Question

Let N be the smallest positive integer whose sum of its digits is 2021. What is the sum of the digits of N + 2021?

Answer 1

`10`.

Explanation:

See below for a proof of why all but the first digit of this
`N` must be "
`9`".

Taking that lemma as a fact, assume that there are
`x` digits in
`N` after the first digit,
`\text{A}`:

`N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}`, where
`x` is a positive integer.

Sum of these digits:

`\text{A} + 9\, x= 2021`.

Since
`\text{A}` is a digit, it must be an integer between
`0` and
`9`. The only possible value that would ensure
`\text{A} + 9\, x= 2021` is
`\text{A} = 5` and
`x = 224`.

Therefore:

`N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}`.

`N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}`.

`N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}`.

Hence, the sum of the digits of
`(N + 2021)` would be
`6 + 2 + 2 = 10`.

Lemma: all digits of this
`N` other than the first digit must be "
`9`".

Proof:

The question assumes that
`N\!` is the smallest positive integer whose sum of digits is
`2021`. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of
`N` is not "
`9`".

For example:
`N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}`, where
`\text{A}`,
`\text{P}`,
`\text{B}`, and
`\text{C}` are digits. (It is easy to show that
`N` contains at least
`5` digits.) Assume that
`\text{B} \!` is one of the non-leading non-"
`9`" digits.

Either of the following must be true:

• 
  `\text{P}`, the digit in front of
  `\text{B}` is a "
  `0`", or
• 
  `\text{P}`, the digit in front of
  `\text{B}` is not a "
  `0`".

If
`\text{P}`, the digit in front of
`\text{B}`, is a "
`0`", then let
`N^(\prime)` be
`N` with that "
`0\!`" digit deleted:
`N^(\prime) :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}`.

The digits of
`N^(\prime)` would still add up to
`2021`:

`\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}`.

However, with one fewer digit,
`N^(\prime) < N`. This observation would contradict the assumption that
`N\!` is the smallest positive integer whose digits add up to
`2021\!`.

On the other hand, if
`\text{P}`, the digit in front of
`\text{B}`, is not "
`0`", then
`(\text{P} - 1)` would still be a digit.

Since
`\text{B}` is not the digit
`9`,
`(\text{B} + 1)` would also be a digit.

let
`N^(\prime)` be
`N` with digit
`\text{P}` replaced with
`(\text{P} - 1)`, and
`\text{B}` replaced with
`(\text{B} + 1)`:
`N^(\prime) :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}`.

The digits of
`N^(\prime)` would still add up to
`2021`:

`\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}`.

However, with a smaller digit in place of
`\text{P}`,
`N^(\prime) < N`. This observation would also contradict the assumption that
`N\!` is the smallest positive integer whose digits add up to
`2021\!`.

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this
`N` other than the first digit must be "
`9`".

Therefore,
`N` would be in the form:
`N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}`, where
`\text{A}`, the leading digit, could also be
`9`.