2021 AMC 10B Problems/Problem 16
Contents
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 , or numbers.
~ilikemath40
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.
Case 1: digit. No numbers work, so numbers.
Case 2: digits. We have the numbers and , but isn't an uphill number, so numbers
Case 3: digits. We have the numbers , so numbers.
Case 4: digits. We have the numbers and , but only satisfies this condition, so number.
Case 5: digits. We have only , so number.
Adding these up, we have .
~JustinLee2017
Solution 3
Like solution 2, we can proceed by using casework. A number is divisible by if is divisible by and In this case, the units digit must be otherwise no number can be formed.
Case 1: sum of digits = 6
There is only one number,
Case 2: sum of digits = 9
There are two numbers: and
Case 3: sum of digits = 12
There are two numbers: and
Case 4: sum of digits = 15
There is only one number,
We can see that we have exhausted all cases, because in order to have a larger sum of digits, then a number greater than needs to be used, breaking the conditions of the problem. The answer is
~coolmath34
Solution 4
An integer is divisible by iff it is divisible by and . Divisibility by means ending in or , but since no digit is less than , the only uphill integer ending in could be , which is not positive. This means the integer must end in .
All uphill integers ending in are formed by picking any subset of the sequence of digits (keeping their order), then appending a . Divisibility by means the sum of the digits is a multiple of , so our choice of digits must add to modulo .
, so the other digits we pick must add to modulo . Since , we can pick either nothing, or one residue (from or ) and one residue (from ), and we can then optionally add a residue (from ). This gives possibilities.
Video Solution by OmegaLearn (Using Divisibility Rules and Casework)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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