the 9 times mystery

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at least it's a mystery to me..............I'm sure some here can explain it
I haven't got a clue
nothing similar happens with any other digits



here we go:



9*5=45.......................add up the digits - 4+5=9

9*8=72.......................add up the digits - 7+2=9

9*7=63.......................add up the digits - 6+3=9

9*17=153...................add up the digits - 1+5+3 =9

9*77=693...................add up the digits - 6+9+3 and get 18............then add up those digits..............1+8=9

9*4,796=43,164..........add up the digits 4+3+1+6+4 and get 18......then add up those digits...............1+8=9

9*104,675 = 942,075.........add up the digits..........9+4+2+0+7+5 and get 27...........then add up those digits..............2+7=9

9*5,327,894 = 47,951,046...........add up the digits .........4+7+9+5+1+0+4+6 and get 36............then add up those digits.................3+6=9



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Quote: lilredrooster

nothing similar happens with any other digits

Any number divisible by 3 will get a single digit sum that is divisible by 3.

Edit:
Any number multiplied by 0 (zero) will have a single digit sum of 0.

It also appears to be a quirk of the base you're using for the calculation.
If I flip to hexadecimal (base 16), 15 (0xf) works this way.
If I flip to octal (base 8), 7 works this way.

The sum of the digits, summed again if needed, of a number divisible by 9 will always be 9.

The easiest way to show it is to consider N and then N+1.

If the last digit of N isn't and 9, then the value goes up by one. If the last digit is 9, then there are a row (1 or more) of 9's at the end of the number and they all turn into 0's with a one carry. Converting the 9's to 0's will not change the sum of the digits mod(9) (this can be proved by induction if needed). The carry will add one to the total of the remaining digits.

I'll try:

When you add 9 to a number whose digits sum to a multiple of 9:
1. If the last digit is 0, it is now 9 and the other digits are the same, so you are adding 9 to the sum of the digits, which means it is still a multiple of 9.
2. If the last digit is not 0, subtract 1 from it and add 1 to the number to its left, unless that number is 9, in which case it becomes 0 and you add 1 to the digit to its left, repeating this until you get to a digit that is not 9. You subtracted 1 from the rightmost digit, added 1 to another digit, and possibly subtracted 9 from other digits, so the sum of the digits is still a multiple of 9.
Except for 1-9, every positive integer > the sum of its digits, so after each sum, you get a number that you know eventually "sums down" to 9.

For other numbers, when you have a "carry," you add 1 to a digit, but subtract more than 1 from the digit to its right.

....................

remember the Beatles track entitled "Revolution 9"

with the lyrics:

"Number nine, number nine, number nine, number nine..."



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