Difference between revisions of "2021 AMC 10B Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
| − | First, note how the number must end in either <math>5</math> or <math>0</math> in order to satisfying being divisible by <math>15</math>. However, the number can't end in <math>0</math> because it's not strictly greater than the previous digits. Thus, our number must end in <math>5</math>. We do casework on the number of digits. \newline | + | First, note how the number must end in either <math>5</math> or <math>0</math> in order to satisfying being divisible by <math>15</math>. However, the number can't end in <math>0</math> because it's not strictly greater than the previous digits. Thus, our number must end in <math>5</math>. We do casework on the number of digits. <math>\newline</math> |
| − | Case <math>1 = 1</math> digit. No numbers work, so <math>0</math> \newline | + | Case <math>1 = 1</math> digit. No numbers work, so <math>0</math> <math>\newline</math> |
| − | Case <math>2 = 2</math> digits. We have the numbers <math>15, 45,</math> and <math>75</math>, but <math>75</math> isn't an uphill number, so <math>2</math> numbers. \newline | + | Case <math>2 = 2</math> digits. We have the numbers <math>15, 45,</math> and <math>75</math>, but <math>75</math> isn't an uphill number, so <math>2</math> numbers. <math>\newline</math> |
| − | Case <math>3 = 3</math> digits. We have the numbers <math>135, 345</math>. So <math>2</math> numbers. \newline | + | Case <math>3 = 3</math> digits. We have the numbers <math>135, 345</math>. So <math>2</math> numbers. <math>\newline</math> |
| − | Case <math>4 = 4</math> digits. We have the numbers <math>1235, 1245</math> and <math>2345</math>, but only <math>1245</math> satisfies this condition, so <math>1</math> number. \newline | + | Case <math>4 = 4</math> digits. We have the numbers <math>1235, 1245</math> and <math>2345</math>, but only <math>1245</math> satisfies this condition, so <math>1</math> number. <math>\newline</math> |
| − | Case <math>5 = 5</math> digits. We have only <math>12345</math>, so <math>1</math> number. \newline | + | Case <math>5 = 5</math> digits. We have only <math>12345</math>, so <math>1</math> number. <math>\newline</math> |
Adding these up, we have <math>2+2+1+1 = 6</math>. <math>\boxed {C}</math> | Adding these up, we have <math>2+2+1+1 = 6</math>. <math>\boxed {C}</math> | ||
~JustinLee2017 | ~JustinLee2017 | ||
Revision as of 23:06, 11 February 2021
==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 
digit. No numbers work, so
Case 
digits. We have the numbers 
and 
, but isn't an uphill number, so 
numbers.
Case 
digits. We have the numbers 
. So numbers.
Case 
digits. We have the numbers 
and 
, but only 
satisfies this condition, so 
number.
Case 
digits. We have only 
, so number.
Adding these up, we have 
. 
~JustinLee2017
