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So let $n$ be a odd integer. Show that $n - 2^k$ is divisible by $3$ if $k$ is SOME SPECIFIC positive integer. $k \ge 0$. So there only has to exist one. For example:

$$7 - 2^2 = 3$$ is divisible by $3$

The approach is modular arithemetic, but it is hard since,

$$2 \equiv 2 \pmod{3}$$

$$n \equiv p \pmod{3}$$

It is hard to combine these? What should I do?

Amad27
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6 Answers6

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Hint $\ {\rm mod}\ 3\!:\ 2^2\equiv 1\,$ so $\,2^k\equiv 2^0\equiv 1$ or $\,2^k\equiv 2^1 \equiv 2$. Thus $\,n\equiv 2^k\iff 3\nmid n$

Bill Dubuque
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  • Actually, isnt, $2^k \equiv -1 \pmod{3}$? – Amad27 Mar 03 '15 at 16:32
  • @Amad27 $\ \ {-}1\equiv 2\pmod{3}.\ $ The point is that the powers of $2$ repeat $, 1,2,1,2\cdots$ so the values taken are precisely the elements $\not\equiv 0\pmod{3}\ \ $ – Bill Dubuque Mar 03 '15 at 16:34
  • I don't understand.

    $$2 \equiv -1 \pmod{3}$$

    $$2^2 \equiv 1 \pmod{3}$$

    $$2^3 \equiv -1 \pmod{3}$$

    ....

    $$2^{n} \equiv (-1)^n \pmod{3}$$

    Isnt it?

     – Amad27 Mar 03 '15 at 16:42
  • @Amad27 Again, $,2\equiv -1\pmod{3},$ so it does not matter which one you use. It is true that $,2\equiv -1,\Rightarrow, 2^n\equiv (-1)^n,,$ so using $-1$ makes the periodicity clearer (but it requires knowledge of the Congruence Power Rule) – Bill Dubuque Mar 03 '15 at 16:45
  • That is what I dont get. How doesn't it make a difference? – Amad27 Mar 03 '15 at 16:46
  • @Amad27 Because $,3\nmid n\iff n \not\equiv 0\pmod3,,$ i.e. the result depends only on the congruence class (or remainder) of $,n,$ modulo $3$. So it doesn't matter which class reps we use. – Bill Dubuque Mar 03 '15 at 16:51
  • $(+1)$ Good answer, but still gray area (from my part). The remainder for $2$ was $-1$. I applied your power rule saying:

    $$2 \equiv -1 \pmod{3}$$

    $$2^n \equiv (-1)^n \pmod{3}$$

    But you say that we can use either of these. So that means it is right to say:

    $$ 2^2 \equiv -1 \pmod{3}$$ ??

     – Amad27 Mar 03 '15 at 16:53
  • @Amad27 $\ \color{#c00}{2\equiv -1},\Rightarrow, \color{#c00}2^2\equiv (\color{#c00}{-1})^2\equiv 1,,$ not $,-1\ \ $ – Bill Dubuque Mar 03 '15 at 16:55
  • Exactly. I seem to have confused this because there werent wordsin the answer. But are you specifically using EVEN $k$? – Amad27 Mar 03 '15 at 16:56
  • @Amad27 The power rule works for both even and odd expts. Here $,2^{2k}\equiv (-1)^{2k}\equiv 1,$ and two times that shows $,2^{2k+1}\equiv 2\equiv -1\ $ – Bill Dubuque Mar 03 '15 at 16:57
  • But if it INCLUDES odd exponents then clearly. It is false that:

    $$2^3 \equiv 1 \pmod{3}$$

     – Amad27 Mar 03 '15 at 16:58
  • @Amad27 Please reread my prior comment. – Bill Dubuque Mar 03 '15 at 17:00
  • So I assume you agree that:

    $$2^k \equiv (-1)^k \pmod{3}$$ Then,

    $$n \equiv p \pmod{3}$$ For some remainder $p$.

    $$n - 2^k \equiv (-1)^k - p \pmod{3}$$?

     – Amad27 Mar 03 '15 at 17:01
  • Would you mind adding words to your answer? It is insanely difficult to understand. Somehow you introduced $n$, in some place suddenly? – Amad27 Mar 03 '15 at 17:02
  • By definition $, 3\mid n-2^k\iff {\rm mod}\ 3!:\ n\equiv 2^k\equiv (-1)^k \iff n\not\equiv 0\iff 3\nmid n.\ $ To master elementary number theory it is essential to be able to understand statements like that. – Bill Dubuque Mar 03 '15 at 17:04
  • I see. You used:

    $$2 \equiv 2 \pmod{3} $$

    $$2^2 \equiv 1 \pmod{3}$$

    Then:

    $$n \equiv 1 \pmod{3}$$ Or $$n \equiv 2 \pmod{3}$$

     – Amad27 Mar 03 '15 at 17:06
  • I seem to have got it. But the conclusion was that as $n$ changes so does the congruence. Since $n=13$ leaves remainder $1$, but $n=11$ leaves remainder $2$ – Amad27 Mar 03 '15 at 17:13
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There are none if $n$ is a multiple of $3$. There are infinitely many otherwise, or a finite amount if you further require $n - 2^k$ to be positive.

If $n = 6m + 1$, then $k = 2$ will do; then $n - 2^k = 6m + 1 - 4 = 6m - 3$.

If $n = 6m + 5$, then $k = 1$ works; then $n - 2^k = 6m + 5 - 2 = 6m + 3$.

John-Luke
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If $n$ is an odd integer not divisible by $3$, then $n $ is $1$ or $2$ mod $3$ and thus $n-2^2$ or $n-2^1$ is $0$ modulo $3$.

On the other hand if $n$ is divisible by $3$ then $n-2^k$ is not divisible by $3$ for every $k$ since otherwise $n-(n-2^k) = 2^k$ would also be divisible by $3$, which is absurd.

(Actually, the "odd" in the first part is irrelevant.)

quid
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  • Unless one redefines "odd" to mean "not divisible by $3$" instead of "not divisible by $2$", but that is absurd, too. – robjohn Mar 03 '15 at 17:10
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    @robjohn yet we might introduce the notion of an oddd number. – quid Mar 03 '15 at 18:54
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The way to combine congruences is by simple arithmetic: if the modulus is the same, you can add, subtract and multiply congruences all day long. For the powers of $2$ we have $2^k \equiv 2 \bmod 3$ if $k$ is odd and $2^k \equiv 1 \bmod 3$ if $k$ is even. Then, there are three possibilities for $n$:

  • If $n \equiv 0 \bmod 3$, for no $k$ can $3 \mid (n - 2^k)$ hold true.
  • If $n \equiv 1 \bmod 3$, for any even $k$ we'll have $3 \mid (n - 2^k)$.
  • If $n \equiv 2 \bmod 3$, for any odd $k$ we'll have $3 \mid (n - 2^k)$.

In your example with $7$, you can also use $2^4$ (which gives $7 - 16 = -9$), $2^6$ (gives $-57$), $2^8$ (gives $-249$), etc.

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Let $[n]_3$ be the equivalence class of $n$ modulo 3. Then $[n-2^k]_3 = [n]_3 - [(-1)^k]_3$. Since $[(-1)^k]_3$ is either $[1]_3$ or $[2]_3$, we see that if $[n]_k = [0]_3$ then $[n-2^k]_3 \neq [0]_3$ for all $k$. If $[n]_3 = [1]_3$, then pick $k=0$, otherwise pick $k=1$.

copper.hat
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0

So if $n$ is an odd integer, you have to prove that there exists some integer exponent $k$ such that $3|(n - 2^k)$? Trouble is, it depends on what $n$ is. There are either infinitely many solutions or none at all.

Let's look at this modulo 6: there are three possibilities: $n \equiv 1, 3 \textrm{ or } 5 \pmod 6$. For the powers of 2, we have $2^k \equiv 2 \textrm{ or } 4 \pmod 6$ for $k > 0$. We want $n - 2^k \equiv 0 \textrm{ or } 3 \pmod 6$.

But how to combine all these? It's actually quite easy: we just subtract! Just be sure to mind the wrap-around.

For example, if $n \equiv 1 \pmod 6$ and $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 1 - 4 \equiv 3 \pmod 6$.

But what if $n \equiv 3 \pmod 6$? No $k$ can work: if $2^k \equiv 2 \pmod 6$, then $n - 2^k \equiv 3 - 2 \equiv 1 \pmod 6$, and if $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 3 - 4 \equiv 5 \pmod 6$. Don't forget $k = 0$, but that's of no help here: $n - 1 \equiv 3 - 1 \equiv 2 \pmod 6$.

The solution for $n \equiv 5 \pmod 6$ should be easy enough for you to figure out now.

One of the other answerers mentioned a requirement for $n - 2^k$ to be positive, but you didn't. I don't know the rules of the contest, so I assume negative values are perfectly acceptable. For example, $-9 \equiv 3 \pmod 6$ since $(-2) \times 6 + 3 = -12 + 3 = -9$.

Robert Soupe
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