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.

Solution 2

First, note how the number must end in either $5$ or $0$ in order to satisfying being divisible by $15$ . However, the number can't end in $0$ because it's not strictly greater than the previous digits. Thus, our number must end in $5$ . We do casework on the number of digits. Case $1 = 1$ digit. No numbers work, so $0$ Case $2 = 2$ digits. We have the numbers $15, 45,$ and $75$ , but $75$ isn't an uphill number, so $2$ numbers. Case $3 = 3$ digits. We have the numbers $135, 345$ . So $2$ numbers. Case $4 = 4$ digits. We have the numbers $1235, 1245$ and $2345$ , but only $1245$ satisfies this condition, so $1$ number. Case $5 = 5$ digits. We have only $12345$ , so $1$ number. Adding these up, we have $2+2+1+1 = 6$ . $\boxed {C}$

~JustinLee2017

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2021 AMC 10B Problems/Problem 16

Solution 1

Solution 2