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

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==Problem==
 
==Problem==
  
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The <math>120</math> permutations of the <math>AHSME</math> are arranged in dictionary order as if each were an ordinary five-letter word.  
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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:
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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
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<math>84th:</math> MESHA
 
<math>84th:</math> MESHA
 
<math>85th:</math> MHAES
 
<math>85th:</math> MHAES
 +
<math>86th:</math> MHASE
  
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We find that the 85th combination ends with the letter S.  
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We find that the <math>86th</math> combination ends with the letter E.  
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So the answer is <math>\textbf{(C)}\ S</math>.
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So the answer is <math>\textbf{(E)}\ E</math>.
  
 
==Solution 2==
 
==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>,
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combinations <math>73</math>-<math>96</math> start with <math>M</math>,
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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

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