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What is the remainder when $4^{96}$ is divided by 6

My approach : When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.

Bill Dubuque
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Sachin
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  • For the above, some inductive argument should be formalized, or just mention that $4=4^2$ in the ring $\Bbb Z/6$. Alternatively, $4^n$ is $0$ mod two, and modulo three $4^n=(3+1)^n=1^n=1$. – dan_fulea Feb 06 '21 at 11:39
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    Since $4^2 \equiv 4 \pmod 6$, you can prove by induction that $$4^N \equiv 4 \pmod 6$$ for all $N$. This is basically the same argument you used, and it's the simplest way to justify that the answer is $4$. – Crostul Feb 06 '21 at 11:39
  • The short cut to solving this kind of problem is modular arithmetic and the chinese remainder theorem. – jMdA Feb 06 '21 at 12:00
  • Hint: $(6a+4)(6b+4)=6(6ab+4a+4b+2)+4$. – J.G. Feb 06 '21 at 12:13
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    Generally the simplest way for these types is to use the mod distributive law as explained in the dupes, i.e. $$\large 4^{n+1}\bmod 6 ,=, 2,\overbrace{(2\cdot \color{#c00}{4^n}\bmod{! 3}}^{\large \color{#c09}{4^n},\equiv, 1^n\pmod{3}}), =, 2(2)\qquad\qquad $$ See the "Linked" sidebar on the prior linked answer for $> 100$ worked examples. – Bill Dubuque Feb 06 '21 at 21:31

8 Answers8

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By induction, we can prove $4^N\equiv4\pmod6$. The base case is $N=1$, which is trivial. Then suppose for $N=k\in\mathbb{N}$, the statement is true. Then for $N=k+1$, we have $$4^N=4^{k+1}=4^k \cdot4\equiv4\cdot4=16\equiv4\pmod6$$ So, we have proved the induction case. So sub $N=96$ into above statement then we are finished with $4^{96}\equiv4\pmod6$.

Culver Kwan
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Your answer is in the right direction. Here is the argument I came up with. (I avoided using modular arithmetic.)

Let's prove that $4^n$ has a remainder of $4$ when divided by $6$. The base case for this argument is $4^1=4$ is obvious. Then assume $4^n$ has a remainder of $4$ when divided by $6$, this means $4^n = 4 + 6*m$ for some integer $m$. This means that $4^{n+1}=4*(4+6*m)=16+6*(4*m)$. Because $6*4*m$ is divisible by $6$ the remainder of $16+6*4*m$ when dividing by $6$ is equal to that of $16$, which is $4$. This proves that $4^{n+1}$ has a remainder of $4$ when divided by $6$. This concludes the proof.

The case for $n=96$ is a special case. That follows from the general case.

jMdA
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$$4^{96}=(6-2)^{96} \equiv 2^{96} \pmod{6}$$

Now, if we find the remainder of $2^{95}=(3-1)^{95}$ by $3$,it is simply $(-1)^{95}=-1$ or $2\pmod{3}$ , so we have $$2^{95}=3k+2$$ Now multiply the above equation by $2$ on both sides to get $$2^{96}=6k+4$$

V.G
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  • I'm just seeing this thread so I didn't downvote, but I can see why this was downvoted. In general $(n-a)^m \mod n = a^m$ is incorrect. It is true here and in general for $m$ an even integer, as $(-1)^2 =1$. And also...there were already a few answers given already that got to the point more consisely than this, what does this really add. – Mike Feb 06 '21 at 15:51
  • @Mike: When did I say that it is true in general? In general, of course $(n-a)^m \equiv (-1)^m \pmod{n}$. And as for your second statement, well I answered at number three if I remember correctly and now there are nine answers to this, are rest of them downvoted? And also, mine was a different answer from the previous ones posted (they were based on induction). AND I know who did this, it was probably just because he has also written answer here (I know that because I saw his reputation getting reduced by $1$ FYI). – V.G Feb 06 '21 at 16:04
  • Well but you did get an upvote too. Two I think. Which was more than even the 1st answer--which is imo the best one here--had gotten. But anyways, the rule of thumb is the "easier" the question the less skipping steps in the proof you can do. It's easier to see that $4^m \mod 6$ stays $4$ for all positive integers $m$ than it is knowing precisely when $(n-a)^m \mod n$ is $a \mod n$. – Mike Feb 06 '21 at 16:18
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No need for induction. We have $$4^{N}=1^{N}=1\mod 3,$$ for all $N\in\mathbb N$, as well as $$4^{N}=0\mod 2.$$ From the former it follows that $4^N$ is either $1$ or $4\mod6$, and from the latter it follows that it's $0$, $2$, or $4\mod 6$. So it must be equal to $4\mod 6$.

Vercassivelaunos
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As was mentioned, since Euler is not directly applicable, you can use CRT. Putting together that $4^{96}\equiv0\bmod2$ and $4^{96}\equiv 1\bmod3$, the latter by Fermat's little theorem, or the binomial theorem, we have using Bezout that $4^{96}\equiv -2\cdot1+0\cdot3=\equiv-2\equiv4\bmod6$.

As it turns out, CRT is not necessary here, as @BillDubuque points out.

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You could proceed this way: $6 = 2\cdot 3$. When you divide $4^{96}$ by $2$ the remainder is $0$. To divide by $3$, note that $4=3+1$ so that $4^{96} = (3+1)^{96}$ which expands to a multiple of $3$ plus $1$, so the remainder is one.

Now you need a number which is $1$ more than a multiple of $3$ and divisible by $2$ and less than $6$. That would be $4$.

B. Goddard
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Just to give a different approach, note that ${n\choose k}$ is even for $1\le k\le n-1$ when $n$ is a power of $2$ (i.e., Pascal's triangle mod $2$ looks like Sierpinski's triangle). Consequently

$$4^{32}=(3+1)^{32}\equiv3^{32}+1\equiv3+1=4\mod 6$$

and thus $4^{96}\equiv4^3=64\equiv4$ mod $6$.

Barry Cipra
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In all cases,

n-odd$\rightarrow 2^n\equiv 2 \mod6$

n-even$\rightarrow 2^n\equiv 4\mod6$

$\implies 2^{96}\equiv 4\mod 6$

poetasis
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