Plato's Theorem

Whizz throgh them mentally discounting anything not divisible by 9, ie those numbers whose digits do not sum to one of 9, 18, 27 etc.
Then discount 4500 because it is not a multiple of 7, because 45 is not. Then discount 6750 because it is not a mutliple of 4 (last 2 digits are not a multiple of 4). 2520 remains.

This question is easy, but not straightforward. Here's why:

What number is divisible by 6 and 3?

$6 \times 3$ = $\boxed {18}$ . Therefore the answer is 18, right?

Yes, but 6 is actually $2 \times 3$ , and because 3 is already in the multiplication, it can be ignored.

Therefore, the smallest number is $\boxed{6}$ .

The same principle can be applied to get a solution.

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To obtain the Lowest Common Multiple of 1, 2, 3, 4, 5, 6, 7, 8, and 9, you must factorise the composite digits.

4 = $2 \times 2$

6 = $2 \times 3$

8 = $2 \times 2 \times 2$

9 = $3 \times 3$

Therefore, substituting each value with its relative absolute minimum factors, the equation is simplified from:

$1 \times 2 \times 3 \times (2 \times 2) \times 5 \times (2 \times 3) \times 7 \times (2 \times 2 \times 2) \times (3 \times 3)$

into

$1 \times 2 \times 3 \times 2 \times 5 \times 7 \times 2 \times 3$ .

Therefore, the actual, hidden question is to calculate the product of $1 \times 2 \times 3 \times 2 \times 5 \times 7 \times 2 \times 3$ .

$1 \times 2 \times 3 \times 2 \times 5 \times 7 \times 2 \times 3$ = 2520.

Therefore, the answer is $\boxed {2520}$ .

According to this question, there is an explanation.

Method : Check if all the options are divisible by numbers 1 to 10.

6750 : not divisible by 4

5090 : not divisible by 8

10000 : not divisible by 3

9240 : not divisible by 9

1030 : not divisible by 4

5050 : not divisible by 3

4500 : not divisible by 6

Therefore 2520 is divisible by ALL the numbers from 1 to 10.