What is the remainder of $17^{{3}^{100}}$ when divided by 60

Question

I'm stuck in this question. I tried to use the calculator to find the remainders up to $17^5$ when they got repeated, but this way seems so daunting that I think it's just not the right way to solve. Can someone provide a solution or give me a hint? Thanks.

Perhaps it might be useless, but:

$17^2=289\equiv 49 \pmod {60} \\ 17^3\equiv 53 \pmod {60} \\ 17^4\equiv 1 \pmod {60}\\ 17^{5}\equiv 17 \pmod {60}$

Your calculationa are not useless. $17^4 \equiv 17^0$ so $17^x$ has period $4$ mod $60$. — GEdgar, Jul 26 '21 at 20:18
It is useful: by mod order reduction $\bmod 60!:\ 17^{\large\color{#0a0}4}\equiv 1\Rightarrow\ 17^{\large\color{#c00}{9^n}}\equiv 17^{\large 9^n\bmod\color{#0a0}4}!\equiv 17^{\color{#c00}{\bf 1}},$ by $\bmod \color{#0a0}4!:\ \color{#c00}{9^n}\equiv 1^n\equiv \color{#c00}{\bf 1}.,$ See the linked dupe for a systematic way to find the exponent $4$ using lcms or Carmichael's lambda function. — Bill Dubuque, Jul 26 '21 at 22:48

score 2 · Answer 1 · answered Jul 26 '21 at 20:13

2

$17^2=289\equiv 49 \pmod {60} \\ 17^3\equiv 53 \pmod {60} \\ 17^4\equiv 1 \pmod {60}\\$

$3^2=9\equiv 1 \pmod {4} \\ 3^{100}\equiv 1 \pmod {4} \\$

$17^{3^{100}}\equiv 17 \pmod {60} \\$

answered Jul 26 '21 at 20:13

Lion Heart

7,073

1

Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 26 '21 at 22:49

What is the remainder of $17^{{3}^{100}}$ when divided by 60

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