Difference between revisions of "2017 AMC 10B Problems/Problem 17"
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Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B) } 1524}</math>. | Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B) } 1524}</math>. | ||
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+ | ==Solution 3== | ||
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+ | Unlike the first two solutions, we can do our casework based off of whether or not the number contains a <math>0</math>. | ||
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+ | If it does, then we know the <math>0</math> must be the last digit in the number (and hence, the number has to be decreasing). Because <math>0</math> is not positive, <math>0</math> is not monotonous. So, we need to pick at least <math>1</math> number in the set <math>[1, 9].</math> After choosing our numbers, there will be just <math>1</math> way to arrange them so that the overall number is monotonous. | ||
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+ | In total, each of the <math>9</math> digits can either be in the monotonous number or not, yielding <math>2^9 = 512</math> total solutions. However, we said earlier that <math>0</math> cannot be by itself so we have to subtract out the case in which we picked none of the numbers <math>1-9</math>. So, this case gives us <math>511</math>. | ||
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+ | Onto the second case, if there are no <math>0</math>s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits <math>1- 9</math>, there are <math>2</math> solutions. This gives <cmath>2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022</cmath> possibilities. Note that we subtracted out the <math>1</math> because we cannot choose none of the numbers. | ||
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+ | However, realize that if we pick just <math>1</math> digit, then there is only <math>1</math> arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out <math>9</math> from there (because there are <math>9</math> possible ways to select one number and for each case, we overcounted by <math>1</math>). | ||
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+ | All in all, that gives <math>511 + 1022 - 9 = \boxed{(B) 1524}</math> monotonous numbers. | ||
==See Also== | ==See Also== |
Revision as of 13:26, 29 December 2020
- 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 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, , , and are monotonous, but , , and are not. How many monotonous positive integers are there?
Solution 1
Case 1: monotonous numbers with digits in ascending order
There are 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, (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to
Case 2: monotonous numbers with digits in descending order
There are 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, (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to We discard the number 0 since it is not positive. Thus there are here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are monotonous numbers.
Solution 2
Like Solution 1, divide the problem into an increasing and decreasing case:
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 ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case.
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 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 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 .
Solution 3
Unlike the first two solutions, we can do our casework based off of whether or not the number contains a .
If it does, then we know the must be the last digit in the number (and hence, the number has to be decreasing). Because is not positive, is not monotonous. So, we need to pick at least number in the set After choosing our numbers, there will be just way to arrange them so that the overall number is monotonous.
In total, each of the digits can either be in the monotonous number or not, yielding total solutions. However, we said earlier that cannot be by itself so we have to subtract out the case in which we picked none of the numbers . So, this case gives us .
Onto the second case, if there are no s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits , there are solutions. This gives possibilities. Note that we subtracted out the because we cannot choose none of the numbers.
However, realize that if we pick just digit, then there is only arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out from there (because there are possible ways to select one number and for each case, we overcounted by ).
All in all, that gives monotonous numbers.
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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