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

Revision as of 13:13, 5 July 2013

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 $\boxed{B}$ .

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

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

@@ Line 14: / Line 14: @@
 == See also ==
 {{AHSME box|year=1997|num-b=23|num-a=25}}
+{{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 13:13, 5 July 2013

Problem

Solution

See also