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Find the remainder when $2^{2^{24}}$ is divided by 9

My approach is as follow $2^{2^{24}}=4^{12}$ Hence we can write is as $2^{2^{24}}=2^{4^{12}}=16^{12}$

$16^{12}=(9+7)^{12}$

We get remainder as $7^{12}$

Now $7^{12}=(9-2)^{12}$

We get remainder as $(-2)^{12}=2^{12}$

Now $2^{12}=8^{4}=(9-1)^4$

Hence final answer is $"1"$ but we are getting different solution.

I presume that I am making some mistake in calculation

Haris
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Samar Imam Zaidi
  • 8,800

1 Answers1

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First note that $$2^{24} = (3 - 1)^{24} = \sum_{n=0}^{24}\binom{24}{n}3^{n}(-1)^{24-n} = (-1)^{24} + 3\sum_{n=1}^{24}\binom{24}{n}3^{n-1}(-1)^{24-n}$$ Then, $2^{24} = 3k + 1$ for some $k \in\mathbb{N}$. Furthermore, $$2^{24} = 3k + 1 \implies 2^{24} - 2k - 1 = k$$ which means $k$ is odd. Now note that, $$8^{k} = (9 - 1)^{k} = \sum_{n=0}^{k}\binom{k}{n}9^{n}(-1)^{k-n} = (-1)^{k} + 9\sum_{n=1}^{k}\binom{k}{n}9^{n-1}(-1)^{k-n}$$ $$ = -1 + 9\sum_{n=1}^{k}\binom{k}{n}9^{n-1}(-1)^{k-n}$$ So that $8^{k} = -1 + 9m$ for some $m \in \mathbb{N}$. We put this altogether like so $$2^{2^{24}} = 2^{3k + 1} = 2\cdot 8^{k} = 2(-1 + 9m) = 2(8 + 9(m-1)) = 16 + 18(m-1) = 7 + 9 + 18(m-1)$$ The remainder after division is then $7$.

Alternatively, we first have that $$2^{24} \equiv_{6} 8^{8} \equiv_{6} 2^{8} \equiv_{6} 4$$ Then, $2^{24} = 6k + 1$ for some $k$. By Eulers Theorem, note $$2^{6} \equiv 1 \mod 9$$ Then, $$2^{2^{24}} = 2^{6k+4} \equiv_{9} 2^{4} \equiv_{9} 7$$

conan
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Feb 20 '24 at 14:39