Sample Question 16:

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 \(15\)?

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

---

Answer Keys

Question 16: C

---

Solutions

Question 16
Step 1: We first notice that for an integer to be divisible by 15, it must be divisible by 3 and 5. From the divisibility rule of 5, the number must end with a 0 or 5.
Step 2: An uphill integer cannot end with a 0 since 0 cannot be greater than any previous digit. So, our integer must end with a 5.
Step 3: From the answer choices, we notice they're all small numbers, so it is possible to generate all numbers that are uphill and divisible by 3.
Step 4: We generate all such numbers, which are 15, 45, 135, 345, 1245, and 12345.
Step 5: We count these numbers and find there are exactly \( \boxed{\textbf{(C) }~6} \) of them. This is our final answer.