Question

If 10 50-74 is written as an integer in base decimal notation, what is the sum of the digits in that integer?

Answer 1

Answer:

The required sum of the digits in that integer is 440.

Step-by-step explanation:

Consider the provided expression.

$10^{50}-74$

We need to find the sum of the digits in that integer.

For $10^2$ = 100 (3 digits)

Now subtract 74 from it.

100-74=26

For $10^3$ = 1000 (4 digits)

Now subtract 74 from it.

1000-74=926

For $10^4$ = 100000 (5 digits)

Now subtract 74 from it.

10000-74=9926

Similarly,

$10^50$ = 1000....[51 digits]

Now subtract 74 from it.

$10^50-74=99999....26$

The number of 9 after subtracting 74 is 3 less than the number of digits.

Therefore, the number of 9 after subtracting 74 from $10^50$ must be: 51-3=48

The sum of the digits is = 9×48 + 2 + 6 = 432 + 2 + 6 = 440.

Hence, the required sum of the digits in that integer is 440.