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>
==Solution: Answer Choices==
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==Solution==
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===Solution: Answer Choices===
 
by e_power_pi_times_i
 
by e_power_pi_times_i
  

Revision as of 18:46, 16 February 2017

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

Solution: Answer Choices

by e_power_pi_times_i


It adds in $9$ at the end, but we know since it is MAA, it is probably a troll question, so we look at the answers. $(C)$ looks likely, as it is just $(B)+9$, but we remember that MAA is trolly so it is probably $\boxed{\textbf{(B) }1524}$.

Legit Solution

The number of one-digit numbers that work is $\binom{10}{1}$, and the number of two-digit integers that work is $\binom{10}{2} + \binom{9}2$. We use similar logic for three-digit integers, four digit integers, etc. Summing, we have $2^{10}+2^9 - 9 - 1 - 1$, and we need to subtract another 1 for the 0 case, so the answer is $2^{10}+2^9 - 9 - 1 - 1 - 1 = \boxed{\textbf{(B) }1524}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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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