Why is the diff of 2 numbers made by a different combinations of the same digits, equal to a multiple of 9?

Question

Sorry for the abreviation in the title, the whole sentence wouldn't fit. Here is the original question:
Why is the difference between two numbers made by a different combination of the same digits, always equal to a multiple of 9? In other words: choose a number at random, then shuffle the digits in it to make a second number. Subtract the smaller number from the higher one. The result is ALWAYS a multiple of 9. (you can find this by adding all the digits in the number, and the sum will also be a multiple of 9). Why does this happen?
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I found this out due to this "mind reader":
http://www.digicc.com/fido/
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Thanks!

Eulercrosser, I don't get your answer, can you please elaborate?

Answer 1

Here is a link to my geocities site. You will find a pdf with a proof of this problem and other yahoo answers answers.
http://www.geocities.com/euleratuo//index.html

Answer 2

THIS DOES NOT NEED TO BE THAT COMPLICATED!!
(eulercrosser, you dont need to go into the sum of the digits being divisible by nine, something that you didnt even prove b.t.w. This solution incorporates the method of that proof anyway.)
HERE IS THE BEST SOLUTION IVE SEEN SO FAR:

a(0), a(1), ... a(n) are the digits of both numbers

n(1) = a(0) + 10a(1) + (10^2)a(2) + ... + (10^n)a(n)

n(2) = a(0)*[10^p(0)] + a(1)*[10^p(1)] + ... + a(n)*[10^p(n)]
where p(0), p(1),...p(n) is some permutation of 0,1,2,...,n

D = n(2)-n(1) = a(0)*[10^(p(0)) -1] + a(1)*10*[10^(p(1)-1)-1] + ... +
a(n)*(10^n)*[10^(p(n)-n)-1]

or

D = Σ a(k)*(10^k)*[10^(p(k)-k) - 1]

where k=0, 1, 2, ..., n

However, if for any term p(k)-k<0, rewrite it as:

-a(k)*[10^p(k)]*[10^(k-p(k)) - 1]

so that z below is non-negative and F below is an integer. *See note below

Therefore each term in the sum (equal to D) has a factor of

F = (10^z - 1) , z ≥ 0 ( z = |p(k)-k| )

if z=0 then F = 10^0 - 1 = 0 and is divisible by 9

If z > 0 factor:

F = (10 - 1)(10^(z-1) + 10^(z-2) + ... + 1)

= 9(10^(z-1) + 10^(z-2) + ... + 1) , divisible by 9

Thus all terms in the sum (equal to D) have a factor F divisible by 9, so the difference D is divisible by 9

Q.E.D.

note: The generalized sum then could be written:

D = Σ a(k)*[10^min(k,p(k)) ]*[10^( |p(k)-k| ) - 1] *sgn(p(k)-k)
Therefore, THIS SOLUTION YIELDS BOTH FACTORS, not just 9!!!!!

D= 9 * Σ {a(k)*[10^min(k,p(k)) ]*[Σ 10^(j-1)]*sgn(p(k)-k) }

where the first sum is taken over k=0,1,2,...,n
min(a,b)=minimum of a and b
z = | p(k)-k |
the second sum is taken over j=1,2,3,...,z
if z=0 then the second sum = 0
and sgn(m) = |m|/m if m ≠ 0; sgn(m) = 0 if m=0

Answer 3

let similar digit (2 digits) huge type be 10a + a = 11a sum of digits = a + a = 2a at the same time as sum of digits is subtracted from the massive type itself we get = 11a -- 2a = 9a, which continuously is a particular of 9, although a be from a million, 2, 3, 4, 5, 6, 7, 8, 9 yet no longer 0. ascertain at the same time as a = 7 huge type is seventy seven = 10*7 + 7 = 11*7 sum of digits = 7 + 7 = 2*7 which at the same time as subtracted from the massive type 11*7 supplies 11*7 -- 2*7 = 9*7 that's a particular of 9.

Answer 4

take any number..if its a 2 digit number then..it can be written as 10X+y where x is 10 th place and y the units place..

if you invert it the new number will be 10Y+x
suppose new number is greater than first one..

then the difference will be

10y+x -(10x+y) = 9y-9x = 9(y-x)
you see this is always divible by 9..
this can also be proved for 3 and more digits number..

Answer 5

If you shuffle two digits next to each other, e.g., you change xy to yx, and y>x
let z=y-x
so you are adding 10z to turn the first x into y, and subtracting 1z to turn the the last y into an x. Net effect is adding 9z. Hence the difference is always divisible by 9.

Answer 6

simply put our basic math system is based on only 9 numbers, Zero is not a number but a place holder or a representation of nothing. Therefore in basic math everything will end up being = to 9 in some form.