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

Revision as of 18:48, 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

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 $2^{10}+2^9 - 9 - 1 - 1 - 1 = \boxed{\textbf{(B) }1524}$ .

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 $2^{10}+2^9 - 9 - 1 - 1 - 1 = \boxed{\textbf{(B) }1524}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution==
 ===Solution: Answer Choices===
-by e_power_pi_times_i
+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>.
-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>.
 ==Legit Solution==

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