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

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==Solution==
 
==Solution==
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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</math> three digit integers without a zero, and <math>9 \cdot 9 \cdot 9 = 1458</math> three-digit integers without a zero. Therefore, the answer is <math>2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}</math>.
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==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 13:21, 16 February 2017

Problem

Placeholder

Solution

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, $9 \cdot 9 \cdot 9$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 1458$ three-digit integers without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{\textbf{(A) }469}$.

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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