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Yes D is correct answer. If you subtract 1 from 10^20 it will be 999(20 times) if you subtract 15 more from that, you will get last digit 4 second last 8 and all others 9. So answer is D.

Yes D is correct answer. If you subtract 1 from 10^20 it will be 999(20 times) if you subtract 15 more from that, you will get last digit 4 second last 8 and all others 9. So answer is D.

I did it the same way. Subtracting 1 first seemed to help me visualize it.

Well, \(10^2-16=84\) and \(10^3-16 = 984\). So, we see the exponent tells us the number of digits involved in our answer (2 in the former case, 3 in the latter).

\(10^2^0-16=999...984\); two numbers are the 8 and the 4, whereas the other eighteen are 9s!

\(10^2^0\) would be a 21 digit number ... subtract 16 and you are left with a 20 digit number with 2 digits as 4 and 8 and the rest 9s. The total is \(8+4+18*9 = 174\) therefore (D) _________________

What is the sum of the digits in decimal notation of number \(10^{20}-16\)?

A. 158 B. 162 C. 165 D. 174 E. 183

\(10^{20}\) has 21 digits: 1 followed by 20 zeros; \(10^{20}-16\) has 20 digits: 18 9's and 84 in the end;

So, the sum of the digits of \(10^{20}-16\) equals to 18*9+8+4=174.

Answer: D.

D can't be the answer. once we subtract 16 from 10^20 we have only 19 numbers left. among 19 numbers 17 are nine and last two are 8 and 4. therefor sum is 17*9+8+4=165 _________________

You have to have the darkness for the dawn to come.

OA is wrong C is the right ans because on subtracting 16 from 10^20, we have only 19 nos left. 17 are 9 and last two are 8 and 4. so sum= 17*9+8+4=165 _________________

You have to have the darkness for the dawn to come.

What is the sum of the digits in decimal notation of number \(10^{20}-16\)?

A. 158 B. 162 C. 165 D. 174 E. 183

\(10^{20}\) has 21 digits: 1 followed by 20 zeros; \(10^{20}-16\) has 20 digits: 18 9's and 84 in the end;

So, the sum of the digits of \(10^{20}-16\) equals to 18*9+8+4=174.

Answer: D.

D can't be the answer. once we subtract 16 from 10^20 we have only 19 numbers left. among 19 numbers 17 are nine and last two are 8 and 4. therefor sum is 17*9+8+4=165

Nope.

\(10^{20}\) has 21 digits: 1 followed by 20 zeros: 100,000,000,000,000,000,000

\(10^{20}-16\) has 20 digits: 18 9's and 84 in the end: 99,999,999,999,999,999,984. _________________