2021 AMC 10B Problems/Problem 16

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 $15, 45, 135, 345, 1245, 12345$ , or $\boxed{\textbf{(C)} ~6}$ numbers.

~ilikemath40

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.

Case 1: $1$ digit. No numbers work, so $0$ numbers.

Case 2: $2$ digits. We have the numbers $15, 45,$ and $75$ , but $75$ isn't an uphill number, so $2$ numbers

Case 3: $3$ digits. We have the numbers $135, 345$ , so $2$ numbers.

Case 4: $4$ digits. We have the numbers $1235, 1245$ and $2345$ , but only $1245$ satisfies this condition, so $1$ number.

Case 5: $5$ digits. We have only $12345$ , so $1$ number.

Adding these up, we have $2+2+1+1=\boxed{\textbf{(C)} ~6}$ .

~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 $\boxed{\textbf{(C)} ~6}.$

~coolmath34

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

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

Page

Toolbox

Search