Why are some numbers divisible by 9?

Question

Take any whole number.
Rearrange the digits to make a new whole number.
Subtract the lower whole number from the higher whole number.
The answer (another whole number) is always divisible by 9. Why?

Answer 1

Any digit of the form ABC... can be written as
10^n A + 10^(n-1) *B + ....

Rearrange the digits and the only thing that changes is the coefficients of A, B, C etc.

Subtract them and you'll get a number whose value is
∑ (10^n - 10^m) * Xi
where Xi = A, B, C ...

You can verify that 10^n - 10^m is divisible by 9 for all n and m, because in both cases the digital sum is 1 so the digital sum of the differerence is 0 or 9.

So the difference between the original two numbers is the sum of a bunch of terms which are individually divisble by 9, hence is divisible by 9.

eg 374 and 437
374 = 3*100 + 7*10 + 4
437 = 4*100 + 3*10 + 7
437 - 374 = 4*(100 -1) + 3*(10 - 100) + 7*(1 - 10)
= 4*99 - 3*90 - 7*9
= 9*k

Answer 2

This is one of the quirks about the number nine. another quirk is that if the digits of any number add up to 9, then it is divisible by 9, An example would be 267408 is divisible by 9, because:

267408 = 27, and 2+7 = 9

Answer 3

Interesting observation.

You can use this information to amaze and astound your friends, such as:

Take your birthdate, as MMDD or MMYY (or any such combination), if the month is 2 digits, remove the first digit, then rearrange the numbers, then subtract the two numbers, then add up the digits. As long as the number is bigger than 10, continue adding the digits until you have a single-digit number. I guess that the special number for your birthdate is (TA-DAH) NINE!

Voila!

Numbers ARE interesting and fun to play with.

Try multiplying 37 times any multiple of 3 (37*3, 37*6, 37*9, etc.)

Answer 4

e.g. 374 and 437
374 = 3*100 + 7*10 + 4
437 = 4*100 + 3*10 + 7
437 - 374 = 4*(100 -1) + 3*(10 - 100) + 7*(1 - 10)
= 4*99 - 3*90 - 7*9
= 9*k
this formula rele does work pick him
and it works because when you do teh subtractions of the seperate parts you end up with three numbers that are the same as taking eaach seperate part of a number

Answer 5

Not a proof, but an example.

Take a three-digit number... abc
Rearrange the digits........... bca

The first number has value 100a + 10b + c
The second number has value 100b + 10c + a

Now look at their difference

100a + 10b + c - 100b - 10c - a
100a - a + 10b - 100b + c - 10c
99a - 90b -9c

Since you can factor a 9 out of the result, the result is divisible by 9.

Answer 6

whenever the digits add up to 9 or anything divisible by nine, the number is divisible by nine. Both of the rearranged numbers will be divisible by 9 for this reason. Also, whenever you take one number divisible by nine from another, it WILL be divisible by nine.

Answer 7

A 3 digit number can be represented as
100h +10t +u
If you reverse the digits, the number is 100u +10t +h.
Subtracting gives 99h -99u = 9(11h -11u) which is clearly divisible by 9.
This can be extended to a number of any number of digits.

Answer 8

Dr D is over my head. I think of it this way: Every number you move from 10s to units (eg, 10 to 01), you subtract 10 and add 1, you've decreased by nine. Go the other way and you add nine. 10s and 1s are replacing each other. Works the same with 100s and 10s, you always change by 90.

Answer 9

I would also add (for completion purpose) to doctor D stuff that for the last part , you should prove it by using modulo mathematics, which I don't have time to do right now :(.