Difference between revisions of "2021 AMC 10B Problems/Problem 16"
m |
Cellsecret (talk | contribs) (→Video Solution by TheBeautyofMath) |
||
| Line 58: | Line 58: | ||
~IceMatrix | ~IceMatrix | ||
| + | |||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/9ZlJTVhtu_s | ||
| + | |||
| + | ~Interstigation | ||
{{AMC10 box|year=2021|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2021|ab=B|num-b=15|num-a=17}} | ||
Revision as of 01:54, 21 February 2021
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 
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. 
Case 1 = 
digit. No numbers work, so
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
Video Solution by OmegaLearn (Using Divisibility Rules and Casework)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
~Interstigation
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
