Art of Problem Solving

2021 AMC 10B Problems/Problem 16: Difference between revisions

Tkfun (talk | contribs)
Created page with "srsly why? u rly thought"
 
Pureswag (talk | contribs)
No edit summary
Line 1: Line 1:
srsly why?
Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, <math>1357, 89, and 5</math> are all uphill integers, but <math>32, 1240, and 466</math> are not. How many uphill integers are divisible by <math>15</math>?
u rly thought
 
<math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8</math>

Revision as of 16:01, 11 February 2021

Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89, and 5$ are all uphill integers, but $32, 1240, and 466$ are not. How many uphill integers are divisible by $15$?

$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$