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, and are all uphill integers, but and are not. How many uphill integers are divisible by ?

Solution 1

The divisibility rule of is that the number must be congruent to mod and congruent to mod . Being divisible by means that it must end with a or a . We can rule out the case when the number ends with a immediately because the only integer that is uphill and ends with a is which is not positive. So now we know that the number ends with a . 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 . These numbers are which are numbers C.

Solution 2

First, note how the number must end in either or in order to satisfying being divisible by . However, the number can't end in because it's not strictly greater than the previous digits. Thus, our number must end in . We do casework on the number of digits. \newline Case digit. No numbers work, so \newline Case digits. We have the numbers and , but isn't an uphill number, so numbers. \newline Case digits. We have the numbers . So numbers. \newline Case digits. We have the numbers and , but only satisfies this condition, so number. \newline Case digits. We have only , so number. \newline Adding these up, we have .

~JustinLee2017