Difference between revisions of "2017 AMC 10B Problems/Problem 16"
The turtle (talk | contribs) |
Slackroadia (talk | contribs) m (→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | We can use complementary counting. There are <math>2017</math> positive integers in total to consider, and there are <math>9</math> one-digit integers, <math>9 \cdot 9 = 81</math> two digit integers without a zero, <math>9 \cdot 9 \cdot 9</math> three digit integers without a zero, and <math>9 \cdot 9 \cdot 9 = 1458</math> | + | We can use complementary counting. There are <math>2017</math> positive integers in total to consider, and there are <math>9</math> one-digit integers, <math>9 \cdot 9 = 81</math> two digit integers without a zero, <math>9 \cdot 9 \cdot 9</math> three digit integers without a zero, and <math>9 \cdot 9 \cdot 9 = 1458</math> four-digit integers without a zero. Therefore, the answer is <math>2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2017|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:09, 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 
four-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. 
