Difference between revisions of "2017 AMC 10B Problems/Problem 16"
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<math>\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481</math> | <math>\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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 = 729</math> three digit integers without a zero, and <math>9 \cdot 9 \cdot 9 = 729</math> four-digit integers starting with a 1 without a zero. Therefore, the answer is <math>2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}</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 = 729</math> three digit integers without a zero, and <math>9 \cdot 9 \cdot 9 = 729</math> four-digit integers starting with a 1 without a zero. Therefore, the answer is <math>2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}</math>. | ||
| + | |||
| + | ==Solution 2 (Casework)== | ||
| + | We can use casework to solve this problem. First, we notice there are no one-digit numbers that contain a zero. There are <math>9</math> two-digit integers and <math>9 \cdot 9 + 9 \cdot 9 + 9 = 171</math> three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are <math>3 \cdot 9 \cdot 9 = 243</math> of these four-digit integers with one zero, <math>\binom3,2 \cdot 9 = 27</math> with two zeros, and <math>1</math> with three zeros <math>(1000)</math>. Finally, we consider the numbers <math>2000</math> to <math>2017</math> which all contain at least one zero. Adding all of these together we get <math>9 + 171 + 243 + 27 + 1 + 18 = \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 16:17, 12 November 2023
Problem
How many of the base-ten numerals for the positive integers less than or equal to 
contain the digit 
?
Solution 1
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 starting with a 1 without a zero. Therefore, the answer is 
.
Solution 2 (Casework)
We can use casework to solve this problem. First, we notice there are no one-digit numbers that contain a zero. There are two-digit integers and 
three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are 
of these four-digit integers with one zero, 
with two zeros, and 
with three zeros 
. Finally, we consider the numbers 
to which all contain at least one zero. Adding all of these together we get 
.
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. 
