If x = 9ab where a is an integer consists of a sequence of 2014 eights and the integer b consists of a sequence of 2014 fives. What is the sum of the digits of x?

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Consider a number x=9ab in which a is an integer consisting of a sequence of 2014 eights andthe integer b consists of a sequence of 2014 fives.Take three of each integer, we get the following number.   x=9 888 555  x=4435560The sum of the digits is given by,    42+3+52+6+0Take five of each integer, we get the following number.   x=9 88888 55555  x=44443555560The sum of the digits is given by,    44+3+54+6+0Take seven of each integer, we get the following number.   x=9 8888888 5555555  x=444444355555560The sum of the digits is given by,    46+3+56+6+0Note that the sum of the digits in the number follows the followingpattern:  4n-1+3+5n-1+6+0Where n is the number of times 8 and 5 are used in the number.This measn that if we are using 2012 eights and fives, then the sum of the resulting number is,   42014-1+3+52014-1+6+0=42013+3+52013+6+0                                                                =8052+3+10065+6+0                                                                =18126So the required sum of digits is 18126.

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