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
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If x=(10^10)-47, what is the sum of all the digit of x?
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Max@Math Revolution
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DavidG@VeritasPrep
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It can be helpful to work with small numbers to establish a pattern.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
(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
