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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>
 
==Solution==
 
==Solution==
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by e_power_pi_times_i
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The number of one-digit numbers that work is <math>\binom{10}{1}</math>, and the number of two-digit integers that work is <math>\binom{10}{2} + \binom{9}2</math>. We use similar logic for three-digit integers, four digit integers, etc. Summing, we have <math>2^{10}+2^9 - 9 - 1 - 1</math>, and we need to subtract another 1 for the 0 case, so the answer is <math>2^{10}+2^9 - 9 - 1 - 1 - 1 = \boxed{\textbf{(B) }1524}</math>.
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It adds in <math>9</math> at the end, but we know since it is MAA, it is probably a troll question, so we look at the answers. <math>(C)</math> looks likely, as it is just <math>(B)+9</math>, but we remember that MAA is trolly so it is probably <math>\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 13:40, 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

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 (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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