When a number is divided by 5, the remainder is 2, when divided by 7, the remainder is 3, when divided by 9, the remainder is 4. The sum of digits of such smallest number is

We are given that when a number is divided by 5, the remainder is 2, when divided by 7, the remainder is 3, and when divided by 9, the remainder is 4. 

Let the number be $$n$$

So, as per the question:

$$n=5a+2$$

$$n=7b+3$$

$$n=9c+4$$

Where $$a,b$$ and $$c$$ are whole numbers.

Now, we have

$$n=5a+2=7b+3$$

$$5a=7b+1$$

This linear equation has a solution when $$a=3 \text{ and } b=2$$

So the next solutions will appear when we increment $$a$$ and $$b$$ by $$7$$ and $$5$$, respectively.

Therefore, the general expression of $$a=3+(i-1)7$$

Let's substitute the value $$a$$ and find the number $$n$$.

$$n= 5a+2 = 5[3+(i-1)7]+2=35i-18$$

$$ $$

Similarly, we have

$$n=5a+2=9c+4$$

$$5a=9c+2$$

This linear equation has a solution when $$a=4 \text{ and } c=2$$

So the next solutions will appear when we increment $$a$$ and $$c$$ by $$9$$ and $$5$$, respectively.

Therefore, the general expression of $$a=4+(k-1)9$$

Let's substitute the value $$a$$ and find the number $$n$$.

$$n= 5a+2 = 5[4+(k-1)9]+2=45k-23$$

$$ $$

So we have: $$n=35i-18=45k-23$$

$$35i=45k-5$$

$$7i=9k-1$$

This linear equation has a smallest solution when $$i=5 \text{ and } k=4$$

Therefore,

$$n=35i-18=35*5-18$$

$$n=175-18=157$$

Hence, the sum of the digits of the smallest number that follows all the given conditions is$$=1+5+7=13$$

Hence, Option A is correct.