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If x=(10^10)-47, what is the sum of all the digit of x?

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by Max@Math Revolution » Sat Aug 27, 2016 4:31 am
If x=(10^10)-47, what is the sum of all the digit of x?
A. 40
B. 45
C. 50
D. 55
E. 80

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by prabsahi » Tue Sep 06, 2016 12:31 am
Option E.

100-47=53
We are left with 8 times 9's
sum of digits=9*8(times)+5+3=80
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by DavidG@VeritasPrep » Tue Sep 06, 2016 7:42 am
Max@Math Revolution wrote:If x=(10^10)-47, what is the sum of all the digit of x?
A. 40
B. 45
C. 50
D. 55
E. 80

*An answer will be posted in 2 days
It can be helpful to work with small numbers to establish a pattern.

(10^2) - 47 = 100-47 = 53. (Sum = 5+3)

(10^3) - 47 = 1000 - 47 = 953. (Sum = 9 + 5 + 3)

(10^4) - 47 = 10,000 - 47 = 9953. (Sum = 9 + 9 + 5 + 3 = 2*9 + 5 + 3)

Now we see a pattern. The number of 9's in the sum will be two less than the exponent value. And we always have the 5 + 3.

(10^10) - 47 --> Sum = 8*9 + 5 + 3 = 80, or E
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