What is a shortcut to doing this kind of problem? I know that 7 and 107 are both prime number; thus, I assume that has something to do with the appropriate approach/solution. But beyond that I am not sure. Any ideas?
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1Do you know Fermat's little theorem? – Thomas Andrews Jan 19 '15 at 03:21
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@ThomasAndrews Since $53$ is a Germain prime it is easy to check that $7$ is a primitive root. Highly improbable this question was assigned without the OP knowing FLT :) – N. S. Jan 19 '15 at 03:25
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I don't see how your first sentence is related to the second, @N.S. – Thomas Andrews Jan 19 '15 at 03:29
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@ThomasAndrews Since the order of $7$ is $106$, calculating the powers of $7$ until they repeat is basically undo-able. And I am not familiar with any other trick which would work, and be taught before FLC. – N. S. Jan 19 '15 at 03:32
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That assumes it was an assigned problem. Back when I was in grade school, I encountered these sort of problems in a math game, without a class attached. Anyway, the method of repeated squaring reduces it to about 100 multiplications :) @N.S. – Thomas Andrews Jan 19 '15 at 03:37
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Hint 1:
Since $107$ is prime $$7^{106} \equiv 1 \pmod{107}$$
Hint 2: $$100 \equiv -6 \pmod{106}$$ Hence $$1,000,000,000,000,000 \equiv 10 \cdot 100^{7} \equiv 10 \cdot (-6)^7 \pmod{106}$$
This is easy to compute.
N. S.
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