Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1986 AHSME Problems/Problem 10"

(Created page with "==Problem== The <math>120</math> permutations of the <math>AHSME</math> are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the <m...")
 
(Fixed the problem statement and adjusted the solutions accordingly)
 
(3 intermediate revisions by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
The <math>120</math> permutations of the <math>AHSME</math> are arranged in dictionary order as if each were an ordinary five-letter word.  
+
The <math>120</math> permutations of <math>AHSME</math> are arranged in dictionary order as if each were an ordinary five-letter word.  
−
The last letter of the <math>85</math>th word in this list is:
+
The last letter of the <math>86</math>th word in this list is:
  
 
<math>\textbf{(A)}\ \text{A} \qquad
 
<math>\textbf{(A)}\ \text{A} \qquad
Line 10: Line 10:
 
\textbf{(E)}\ \text{E}  </math>   
 
\textbf{(E)}\ \text{E}  </math>   
  
−
==Solution==
+
==Solution 1==
  
 +
We could list out all of the possible combinations in dictionary order.
 +
 +
<math>73rd:</math> MAEHS                 
 +
<math>74th:</math> MAESH
 +
<math>75th:</math> MAHES
 +
<math>76th:</math> MAHSE
 +
<math>77th:</math> MASEH
 +
<math>78th:</math> MASHE
 +
<math>79th:</math> MEAHS
 +
<math>80th:</math> MEASH
 +
<math>81th:</math> MEHAS
 +
<math>82th:</math> MEHSA
 +
<math>83th:</math> MESAH
 +
<math>84th:</math> MESHA
 +
<math>85th:</math> MHAES
 +
<math>86th:</math> MHASE
 +
 +
We find that the <math>86th</math> combination ends with the letter E.
 +
So the answer is <math>\textbf{(E)}\ E</math>.
 +
 +
==Solution 2==
 +
We can do this problem without having to list out every single combination.
 +
There are <math>5</math> distinct letters, so therefore there are <math>5!=120</math> ways to rearrange the letters.
 +
We can divide the <math>120</math> different combinations into 5 groups. Words that start with <math>A</math>, words that start with <math>E</math> and so on...
 +
Combinations <math>1</math>-<math>24</math> start with <math>A</math>,
 +
combinations <math>25</math>-<math>48</math> start with <math>E</math>,
 +
combinations <math>49</math>-<math>72</math> start with <math>H</math>,
 +
combinations <math>73</math>-<math>96</math> start with <math>M</math>,
 +
and combinations <math>97</math>-<math>120</math> start with <math>S</math>.
 +
We are only concerned with combination <math>86</math>, so we focus on combinations <math>73</math>-<math>96</math>.
 +
We can divide the remaining 24 combinations into 4 groups of 6, based upon the second letter.
 +
Combinations <math>73</math>-<math>78</math> begin with <math>MA</math>,
 +
combinations <math>79</math>-<math>84</math> begin with <math>ME</math>,
 +
combinations <math>85</math>-<math>90</math> begin with <math>MH</math>,
 +
and combinations <math>91</math>-<math>96</math> begin with <math>MS</math>.
 +
Combination <math>86</math> begins with <math>MH</math>. Now we can fill in the rest of the letters in alphabetical order and get <math>MHASE</math> (as <math>85</math> is <math>MHAES</math>). The last letter of the word is <math>E</math>, so the answer is <math> \textbf{(E)}\ E</math> .
  
 
== See also ==
 
== See also ==

Latest revision as of 17:29, 1 April 2018

Problem

The $120$ permutations of $AHSME$ are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the $86$th word in this list is:

$\textbf{(A)}\ \text{A} \qquad \textbf{(B)}\ \text{H} \qquad \textbf{(C)}\ \text{S} \qquad \textbf{(D)}\ \text{M}\qquad \textbf{(E)}\ \text{E}$

Solution 1

We could list out all of the possible combinations in dictionary order.

$73rd:$ MAEHS $74th:$ MAESH $75th:$ MAHES $76th:$ MAHSE $77th:$ MASEH $78th:$ MASHE $79th:$ MEAHS $80th:$ MEASH $81th:$ MEHAS $82th:$ MEHSA $83th:$ MESAH $84th:$ MESHA $85th:$ MHAES $86th:$ MHASE

We find that the $86th$ combination ends with the letter E. So the answer is $\textbf{(E)}\ E$.

Solution 2

We can do this problem without having to list out every single combination. There are $5$ distinct letters, so therefore there are $5!=120$ ways to rearrange the letters. We can divide the $120$ different combinations into 5 groups. Words that start with $A$, words that start with $E$ and so on... Combinations $1$-$24$ start with $A$, combinations $25$-$48$ start with $E$, combinations $49$-$72$ start with $H$, combinations $73$-$96$ start with $M$, and combinations $97$-$120$ start with $S$. We are only concerned with combination $86$, so we focus on combinations $73$-$96$. We can divide the remaining 24 combinations into 4 groups of 6, based upon the second letter. Combinations $73$-$78$ begin with $MA$, combinations $79$-$84$ begin with $ME$, combinations $85$-$90$ begin with $MH$, and combinations $91$-$96$ begin with $MS$. Combination $86$ begins with $MH$. Now we can fill in the rest of the letters in alphabetical order and get $MHASE$ (as $85$ is $MHAES$). The last letter of the word is $E$, so the answer is $\textbf{(E)}\ E$ .

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_10&oldid=93697"