Difference between revisions of "2017 AMC 10B Problems/Problem 16"
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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 | + | ==Solution 2== |
− | + | 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>. | |
+ | |||
+ | ~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 11:03, 20 November 2023
Contents
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
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
.
~vsinghminhas
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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