Areteem: MCIII Number Theory 3.21

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MCIII Number Theory 3.21

Hello,

Hello! For this one I added (sqrt(29)-sqrt(21))^1984, but I don't know how to find the last two digits. Am I just supposed to keep multiplying it until it repeats?

Thanks,

Tina Jin

Before doing 3.21, can you work on 3.20? These two questions use almost exactly the same methods, and the numbers in 3.20 are easier to manipulate.

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I figured out how to do 3.20, just by adding the same power of the conjugate, and since its really small (<1), it would just be 8-1=7.

(the last digit is 8 because the last digits are 2(9+6+4+6+4+6+4+...+6+4) congruent to 8 (mod 10) then subtract 1 because of the power of the conjugate.

I tried doing the same thing for 3.21, but I couldn't figure out how to get the last two digits, can you help me with that? Like maybe a hint?

Thank you!

Tina Jin

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Assuming things started off correctly, you need to find the last two digits of $$(\sqrt{29} + \sqrt{21})^{1984} + (\sqrt{29} - \sqrt{21})^{1984} = (50+2\sqrt{609})^{992} +(50-2\sqrt{609})^{992}.$$Hint: As in 3.20, the expansions (using the binomial theorem) will have lots of terms that cancel. Which terms are left? Which ones matter for the last two digits?

Note: If this doesn't seem like something you did in 3.20, I would recommend (similar to Dr. Wang) looking at that one again and trying this method.

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Oh, I see, I didn't square (sqrt(29)±sqrt(21)). Now I only need to calculate some of the terms (the ones ending in 00 won't effect the final answer).

Thank you!

Tina Jin

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That is correct! Just want to point out that for 3.20, getting $(\sqrt{3} + \sqrt{2})^2 = 5 + 2\sqrt{6}$ first will be really helpful.

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