Difference between revisions of "1997 AHSME Problems/Problem 24"

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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
  
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==Solution==
  
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The list starts with <math>12345</math>.  There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.
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Thus, the <math>71^{st}</math> number is <math>23456</math>.  There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.
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Thus, the <math>91^{st}</math> number is <math>24567</math>.  Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{th}</math> number, which does not contain the digit <math>5</math>.  The answer is <math>\boxed{B}</math>.
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=23|num-a=25}}
 
{{AHSME box|year=1997|num-b=23|num-a=25}}

Revision as of 20:58, 9 August 2011

Problem

A rising number, such as $34689$, is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{th}$ number in the list does not contain the digit

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

The list starts with $12345$. There are $\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$, and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.

Thus, the $71^{st}$ number is $23456$. There are $\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$, and thus $20$ five digit rising numbers that begin with a $23$.

Thus, the $91^{st}$ number is $24567$. Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{th}$ number, which does not contain the digit $5$. The answer is $\boxed{B}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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