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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==Video Solution==
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https://youtu.be/nctNL-xLImI Soo, DRMS, NM
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https://www.youtube.com/watch?v=eAeVBrQ1PQI  ~David
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==Video Solution by WhyMath==
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https://youtu.be/3FikBB_Lx7c
  
 
==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}}

Latest revision as of 13:33, 29 October 2024

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

Numbers with exactly one zero have the form $\overline{a0b}$ or $\overline{ab0}$, where the $a,b \neq 0$. There are $(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}$ such numbers, hence our answer is $\fbox{D}$.

Video Solution

https://youtu.be/nctNL-xLImI Soo, DRMS, NM

https://www.youtube.com/watch?v=eAeVBrQ1PQI ~David

Video Solution by WhyMath

https://youtu.be/3FikBB_Lx7c

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

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