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3 years ago

when a number is subtracted from 6, the result is two times five less than the number. find the number?

Mathematics

2 answers:

kolbaska11 [484]3 years ago

6 0

Answer:

16/3

Step-by-step explanation:

Let that number be n

6 - n = 2(n - 5)

6 - n = 2n - 10

16 = 3n

n = 16/3

Kryger [21]3 years ago

5 0

Answer:

let number=x

According to given condition

6-x=x-2(5)

6-x=x-10

6+10=x+x

16=2x

16/2=x

8=x

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the hypotenuse of a 45-45-90 triangle has a length of 10 units. what is the length of one of its legs?

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When a triangle is a 45-45-90 triangle, the following rules apply:

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if the hypotenuse is 10, then 10 divided by the square root of 2 is how you find the leg.
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3 years ago

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4 years ago

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

kondor19780726 [428]

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 $9$ .