2017 AMC 12B Problem 11

Attempt Problem 11 of the 2017 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12B solutions, or check the answer key.

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11.

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, 3,3, 23578,23578, and 987620987620 are monotonous, but 88,88, 7434,7434, and 2355723557 are not. How many monotonous positive integers are there?
10241024
15241524
15331533
15361536
20482048
Answer: B
Concepts:subsetsbijectioninclusion-exclusion
Small hint:
Each nonempty subset of {1,2,…,9}\{1, 2, \ldots, 9\} gives exactly one strictly increasing number
Big hint:
Decreasing numbers correspond to subsets of {0,1,…,9}\{0, 1, \ldots, 9\} except ∅\varnothing and {0};\{0\}; single digits are counted twice
Solution:
Strictly increasing monotonous numbers correspond to nonempty subsets of {1,…,9},\{1, \ldots, 9\}, giving 29−1=511.2^9 - 1 = 511. Strictly decreasing ones correspond to subsets of {0,1,…,9}\{0, 1, \ldots, 9\} other than ∅\varnothing and {0}\{0\} (a leading 00 is not allowed), giving 210−2=1022.2^{10} - 2 = 1022. The nine single-digit numbers are counted in both, so the total is 511+1022−9=1524.511 + 1022 - 9 = 1524.
Thus, the correct answer is B.
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