soroban wrote:

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I tested this in Excel...which found that it is true for 3645 of the 9000 numbers in the range 1000 to 9999.

I didn't try to weed out multiples of 1111 (or anything else).

Found by Excel:**5**-digit results*10890*: 3645 times*10989*: 640 times

Example: 8991 - 1998 = 6993: 6993 + 3996 = 10989

I haven't tried to work out why, but of the numbers that I checked the middle two digits were always 99.

**4**-digit results*9999*: 2880 times*1170, 1251, 1332, 1413, 1494, 1575, 1656, 1737, 1818 & 1998*: a total of 3815 times

**3**-digit results*261, 342, 423, 504, 585, 666, 747 & 828*: a total of 648 times

**2**-digit results*99*: 162

**1**-digit results*Zero*: 90

The digit-sum of the above multiple-digit numbers always reduces to 9: eg,

(a) *10989*'s digit-sum is 27, whose digit-sum is 9

(b) *9999*'s digit sum is 36, whose digit-sum is 9

(c) *747*'s digit sum is 18, whose digit-sum is 9

It also works on 5 digit numbers:

Again the right hands side is divisible by 9.

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The answer will always be 9. Can someone prove how?

1111-1111 = 0

]]>represent the 2 numbers like this.

As you can see the expression on the right hand side is divisible by 9.

Any number that is divisible by nine when you add its digits they will sum to 9.

7694-4967=2727

Now add all the digits:

2+7+2+7=18

1+8=9

The answer will always be 9. Can someone prove how?

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