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

Answer:

$10$.

Step-by-step 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 $9$.