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Q.
Sum of all three digit numbers (no digit
being zero) having the property that all digits are perfect
squares, is
Permutations and Combinations
Solution:
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
$(1+4+9)(1+10+100)(9)=13986$