Difference between revisions of "2002 AMC 8 Problems/Problem 19"

Line 10: Line 10:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=18|num-a=20}}
 
{{AMC8 box|year=2002|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Revision as of 23:45, 4 July 2013

Problem

How many whole numbers between 99 and 999 contain exactly one 0?

$\text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180$

Solution

This list includes all the three digit whole numbers except 999. Because the hundreds digit cannot be 0, there are $2$ ways to choose whether the tens digit or the ones digit is equal to 0. Then for the two remaining places, there are $9$ ways to choose each digit. This gives a total of $(2)(9)(9)=\boxed{\text{(D)}\ 162}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
All AJHSME/AMC 8 Problems and Solutions

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