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

Revision as of 23:38, 7 December 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 $\fbox{D}$ .

Solution 2 (Complementary Counting)

(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). $1)$

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

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.

@@ Line 9: / Line 9: @@
 ==Solution 2 (Complementary Counting)==
-(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)
+(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>1)</math>
 ==Video Solution==

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