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

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.

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

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 $$5$$?

• So the remainders are 3, 4, 2 and 1 – Jamminermit Nov 10 '19 at 9:24
• 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)? – Jamminermit Nov 10 '19 at 9:31
• @Jamminermit that is perfectly correct. – YiFan Nov 10 '19 at 9:45
• which is the same as saying $12-2$ is divisible by 5. – Roddy MacPhee Nov 10 '19 at 11:57
• @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. – Bill Dubuque Nov 10 '19 at 23:35

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?

• 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. – Bill Dubuque Nov 11 '19 at 2:23
• @BillDubuque Thanks. This is what I should have said. – Allawonder Nov 11 '19 at 7:20