Difference between revisions of "2021 AMC 10B Problems/Problem 16"
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<math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8</math> | <math>\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8</math> | ||
| − | ==Solution== | + | ==Solution 1== |
The divisibility rule of <math>15</math> is that the number must be congruent to <math>0</math> mod <math>3</math> and congruent to <math>0</math> mod <math>5</math>. Being divisible by <math>5</math> means that it must end with a <math>5</math> or a <math>0</math>. We can rule out the case when the number ends with a <math>0</math> immediately because the only integer that is uphill and ends with a <math>0</math> is <math>0</math> which is not positive. So now we know that the number ends with a <math>5</math>. 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 <math>3</math>. These numbers are <math>15, 45, 135, 345, 1245, 12345</math> which are <math>6</math> numbers C. | The divisibility rule of <math>15</math> is that the number must be congruent to <math>0</math> mod <math>3</math> and congruent to <math>0</math> mod <math>5</math>. Being divisible by <math>5</math> means that it must end with a <math>5</math> or a <math>0</math>. We can rule out the case when the number ends with a <math>0</math> immediately because the only integer that is uphill and ends with a <math>0</math> is <math>0</math> which is not positive. So now we know that the number ends with a <math>5</math>. 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 <math>3</math>. These numbers are <math>15, 45, 135, 345, 1245, 12345</math> which are <math>6</math> numbers C. | ||
Revision as of 22:46, 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.
