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

User Thallius
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5 votes

Answer:

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

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.

User Pzijd
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