2021 AMC 10B Problems/Problem 16

==Problem==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 $15$ ?

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

Solution 1

The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$ . Being divisible by $5$ means that it must end with a $5$ or a $0$ . We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0$ which is not positive. So now we know that the number ends with a $5$ . Looking at the answer choices, the answer choices are all pretty small, so we can generate all of the numbers that are uphill and are divisible by $3$ . These numbers are $15, 45, 135, 345, 1245, 12345$ which are $6$ numbers C.

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2021 AMC 10B Problems/Problem 16

Solution 1