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Difference between revisions of "2017 AMC 10B Problems/Problem 17"

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<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math>
 
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math>
  
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==Solution==
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==Solution 1==
  
 
Case 1: monotonous numbers with digits in ascending order
 
Case 1: monotonous numbers with digits in ascending order
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Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are <math>511+1022-9=\boxed{\textbf{B} \ 1524}</math> monotonous numbers.
 
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are <math>511+1022-9=\boxed{\textbf{B} \ 1524}</math> monotonous numbers.
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==Solution 2==
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Like Solution 1, divide the problem into an increasing and decreasing case:
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<math>\bullet</math> Case 1: Monotonous numbers with digits in ascending order.
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 +
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
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To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are <math>2^9 = 512</math> ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get <math>512-1=511</math> monotonous numbers for this case.
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<math>\bullet</math> Case 2: Monotonous numbers with digits in descending order.
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 +
This time, we arrange all 10 digits in decreasing order and repeat the process to find <math>2^{10} = 1024</math> ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get <math>1024-2=1022</math> monotonous numbers for this case.
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At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
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Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B)} 1524}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2017|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:58, 17 February 2017

The following problem is from both the 2017 AMC 12B #11 and 2017 AMC 10B #17, so both problems redirect to this page.

Problem

Call a positive integer $monotonous$ if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$, $23578$, and $987620$ are monotonous, but $88$, $7434$, and $23557$ are not. How many monotonous positive integers are there?

$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution 1

Case 1: monotonous numbers with digits in ascending order

There are $\Sigma_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. The sum is equivalent to $\Sigma_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$

Case 2: monotonous numbers with digits in descending order

There are $\Sigma_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. The sum is equivalent to $\Sigma_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.

Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{\textbf{B} \ 1524}$ monotonous numbers.

Solution 2

Like Solution 1, divide the problem into an increasing and decreasing case:

$\bullet$ Case 1: Monotonous numbers with digits in ascending order.

Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.

To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.

$\bullet$ Case 2: Monotonous numbers with digits in descending order.

This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.

At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.

Thus our final answer is $511+1022-9 = \boxed{\textbf{(B)} 1524}$.

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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