Difference between revisions of "2017 AMC 10B Problems/Problem 16"
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==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 = | + | 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 = 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>. |
==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 23:56, 21 January 2018
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 starting with a 1 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 |
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