Floor functions encode intervals. Introducing the shared integer value can turn a difficult-looking equality into two simple range conditions, while keeping endpoints explicit.
Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 FLOORS 1027 and tell us your school year and question.
Name the common integer value
Find positive integers n ≤ 30 such that floor(n/3) = floor(n/4). Let their common value be k. Then 3k ≤ n < 3k + 3 and 4k ≤ n < 4k + 4.
Since n is positive, k is nonnegative. For k ≥ 0, the intersection has lower bound 4k and upper bound 3k + 3.
Bound the parameter before listing
A nonempty intersection requires 4k < 3k + 3, so k < 3. Only k = 0, 1 and 2 need checking.
For k = 0, the allowed positive integers are 1 and 2. For k = 1, they are 4 and 5. For k = 2, only 8 remains. All lie within the requested upper limit.
Check the endpoints
For n = 8, both floors are 2. For n = 9, the floors are 3 and 2, so 9 is excluded by the strict upper endpoint.
The complete solution list is 1, 2, 4, 5 and 8, giving five solutions. The parameter bound proves no later values were missed.
Use the definition rather than a graph guess
The defining relationship k ≤ x < k + 1 explains why one endpoint is included and the other excluded. A rough sketch should not replace that convention.
This original foundation exercise is not a released AMC or AIME question. Preserve integer and positivity conditions throughout the transformation.
Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.
Contact Hanlin
Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 FLOORS 1027. Include your school year, country or competition region, experience and the question you want to discuss.
On your phone, use the button or save the QR image and scan it within WeChat.
Official information and further reading: Source 1
