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Show that 9453(6824)$\equiv$6782(5675341)$\equiv$2 (mod 5)

I am very new to modular arithmetic and I am not entirely sure what this question is asking me to do, or how you would go about showing what it is looking for. I apologise if this is very simple, but I am looking for some clarification on what this actually means. Thanks.

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  • $\begingroup$ Do you know that $a \equiv b$ (mod $p$) and $c \equiv d$ (mod $p$) implies $ac \equiv bd$ (mod $p$)? $\endgroup$ – Jerry Nov 10 '19 at 9:07
  • $\begingroup$ Yeah, I think I have seen this before, but how do you apply it to this problem? $\endgroup$ – Jamminermit Nov 10 '19 at 9:12
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Hint: $a\equiv r$ (mod $p$), where $r$ is the remainder when $a$ is divided by $p$.

What are the remainders when $9453, 6824, 6782$ and $5675341$ are divided by $5$?

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  • $\begingroup$ So the remainders are 3, 4, 2 and 1 $\endgroup$ – Jamminermit Nov 10 '19 at 9:24
  • $\begingroup$ So can I say that 9453$\equiv$3 (mod 5) and 6824$\equiv$4 (mod 5) and 6782 $\equiv$2 (mod 5) and 5675341 $\equiv$1 (mod 5)? And using the rule above, then 3(4)$\equiv$2(1) (mod 5) so 12$\equiv$2 (mod 5)? $\endgroup$ – Jamminermit Nov 10 '19 at 9:31
  • $\begingroup$ @Jamminermit that is perfectly correct. $\endgroup$ – YiFan Nov 10 '19 at 9:45
  • $\begingroup$ which is the same as saying $12-2$ is divisible by 5. $\endgroup$ – Roddy MacPhee Nov 10 '19 at 11:57
  • $\begingroup$ @Jamminermit Another way is to compute the decimal units digits then reduce that $\bmod 5,\,$ i.e. $\, n\bmod 5 = (n\bmod 10)\bmod 5,\,$ a special case of the method of simpler multiples. $\endgroup$ – Bill Dubuque Nov 10 '19 at 23:35
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Hint. It is relatively easy to reduce numbers in decimal form modulo $5.$ Just expand out and note that any positive power of $10$ vanishes modulo $5.$ Thus, the residues are just the residues of the units digits, which are easy to do. Can you now proceed?

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  • $\begingroup$ The residue $\bmod 5\,$ is the units digits reduced $\bmod 5,\,$ i.e. $\, n\bmod 5 = (n\bmod 10)\bmod 5,\,$ see my comment in Jerry's answer. $\endgroup$ – Bill Dubuque Nov 11 '19 at 2:23
  • $\begingroup$ @BillDubuque Thanks. This is what I should have said. $\endgroup$ – Allawonder Nov 11 '19 at 7:20

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