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$(12\cdot 53 ) \bmod 32$

In the above case, if I multiple 12 by 53 and then take mod then what is the use of modular arithmetic. The other way is:

$(12\cdot 53 ) mod 32 = (12 mod 32 = 12) + (53 mod 32 = 21)$ but the sum is 33 which is not correct.

Somebody please guide me.

Zulfi.

zak100
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  • Do you mean $12 \times 53 \mod 32$ or $12^{53} \mod 32 ? $ If the former , you can simply multiply the terms and find the remainder. – The Demonix _ Hermit Dec 01 '19 at 14:27
  • Writing run on "equations" like "(12* 53 ) mod 32 = 12 mod 32 = 12 + 53 mod 32 = 21" makes no sense and it is very confusing. In this case, I think every one of those equal signs is false. – lulu Dec 01 '19 at 14:30
  • I found the property: (12* 53 ) mod 32] i.e.(12 mod 32) * (53 mod 32)] mod 32 =[12 + 21] mod 32 = 1. – zak100 Dec 01 '19 at 14:40
  • Answer is not correct, somebody please guide me why is the property: (a* b) mod n = [(a mod n) * (b mod n)] mod n : not working? – zak100 Dec 01 '19 at 14:51
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    you multiply the remainders not add. –  Dec 01 '19 at 14:58
  • If you have further questions after reading the linked answers then reply here (or there). – Bill Dubuque Dec 01 '19 at 15:15
  • [(12* 53 ) mod 32 ]mod 32= [(12mod 32) * (53 mod 32)] mod 32 = [12 * 21] mod 32 = 252 mod 32 = 28. @Bill: I have questions with multiplication, I would read your post first. – zak100 Dec 01 '19 at 16:30
  • @zak Yes, that is correct. Below is another way, using the mod Distributive Law to simplify arithmetic $$\begin{align} \bmod 8!:&\ \ \ \ \overbrace{3\cdot 53}^{\large!!!!!! \equiv\ 3\ \cdot\ 5}\equiv 7\ \overset{\large \times\ 4}\Rightarrow,\bmod 32!:&\ \ 12\cdot 53\equiv 4\cdot 7\end{align}\qquad\qquad\qquad$$ – Bill Dubuque Dec 01 '19 at 16:54

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