Can someone me ! let n be the least postive integer whose digits sum to 2018. compute the sum of the digits of n+1.

3

Explanation:

Since we want the least number of integers, divide by the largest integer (9).

2018 ÷ 9 = 224 remainder 2

So, N = 2999...999 (there are 224 of the 9's)

Thus, N + 1 = 2999...999 + 1 (there are 224 of the 9's)

                  = 3000...000     (there are 224 of the 0's)

The sum of the digits is: 3 + 0 + ... + 0  (there are 224 of the 0's)

                                      = 3

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get