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I'm a very slow learner in math. Would someone be kind enough to explain in detail how to solve the above question?

Thanks in advance.

Bernard
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Rogs
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    Being a slow learner is not a problem. So you should tell us how and from where you got this question. What are your thoughts on this question or even part of it. Please let us know what you don't understand and also let us know the theorems/formulae/rules you are familiar with that are related to this question. – paulplusx Sep 26 '18 at 08:30

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You have addition and multiplication modulo $n$. To calculate $2n-1\mod n$, you may calculate $2n\mod n$ and $-1\mod n$ separately and then add the results (and take the result $\mod n$ again, if necessary).

Thus, $2n-1\mod n = ((2n\mod n)+(-1\mod n))\mod n= (0-1)\mod n \equiv -1$.

Does this answer your question?

  • Rogs.Have a look at https://math.stackexchange.com/questions/2883235/modulus-of-negative-numbers. – Peter Szilas Sep 26 '18 at 08:33
  • @James thank you so much bro. Yes it does! Been stuck on this question for hours and you cleared it. Thank you again for the clear explanation. – Rogs Sep 26 '18 at 09:05
  • You're welcome. If there are no further questions, please accept the answer. –  Sep 26 '18 at 09:36
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Consider basic definition of congruence:

If $A= 2\times n +r$

Then:

$A ≡ r\mod n$

Where r is remainder which in you question is $r=-1$

So $A=2\times n -1 ≡ -1\mod n$

sirous
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