Question

6. If the sum of the digits of a number
10 - 1, where 'n' is a natural number, is equal to 3798,
then what is the value of n?

let n=1
(10^1)-1=3798
10-1=3798
n=3798/9
n=422

Note*=rest other answers are wrong...this is a right answer ...
(eg...379.9,379,423...all are wrong..
422 is a right answer)
​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The value of n is 422.

Step-by-step explanation:

It is given that the sum of digits of a number 10ⁿ-1 is 3798.

We have to find the value of n.

Let n = 1:

10¹ - 1 = 10 - 1

10¹ - 1 = 9

sum of digits = 9 = 9 × 1

Let n = 2:

10² - 1 = 100 - 1

10² - 1 = 99

sum of digits = 9 + 9 = 18

sum of digits = 9 × 2

Let n = 3:

10³ - 1 = 1000 - 1

10³ - 1 = 999

sum of digits = 9 + 9 + 9 = 27

sum of digits = 9 × 3

That is, the sum of digits is given as 9n. Therefore,

3798 = 9n

n = 3798/9

n = 422

Therefore, the value of n is 422.

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