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Home » \[ = rac12 \left| 3 + 4 - 20 - Blask

\[ = rac12 \left| 3 + 4 - 20 - Blask

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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We test small values of $ n $:
$ n = 1 $: $ 1^2 = 1 \equiv 1 \pmod{12} $ ✅
Thus, the smallest such $ n $ is $ \boxed{1} $.
Solution: Consider four consecutive integers: $ n, n+1, n+2, n+3 $.
The product is $ P = n(n+1)(n+2)(n+3) $.
Among any four consecutive integers, one is divisible by 4.
At least two are even, so the product is divisible by $ 2^3 = 8 $.
Thus, $ P $ is divisible by $ \text{lcm}(8, 3) = 24 $.
We check if a larger fixed divisor is guaranteed:
Not always divisible by 5 (e.g., $ n = 1 $: product = 24, not divisibl...

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eta^4 + 3 = -1 = a(1 - i) + b = a - ai + b
Imaginary part: \( -a = 0 \Rightarrow a = 0 \), then \( b = -1 \)
So both give \( a = 0 \), \( b = -1 \)
Conclusion: the remainder is constant \( -1 \)
Alternatively, polynomial division confirms:
Divide \( u^4 + 0u^3 + 0u^2 + 0u + 3 \) by \( u^2 - 2u + 2 \)
First: \( u^4 \div u^2 = u^2 \), multiply: \( u^2(u^2 - 2u + 2) = u^4...
Subtract: \( (u^4 + 0u^3 + 0u^2) - (u^4 - 2u^3 + 2u^2) = 2u^3 - 2u^2 \...
Bring down: \( +0u + 3 \), now divide \( 2u^3 \div u^2 = 2u \)
Multiply: \( 2u(u^2 - 2u + 2) = 2u^3 - 4u^2 + 4u \)

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