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

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(→‎Solution 4: Will come back later.)
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<math>\quad\bullet</math> It is divisible by <math>15</math> if and only if it is divisible by both <math>3</math> and <math>5.</math>
 
<math>\quad\bullet</math> It is divisible by <math>15</math> if and only if it is divisible by both <math>3</math> and <math>5.</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>
+
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 deleting its digits. Clearly, we cannot delete 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 uphill integers)</b>
+
Note that the sum of the deleted digits must be a multiple of <math>3,</math> so that the difference between <math>15</math> and this sum is also divisible by <math>3</math> (Quick Proof: Suppose the sum of the deleted digits is <math>3k.</math> It follows that <math>15-3k=3(5-k)</math> is also divisible by <math>3.</math>). Two solutions follow from here:
 +
 
 +
===Solution 4.1 (Casework on the Number of Digits Deleted)===
 +
<b>Case (1): Delete 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>
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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 uphill integers)</b>
+
<b>Case (2): Delete 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 delete the digit <math>3.</math>
  
 
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 uphill integers)</b>
+
<b>Case (3): Delete 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> (as the difference between <math>15</math> and this sum must be divisible by <math>3</math>). This sum must be <math>3</math> or <math>6</math> here.
+
We can only delete the digits that sum to either <math>3</math> or <math>6.</math>
  
 
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 uphill integers)</b>
+
<b>Case (4): Delete 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> (same reasoning as Case (3)). This sum must be <math>6</math> or <math>9</math> here.
+
We can only delete the digits that sum to either <math>6</math> or <math>9.</math>
  
 
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 uphill integers)</b>
+
<b>Case (5): Delete 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> However, since <math>5</math> is not divisible by <math>3,</math> this case is impossible.
  
 
<b>Total</b>
 
<b>Total</b>
  
 
Together, our answer is <math>1+1+2+2=\boxed{\textbf{(C)} ~6}.</math>
 
Together, our answer is <math>1+1+2+2=\boxed{\textbf{(C)} ~6}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
===Solution 4.2 (Casework on the Sum of Digits Deleted)===
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 22:53, 5 March 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

Solution 3

Like solution 2, we can proceed by using casework. A number is divisible by $15$ if is divisible by $3$ and $5.$ In this case, the units digit must be $5,$ otherwise no number can be formed.

Case 1: sum of digits = 6

There is only one number, $15.$

Case 2: sum of digits = 9

There are two numbers: $45$ and $135.$

Case 3: sum of digits = 12

There are two numbers: $345$ and $1245.$

Case 4: sum of digits = 15

There is only one number, $12345.$

We can see that we have exhausted all cases, because in order to have a larger sum of digits, then a number greater than $5$ needs to be used, breaking the conditions of the problem. The answer is $\textbf{(C)}.$

~coolmath34

Solution 4

For every positive integer:

$\quad\bullet$ It is divisible by $3$ if and only if its digit-sum is divisible by $3.$

$\quad\bullet$ It is divisible by $5$ if and only if its units digit is $0$ or $5.$

$\quad\bullet$ It is divisible by $15$ if and only if it is divisible by both $3$ and $5.$

Since the desired positive integers are going uphill, their units digits must be $5$s. We start with $12345,$ the largest of such positive integers, and perform casework on deleting its digits. Clearly, we cannot delete the digit $5,$ as that is the only way to satisfy the divisibility rule of $5.$ Now, we focus on the divisibility rule of $3.$

Note that the sum of the deleted digits must be a multiple of $3,$ so that the difference between $15$ and this sum is also divisible by $3$ (Quick Proof: Suppose the sum of the deleted digits is $3k.$ It follows that $15-3k=3(5-k)$ is also divisible by $3.$). Two solutions follow from here:

Solution 4.1 (Casework on the Number of Digits Deleted)

Case (1): Delete exactly $\boldsymbol{0}$ digits. ($\boldsymbol{5}$-digit uphill integers)

The number $12345$ has a digit-sum of $15.$ So, it is divisible by $3.$

There is $1$ uphill integer in this case: $12345.$

Case (2): Delete exactly $\boldsymbol{1}$ digit. ($\boldsymbol{4}$-digit uphill integers)

We can only delete the digit $3.$

There is $1$ uphill integer in this case: $1245.$

Case (3): Delete exactly $\boldsymbol{2}$ digits. ($\boldsymbol{3}$-digit uphill integers)

We can only delete the digits that sum to either $3$ or $6.$

There are $2$ uphill integers in this case: $345,135.$

Case (4): Delete exactly $\boldsymbol{3}$ digits. ($\boldsymbol{2}$-digit uphill integers)

We can only delete the digits that sum to either $6$ or $9.$

There are $2$ uphill integers in this case: $45,15.$

Case (5): Delete exactly $\boldsymbol{4}$ digits. ($\boldsymbol{1}$-digit uphill integers)

As discussed above, we must keep the digit $5.$ However, since $5$ is not divisible by $3,$ this case is impossible.

Total

Together, our answer is $1+1+2+2=\boxed{\textbf{(C)} ~6}.$

~MRENTHUSIASM

Solution 4.2 (Casework on the Sum of Digits Deleted)

~MRENTHUSIASM

Video Solution by OmegaLearn (Using Divisibility Rules and Casework)

https://youtu.be/n2FnKxFSW94

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ

~IceMatrix

Video Solution by Interstigation

https://youtu.be/9ZlJTVhtu_s

~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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