Mathematics, 03.07.2021 15:50 galarzachristopher

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

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Answer from: bryan1528489

Step-by-step explanation:

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

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

$N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}$ , where 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 and . 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 other than the first digit must be "".

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 is not "".

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 contains at least digits.) Assume that $\text{B} \!$ is one of the non-leading non-"" digits.

Either of the following must be true:

$\text{P}$ , the digit in front of $\text{B}$ is a "

", or $\text{P}$ , the digit in front of $\text{B}$ is not a "

If $\text{P}$ , the digit in front of $\text{B}$ , is a "", then let $N^{\prime}$ be 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 "", then $(\text{P} - 1)$ would still be a digit.

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

let $N^{\prime}$ be 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 other than the first digit must be "".

Therefore, 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 .