Question

The sum of all $3$ digit numbers that can be formed from the digits $1$ to $9$ and when the middle digit is a perfect square is (repetitions are allowed).

• A. $1,34,055$
• B. $2,70,540$
• C. $1,70,055$
• D. $2,34,520$

Answer 1

Answer: (A)

$1,34,055$

The units place can take any of the $9$ digits
The hundreds place can take any of the $9$ digits
The tens place can take $3$ digits ($1$, $4$ and $9$)
Total number of these numbers is $243$
When we fix the units digit the total number of possible permutations of the tens and hundreds digits is $27$
Therefore each units digit occurs $27$ times
Sum of all units digits is therefore $27\times (1+...9)$
Similarly, sum of all tens digits is therefore $81\times (1+4+9)$
And sum of all hundreds digits is therefore $27\times (1+2+....+9)$
The sum of all the numbers is thus
$27\times (0+1+...9)+10\times (81\times (1+4+9\left)\right)+100\times (27\times (1+2+....+9\left)\right)=134055$
Hence, the answer is A