0

I had to find the remainder when $5^{30009}$ is divided by $7.$

  1. I wrote the expression like this: $$(A)\ x \equiv 5^{30009} \pmod 7$$

  2. Found that: $$ 5^6 \equiv 1 \pmod 7 $$

  3. Then wrote $(A)$ like this: $$ (5^6)^{50001}\cdot 5^3 \equiv 5^{30009} \pmod 7$$

  4. Replacing $5^6$ with $1$, the expression becomes, $$5^3 \equiv 5^{30009} \pmod 7 $$

Now I'm not sure how to continue from there. I can easily verify that $$125 = 17\cdot7 + 6$$ So, the remainder is $6.$ But is showing that enough or I should keep using congruence notation to solve this?

Sathvik
  • 3,609
Jonas
  • 307
  • 2
    Looks like you've done all the work (in a good way, well done!). The last step would be to note that the identity $125=17\cdot7+6$ implies $5^3\equiv6\pmod7$. There's nothing new that you need to add, other than to show that your calculations actually demonstrate what you've been asked to demonstrate. – Greg Martin Dec 11 '22 at 00:48
  • 1
    \pmod{7} will produce the correct spacing, typeface, and the parentheses around "mod 7". – Arturo Magidin Dec 11 '22 at 00:50
  • 1
    Did you mean $5001$ where you typed $50001$? – J. W. Tanner Dec 11 '22 at 01:31
  • 1
    Cf. this question – J. W. Tanner Dec 11 '22 at 01:36
  • 1
    $!\bmod 7!:\ \color{#c00}{5^3\equiv -1}\Rightarrow 5^{30009}\equiv (\color{#c00}{5^3})^{10003}\equiv (\color{#c00}{-1})^{10003}\equiv -1,$ by the Congruence Power Rule. $ $ Or use $,5^{\color{#c00}6}\equiv 1\pmod{!7},\ \color{#0a0}{30009\equiv 3\pmod{!\color{#c00}{6}}}\Rightarrow 5^{\color{#0a0}{30009}}\equiv 5^{\color{#0a0}3}\pmod{!7},$ by mod order reduction in the linked dupe (whose proof also uses the Congruence Power Rule as above, using $,\color{#c00}{5^6\equiv 1}),,$ as you did. – Bill Dubuque Dec 11 '22 at 07:47

1 Answers1

1

As said in the comments, your proof is fine.

Here is one that involves even less calculations, exploiting negative values of congruences:

$5^{30009}\equiv(-2)^{30009}\equiv (-8)^{10003}\equiv(-1)^{10003}\equiv -1\equiv 6\pmod{7}$

zwim
  • 28,563