26's power!

Given the last two digit is 26 and the number is also divisible by 26, we can put it into the form of $26(100x)+26 = 26(100x+1)$ . Since the sum of the remaining digit is 18, we will have the find the number $x$ which produce the sum of digit in $26x$ as 18

As $26x\neq99$ , we have to look for combination of three-digit number of $26x$ that is a multiple of 9 (as the remaining digit has a sum of 18 which means the number is divisible by 9). Given that 26 does not have any multiple of $3$ , it is safe we can try $x = 9j$ when $j = 1, 2, 3, \dots$ . The least $j$ which produce sum of the digit in $234j$ as 18 is $j = 2$ and the digit is $468$

Therefore, the least number possible is $46826$

Since the last two digits are 26, we only need a sum of 18 for the missing digits that is divisible by 26.. the number must be _ 26... just find a multiplier of 26 whose sum of the digits is 18.. 18x26=468 then the number is 46826...