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Problem 246

Algebra Difficulty 2.0 Prove it CEMC Galois · Canada · 2021

The operation \triangle is defined by ab=a(2b+4)a \triangle b = a(2b +4) for integers aa and bb. For example, 36=3(2×6+4)=3(16)=48.3\triangle 6=3(2\times 6+4)=3(16)=48.

What is the value of 515 \triangle 1?
If k2=24k \triangle 2 = 24, what is the value of kk?
Determine all values of pp for which p3=3pp \triangle 3 = 3\triangle p.
Determine all values of mm for which m(m+1)=0m \triangle (m+1) = 0.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Substituting a=5a=5 and b=1b=1, we get 51=5(2×1+4)=5(6)=305 \triangle 1 = 5(2\times1 +4)=5(6)=30.
If k2=24k \triangle 2 = 24, then k(2×2+4)=24k(2\times2 +4)=24 or 8k=248k=24, and so k=3k=3.
Solving the given equation for pp, we get p3=3pp(2×3+4)=3(2p+4)p(10)=6p+1210p6p=124p=12p=3\begin{aligned} p \triangle 3 & = 3 \triangle p \\ p(2\times3+4) & = 3(2p+4)\\ p(10) & = 6p+12\\ 10p-6p & = 12\\ 4p & = 12\\ p & = 3\end{aligned} The only value of pp for which p3=3pp \triangle 3 = 3 \triangle p is p=3p=3.
Simplifying the given equation, we get m(m+1)=0m(2(m+1)+4)=0m(2m+2+4)=0m(2m+6)=0\begin{aligned} m \triangle (m+1) & = 0 \\ m(2(m+1)+4) & = 0\\ m(2m+2+4) & = 0\\ m(2m+6) & = 0\end{aligned} Thus, m=0m=0 or 2m+6=02m+6=0 which gives m=3m=-3.

The values of mm for which m(m+1)=0m \triangle (m+1) = 0 are m=0m=0 and m=3m=-3.

(Substituting each of these values of mm, we may check that 01=0(2×1+4)=0(6)=00 \triangle 1 = 0(2\times1 +4)=0(6)=0, and that (3)(2)=3(2×(2)+4)=3(0)=0(-3)\triangle(-2) = -3(2\times(-2) +4)=-3(0)=0.)

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.