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

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Thus, the <math>71^{\text{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>.
 
Thus, the <math>71^{\text{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^{\text{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>\boxed{\textbf{(B)} \ 5}</math>.
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Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>.  Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{\text{th}}</math> number, which does not contain the digit <math>\boxed{\textbf{(B)} \ 5}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:52, 23 March 2020

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 $97^{\text{th}}$ 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 $71^{\text{st}}$ 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 $91^{\text{st}}$ number is $24567$. Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\text{th}}$ number, which does not contain the digit $\boxed{\textbf{(B)} \ 5}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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