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==
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The divisibility rule of 15 is that the number must be congruent to 0 mod 3 and congruent to 0 mod 5. Being divisible by 5 means that it must end with a 5 or a 0. We can rule out the case when the number ends with a 0 immediately because the only integer that is uphill and ends with a 0 is 0 which is not positive. So now we know that the number ends with a 5. 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 3. These numbers are <math>15, 45, 135, 345, 1245, 12345</math> which is 6 numbers C.

Revision as of 22:41, 11 February 2021

==Problem==Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$?

$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$

Solution

The divisibility rule of 15 is that the number must be congruent to 0 mod 3 and congruent to 0 mod 5. Being divisible by 5 means that it must end with a 5 or a 0. We can rule out the case when the number ends with a 0 immediately because the only integer that is uphill and ends with a 0 is 0 which is not positive. So now we know that the number ends with a 5. 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 3. These numbers are $15, 45, 135, 345, 1245, 12345$ which is 6 numbers C.