Question

Total number less than $3\times {10}^8$ and can be formed using the digits $1,2,3$ is equal to $\frac{a}{b}\left(\right[a+2b]\cdot a^7-1)$, then

• A. $a=2$
• B. $b=2$
• C. $a=3$
• D. $b=3$

Answer 1

Answer: (B), (C)

$a=3$

​Formed number can be utmost of nine digits

number of one digit numbers which are less than $3\times {10}^8$ are $3$

number of two digit numbers which are less than $3\times {10}^8$ are $3\times 3=3^2$

number of three digit numbers which are less than $3\times {10}^8$ are $3\times 3\times 3=3^3$
$⋮⋮⋮$
number of eight digit numbers which are less than $3\times {10}^8$ are $3^8$

For, nine digit numbers to be less than $3\times {10}^8$ leftmost digit should be $1,2$ only two options

hence, $2\times 3^8$ nine digit numbers are less than $3\times {10}^8$

Total number of such numbers is

=$3+3^2+3^3+\dots \dots +3^8+2\times 3^8$
$=\frac{3(3^8-1)}{3-1}+2\times 3^8=\frac{3^9-3+4\times 3^8}{2}=\frac{7\times 3^8-3}{2}=\frac{3}{2}\left(\right[2\cdot 2+3]\cdot 3^7-1)$
On comparing we get
$a=3,b=2$