I don't know why I am wrong. According to the problem, since $de$ is a multiple of $3$, eith $d$ or $e$ has to be $6$. But my answer is wrong. The correct answer is $\frac13$.
Digit $d$ is randomly selected from the set $\{4,5,6,7\}$. Without replacement of $d$, another digit $e$ is selected. What is the probability that the two-digit number $de$ is a multiple of $3$? Express your answer as a common fraction.
Best Answer
"Either $d$ or $e$ has to be $6$", this is incorrect.
In order for the number $de$ to be divisible by $3$, the sum of the two digits $(d+e)$ should be a multiple of $3$. Therefore, $(d+e)=(4+5)$ or $(5+7)$
Proof: Suppose you have a three-digit number $abc$
Then$$abc=100a+10b+c=99a+9b+0c+(a+b+c)$$ if $abc$ is divisible by $3$, then $99a+9b+0c+(a+b+c)$ must also be divisible by 3. Now because $99a,9b,0c$ can clearly be divided by $3$ (because of $99$), so $(a+b+c)$ must be a multiple by $3$.
For higher-digit numbers this proof also holds.