Four-Digit Dilemma

Let the number be $x$

If $x$ is divisible by $5$ then it's unit digit is either $0$ or $5$ . Since $x$ is divisible by $2$ , the unit digit can only be $0$ .

Now largest 4 digit number, ending with 0, is $9990$ . Let's now check the divisibility rule for $3$ .

$9 + 9 + 9 + 0 = 27 \implies 2 + 7 = 9 \implies 3|9990$

Therefore, the largest 4 digit number is $9990$

2x3x5=30

number is _ _ _ 0

999 is biggest 3 digit number that is divisible by 3

therefore, the answer is 9990

The number must be divisible by $2 \times 5 = 10$ so must end with a $0.$

The number must be divisible by $3$ so the sum of the digits must be divisible by $3;$ $9 + 9 + 9 = 27$ is the largest possible sum out of the three remaining digits (which happens to also be divisible by 3).

Hence the number must be $9990.$