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Missing 3, 6 and 9's

by Hippasos
Tags: power of 3 6 and 9
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Feb8-13, 03:35 PM
P: 72









Missing decimal numbers 3 - why are these"missing"? Not needed even during calculations - at all?

Can You please explain?
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Feb8-13, 03:38 PM
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phinds's Avatar
P: 6,089
what's to explain? it's just the result of particular calculations.

Hey, try 1/3 = .333333 you don't need 1,2,4,5,6,7,8, or 9 Explain that !
Feb9-13, 01:58 AM
P: 72
Quote Quote by phinds View Post
what's to explain? it's just the result of particular calculations.

Hey, try 1/3 = .333333 you don't need 1,2,4,5,6,7,8, or 9 Explain that !
Yes of course my stupid mistake. I was *runk and it was late. Still interesting ...142857.. cycles there.

Feb11-13, 12:38 PM
Integral's Avatar
P: 7,315
Missing 3, 6 and 9's

Hey! don't dismiss your observation so quickly. It is very interesting and if you look at the long division process you can see just why there are no 3 or multiples of, in the decimal expansion.

First of all observe that there are at most 6 digits possible. There is a limited set of possible remainders in the long division process{1,2,3,4,5,6}. A zero would terminate the process, 7 and greater means that you have not made a correct choice of a multiplier. This implies that there can be at most 6 digits in the expansion. 7 is unique in that it is the only integer with the max possible number of unique digits in the decimal expansion.

Now lets go through the long division process and see if we can see why 3 does not appear.

The first step:

10/7 = 1 r 3 : so 1 is the first digit the remainder sets up the second step, r*10
30/7 = 4 t 2 : so 4 is the 2nd digit.
20/7= 2 r 6 : so 2 is the 3rd digit.

now we can see why 3 does not appear, each of the steps involves a multiple of 10 divided by 7. But the only way to get a 3 would be if you could have a non zero as the last digit, this cannot happen. Since 7*3=21 and 7*4=28.

I also find this an interesting little tidbit. Hope I helped throw some light on it for you.

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