Your md is simpler fr this sum.. But this formula enhances our capability ot solvesuch sums even with more no of digits… and generic formulae always help .. thanks atul for the formula and explanation.. very helpful

Great Explanation !!! Thanks!!

@Charanya / manisha,
Nice formula .. there is a slight change is the formula to make it generic:

suppose n is the no. of digits given. Our scenario it is 3 (digits given 3,4,5)

Formula will be:

(n times 1) * {fact (n-1)} * (sum of digits)

explaination:
First term : ‘n times 1′

for n = 3 this means ’3 times 1′ i.e. 111
for n = 4 this will be ’4 times 1′ i.e. 1111

Second term : fact(n-1) ……..(factorial of (n-1)

for n = 3 this means fact(3-1) = fact (2) = 2
for n = 4 it will be !(4-1) = !3 = 6

Third term : Sum of the digits

in our case = 3+4+5 = 12

so the solution of our problem with this formula will be as Charanya explained:

= (3 times 1) (fact(3-1) (3+4+5)
= (111)(!2)(12) = 111*2*12
= 2664

This formula can be used in the scenarios were ’0′ is also present in the given digits. Hope it will help.

(111..n times)(n – 1) (Sum of digits)

The formula varies when one of the digits is zero.

I’ll post it ASAP.

Can this method be used for any kind of problem similar to give here?
Nice concept..do you have some more tricks/ logical solutions…i would appreciate if you would share it!

What is the basic concept behind this formula? whr all it can be actually used?

(111)(n-1)(sum of the digits)
= (111)(2)(12)
= 2664