Difference between revisions of "2002 AMC 8 Problems/Problem 19"
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==Solution== | ==Solution== | ||
Numbers with exactly one zero have the form <math>\overline{a0b}</math> or <math>\overline{ab0}</math>, where the <math>a,b \neq 0</math>. There are <math>(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}</math> such numbers, hence our answer is <math>\fbox{D}</math>. | Numbers with exactly one zero have the form <math>\overline{a0b}</math> or <math>\overline{ab0}</math>, where the <math>a,b \neq 0</math>. There are <math>(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}</math> such numbers, hence our answer is <math>\fbox{D}</math>. | ||
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| + | ==Solution 2 (Complementary Counting)== | ||
| + | <math>(whole numbers between 99 and 999)-(whole numbers which do not contain exactly one 0)= (how many whole numbers between 99 and 999 contain exactly one 0)</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 23:35, 7 December 2024
Contents
Problem
How many whole numbers between 99 and 999 contain exactly one 0?
Solution
Numbers with exactly one zero have the form 
or 
, where the 
. There are 
such numbers, hence our answer is 
.
Solution 2 (Complementary Counting)
Video Solution
https://youtu.be/nctNL-xLImI Soo, DRMS, NM
https://www.youtube.com/watch?v=eAeVBrQ1PQI ~David
Video Solution by WhyMath
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. 
