Difference between revisions of "2017 AMC 10B Problems/Problem 16"

(Solution 2)
 
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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)==
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==Solution 2==
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>{3 \choose 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>.
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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>{3 \choose 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>.
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~vsinghminhas
  
 
==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}}

Latest revision as of 10:03, 20 November 2023

Problem

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$?

$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$

Solution 1

We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 \cdot 9 = 81$ two digit integers without a zero, $9 \cdot 9 \cdot 9 = 729$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 729$ four-digit integers starting with a 1 without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}$.

Solution 2

First, we notice there are no one-digit numbers that contain a zero. There are $9$ two-digit integers and $9 \cdot 9 + 9 \cdot 9 + 9 = 171$ three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are $3 \cdot 9 \cdot 9 = 243$ of these four-digit integers with one zero, ${3 \choose 2} \cdot 9 = 27$ with two zeros, and $1$ with three zeros $(1000)$. Finally, we consider the numbers $2000$ to $2017$ which all contain at least one zero. Adding all of these together we get $9 + 171 + 243 + 27 + 1 + 18 = \boxed{\textbf{(A) }469}$.

~vsinghminhas

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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

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