Difference between revisions of "2017 AMC 10B Problems/Problem 16"
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==Problem== | ==Problem== | ||
| − | + | How many of the base-ten numerals for the positive integers less than or equal to <math>2017</math> contain the digit <math>0</math>? | |
| + | |||
| + | <math>\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481</math> | ||
==Solution== | ==Solution== | ||
Revision as of 12:29, 16 February 2017
Problem
How many of the base-ten numerals for the positive integers less than or equal to 
contain the digit 
?
Solution
We can use complementary counting. There are positive integers in total to consider, and there are 
one-digit integers, 
two digit integers without a zero, 
three digit integers without a zero, and 
three-digit integers without a zero. Therefore, the answer is 
.
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
