Finding the remainder of $2^{2021}$ divided by $2700.$ I can't solve this because $2700= 3\cdot 9\cdot 100$ and we cannot use Little Fermat.
Asked
Active
Viewed 258 times
0
-
1You'll probably need to use the Chinese Remainder Theorem. First find the remainder when you divide by $27$, then when you divide by $4$, then we you divide by $25$. – ndhanson3 Mar 11 '21 at 00:57
-
1Are you allowed Chinese Remainder Theorem? I'd hate to do it be successive squaring. But if we had to it'd be easier to do it withh small moduli. – fleablood Mar 11 '21 at 01:07
-
Does this answer your question? How do I compute $a^b,\bmod c$ by hand? – Jyrki Lahtonen Mar 11 '21 at 06:15
2 Answers
3
We have $2700 = 3^3 2^2 5^2$. We will evaluate $2^{2021}$ modulo each of $3^3$, $2^2$, and $5^2$ and use the Chinese remainder theorem to piece the result back together.
Note that $totient(3^3) = 3^2 \cdot 2 = 18$ and $2021 \equiv 5$ modulo 18, so $2^{2021} \equiv 2^5 = 32 \equiv 5$ modulo 27.
Note that $2^2 | 2^{2021}$, so $2^{2021} \equiv 0$ modulo $4$.
Note that $totient(5^2) = 5 \cdot 4 = 20$ and $2021 \equiv 1$ modulo $20$, so $2^{2021} \equiv 2^1 = 2$ modulo 25.
Now, we must piece these puzzle pieces together. Using the Euclidean algorithm, we compute that
$7 \cdot 4 - 27 = 1$
$13 \cdot 25 - 12 \cdot 27 = 1$
$25 - 6 \cdot 4 = 1$
So to compute a number which is $5$ mod 27 and 0 mod 4 and 2 mod 25, we take $5 \cdot (1 + 27) \cdot (1 + 12 \cdot 27) + (1 - 25) \cdot (1 - 13 \cdot 25)$, which evaluates to $47552 = 1652$ mod $2700$.
The final answer is therefore 1652.
Mark Saving
- 31,855
-
@user2661923 I am using the Chinese Remainder Theorem to show that because $2^{2021}$ and $1652$ agree modulo $4$, $27$, and $25$, they must agree modulo $2700$. So I'm at least using the "uniqueness" part. I'm using the algorithm contained in the constructive proof of the Chinese Remainder Theorem to compute the number - see my answer here - https://math.stackexchange.com/a/4051746/798694 – Mark Saving Mar 11 '21 at 01:24
-
1excellent rebuttal. I went back and checked the pertinent Wikipedia article. While the article does contain an existence proof, I had overlooked that it also contains a constructive proof. I have deleted my previous comment. – user2661923 Mar 11 '21 at 01:31
-
-
1
As $(2^{2021},2700)=4$
let us find $x=2^{2021-2}\pmod{\dfrac{2700}4}$
Now as $\phi(27)=18,2019\equiv3\pmod{18}, x\equiv2^3\pmod{27}$
and $\phi(25)=20,2019\equiv-1\pmod{20}, x\equiv2^{-1}\pmod{25}\equiv13$
We need $x=8+27a=13+25b\iff27(a+1)-25b=5+27=16(27-25)$
$$\iff27(a+1-16)=25(b-16), \dfrac{25(b-16)}{27}=a+1-16\text{ which is an integer}$$
So, $27$ must divide $b-16,b=16+27d$(say)
$x=13+25(16+27d)\equiv13+400\pmod{27\cdot25}$
Like How to find last two digits of $2^{2016}$,
multiply all sides by $2^2$ to find $$2^{2019+2}\equiv2^2\cdot413\pmod{2^2\cdot27\cdot25}$$
lab bhattacharjee
- 274,582