Nine Digit Numbers

  • Thread starter Ian Rumsey
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  • #1
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The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. are placed in a bag and then individually removed, in no particular order, and placed in line to form a nine digit number.
What is the probability that this random nine digit number is divisible by 9.
 

Answers and Replies

  • #2
403
1
100% .
The digits sum 45, which is divisible by 9.
 
  • #3
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Well Done, Will have to think of something harder.
 
  • #4
Wrong. If you line up 1 through 9 in a random order, they form a number like 142385976 or 5829345167 etc. For you to say that 9 fits into every one of these numbers is ludacrous.

Unless you specify in the question that the numbers are to be added together which you clearly did not.
 
  • #5
chroot
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Tau,

No, you're wrong. Please consider that you don't know everything yet before telling other people they are wrong.

http://www.mathforum.org/k12/mathtips/division.tips.html

It can be proven that every number whose digits sum to a number divisible by nine is itself divisible by nine -- rather simply, in fact.

Since any combination of all the digits 1-9 has a sum of 45, any combination is divisible by nine. If you would like to find a counterexample, a nine-digit number with all the digits 1-9 occuring once each that is not divisible by nine, be my guest.

It's not ludicrous, it's number theory.

- Warren
 
  • #6
Gokul43201
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It's not at all ludicrous.

For a number to be divisible by 9, the sum of its digits must be divisible by 9. No matter what order you place the numbers in, the sum will always be 45, as Rog pointed out.

If you don't believe me, check to see that 142385976 = 9* 15820664 and 582934167 = 9* 64770463 - I changed your number by removing the repeated '5'.


Edit : Chroot got in before me, so ignore this.
 
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  • #7
Sorry, I am both a moron and a jerk. lol. Not a healthy combo.
 
  • #8
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Tau_Muon_PlanetEater said:
Sorry, I am both a moron and a jerk. lol. Not a healthy combo.
No. You made a mistake, forgot to check it, and agrued a mistake. Something we all do from time to time.

The Bob (2004 ©)
 
  • #9
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I'd hate to gravedig, but there is a neat proof for this.

It uses modular arithmetic which is one of the simplest and most useful little tricks in mathematics. This is basically a method of finding a remainder. I use it for puzzling out dates all the time. Lets say that today is a Monday. Seven days from now it will be a Monday again. 21 days from now...Monday. Lets say that we wish to find what day it will be 50 days from Monday. 49 is a factor of seven, so we find what day it will be a Monday in 49 days from Monday. 49 + 1 is 50 and Monday plus a day is Tuesday.

We can use this in this case (believe it or not).

Lets use this for a three digit number- we'll give a variable for each digit. We can write that as:

100a + 10b + c (if a=1 b=3 c=5 then our number would be 135)

Now modular arithmetic says that we can simplify this by subtracting a factor of 9. The closest multiple of 9 to 100 is 99; 9 to 10; and 0 to 1.

We go ahead and subtract:

100a + 10b +c
- 99a - 9b -0c
-------------------
a + b + c

So it stands to reason that adding the digits of any three digit number divisible by 9 will result in a new number that is divisible by 9.

Since every multiple of 10 will fall short by 1, this will work for a number with n digits.

This method works for all numbers (just not always as cleanly). Just keep in mind that for most numbers, the number of digits matters. Let's try it for 8 and a four digit number.

1000a + 100b + 10c + d
-1000a - 96b - 8c - 0d
---------------------------
4b + 2c + d

This works nicely for the last three digits of any number. We are also allowed to overshoot. So, substituting 104 instead of 96 will also work.
 
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