Difference between revisions of "2017 AMC 10B Problems/Problem 17"
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There are <math>\Sigma_{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. The sum is equivalent to <math>\Sigma_{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>\Sigma_{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. The sum is equivalent to <math>\Sigma_{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. | ||
− | 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} | + | 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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At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total. | 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 <math>511+1022-9 = \boxed{\textbf{(B)} 1524}</math>. | + | Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B) } 1524}</math>. |
==Solution 3: Answer Choices== | ==Solution 3: Answer Choices== |
Revision as of 23:20, 22 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 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. 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. 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: Answer Choices
by e_power_pi_times_i
It adds in at the end, but we know since it is MAA, it is probably a troll question, so we look at the answers. looks likely, as it is just , but we remember that MAA is trolly so it is probably because they would put the answer as the one that looks the most unlikely.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.