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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.
]]>. .
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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