Difference between revisions of "2002 AMC 8 Problems/Problem 19"
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==Solution== | ==Solution== | ||
This list includes all the three digit whole numbers except 999. Because the hundreds digit cannot be 0, there are <math>2</math> ways to choose whether the tens digit or the ones digit is equal to 0. Then for the two remaining places, there are <math>9</math> ways to choose each digit. This gives a total of <math>(2)(9)(9)=\boxed{\text{(D)}\ 162}</math>. | This list includes all the three digit whole numbers except 999. Because the hundreds digit cannot be 0, there are <math>2</math> ways to choose whether the tens digit or the ones digit is equal to 0. Then for the two remaining places, there are <math>9</math> ways to choose each digit. This gives a total of <math>(2)(9)(9)=\boxed{\text{(D)}\ 162}</math>. | ||
| + | or we can Numbers with exactly one zero have the form <math>\_ 0 \_</math> or <math>\_ \_ 0</math>, where the blanks are not zeros. There are <math>(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}</math> such numbers.-Alcumus(all credit to Alcumus, not me) | ||
==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}} | {{MAA Notice}} | ||
Revision as of 20:17, 6 July 2018
Problem
How many whole numbers between 99 and 999 contain exactly one 0?
Solution
This list includes all the three digit whole numbers except 999. Because the hundreds digit cannot be 0, there are 
ways to choose whether the tens digit or the ones digit is equal to 0. Then for the two remaining places, there are 
ways to choose each digit. This gives a total of 
.
or we can Numbers with exactly one zero have the form 
or 
, where the blanks are not zeros. There are 
such numbers.-Alcumus(all credit to Alcumus, not me)
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
