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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



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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

3 4 3
We can cross the small numbers(four 1 and one 2)
the large number =44443333222
0 0 0
thanks kvnmurty for the answer
jithus thanks for trying
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