2002 AMC 8 Problems/Problem 19

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).

$1)$ How many whole numbers are between 99 and 999, $999-99=900$ , so there are 900 numbers between 99 and 999.

$2a)$ How many whole numbers in this range do not contain the digit $0$ , there are $10-1=9$ possible digits for each digit in this three digit number, $9^3=729$ . So there are $729$ numbers in this range which do not contain the digit $0$ .

$2b)$ How many whole numbers in this range contain the digit $0$ 2 times, there are $10-1=9$ possible digits for the first digit and the other two digits have to be the digit $0$ . So there are $9$ of these numbers.

So there are $900-(729+9)$ whole numbers between 99 and 999 that contain exactly one 0.

Simplifying the expression we get that there are $162$ whole numbers between 99 and 999 that contain exactly one 0. So the answer is $\fbox{D}$ .

~blankbox

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.

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