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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 » Tue Aug 30, 2016 7:22 pm
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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by dustystormy » Wed Aug 31, 2016 5:43 am
IMO E
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by Brent@GMATPrepNow » Wed Aug 31, 2016 6:34 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
First recognize that 10^10 equals 1 followed by 10 zeros.
That is, 10^10 = 10,000,000,000
In other words, 10^10 is an 11-digit number.

So, 10^10 - 47 will be a 10-digit number with a lot of 9's with 53 at the end (since 100 - 47 = 53)
If 10^10 - 47 has 10 digits, and the last 2 digits are 53, then 10^10 - 47 must equal eight 9's followed by 53
(8)(9) + 5 + 3 = 80

Alternatively, 10^10 - 47 = 10,000,000,000 - 47
= 9,999,999,953
The sum of the digits is 80

Answer: E

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Brent
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by Max@Math Revolution » Wed Aug 31, 2016 8:08 pm
From (10^10)-47=9,999,999,953, the sum of all digit is 9(8)+6+2=80. The answer is E.
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by jawwadali » Mon Oct 17, 2016 11:16 am
I didn't understand the answer. Can you please simplify it.

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by GMATGuruNY » Mon Oct 17, 2016 11:21 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.
Test easy cases and look for a pattern.

10³ - 47 = 953.
10� - 47 = 9953.
10� - 47 = 99953.

In every case:
The number of 9's is equal to TWO LESS THAN THE EXPONENT.
The last two digits are 5 and 3.
Thus, 10¹� - 47 will be composed of eight 9's, one 5 and one 3, yielding the following sum:
(8*9) + 5 + 3 = 80.

The correct answer is E.
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by Brent@GMATPrepNow » Wed Jan 10, 2018 8:16 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
First recognize that 10^10 equals 1 followed by 10 zeros.
That is, 10^10 = 10,000,000,000
In other words, 10^10 is an 11-digit number.

So, 10^10 - 47 will be a 10-digit number with a lot of 9's with 53 at the end (since 100 - 47 = 53)
If 10^10 - 47 has 10 digits, and the last 2 digits are 53, then 10^10 - 47 must equal eight 9's followed by 53
(8)(9) + 5 + 3 = 80

Alternatively, 10^10 - 47 = 10,000,000,000 - 47
= 9,999,999,953
The sum of the digits is 80

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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