How many 2-digit numbers can be formed using only the digits 2, 3,5 and 6

\begin{equation} 1 = 1 \mod 6\\ 10 = 4 \mod 6\\ 100 = 4*10 = 4 \mod 6,\\ \text{and so on for higher powers of 10} \end{equation}

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How many two
How many 3 digit numbers can be formed from the digits 2 3 5 6 and 7 which are divisible by 5 and none of the digits is repeated?

Thus, we find: a number X is divisable by 6 iff, cutting of the last digit, taking the sum of the other digits times 4 and adding the last digit the result is divisable by 6.

You are asked for a 6 digit number using only $2,3,9$. We are thus asked to find $a,b,c,d,e,f \in {2, 3, 9}$ such that $4 * (a + b + c + d + e) + f = 0 \mod 6$. As you noted, the last digit must be $2$, which you can conclude from the equation above quite easily by noticing that f must be even. So we conclude $4 * (a + b + c + d + e) = 4 \mod 6$ thus $a + b + c + d + e = {1,4} \mod 6$ and $a + b + c + d + e = 1 \mod 3$ follows. Since both $6$ and $9$ reduce modulo 6, either 2 or 5 higher digits must be equal to $2$, the rest can be chosen freely.