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Difference between revisions of "2021 AMC 10B Problems/Problem 16"

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==Solution 2==
 
==Solution 2==
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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
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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>
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Case <math>1 = 1</math> digit. No numbers work, so <math>0</math> \newline
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Case <math>1 = 1</math> digit. No numbers work, so <math>0</math> <math>\newline</math>
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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
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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>
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Case <math>3 = 3</math> digits. We have the numbers <math>135, 345</math>. So <math>2</math> numbers.  \newline
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Case <math>3 = 3</math> digits. We have the numbers <math>135, 345</math>. So <math>2</math> numbers.  <math>\newline</math>
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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
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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>
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Case <math>5 = 5</math> digits. We have only <math>12345</math>, so <math>1</math> number. \newline
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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, $1357, 89,$ 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 1

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 are $6$ numbers C.


Solution 2

First, note how the number must end in either $5$ or $0$ in order to satisfying being divisible by $15$. However, the number can't end in $0$ because it's not strictly greater than the previous digits. Thus, our number must end in $5$. We do casework on the number of digits. $\newline$ Case $1 = 1$ digit. No numbers work, so $0$ $\newline$ Case $2 = 2$ digits. We have the numbers $15, 45,$ and $75$, but $75$ isn't an uphill number, so $2$ numbers. $\newline$ Case $3 = 3$ digits. We have the numbers $135, 345$. So $2$ numbers. $\newline$ Case $4 = 4$ digits. We have the numbers $1235, 1245$ and $2345$, but only $1245$ satisfies this condition, so $1$ number. $\newline$ Case $5 = 5$ digits. We have only $12345$, so $1$ number. $\newline$ Adding these up, we have $2+2+1+1 = 6$. $\boxed {C}$

~JustinLee2017

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