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I worked through problem 7 in this GRE prep material (https://math.uchicago.edu/~min/GRE/files/week1.pdf), and I got the right answer, but I'm wondering if there is a more efficient way to do this, particularly considering that the GRE math subject test doesn't permit using a calculator. The question is to find the remainder upon dividing $ 13 ^{2019} $ by $95$. So we must solve for $x$ in $ 13^{2019} \equiv_{95} x.$ Since 95 factors into 5*19, I first found each of those residues.

$ 13^{2019} \equiv_{5} {{(13 \pmod 5})}^{2019 \pmod{\phi(5)}}\equiv_{5} 3^{2019 \pmod{4}}\equiv_{5} 3^3=27\equiv_{5}2.$ $ 13^{2019} \equiv_{19} {13}^{2019 \pmod{\phi(19)}}\equiv_{19} {13}^{2019 \pmod{18}}= 13^3\equiv_{19}(-6)^3=36*(-6)\equiv_{19}(-2)*(-6)\equiv_{19}12.$
Since $2 \equiv_5 12$, that means $13^{2019}$ is equivalent 12 with respect to each modulus, which means that $13^{2019} \equiv_{95} 12.$

Are there any other techniques I could have used to get the answer more quickly?

Bill Dubuque
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Hank Igoe
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    This looks pretty fast to me, you've used most of the tricks it would occur to me to use. You can get the remainder $\bmod 5$ slightly faster by writing $3^{2019} \equiv (-2)^3 \equiv -8 \equiv 2 \bmod 5$ but it's not much faster. You haven't written out how you computed $2019 \bmod 18$ but you can do it by writing $2019 = 1800 + 180 + 39$. – Qiaochu Yuan Dec 22 '20 at 22:54
  • For that last one (2019 mod 18), I cheated by using python, but your approach is good for doing by hand, thanks. – Hank Igoe Dec 22 '20 at 23:50
  • Or: $,2020\bmod 18 = 2(\color{#c00}10\color{#c00}10\bmod 9) = 2(\color{#c00}{1!+!1}),$ by mod Distributive Law and casting out nines. – Bill Dubuque Dec 23 '20 at 10:26

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$\phi(95)=\phi(5) \cdot \phi(19) = 4 \cdot 18=72$.

$2019 \equiv 3 \pmod{72}, 13^{72} \equiv 1 \pmod{95}$, therefore

$$13^{2019} \equiv 13^3 \equiv 169 \cdot 13 \equiv (-21) \cdot 13 \equiv 12 \pmod{95}$$

Neat Math
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