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

The answer is 20^3 because by the end of 6 hours the number of bacteria has doubled every two hours meaning that after the first two hours you have 2000, then another 2000 after two hours, then 4000 after the next two hours, ans finally at the end of 6 hours, you have 8000 bacteria aka 20^3Step-by-step explanation:...

Let N be the smallest positive integer whose sum of

Ответ:

kylea97

02.08.2021

Mathematics

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 .

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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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