Difference between revisions of "2017 AMC 10B Problems/Problem 17"
m (→Solution) |
m (→Solution) |
||
| Line 4: | Line 4: | ||
<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== | ||
| − | + | ||
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>. | 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>. | ||
Revision as of 18:49, 16 February 2017
Contents
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
The number of one-digit numbers that work is 
, and the number of two-digit integers that work is 
. We use similar logic for three-digit integers, four digit integers, etc. Summing, we have 
, and we need to subtract another 1 for the 0 case, so the answer is 
.
Legit Solution
The number of one-digit numbers that work is , and the number of two-digit integers that work is . We use similar logic for three-digit integers, four digit integers, etc. Summing, we have , and we need to subtract another 1 for the 0 case, so the answer is .
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. 
