-1

I'm very confused at how this math problem is being set up... I just came across a page on my study on congruences. I've studied the theorems on it but I seem to be lagging behind in this direction. Can I get help with these two questions and I would love to know more about the answer with steps to how the question was solved. Thanks in advance.

(1) Find $2^{501} \pmod {17}$

(2) Solve for $41x \equiv 5\pmod {51}$

Alessio K
  • 10,599
Dave Kent
  • 27
  • Question $(1)$ asks you to find the remainder when $2^{501}$ is divided by $17$. $(mod\ n)$ is an environment where you denote any integer $m$ by it's remainder when divided by $n$, so that if $m_1,m_2$ have same remainders on division by $n$, they are equal(or congruent, like geometric congruency) modulo n i.e. equal in the (mod n) environment. Where are you studying from? Do you know properties like $a\equiv b (mod \ n) \implies a^k\equiv b^k (mod\ n)$ for any positive integer $k$? – Fawkes4494d3 Aug 30 '20 at 01:01
  • I know that... I'm actually studying from a document and the property you stated is in the document. – Dave Kent Aug 30 '20 at 01:15
  • for (2), $x\equiv \frac{5}{-10} \equiv \frac{-1}{2} \equiv \frac{50}{2} \equiv 25 \pmod{51}$ – Evariste Aug 30 '20 at 01:37

1 Answers1

1

For (1), note that $2^4\equiv-1\bmod17$, so $(2^4)^{125}\equiv-1\bmod17$.

For (2), use the extended Euclidean algorithm:

$$51=41+10$$

$$41=4\times10+1.$$

Therefore $1=41-4\times10=41-4(51-41)=5\times41-4\times51$,

so $5=25\times41-20\times51$.

Can you take it from here?

J. W. Tanner
  • 60,406