3
$\begingroup$

This is what I have : By induction

Base case: $n=1$ where it is true Inductive hypothesis: Assume there exists a natural number k where this is true for $k<n$ Show for $k+1$, $11^{k+1} ≡ 6 \pmod{5}$ implies $11^{k}\cdot 11 ≡ 6\cdot 6 \pmod{5} =36\pmod{5}$ Therefore it is true for all natural numbers. Is this true. I find this too easy

$\endgroup$
  • $\begingroup$ Hint: what is $11$ congruent to mod $5$? Now use your induction hypothesis. $\endgroup$ – Alex Wertheim Nov 12 '14 at 2:40
  • $\begingroup$ I'm sorry it won't let me post my solution. But I wanna know if what I have is correct, please check . $\endgroup$ – Jessy White Nov 12 '14 at 2:41
  • 1
    $\begingroup$ Base case: n=1 where it is true Inductive hypothesis: Assume there exists a natural number k where this is true for k<n Show for k+1, 11^(k+1) ≡ 6 mod 5 implies 11^(k)*11 ≡ 6 *6 mod 5 =36 mod 5 Therefor it is true for all natural numbers. Is this true. I find this too easy $\endgroup$ – Jessy White Nov 12 '14 at 2:41
  • 1
    $\begingroup$ Looks just fine to me! (Although just for clarity's sake, you may want to remark at the end that $36 \equiv 6 \pmod 5$.) $\endgroup$ – Alex Wertheim Nov 12 '14 at 2:44
  • $\begingroup$ Is it stated anywhere that you need to use induction? Because it's as simple as $11^n \equiv 1^n \equiv 1 \equiv 6 \pmod 5$. A simple (almost trivial) direct proof. Why use induction? $\endgroup$ – Deepak Nov 12 '14 at 4:09
2
$\begingroup$

Since $6 \cong 1\pmod{5}$, thus we prove: $11^n \cong 1\pmod{5}$, but $11^n - 1 = (11-1)(10^{n-1} + 10^{n-2} + .... + 1) \cong 0\pmod{5}$

$\endgroup$
  • $\begingroup$ Notice the difference of pmod and (mod ), pmod is adjusted correctly. $\endgroup$ – dustin Nov 12 '14 at 2:48
0
$\begingroup$

You have $$11^n = 6^n = 1^n = 1 \text{mod} 5.$$

$\endgroup$
  • $\begingroup$ Does this seem right Base case: n=1 where it is true Inductive hypothesis: Assume there exists a natural number k where this is true for k<n Show for k+1, 11^(k+1) ≡ 6 mod 5 implies 11^(k)*11 ≡ 6 *6 mod 5 =36 mod 5 Therefor it is true for all natural numbers. Is this true. I find this too easy $\endgroup$ – Jessy White Nov 12 '14 at 2:51
0
$\begingroup$

Hints:

$$11^n=(10+1)^n= 1 \mod 5$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.