribution is<br />
<br />
<asy><br />
unitsize(12);<br />
for(int i = 1; i <= 7; ++i)<br />
{<br />
draw((0,i)--(19,i),dotted);<br />
draw((-0.5,i)--(0.5,i));<br />
}<br />
for(int i = 0; i <= 5; ++i)<br />
{<br />
draw((3*i+2,0)--(3*i+2,-0.5));<br />
}<br />
<br />
fill((1,0)--(1,2)--(3,2)--(3,0)--cycle,white);<br />
fill((4,0)--(4,1)--(6,1)--(6,0)--cycle,white);<br />
fill((7,0)--(7,2)--(9,2)--(9,0)--cycle,white);<br />
fill((10,0)--(10,2)--(12,2)--(12,0)--cycle,white);<br />
fill((13,0)--(13,6)--(15,6)--(15,0)--cycle,white);<br />
<br />
draw((0,9)--(0,0)--(19,0));<br />
draw((1,0)--(1,2)--(3,2)--(3,0));<br />
draw((4,0)--(4,1)--(6,1)--(6,0));<br />
draw((7,0)--(7,2)--(9,2)--(9,0));<br />
draw((10,0)--(10,2)--(12,2)--(12,0));<br />
draw((13,0)--(13,6)--(15,6)--(15,0));<br />
<br />
label("$1$",(2,-0.5),S);<br />
label("$2$",(5,-0.5),S);<br />
label("$3$",(8,-0.5),S);<br />
label("$4$",(11,-0.5),S);<br />
label("$5$",(14,-0.5),S);<br />
label("$6$",(17,-0.5),S);<br />
<br />
label("$2$",(-0.5,2),W);<br />
label("$4$",(-0.5,4),W);<br />
label("$6$",(-0.5,6),W);<br />
<br />
label("$\textbf{Number of Children}$",(9,-1.5),S);<br />
label("$\textbf{in the Family}$",(9,-2.5),S);<br />
<br />
label("$\textbf{Number}$",(-1.5,6),W);<br />
label("$\textbf{of}$",(-3,5),W);<br />
label("$\textbf{Families}$",(-1.5,4),W);<br />
</asy><br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5</math><br />
<br />
[[1995 AJHSME Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Diana and Apollo each roll a standard die obtaining a number at random from 1 to 6. What is the probability that Diana's number is larger than Apollo's number?<br />
<br />
<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{5}{12} \qquad \text{(C)}\ \dfrac{4}{9} \qquad \text{(D)}\ \dfrac{17}{36} \qquad \text{(E)}\ \dfrac{1}{2}</math><br />
<br />
[[1995 AJHSME Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides as shown. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?<br />
<br />
<asy><br />
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br />
draw(circle((2,2),1));<br />
draw((4,0)--(6,1)--(6,5)--(4,4));<br />
draw((6,5)--(2,5)--(0,4));<br />
draw(ellipse((5,2.5),0.5,1));<br />
fill(ellipse((3,4.5),1,0.25),black);<br />
fill((2,4.5)--(2,5.25)--(4,5.25)--(4,4.5)--cycle,black);<br />
fill(ellipse((3,5.25),1,0.25),black);<br />
</asy><br />
<br />
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math><br />
<br />
[[1995 AJHSME Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?<br />
<br />
<math>\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222</math><br />
<br />
[[1995 AJHSME Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?<br />
<br />
<math>\text{(A)}\ 1120 \qquad \text{(B)}\ 1400 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 2025 \qquad \text{(E)}\ 2500</math><br />
<br />
[[1995 AJHSME Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
In parallelogram <math>ABCD</math>, <math>\overline{DE}</math> is the altitude to the base <math>\overline{AB}</math> and <math>\overline{DF}</math> is the altitude to the base <math>\overline{BC}</math>. ['''Note:''' ''Both pictures represent the same parallelogram.''] If <math>DC=12</math>, <math>EB=4</math>, and <math>DE=6</math>, then <math>DF=</math><br />
<br />
<asy><br />
unitsize(12);<br />
pair A,B,C,D,P,Q,W,X,Y,Z;<br />
A = (0,0); B = (12,0); C = (20,6); D = (8,6);<br />
W = (18,0); X = (30,0); Y = (38,6); Z = (26,6);<br />
draw(A--B--C--D--cycle);<br />
draw(W--X--Y--Z--cycle);<br />
P = (8,0); Q = (758/25,6/25);<br />
dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot(P); dot(Q);<br />
draw(A--B--C--D--cycle);<br />
draw(W--X--Y--Z--cycle);<br />
draw(D--P);<br />
draw(Z--Q);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
label("$E$",P,S);<br />
label("$A$",W,SW);<br />
label("$B$",X,S);<br />
label("$C$",Y,NE);<br />
label("$D$",Z,NW);<br />
label("$F$",Q,E);<br />
</asy><br />
<br />
<math>\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math><br />
<br />
[[1995 AJHSME Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes 5 hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass in the highway (not in the station)?<br />
<br />
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math><br />
<br />
[[1995 AJHSME Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AJHSME box|year=1995|before=[[1994 AJHSME Problems|1994 AJHSME]]|after=[[1996 AJHSME Problems|1996 AJHSME]]}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_15&diff=1107412015 AMC 8 Problems/Problem 152019-10-31T22:37:31Z<p>Prestonh: /* Solution 3 */</p>
<hr />
<div>At Euler Middle School, <math>198</math> students voted on two issues in a school referendum with the following results: <math>149</math> voted in favor of the first issue and <math>119</math> voted in favor of the second issue. If there were exactly <math>29</math> students who voted against both issues, how many students voted in favor of both issues?<br />
<br />
<math>\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149</math><br />
<br />
==Solution 1==<br />
We can see that this is a Venn Diagram Problem.[SOMEBODY DRAW IT PLEASE !!!] <br />
<br />
First, we analyze the information given. There are <math>198</math> students. Let's use A as the first issue and B as the second issue. <br />
<br />
<math>149</math> students were for the A, and <math>119</math> students were for B. There were also <math>29</math> students against both A and B. <br />
<br />
Solving this without a Venn Diagram, we subtract <math>29</math> away from the total, <math>198</math>. Out of the remaining <math>169</math> , we have <math>149</math> people for A and <br />
<br />
<math>119</math> people for B. We add this up to get <math>268</math> . Since that is more than what we need, we subtract <math>169</math> from <math>268</math> to get <br />
<br />
<math>\boxed{\textbf{(D)}~99}</math><br />
<br />
<br />
<br />
<br />
Venn Diagram (I couldn't make circles)<br />
<br />
We need to know how many voted in favor for both<br />
<br />
<br />
Issue A Against both issues Issue B<br />
149 students 29 students 119 students<br />
<br />
149+29+119=297<br />
297-198=99 students in favor for both<br />
<br />
==Solution 2==<br />
There are <math>198</math> people. We know that <math>29</math> people voted against both the first issue and the second issue. That leaves us with <math>169</math> people that voted for at least one of them. If <math>119</math> people voted for both of them, then that would leave <math>20</math> people out of the vote, because <math>149</math> is less than <math>169</math> people. <math>169-149</math> is <math>20</math>, so to make it even, we have to take <math>20</math> away from the <math>119</math> people, which leaves us with <math>\boxed{\textbf{(D)}~99}</math><br />
<br />
==Solution 3==<br />
Divide the students into four categories:<br />
* A. Students who voted in favor of both issues.<br />
* B. Students who voted against both issues.<br />
* C. Students who voted in favor of the first issue, and against the second issue.<br />
* D. Students who voted in favor of the second issue, and against the first issue.<br />
<br />
We are given that:<br />
* <math>A + B + C + D = 198.</math><br />
* <math>B = 29.</math><br />
* <math>A + C = 149</math> students voted in favor of the first issue.<br />
* <math>A + D = 119</math> students voted in favor of the second issue.<br />
<br />
We can quickly find that:<br />
* <math>198 - 119 = 79</math> students voted against the second issue.<br />
* <math>198 - 149 = 49</math> students voted against the first issue. <br />
* <math>B + C = 79, B + D = 49, \text{so} C = 50, D = 20, A = 99.</math><br />
<br />
The answer is <math>\boxed{\textbf{(D)}~99}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_20&diff=1102541991 AJHSME Problems/Problem 202019-10-12T20:58:52Z<p>Prestonh: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then <math>C=</math><br />
<br />
<asy><br />
unitsize(18);<br />
draw((-1,0)--(3,0));<br />
draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4));<br />
label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N);<br />
label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N);<br />
label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
From this we have <br />
<cmath>111A+11B+C=300.</cmath><br />
Clearly, <math>A<3</math>. Since <math>B,C\leq 9</math>, <br />
<cmath>111A > 201 \Rightarrow A\geq 2.</cmath><br />
Thus, <math>A=2</math> and <math>11B+C=78</math>. From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.<br />
<br />
==Solution==<br />
<br />
Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.<br />
So <math>b+a=9</math><br />
<math>A+1=3</math><br />
<br />
Thus, <math>A=2</math> and <math>11B+C=78</math>. From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1991|num-b=19|num-a=21}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_20&diff=1102531991 AJHSME Problems/Problem 202019-10-12T20:58:33Z<p>Prestonh: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then <math>C=</math><br />
<br />
<asy><br />
unitsize(18);<br />
draw((-1,0)--(3,0));<br />
draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4));<br />
label("$A$",(0.5,2.1),N); label("$B$",(1.5,2.1),N); label("$C$",(2.5,2.1),N);<br />
label("$A$",(1.5,1.1),N); label("$B$",(2.5,1.1),N); label("$A$",(2.5,0.1),N);<br />
label("$3$",(0.5,-.1),S); label("$0$",(1.5,-.1),S); label("$0$",(2.5,-.1),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
From this we have <br />
<cmath>111A+11B+C=300.</cmath><br />
Clearly, <math>A<3</math>. Since <math>B,C\leq 9</math>, <br />
<cmath>111A > 201 \Rightarrow A\geq 2.</cmath><br />
Thus, <math>A=2</math> and <math>11B+C=78</math>. From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.<br />
<br />
==Solution==<br />
<br />
Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.<br />
so <math>b+a=9</math><br />
<math>A+1=3</math><br />
<br />
Thus, <math>A=2</math> and <math>11B+C=78</math>. From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1991|num-b=19|num-a=21}}<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_18&diff=1102462004 AMC 8 Problems/Problem 182019-10-12T15:55:50Z<p>Prestonh: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers <math>1</math> through <math>10</math>. Each throw hits the target in a region with a different value. The scores are: Alice <math>16</math> points, Ben <math>4</math> points, Cindy <math>7</math> points, Dave <math>11</math> points, and Ellen <math>17</math> points. Who hits the region worth <math>6</math> points?<br />
<br />
<math>\textbf{(A)}\ \text{Alice}\qquad \textbf{(B)}\ \text{Ben}\qquad \textbf{(C)}\ \text{Cindy}\qquad \textbf{(D)}\ \text{Dave} \qquad \textbf{(E)}\ \text{Ellen}</math><br />
<br />
==Solution 1==<br />
The only way to get Ben's score is with <math>1+3=4</math>. Cindy's score can be made of <math>3+4</math> or <math>2+5</math>, but since Ben already hit the <math>3</math>, Cindy hit <math>2+5=7</math>. Similarly, Dave's darts were in the region <math>4+7=11</math>. Lastly, because there is no <math>7</math> left, <math>\boxed{\textbf{(A)}\ \text{Alice}}</math> must have hit the regions <math>6+10=16</math> and Ellen <math>8+9=17</math>.<br />
<br />
==Solution 2==<br />
Taking the 2 largest scores to narrow out the points we get<br />
<br />
Ellen with either <math>10,7</math> or <math>9,8</math> and Alice with either <math>10,6</math> or <math>9,7</math>, the 2 pairs that work are<br />
Alice=<math>10,6</math><br />
Ellen=<math>9,8</math>, therefore Alice scored the <math>6</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2004|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_24&diff=1101382010 AMC 8 Problems/Problem 242019-10-05T21:36:43Z<p>Prestonh: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?<br />
<br />
<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\\<br />
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \\<br />
\textbf{(C)}\ 5^{12}<2^{24}<10^8 \\<br />
\textbf{(D)}\ 10^8<5^{12}<2^{24} \\<br />
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math><br />
<br />
==Solution 1==<br />
Use brute force.<br />
<math>10^8=100,000,000</math>, <br />
<math>5^{12}=244,140,625</math>, and<br />
<math>2^{24}=16,777,216</math>.<br />
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for this contest)<br />
<br />
== Solution 2==<br />
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.<br />
<br />
== Solution 3==<br />
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.<br />
<math>10^8</math> is fine as is.<br />
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.<br />
We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>.<br />
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.<br />
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.<br />
Solution by coolak<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=User:Prestonh&diff=109691User:Prestonh2019-09-06T04:05:11Z<p>Prestonh: Created page with "This is a person that is a user in AoPS."</p>
<hr />
<div>This is a person that is a user in AoPS.</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_15&diff=1096752009 AMC 8 Problems/Problem 152019-09-04T03:10:42Z<p>Prestonh: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A recipe that makes <math> 5</math> servings of hot chocolate requires <math> 2</math> squares of chocolate, <math> \frac{1}{4}</math> cup sugar, <math> 1</math> cup water and <math> 4</math> cups milk. Jordan has <math> 5</math> squares of chocolate, <math> 2</math> cups of sugar, lots of water, and <math> 7</math> cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?<br />
<br />
<br />
<math> \textbf{(A)}\ 5 \frac18 \qquad<br />
\textbf{(B)}\ 6\frac14 \qquad<br />
\textbf{(C)}\ 7\frac12 \qquad<br />
\textbf{(D)}\ 8 \frac34 \qquad<br />
\textbf{(E)}\ 9\frac78</math><br />
<br />
==Solution==<br />
Assuming excesses of the other ingredients, the chocolate can make <math>\frac52 \cdot 5=12.5</math> servings, the sugar can make <math>\frac{2}{1/4} \cdot 5 = 40</math> servings, the water can make unlimited servings, and the milk can make <math>\frac74 \cdot 5 = 8.75</math> servings. Limited by the amount of milk, Jordan can make at most <math>\boxed{\textbf{(D)}\ 8 \frac34}</math> servings.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2009|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_18&diff=1095542007 AMC 8 Problems/Problem 182019-08-31T00:20:02Z<p>Prestonh: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The product of the two <math>99</math>-digit numbers<br />
<br />
<math>303,030,303,...,030,303</math> and <math>505,050,505,...,050,505</math><br />
<br />
has thousands digit <math>A</math> and units digit <math>B</math>. What is the sum of <math>A</math> and <math>B</math>?<br />
<br />
<math>\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10</math><br />
<br />
==Solution==<br />
We can first make a small example to find out <math>A</math> and <math>B</math>. So, <br />
<br />
<math>303\times505=153015 </math><br />
<br />
The ones digit plus thousands digit is <math>5+3=8</math>.<br />
<br />
Note that the ones and thousands digits are, added together, <math>8</math>. (and so on...) So the answer is <math>\boxed{\textbf{(D)}\ 8}</math><br />
This is a direct multlipication way.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2007|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Prestonhhttps://artofproblemsolving.com/wiki/index.php?title=Base_numbers/Conversion&diff=109353Base numbers/Conversion2019-08-25T18:05:31Z<p>Prestonh: /* Converting from base 10 to base b */</p>
<hr />
<div>== Converting from base b to base 10 ==<br />
The next natural question is: how do we convert a number from another base into base 10? For example, what does <math>4201_5</math> mean? Just like base 10, the first digit to the left of the decimal place tells us how many <math>5^0</math>'s we have, the second tells us how many <math>5^1</math>'s we have, and so forth. Therefore:<br />
<br />
<center><math>4201_5 = (4\cdot 5^3 + 2\cdot 5^2 + 0\cdot 5^1 + 1\cdot 5^0)_{10}</math></center><br />
<center><math>=4\cdot 125 + 2\cdot 25 + 1</math></center><br />
<center><math>= 551_{10}</math></center><br />
<br />
From here, we can generalize. Let <math>x=(a_na_{n-1}\cdots a_1a_0)_b</math> be an <math>n+1</math>-digit number in base <math>b</math>. In our example (<math> 2746_{10}</math>) <math>a_3 = 2, a_2 = 7, a_1 = 4</math> and <math>a_0 = 6 </math>. We convert this to base 10 as follows:<br />
<br />
<center><math> x = (a_na_{n-1}\cdots a_1a_0)_b</math></center><br />
<center><math> = (b^n\cdot a_n + b^{n-1}\cdot a_{n-1}+\cdots + b\cdot a_1 + a_0)_{10}</math></center><br />
<br />
<br />
== Converting from base 10 to base b ==<br />
<br />
It turns out that converting from base 10 to other bases is far harder for us than converting from other bases to base 10. This shouldn't be a surprise, though. We work in base 10 ''all the time'' so we are naturally less comfortable with other bases. Nonetheless, it is important to understand how to convert from base 10 into other bases.<br />
<br />
We'll look at two methods for converting from base 10 to other bases.<br />
<br />
=== Method 1 ===<br />
<br />
Let's try converting 1000 base 10 into base 7. Basically, we are trying to find the solution to the equation<br />
<br />
<center><math> 1000 = a_0 + 7a_1 + 49a_2 + 343a_3+2401a_4+\cdots</math></center><br />
<br />
where all the <math>a_i</math> are digits from 0 to 6. Obviously, all the <math>a_i</math> from <math>a_4</math> and up are 0 since otherwise they will add in a number greater than 1000, and all the terms in the sum are nonnegative. Then, we wish to find the largest <math>a_3</math> such that <math>343a_3</math> does not exceed 1000. Thus, <math> a_3= 2</math> since <math>2a_3=686</math> and <math>3a_3=1029</math>. This leaves us with<br />
<br />
<center><math> 1000 = a_0 + 7a_1 + 49 a_2 + 343(2)\Leftrightarrow 314 = a_0 + 7a_1 + 49 a_2.</math></center><br />
<br />
Using similar reasoning, we find that <math>a_2 = 6</math>, leaving us with<br />
<br />
<center><math>20 = a_0 + 7a_1.</math></center><br />
<br />
We use the same procedure twice more to get that <math>a_1=2</math> and <math>a_0=6</math>.<br />
<br />
Finally, we have that <math>1000_{10}=2626_7</math>.<br />
<br />
An alternative version of method 1 is to find the "digits" <math>a_0,a_1,\dots</math> starting with <math>a_0</math>. Note that <math>a_0</math> is just the [[remainder]] of division of <math>1000</math><br />
by <math>7</math>. So, to find it, all we need to do is to carry out one division with remainder. We have <math>1000:7=142(R6)</math>. How do we find <math>a_1</math>, now? It turns out that all we need to do is to find the remainder of the division of the quotient <math>142</math> by <math>7</math>:<br />
<math>142:7=20(R2)</math>, so <math>a_1=2</math>. Now, <math>20:7=2(R6)</math>, so <math>a_3=6</math>. Finally, <math>2:7=0(R2)</math>, so <math>a_4=2</math>. We may continue to divide beyond this point, of course, but it is clear that we will just get <math>0:7=0(R0)</math> during each step.<br />
<br />
Note that both versions of this method use computations in base <math>10</math>. <br />
<br />
It's often a good idea to double check by converting your answer back into base 10, since this conversion is easier to do. We know that <math>2626_7=343\cdot 2 + 6\cdot 49 + 2\cdot 7 + 6=1000</math>, so we can rest assured we got the right answer.<br />
<br />
=== Method 2 ===<br />
We'll exhibit the second method with the same problem used to exhibit the first method.<br />
<br />
The second method is just like how we converted from other bases into base 10. To do this, we pretend that our standard number system is base 7. In base 7, however, there is no digit 7. So 7 is actually represented as "10." Also, the multiplication rules we know do not hold. For example, <math>3\cdot 3\neq 9</math> (in base 7). For one, there is no 9 in base 7. Second, we need to go back to the definition of multiplication to fully understand what's happening. Multiplication is a shorthand for repeated addition. So, <math>3\cdot 3 = 3 + 3 + 3 = 12_7</math>.<br />
<br />
In base 7, we have that 10 (the decimal number 10) is 13. Thus, if we view everything from base 7, we are actually converting <math>1000_{13}</math> to base 10. So, this is just <math>13^{3}</math>. Remember that we aren't doing this in our regular decimal system, so <math>13^3\neq 2197</math>. Instead, we have to compute <math>13\times 13\times 13</math> as <math>(13\times 13)\times 13=202\times 13=2626</math>.<br />
<br />
This method can be ''very'' confusing unless you have a very firm grasp on the notion of number systems.<br />
<br />
== Binary and Hexadecimal ==<br />
Two of the most common number bases are binary (base 2) and hexadecimal (base 16). Both of these are used in computer science.<br />
A binary number consists entirely of 0s and 1s. Each binary digit is called a bit.<br />
A base 16 number requires additional symbols after 9 since any positive integer less than 16 is a single digit. The standard notation is to use the letters a=10,b=11,c=12,d=13,e=14,f=15.<br />
In order to indicate that a number is written in hexadecimal, the prefix 0x is used. For instance, 0x95 is the same as <math>95_{16}</math> and is equivalent to <math>9 \times 16^1 + 5 = 149</math><br />
<br />
=== Examples ===<br />
1. Convert the binary number 1111 1110 to decimal. (A space is often inserted every 4 digits to improve readability of binary numbers. A group of 4 digits (bits) is called a nible and a group of 8 bits is called a byte)<br />
<br />
We could simply add all of the digit values: <math>0 \times 1 + 1 \times 2 + 1 \times 4 + 1 \times 8 + 1 \times 16 + 1 \times 32 + 1 \times 64 + 1 \times 128 </math>.<br />
But, there is a much shorter way. Note that in base 10, 99 = 10<sup>2</sup> - 1 and 999 = 10<sup>3</sup> - 1. Similarly, in base 2, 11 = 2<sup>2</sup> - 1, 111 = 2<sup>3</sup> - 1, and so on.<br />
Therefore, 1111 1111 = 2<sup>8</sup> -1 = 255.<br />
The number we are trying to convert is one less, so it must be <b>254</b>.<br />
<br />
<br />
2. Convert the hexadecimal number 0xA7 into decimal.<br />
<br />
Since A means ten and the A is in the "16s place," we have 0xA7<math> = 10 \times 16 + 7 = 167</math><br />
<br />
<br />
3. Convert 700 into hexadecimal.<br />
<math>16^2 = 256</math>, so 700 will require three digits.<br />
Using Method 1 above, we want to solve:<br />
<center><math> 700 = a_0 + 16a_1 + 256a_2</math></center><br />
<br />
<math>\frac{700}{256} = 2 </math> remainder 188, so <math> a_2 = 2</math><br />
<center><math> 188 = a_0 + 16a_1</math></center><br />
<math>\frac{188}{16} = 11</math> remainder 12, so <math> a_1 = 11</math> and <math>a0=12</math><br />
We have<br />
<center><math> 700 = 12 + 16 \times 11 + 256 \times 2</math></center><br />
Using standard hexadecimal notation, we can write<br />
<center>'''700 = 0x2BC'''</center><br />
<br />
<br />
4. Convert 0xF7B9 into binary. (hint: You can do this directly without first converting into decimal)<br />
<br />
== Resources ==<br />
==== Books ====<br />
* The AoPS [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10 Introduction to Number Theory] by [[Mathew Crawford]].<br />
<br />
* The [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=1 Art of Problem Solving Volume 1: the BASICS], pgs. 52-54 by Sandor Lehoczky and Richard Rusczyk.<br />
<br />
==== Classes ====<br />
* [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#begnum AoPS Introduction to Number Theory Course]<br />
<br />
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== See also ==<br />
* [[base numbers]]<br />
* [[number theory]]</div>Prestonh