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

Latest revision as of 14:33, 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

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 6: / Line 6: @@
 ==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>.
+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>.
+==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==
 {{AMC8 box|year=2002|num-b=18|num-a=20}}
+{{MAA Notice}}

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