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


$5^{17} \equiv 5 \pmod{10}$ 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 $5^{17} \equiv 45 \pmod{70}$.

  • $\begingroup$ 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

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

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


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