Digit sums problem

Number Theory Level 3

The digit-sum of a number is the sum of its digits till a single digit is left.

For example: 39 = 3+9 = 12 = 1+2 =3. So 3 is the digit-sum of 39.

If the digit-sums of all the integers starting from 1 are added, at which integer would the sum of the digit-sums cross 2017?

405 406 403 404

2 solutions

Fletcher Mattox
Sep 24, 2020

sumsum = 0
n = 0
while sumsum < 2017:
    n += 1
    dsum = sum(list(map(int, str(n))))
    while dsum > 9:
        dsum = sum(list(map(int, str(dsum))))
    sumsum += dsum
print(n)

Chris Lewis
Oct 31, 2019

The digit sum of $n$ is the same as its remainder when divided by $9$ , unless $n$ is a multiple of $9$ , in which case the digit sum is $9$ .

So the sum of the digit sums of every nine consecutive numbers is $45$ .

$45\times 45=2025$ ; so the sum of the first $45$ sets of $9$ consecutive numbers (ie the sum of digit sums from $1$ to $45\times 9=405$ is $2025$ . The sum up to $404$ is therefore $2016$ , so we cross $2017$ at $n=405$ .

Digit sums problem

2 solutions

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