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Modulo $12$ multiplication table and I got $3$, but I don't know if it's true or not. I don't know whether it's correct place. Please reopen it.

Mohsen Shahriari
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Damilare Sheu
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    It is true that $(3 \cdot 5)(5 \cdot 9) \equiv 3 \pmod{12}$. Did you pose your question to check your answer, or do you have a more general question? – N. F. Taussig Aug 08 '20 at 13:50
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    Welcome to MSE! Firstly, you are asking a Maths related question at Maths' stackexchange. So, you are definitely in the right place. Secondly, although there will be many people who will be able to explain it you from scratch, it helps to lay out the steps, so we can help you locate the error step and help specifically. – Jaspreet Aug 08 '20 at 13:53

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$(3\cdot5)(5\cdot9)=675$. Now, since we are taking modulo $12$, we will take $675-12k$, where $k$ is some positive integer, until we get a "least positive residue" (i.e. the smallest positive number where if we subtract $12$ one more time, then we would get a negative number). So, plug in different values of $k$ and see what you get...

User7238
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The straightforward solution: calculate $3\cdot 5\cdot 5\cdot 9=675$, divide $675$ by $12$ (e.g. long division), and you get the quotient $56$ and remainder $3$. So $3\cdot 5\cdot 5\cdot 9\equiv 3\pmod{12}$.

However, you may be expected to know that you are allowed to reduce (modulo any whole number) at each multiplication step. For example: $3\cdot 5=15\equiv 3\pmod{12}$. Also, $5\cdot 9=45\equiv 9\pmod{12}$. Therefore:

$$(3\cdot 5)\cdot(5\cdot 9)\equiv 3\cdot 9=27\equiv 3\pmod{12}$$

This technique is useful to avoid the numbers growing too much as you are calculating the remainders - in each step the result is reduced to something smaller than the modulus (in this case - smaller than $12$). You may be given a problem with a bigger expression and no calculator allowed - in which case you would be compelled to use this idea to shortcut the calculations.