Hint: Largest number of n digits is 9999…9 (n times) and the smallest number of n-digits = 10….0(n-1 times).Hence find the largest number of six digits and the smallest number of 4 digits. Subtract the two to get the difference.
Complete step-by-step answer:
We know the largest n-digit number is 9999…99 (n times).
Hence the largest six-digit number is 999999.
The smallest n – digits number is 1000….00(n-1 times).
Hence the smallest 4-digit number is 1000.
Hence the difference between the largest 6-digit number and the smallest 4-digit number is 999999-1000 = 998999.
Hence option [b] is correct.
Note: [1] Largest number of n digits is 9999…9 (n times).
Proof: Consider an arbitrary n -digit number $a={{a}_{1}}{{a}_{2}}...{{a}_{n}},a\ne 999\ldots 99$
Since the numbers are not equal, they differ by at least one decimal place.
Let i=k be the smallest value of i, such that ${{a}_{i}}\ne 9$.
So we have $a=999..{{a}_{k}}{{a}_{k+1}}...{{a}_{n}}$.
Since $0\le {{a}_{k}}\le 9$we have ${{a}_{k}}<9$
So, we have a<999…99
Hence 999..99 is the largest n-digit number
[2] Smallest number of n-digits is 1000…0 (n-1 times).
Proof: Consider an arbitrary n -digit number $a={{a}_{1}}{{a}_{2}}...{{a}_{n}},a\ne 100...00$
Since a is an n-digit number we have ${{a}_{1}}\ge 1$
If ${{a}_{1}}>1$, then clearly a>100…00 and we are done.
If ${{a}_{1}}=1$, then since a is not equal to 100…00, they differ by at least one decimal place.
Let i=k be the smallest value of i such that $i>1,{{a}_{i}}\ne 0$.
So, we have $a=100...{{a}_{k}}{{a}_{k+1}}\ldots {{a}_{n}}$
Since $0\le {{a}_{k}}\le 9$we have ${{a}_{k}}>0$
So a>100…00.
Hence proved.
