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10's complement of a positive decimal integer



 
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Scattershot
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Joined: 05 Feb 2004
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PostPosted: Fri Feb 06, 2004 11:31 am    Post subject: 10's complement of a positive decimal integer
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Hi Chris,
This is an exercise taken from a book that I am currently studying from. I don't understand this question very well, among others, what the d is supposed to represent. It also doesn't provide the answer in the back of the book. Could you please shed some light on this?
And explain how I can calculate the first question, I'll try the second myself.

The 10's complement of a positive decimal integer n is 10 to the power of k minus n, where k is the number of digits in the decimal representation of n. It can be calculated in the following way:

1. All the zeros at the right-hand end of the number remain as zeros in the answer.

2. The rightmost non-zero digit d of the number is replaced by 10 - d in the answer.

3. Each other digit d is replaced by 9 - d.

Find the 10's complements of the following decimal numbers using the rules given above, and check your answers by evaluating 10 to the power of k minus n on a calculator:

(a) 3296
(b) 10350
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chris
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PostPosted: Fri Feb 06, 2004 12:49 pm    Post subject: Re: 10's complement of a positive decimal integer
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Scattershot wrote:
Open quoteHi Chris,
what the d is supposed to represent. Close quote


It tells you in 2. what d is: the rightmost non-zero digit d of the number. In other words, d is the non-zero coefficient of the smallest power of 10 in the decimal representation of the number.

To take (a) as an example:

3296

Rule 1. does not apply, since we don't have any zeroes in that number.

Rule 2: we must find the rightmost non-zero digit. What are the non-zero digits of 3296?

3, 2, 9 and 6 of course!

What are their positions?

3 is the leftmost, 2 is to the right of 3 and 9 to the right of 2 and 6 to the right of 9. What is the "rightmost" digist of those?

6 of course. What does rule 2 say? "The rightmost non-zero digit d of the number is replaced by 10 - d in the answer.". So our d is 6 and we replace 6 with 10-6 = 4.

Now comes rule 3 into play: "Each other digit d is replaced by 9 - d.". What are our "other digits"? What remained of course! That's 3, 2 and 9. Now take each one of them, let d be the digit and replace d by 9-d:

  • For d = 3: we replace the digit 3 with 9 - 3 = 6
  • For d = 2: we replace the digit 2 with 9 - 2 = 7
  • For d = 9: we replace the digit 9 with 9 - 9 = 0


So our 10's complement is 6704. Indeed 3296 + 6704 = 10000 = 10^5.
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dhwani
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PostPosted: Tue Apr 28, 2009 5:04 am    Post subject: Re: 10's complement of a positive decimal integer
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Hi Chris,

Thanks a lot as the way you have explained 10's complement is really simple.
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