1
$\begingroup$

I'm currently studying for my summer maths exam and I've come across a problem that has appeared in some form in all of the previous years' papers.

Unfortunately, our Maths teacher wasn't very good at explaining things so I have no idea how to approach the problem and the notes have only increased my frustration.

The question is as follows: Evaluate $5^{17}$ modulo $70$ i.e. Determine the smallest positive remainder $b$ such that $5^{17} \equiv b \pmod {70}$

Rather than a solution to the problem, what I'm more concerned about is a method to solve the problem

$\endgroup$
0
$\begingroup$

$5^{17} \equiv 5 \pmod{10}$ while $5^{17} = 125^5 \cdot 5^2 \equiv (-1)^5 \cdot 25 \equiv -25 \equiv 3 \pmod{7}$. Thus, by Chinese Remainder's Theorem, we have $5^{17} \equiv 45 \pmod{70}$.

$\endgroup$
  • $\begingroup$ I have no idea where you're getting any of that from. Could you explain your answer a bit better please? $\endgroup$ – Declan May 6 '14 at 12:41
  • $\begingroup$ Which part do you want me to clarify $\endgroup$ – zscoder May 6 '14 at 12:44
  • $\begingroup$ As awful as I feel saying it: All of it? I feel like I'm missing something crucial in understanding congruences so following along with the solutions is beyond my right now. $\endgroup$ – Declan May 6 '14 at 12:48
  • $\begingroup$ First one : Since it is a multiple of 5 and not a multiple of 10 $\endgroup$ – zscoder May 6 '14 at 12:49
  • $\begingroup$ Ok, It's not a multiple of 10 because it's to the power of an odd number? $\endgroup$ – Declan May 6 '14 at 12:56
3
$\begingroup$

$5^4\equiv -5 \mod 70 \implies 5^{16}\equiv 5^4\equiv -5\mod 70 \implies 5^{17}\equiv -25 \mod 70$

Hence, $5^{17}\equiv 45 \mod 70$

$\endgroup$

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.