$4^{2020} - 2020$ mod $48$

Question

$4^{2020} - 2020$ mod $48$

My Approach:

I basically tried to separate $4^{2020}$ into multiple entities and tried to break them down but at one point it just wasn't computational.

Try separating $48$ with the Chinese Remainder Theorem instead. — player3236, Jan 16 '21 at 11:13
$4^{\large 2+N}! \bmod 48 = 4^{\large 2}(4^{\large N} \bmod 3) = 4^2(1),$ by the mod Distributive Law, and by $\bmod 3!:\ 4\equiv 1\Rightarrow 4^N \equiv 1^N\equiv 1,$ by the Congruence Power Rule. $\ \ $ — Bill Dubuque, Jan 16 '21 at 13:31

score 1 · Accepted Answer · answered Jan 16 '21 at 11:17

1

Hint: $4^{2018}\equiv1\pmod{3}$ implies $4^{2020} \equiv 16\pmod {48}$

answered Jan 16 '21 at 11:17

cr001

wouldn't $4^{2020}$ mod $3$ be equal to $1$ as well? – Patrick Schick Jan 16 '21 at 11:22
Yes, it would. But that doesn't help when mod 48. Indeed $16\equiv 1 \pmod{3}$. – cr001 Jan 16 '21 at 11:23
right. I think this answers the question so thank you! – Patrick Schick Jan 16 '21 at 11:27

score 1 · Answer 2 · answered Jan 16 '21 at 11:31

1

Hint :

$4^{3}\pmod{48}\equiv4^{2}\pmod{48}$

answered Jan 16 '21 at 11:31

Avi

2 Answers2