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1997 AHSME Problems/Problem 24

Revision as of 17:52, 23 March 2020 by Ryjs (talk | contribs) (Solution)

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 $97^{th}$ number, which does not contain the digit $\boxed{\textbf{(B)} \ 5}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AHSME Problems and Solutions

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