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Cant get this concept

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by [email protected] » Sun Oct 13, 2013 8:20 am
The sum of the digits of integer z is 186 and z=(10^n)-4, what is the value of positive integer n

Answer is 21
Last edited by [email protected] on Sun Oct 13, 2013 4:14 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Sun Oct 13, 2013 8:43 am
[email protected] wrote:The sum of digits of integer z is 186 and z -10^n-4 what is the value of positive integer n
This question doesn't make any sense. Namely, the part in green.
Also, please make the exponents clear.
For example, 10^n-4 can be interpreted as EITHER (10^n)-4 OR 10^(n-4)

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by [email protected] » Sun Oct 13, 2013 9:02 am
Sorry for the typo it is the sum of the digits of integer z is 186 and z=(10^n)-4, what is the value of positive integer n

Brent@GMATPrepNow wrote:
[email protected] wrote:The sum of digits of integer z is 186 and z -10^n-4 what is the value of positive integer n
This question doesn't make any sense. Namely, the part in green.
Also, please make the exponents clear.
For example, 10^n-4 can be interpreted as EITHER (10^n)-4 OR 10^(n-4)

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Oct 13, 2013 9:08 am
Please go back and edit the original post so that others see the correct question.

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by Brent@GMATPrepNow » Sun Oct 13, 2013 9:15 am
[email protected] wrote:The sum of the digits of integer z is 186 and z=(10^n)-4, what is the value of positive integer n
Since 10^n is a 1 followed by n zeros, we know that z = 999....996
So, how many 9's are there?

The sum of the digits of integer z is 186
So, 9+9+9+9+9+....+9+9+6 = 186
This means that 9+9+9+9+9+....+9+9 = 180
So, there must be 20 9's in the above sum.
So, z = 9999999....9996 (where there are twenty 9's and one 6)
In other words, z is a 21-digit number.
This means that 10^n must be a 22-digit number.
In other words, 10^n must be 1 followed by 21 zeros.
So, n = 21

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