Question

Sum of all three digit numbers (no digit being zero) having the property that all digits are perfect squares, is

• A. 3108
• B. 6216
• C. 13986
• D. none of these

Answer 1

Answer: (C)

The non-zero perfect square digits are 1,4 and 9 . 1 can occur at units place in $3\times 3=9$ ways.
$\therefore$ Sum due to 1 at units place is $1\times 9$. Similarly, sum due to 1 at tens place is $1\times 10\times 9$ and sum due to 1 at hundreds place is $1\times 100\times 9$. We can deal with the digits 4 and 9 in a similar way.
Thus, sum of the desired number is
$(1+4+9)(1+10+100)(9)=13986$