The greatest number that leaves the same remainder when it divides 30,53 and 99 is

Let the required number be $$d$$.

Since $$d$$ leaves the same remainder when dividing $$30,53,$$ and $$99$$, we can write

$$30=dq_1+r$$

$$53=dq_2+r$$

$$99=dq_3+r$$

where $$r$$ is the same remainder in all three cases.

Now, if two numbers leave the same remainder when divided by the same number, their difference will be exactly divisible by that number. 

$$53-30=(dq_2+r)-(dq_1+r)$$

The remainders cancel:

$$53-30=d(q_2-q_1)$$

Therefore, $$53-30$$ is exactly divisible by $$d$$.

Similarly,

$$99-53$$ is also exactly divisible by $$d$$.

$$53-30=23$$

$$99-53=46.$$

Therefore, $$d$$ must be a common divisor of $$23$$ and $$46$$.

The greatest common divisor is

$$\gcd(23,46)=23.$$

Hence, the greatest possible value of $$d$$ is $$ 23$$