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

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==Problem==
 
==Problem==
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Call a positive integer <math>\textbf{monotonous}</math> 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, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?
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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, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?
  
 
<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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There are <math>\sum_{n=1}^{10} \binom{10}{n}</math> 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. However, <math>\emptyset</math> (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to <math>\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.</math> We discard the number 0 since it is not positive. Thus there are <math>1022</math> here.  
 
There are <math>\sum_{n=1}^{10} \binom{10}{n}</math> 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. However, <math>\emptyset</math> (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to <math>\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.</math> We discard the number 0 since it is not positive. Thus there are <math>1022</math> here.  
  
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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.
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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.
  
 
==Solution 2==
 
==Solution 2==
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All in all, that gives <math>511 + 1022 - 9 = \boxed{\textbf{(B)  } 1524}</math> monotonous numbers.
 
All in all, that gives <math>511 + 1022 - 9 = \boxed{\textbf{(B)  } 1524}</math> monotonous numbers.
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==Solution 4==
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Let <math>n</math> be the number of digits of a monotonous number. Notice for an increasing monotonous number with <math>n \ge 2</math>, we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a <math>0</math> to the end of the reversed digits. Whenever <math>n</math> digits are chosen, the order is fixed. There are <math>\binom{9}{n}</math> ways to obtain an increasing monotonous number with <math>n</math> digits. So, there are <math>3\cdot \sum_{n=2}^{9} \binom{9}{n}</math> monotonous numbers when <math>n \ge 2</math>. When <math>n=1</math>, there is no reverse but we could add <math>0</math> to the end, so there are <math>2 \cdot \binom{9}{1}</math> monotonous numbers.
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The answer is:
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<math>3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}</math> <math>=3\cdot \sum_{n=1}^{9} \binom{9}{n}  -  \binom{9}{1}</math> <math>=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) -  \binom{9}{1}</math> <math>= 3 \cdot (2^9-1) - 9</math> <math>=\boxed{\textbf{(B)  } 1524}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 5 (Casework / 1-1 Correspondence)==
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We have 2 cases.
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'''Case 1: Ascending order'''
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We can set up a 1-1 Correspondence. For any subset of all the digits <math>1</math> to <math>9</math> (<math>0</math> cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers <math>1</math> to <math>9</math> is equivalent to the number of ascending integers. So, <math>2^9=512</math>. However, the empty set (<math>\emptyset</math>) is not an integer, so we must subtract 1. Thus, <math>512-1=511</math>.
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'''Case 2: Descending order'''
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Similarly, any subset of the digits <math>0</math> to <math>9</math> can be rearranged into a descending monotonous number. So, <math>2^{10}=1024</math>. However, <math>\emptyset</math> and <math>0</math> are not ''positive'' integers, so we must subtract 2. Thus, <math>1024-2=1022</math>.
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We have covered all the cases. We add <math>511</math> to <math>1022</math>, giving us <math>1533</math>. So now we just innocently go ahead and choose <math>\textbf{(C) } 1533</math> as our answer, right? No! We overcounted the <math>9</math> ''single-digit integers''. The answer is actually <math>1533-9=\boxed{\textbf{(B)  } 1524}</math>.
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~MrThinker
  
 
==See Also==
 
==See Also==

Latest revision as of 18:34, 27 March 2024

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 $\sum_{n=1}^{9} \binom{9}{n}$ 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. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$

Case 2: monotonous numbers with digits in descending order

There are $\sum_{n=1}^{10} \binom{10}{n}$ 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. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ 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 $511+1022-9 = \boxed{\textbf{(B) } 1524}$.

Solution 3

Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$.

If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.

In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$. So, this case gives us $511$.


Onto the second case, if there are no $0$s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$, there are $2$ solutions. This gives \[2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022\] possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.

However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$).

All in all, that gives $511 + 1022 - 9 = \boxed{\textbf{(B) } 1524}$ monotonous numbers.

Solution 4

Let $n$ be the number of digits of a monotonous number. Notice for an increasing monotonous number with $n \ge 2$, we can obtain 2 more monotonous numbers that are decreasing by reversing its digits and adding a $0$ to the end of the reversed digits. Whenever $n$ digits are chosen, the order is fixed. There are $\binom{9}{n}$ ways to obtain an increasing monotonous number with $n$ digits. So, there are $3\cdot \sum_{n=2}^{9} \binom{9}{n}$ monotonous numbers when $n \ge 2$. When $n=1$, there is no reverse but we could add $0$ to the end, so there are $2 \cdot \binom{9}{1}$ monotonous numbers.

The answer is:

$3\cdot \sum_{n=2}^{9} \binom{9}{n} + 2 \cdot \binom{9}{1}$ $=3\cdot \sum_{n=1}^{9} \binom{9}{n} - \binom{9}{1}$ $=3\cdot \left( \sum_{n=0}^{9} \binom{9}{n} - \binom{9}{0} \right) - \binom{9}{1}$ $= 3 \cdot (2^9-1) - 9$ $=\boxed{\textbf{(B) } 1524}$

~isabelchen

Solution 5 (Casework / 1-1 Correspondence)

We have 2 cases.

Case 1: Ascending order

We can set up a 1-1 Correspondence. For any subset of all the digits $1$ to $9$ ($0$ cannot be a digit for ascending order), we can always rearrange them into an ascending monotonous number. Therefore, the number of subsets of the integers $1$ to $9$ is equivalent to the number of ascending integers. So, $2^9=512$. However, the empty set ($\emptyset$) is not an integer, so we must subtract 1. Thus, $512-1=511$.

Case 2: Descending order

Similarly, any subset of the digits $0$ to $9$ can be rearranged into a descending monotonous number. So, $2^{10}=1024$. However, $\emptyset$ and $0$ are not positive integers, so we must subtract 2. Thus, $1024-2=1022$.


We have covered all the cases. We add $511$ to $1022$, giving us $1533$. So now we just innocently go ahead and choose $\textbf{(C) } 1533$ as our answer, right? No! We overcounted the $9$ single-digit integers. The answer is actually $1533-9=\boxed{\textbf{(B) } 1524}$.

~MrThinker

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
2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 12 Problems and Solutions

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