To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

Question

$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.

$199x \equiv 1\pmod {10}$ has a solution $x =-1$ and $10x \equiv 1\pmod{199}$ has a solution $x = 20$.

Thus using Chinese Remainder Theorem we have,

$a \equiv 4(-1)199 + 29(20)10 \equiv 5004 \equiv 1024\pmod {1990}$.

Thus our $a = 1024$.

Is the solution correct? Is there any better proof?

Us pmod for better formatting. a\equiv b\pmod{1900} becomes $a\equiv b\pmod{1900}$ — Thomas Andrews, Mar 29 '15 at 02:01
Smartass answer: $a=2^{1990}$. :) Presumably, you want the smallest positive $a$ :) — Thomas Andrews, Mar 29 '15 at 02:01
It's not clear how you jump from modulo $10$ to modulo $100$. As it turns out, your answer is correct, but it might be an accident. — Thomas Andrews, Mar 29 '15 at 02:09
You don't know $a\equiv 29\pmod{199}$. $199$ is not a divisor of $1900$. $1900=19\cdot 2^2\cdot 5^2$. You don't want to know $a\pmod {1990}$. — Thomas Andrews, Mar 29 '15 at 02:14
If $a\equiv 4\pmod{10}$ it is not possible that $199a\equiv 1\pmod{10}$ because $a$ is even. I think you want $199x\equiv 1\pmod{10}$ - a new variable. — Thomas Andrews, Mar 29 '15 at 02:19
Now it looks fine. Amusing that the answer to the wrong question ($\pmod 1900$) was the same as the answer to the corrected question. — Thomas Andrews, Mar 29 '15 at 02:22
$\varphi(1990)=792$ and $1990$ mod $792$ is $406$ which brings it down to finding $2^{406}$ mod $1990$ which is indeed $1024$ so you're right. — coffeemath, Mar 29 '15 at 02:32

Thomas Andrews · Answer 1 · 2015-03-29T02:25:06.667

1

Checking your answer of $a=1024=2^{10}$: $$\begin{align}2^{1990}&\equiv 0\equiv 2^{10}&\pmod 2\\ 2^{1990}&\equiv \left(2^{4}\right)^{495}2^{10}\equiv 2^{10}&\pmod{5}\\ 2^{1990}&\equiv \left(2^{198}\right)^{10}2^{10}\equiv 2^{10}&\pmod{199} \end{align}$$

After your edits, your answer is correct.

edited Mar 29 '15 at 02:25

answered Mar 29 '15 at 02:06

Thomas Andrews

177,126

score 0 · Answer 2 · answered Mar 29 '15 at 07:21

Yes, it is correct. More simply, by little Fermat $\,2^4\equiv 1\pmod 5\,$ and $\,2^{198}\equiv 1\pmod{199}\,$ so $\,4,198\mid 1980\,\Rightarrow\,2^{1980}\equiv 1\,\Rightarrow\,2^{1990}\equiv 2^{10}\,$ mod $5$ and mod $199$. Finally, combining these we conclude that $\,2^{1990}-2^{10}\,$ is divisible by $\,2,5,199,\,$ so also by their lcm = product $=1990$.

To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

2 Answers2