Question

If 5 digits from the number 1234123412341234 are crossed out so that the remaining number is as large as possible, then sum of the digits of the remaining number is

Answer 1

123 412 341  234 123 4

we delete the first three digits. 
412 341 234 123 4
now delete the 2nd and 3rd digits.

434 123 412 34      - this is the largest number remaining after deleting 5 digits.

sum of digits: 31

Answer 2

we can cross the small numbers(four 1 and one 2)
the large number =44443333222
sum=16+12+6=34