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Fast Modular Arithmetic

What is the fastest way to show that \(102^2 \equiv 30\mod {247}\)?

Note by Axas Bit
9 months, 3 weeks ago

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I suppose one way would be to first note that \(102 = 124 - 22\), and so

\(A = 102^{2} = (124 - 22)^{2} = 124^{2} - 2*22*124 + 22^{2} = 62*248 - 22*248 + 484.\)

Now \(248 \equiv 1 \pmod{247}\) and \(484 = 2*248 - 10 \equiv -10 \pmod{247}\), so

\(A \equiv (62*1 - 22*1 - 10) \pmod{247} \equiv 30 \pmod{247}\). Brian Charlesworth · 9 months, 3 weeks ago

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Define "fastest".

In terms of no need to think about what to do: \( 102^2 = 42 \times 247 + 30 \). Calvin Lin Staff · 9 months, 3 weeks ago

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'Fastest way' should definitely be checking if 247 divides \(102^2-30\). We find that it is true very fast. Svatejas Shivakumar · 9 months, 3 weeks ago

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Brian Charlesworth gave a standard answer. Here's an alternative approach:

Notice that \(50^2 = 2500 = 30 + 2470 = 30 + 247(10) \), then \[ 102^2 - 50^2 = (102 - 50)(102 + 50) = 52 \times 152 = 247 \times (\ldots ). \]

This tells us that \(102^2 - 50^2 \equiv 0 \pmod {247} \Rightarrow 102^2 \equiv 50^2 \pmod{247} \Rightarrow 10^2 \equiv 30 \pmod{247} \). Pi Han Goh · 9 months, 3 weeks ago

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@Calvin Lin @Pi Han Goh @Otto Bretscher Axas Bit · 9 months, 3 weeks ago

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@Axas Bit @Brian Charlesworth Any thoughts? Axas Bit · 9 months, 3 weeks ago

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