Divisible by 3

The digits that work are 2, 5, and 8. So, $\frac{3}{10}$ is the answer.

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Explanation:

A number is divisible by 3 if the sum of its digits is divisible by 3. (proof of this is left to the reader)

Step 1: group sets of existing digits into sets that are divisible by 3 and see what the current remainder when dividing by 3 would be. $(1+2)+(3)+(6)+(7+8)+(9) ... +\hspace{2mm} \color{#D61F06}{ (4+A) \hspace{2mm} extra}$

Step 2: The goal is to make the remainder of this last group, $(4+A)$ , equal to a multiple of 3, up to 0, 3, 6, 9, 12, 15, or 18. What values of $A$ accomplish this?

• $\fbox{2}$ would bring the sum of that last group $(4+A)$ up to 6
• $\fbox{5}$ would bring $(4+A)$ up to 9, and
• $\fbox{8}$ would bring $(4+A)$ up to 12.

Any other choices for F result in numbers that aren't divisible by 3. So, $\frac{3}{10}$ of the digits work.

If $N$ is divisible by 3, then sum of its digit is divisible by 3. In this case:

$1+2+3+4+F+6+7+8+9$

$= 40 + F \equiv 0 \quad(mod \quad3)$

This suggests that $F \in \{2,5,8\}$ as $F$ is a digit, hence opens options not to $\{11,14,...\}$

Since from $0 \rightarrow 9$ there are 10 numbers, therefore;

$P(N\vdots 3) = \boxed{\frac{3}{10}}$

123406789/3 = 41135596.33

123416789/3 = 41138929.67

123426789/3 = 41142263

123436789/3 = 41145596.33

123446789/3 = 41148929.67

123456789/3 = 41152263

123466789/3 = 41155596.33

123476789/3 = 41158929.67

123486789/3 = 41162263

123496789/3 = 41165596.33

If N= {2,5,8}, then number will be divisible by 3. Therefore there are 3 possible numbers out of 10 that the number will be divisible by 3.

$1+2+3+4+6+7+8+9 = 40$ . The only numbers in the range from 40 to 50 that are divisible by 3 are $42$ , $45$ and $48$ , meaning that only $\frac{3}{8}$ of the numbers from 0 to 9 satisfy.

Then $10+F$ divisible by 9 which is $2,5,8$ in the set & that fraction is $\frac{3}{10}$

By symmetry, you can already see 5 works. 9 * 5 = 45 which is divisible by 9. Then you can add and subtract 3 to get 2 and 8. 3/10

Relevant wiki: Divisibility Rules

Nice question. The answer is $\frac{3}{10}$ . Here is why:

For a number to be divisible by 3, the sum of its digits must be divisible by 3.

Using the divisibility rules of 3, we have:

$\frac { 1+2+3+4+F+6+7+8+9 }{ 3 } =\frac { 40+F }{ 3 }$ must be an integer for N to be divisible by 3.

The only solutions of F that leads to $\frac { 40+F }{ 3 }=x,$ where $x$ is an integer are $F=2,5,8$ because $42$ , $45$ and $48$ are divisible by 3.

The set consists of 10 numbers: $(0,1,2,3,4,5,6,7,8,9)$

Hence, the answer is $\boxed{\frac{3}{10}}$

cant it also be all of them coz technically u can divide any number by any number its only that those 3 digits provide an answer that has no remander?