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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.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1

A large power can become simple when the base satisfies a short cycle. For roots of unity, combine that cycle with a relationship between the roots rather than expanding repeated complex products.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 UNITY 1026 and tell us your school year and question.

Identify the three roots

The roots of z³ = 1 are 1, ω and ω², where ω is a nonreal cube root. They satisfy ω³ = 1 and 1 + ω + ω² = 0.

The second relationship follows from z³ - 1 = (z - 1)(z² + z + 1), evaluated at ω ≠ 1.

Reduce the exponent

Since 2026 leaves remainder 1 on division by 3, the 2026th powers of the three roots are 1, ω and ω². Their sum is zero.

By contrast, 2025 is divisible by three, so every root raised to that power equals 1. The sum is then 3.

Distinguish one root from the complete set

For an unspecified nonreal root, its 2026th power equals that same root; the information does not choose between the two nonreal values.

A symmetric sum over all roots can be determined even when an individually labelled value is not fixed.

Check the polynomial before generalising

This short cycle depends on the equation z³ = 1. Roots of another polynomial may have a different structure.

This is original algebra practice for AMC 12 preparation, not an official paper prediction. Derive the root relationship once, then explain precisely which remainder rule you are using.

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 UNITY 1026. Include your school year, country or competition region, experience and the question you want to discuss.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1