Difference between revisions of "2021 AMC 10B Problems/Problem 16"
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Since the desired positive integers are going uphill, their units digits must be <math>5</math>s. We start with <math>12345,</math> the largest of such positive integers, and perform casework on removing the number of its digits. Clearly, we cannot remove the digit <math>5,</math> as that is the only way to satisfy the divisibility rule of <math>5.</math> Now, we focus on the divisibility rule of <math>3.</math> | Since the desired positive integers are going uphill, their units digits must be <math>5</math>s. We start with <math>12345,</math> the largest of such positive integers, and perform casework on removing the number of its digits. Clearly, we cannot remove the digit <math>5,</math> as that is the only way to satisfy the divisibility rule of <math>5.</math> Now, we focus on the divisibility rule of <math>3.</math> | ||
| − | <b>Case (1): Remove exactly <math>\boldsymbol{0}</math> digits. (<math>\boldsymbol{5}</math>-digit | + | <b>Case (1): Remove exactly <math>\boldsymbol{0}</math> digits. (<math>\boldsymbol{5}</math>-digit uphill integers)</b> |
The number <math>12345</math> has a digit-sum of <math>15.</math> So, it is divisible by <math>3.</math> | The number <math>12345</math> has a digit-sum of <math>15.</math> So, it is divisible by <math>3.</math> | ||
| Line 65: | Line 65: | ||
There is <math>1</math> uphill integer in this case: <math>12345.</math> | There is <math>1</math> uphill integer in this case: <math>12345.</math> | ||
| − | <b>Case (2): Remove exactly <math>\boldsymbol{1}</math> digit. (<math>\boldsymbol{4}</math>-digit | + | <b>Case (2): Remove exactly <math>\boldsymbol{1}</math> digit. (<math>\boldsymbol{4}</math>-digit uphill integers)</b> |
We can only remove the digit <math>3.</math> | We can only remove the digit <math>3.</math> | ||
| Line 71: | Line 71: | ||
There is <math>1</math> uphill integer in this case: <math>1245.</math> | There is <math>1</math> uphill integer in this case: <math>1245.</math> | ||
| − | <b>Case (3): Remove exactly <math>\boldsymbol{2}</math> digits. (<math>\boldsymbol{3}</math>-digit | + | <b>Case (3): Remove exactly <math>\boldsymbol{2}</math> digits. (<math>\boldsymbol{3}</math>-digit uphill integers)</b> |
We can only remove the digits that sum to a multiple of <math>3.</math> This sum can only be <math>3</math> or <math>6</math> here. | We can only remove the digits that sum to a multiple of <math>3.</math> This sum can only be <math>3</math> or <math>6</math> here. | ||
| Line 77: | Line 77: | ||
There are <math>2</math> uphill integers in this case: <math>345,135.</math> | There are <math>2</math> uphill integers in this case: <math>345,135.</math> | ||
| − | <b>Case (4): Remove exactly <math>\boldsymbol{3}</math> digits. (<math>\boldsymbol{2}</math>-digit | + | <b>Case (4): Remove exactly <math>\boldsymbol{3}</math> digits. (<math>\boldsymbol{2}</math>-digit uphill integers)</b> |
We can only remove the digits that sum to a multiple of <math>3.</math> This sum can only be <math>6</math> or <math>9</math> here. | We can only remove the digits that sum to a multiple of <math>3.</math> This sum can only be <math>6</math> or <math>9</math> here. | ||
| Line 83: | Line 83: | ||
There are <math>2</math> uphill integers in this case: <math>45,15.</math> | There are <math>2</math> uphill integers in this case: <math>45,15.</math> | ||
| − | <b>Case (5): Remove exactly <math>\boldsymbol{4}</math> digits. (<math>\boldsymbol{1}</math>-digit | + | <b>Case (5): Remove exactly <math>\boldsymbol{4}</math> digits. (<math>\boldsymbol{1}</math>-digit uphill integers)</b> |
As discussed above, we must keep the digit <math>5.</math> Since <math>5</math> is not divisible by <math>3,</math> this case is impossible. | As discussed above, we must keep the digit <math>5.</math> Since <math>5</math> is not divisible by <math>3,</math> this case is impossible. | ||
Revision as of 15:21, 5 March 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
Solution 4
For every positive integer:
It is divisible by if and only if its digit-sum is divisible by 
It is divisible by if and only if its units digit is or
It is divisible by if and only if it is divisible by both and
Since the desired positive integers are going uphill, their units digits must be s. We start with 
the largest of such positive integers, and perform casework on removing the number of its digits. Clearly, we cannot remove the digit as that is the only way to satisfy the divisibility rule of Now, we focus on the divisibility rule of
Case (1): Remove exactly 
digits. (
-digit uphill integers)
The number has a digit-sum of So, it is divisible by
There is uphill integer in this case:
Case (2): Remove exactly 
digit. (
-digit uphill integers)
We can only remove the digit
There is uphill integer in this case:
Case (3): Remove exactly 
digits. (
-digit uphill integers)
We can only remove the digits that sum to a multiple of This sum can only be or here.
There are uphill integers in this case: 
Case (4): Remove exactly digits. (-digit uphill integers)
We can only remove the digits that sum to a multiple of This sum can only be or 
here.
There are uphill integers in this case: 
Case (5): Remove exactly digits. (-digit uphill integers)
As discussed above, we must keep the digit Since is not divisible by 
this case is impossible.
Total
Together, our answer is 
~MRENTHUSIASM
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 | |
| 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 | ||
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