# modular arithmetic congruence class simple proof

Hello I have the following question but I'm unsure of how it can be approached by a method of proof. I'm new to modular arithmetic and any information on how to solve this would be great for me.

http://imgur.com/3iHvQ

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Imagine trying everything. Clearly $5\cdot 5$ is no good. All the other choices give (ordinary) product $m$ between $0$ and $20$, so it is clear that $25$ cannot divide $24-m$.

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First note that $24\equiv -1\pmod{25}$ and hence we are trying to show that $ts\not\equiv-1\pmod{25}$. Suppose for contradiction that $ts\equiv -1\pmod{25}$, then multiplying through by $-1$ we get $-ts\equiv 1\pmod{25}$ so $t$ or $s$ is invertible, say $t$ with inverse $-s$. Therefore $\gcd(t,25) = 1$ (why?), and hence $t = 1,2,3,4$. There is a unique value of $s$ corresponding to each of these (why?), each of which should give you a contradiction.
$[t][s]$ refers to the product of the equivalence classes of $[t]$ and $[s]$ taken modulo $25$ –  John Martin Oct 8 '12 at 15:32
Hint $\$ If $\rm\:a\:|\:b\:$ in $\Bbb Z$ then so too in every ring, and the quotient is unique if $\rm\:a\:$ is not a zero divisor. Also, divisors of units are units, so $\rm\: mod\ 25\:$ the divisors of $\:24\equiv -1\:$ are units, so coprime to $5$.