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Sagot :
Answer:
[tex]10[/tex].
Step-by-step explanation:
See below for a proof of why all but the first digit of this [tex]N[/tex] must be "[tex]9[/tex]".
Taking that lemma as a fact, assume that there are [tex]x[/tex] digits in [tex]N[/tex] after the first digit, [tex]\text{A}[/tex]:
[tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}[/tex], where [tex]x[/tex] is a positive integer.
Sum of these digits:
[tex]\text{A} + 9\, x= 2021[/tex].
Since [tex]\text{A}[/tex] is a digit, it must be an integer between [tex]0[/tex] and [tex]9[/tex]. The only possible value that would ensure [tex]\text{A} + 9\, x= 2021[/tex] is [tex]\text{A} = 5[/tex] and [tex]x = 224[/tex].
Therefore:
[tex]N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}[/tex].
[tex]N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}[/tex].
[tex]N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}[/tex].
Hence, the sum of the digits of [tex](N + 2021)[/tex] would be [tex]6 + 2 + 2 = 10[/tex].
Lemma: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".
Proof:
The question assumes that [tex]N\![/tex] is the smallest positive integer whose sum of digits is [tex]2021[/tex]. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of [tex]N[/tex] is not "[tex]9[/tex]".
For example: [tex]N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}[/tex], where [tex]\text{A}[/tex], [tex]\text{P}[/tex], [tex]\text{B}[/tex], and [tex]\text{C}[/tex] are digits. (It is easy to show that [tex]N[/tex] contains at least [tex]5[/tex] digits.) Assume that [tex]\text{B} \![/tex] is one of the non-leading non-"[tex]9[/tex]" digits.
Either of the following must be true:
- [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex] is a "[tex]0[/tex]", or
- [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex] is not a "[tex]0[/tex]".
If [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is a "[tex]0[/tex]", then let [tex]N^{\prime}[/tex] be [tex]N[/tex] with that "[tex]0\![/tex]" digit deleted: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}[/tex].
The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:
[tex]\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}[/tex].
However, with one fewer digit, [tex]N^{\prime} < N[/tex]. This observation would contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].
On the other hand, if [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is not "[tex]0[/tex]", then [tex](\text{P} - 1)[/tex] would still be a digit.
Since [tex]\text{B}[/tex] is not the digit [tex]9[/tex], [tex](\text{B} + 1)[/tex] would also be a digit.
let [tex]N^{\prime}[/tex] be [tex]N[/tex] with digit [tex]\text{P}[/tex] replaced with [tex](\text{P} - 1)[/tex], and [tex]\text{B}[/tex] replaced with [tex](\text{B} + 1)[/tex]: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}[/tex].
The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:
[tex]\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}[/tex].
However, with a smaller digit in place of [tex]\text{P}[/tex], [tex]N^{\prime} < N[/tex]. This observation would also contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].
Either way, there would be a contradiction. Hence, the claim is verified: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".
Therefore, [tex]N[/tex] would be in the form: [tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}[/tex], where [tex]\text{A}[/tex], the leading digit, could also be [tex]9[/tex].
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