Thread: Sums of digits in multiples

In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

9 = 9 = 9
18 -> 1+8 = 9
27 -> 2+7 = 9
909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

Also, is there a way to prove this for every multiple of 9?

Its true. The gneral idea of a proof is that every power of ten is 1 + multiple of 9.
For example a 3 digit number abc is (99 + 1)a + (9 +1)b +c. Since 99a + 9b is obviously divisible by 9, then a+b+c must be divisible by 9.

Okay, I see now. Thank you. I was at first confused with your statement, considering 28 (9*3+1) is not a power of ten. But it must have had that "all m's are n's, but not all n's are m's" quality.

Alternatively, it's called "modulo 9 arithmetic", which is just another way of saying "base 10 arithmetic". If you want the full works, the modulo n relation partitions the integers into disjoint modulo classes, for which each member of one class differs from some member of another class by n or a multiple thereof.

Try this party trick:

Invite your victim to chose a number with more than one digit, add them (and add again if needed) to obtain a single digit.

Now say to subtract this number from the starting number, and repeat the adding digits process until he arrives again at a single digit.

Close your eyes, pretend to think really hard and state "the answer is 9".

For any cheats who start with a single digit, the answer is of course zero.

Originally Posted by brody

In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

9 = 9 = 9
18 -> 1+8 = 9
27 -> 2+7 = 9
909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

Also, is there a way to prove this for every multiple of 9?

Consider the number where each of the is a digit from 0 to 9.

Then where .

Notice that for each non-negative integer , is divisible by 9:

This can be rigorously established by mathematical induction (or by simply noting that ).

It follows that is divisible by 9. Hence we have , i.e. is divisible by 9 if and only if the sum of its digits is divisible by 9.

the 9s repeat in the order of 1/9 until inf

Originally Posted by brody

So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

Also, is there a way to prove this for every multiple of 9?

You could also look at a generator function. and write the rules about the digits

Multiples of 9 are just 9+9+9 for n 9's.
So use addition.

When will adding 9 break the rule that the sum of digits is a multiple of 9, and recursive summing of digits will not end at 9 ?


If the right hand digit is 0, then you change the 0 to 9, which does add 9 to the sum of the digits.
Is this leaving the sum of digits as a multiple of 9 ? yes.


Every time you add 9, you take 1 off the right hand digit, and add 1 to the digit to the left.
If you have not caused a carry, then there is no change to the sum of the digits, since you took 1 off one digit and added 1 to another.
So this is leaving the sum of the digits to be a multiple of 9.

If you have caused a carry, this is because 9+1=10, which means you remove a 9 from one column, and and taken 1 off one column and added 1 to another.. The total change in the sum of digits, but the sum of digits must remain a multiple of 9.

Even if there are multiple 9's eg "xxx999992" , the carry will eventually come to a column where the 1 can you still do a carry of the to the next digit left.

Now since you increased through 2 digits, 3 digits, etc, its sure that the smaller ones were always meeting the rules, because they were generated according to the rules.

Also, since its a number larger than 0, the digits cannot sum to exactly 0. The removal of the 9 could never reduce the sum of digits to be 0.


Therefore, its not possible that the digits of a multiple of 9 do not sum to a multiple of 9.
And recursive summing of the digits must always produce lesser and lesser sums, which much be multiples of 9, and so when it reaches a single digit, that digit must be 9

I agree if an only if it is a whole number multiple of 9. Given a fractional multiple of 9 say z/9 multiplied by 9 then an identity is created and any given number z is returned even ones that dont add to nine. Even then I havent tested every whole number multiple of nine greater that zero to prove that a whole number multiple which doesnt add to nine doesnt exsist.

Originally Posted by brody

In positive multiples of 9 (e.g. 18, 72, 909) why do the sum of the digits (no matter what order or combination) make 9 or another multiple of 9?

9 = 9 = 9
18 -> 1+8 = 9
27 -> 2+7 = 9
909 -> 9+0+9 = 18 (1+8=9) or... 90+9=99 (9+9=18, 1+8=9)

So no matter how large the multiple is, you can deduce it by summing its digits until you eventually reach 9.

Also, is there a way to prove this for every multiple of 9?

All the multiples of 9 will end up reaching to 9 if the result is added as per the example given.

108 = 1+0+8=9
117=1+1+7=9 and so on.

the rule applies for all the multiples of 9.