Why must 2 digits of any 2-digit number that is multiple of 9 (eg 27) add up to 9?

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One way to explain why two digits of any two-digit number that is multiple of 9 add up to 9 is to think about how we write numbers in base-10 system. That means we use 10 symbols (0, 1, 2, …, 9) to represent any number. For example, when we write 27, we mean 2 tens and 7 ones.

Now, when we multiply a number by 9, we are making it a multiple of 9. That means it can be divided by 9 without any remainder. For example, 27 is a multiple of 9 because it can be divided by 9 exactly three times.

But how do we know if a number is a multiple of 9? One way is to add up its digits and see if the result is also a multiple of 9. For example, if we add up the digits of 27 (2 + 7), we get 9, which is also a multiple of 9.

This works because every time we add 9 to a number, we are increasing the value of the tens place by 1, but decreasing the value of the ones place by 11. For example:

If we start with 0 and add 9, we get 09, which has 0 tens and 9 ones. The sum of the digits is 0 + 9 = 9.

If we start with 09 and add another 9, we get 18, which has 1 ten and 8 ones. The sum of the digits is 1 + 8 = 9.

If we start with 27 and add another 9, we get 36, which has 3 tens and 6 ones. The sum of the digits is 3 + 6 = 9.

And so on. You can see that every time we add 9 to a number, the sum of the digits stays the same as 9. That’s why two digits of any two-digit number that is multiple of 9 add up to 9.

Does that make sense?

Because 9 is a power of 3, and as such carries some of the properties of 3, such as the digits of a product of 9 adding up to a number divisible by 9

People are talking about addition, which is true, but this is actually easier by subtraction.

9 is 10 – 1. Additionally, 99 is 100 – 1, 999 is 1000 – 1, and so on… or “any power of ten is always one more than a multiple of nine.”

What a number actually means (ie 268) is: 2×10^2 + 6×10^1 + 8×10^0. This is what we call *base 10* because every digit represents an order of magnitude of 10. This is why zeros don’t matter.

Anyway, back to the question. 268 is two 100s (99+99+2), six 10s (9+9+9+9+9+9+6), eight 1s. Notice how the remainders pile up. If they sum to any multiple of nine, then the number must be divisible by nine. Try writing it out with something like 20412.

The tricky part is that nine only has this special property in base 10. In base 5 (where twelve is 22), the magic number is 4. Always one less than the base (or a divisor of the magic number, like 3 in base 10). The key here is the infinite -1 multiples at the top of this comment.

Congratulations, you get to discover Square One today!

https://www.youtube.com/watch?v=Q53GmMCqmAM

As specified by others, when you add 9 the tens digit goes up by 1, then the ones digit goes down by 1, so the sum is the same. And it continues working for any number of digits.

Because 9 is exactly 1 less than the number at wich we roll over digits. That means whenever you add 9 to any number the digital sum doesn’t change because the last digit reduces by one and the second last digit increases by one.

Since a multiple of 9 is 9+9+…+9 each addition keeps the digital sum at 9

The exception is when your number ends in 0, wich also applies in your example: 99 is a multiple of 9 and it’s digital sum is 18. But the next exception corrects that again, if your second last digits rolls over it reduces the sum again. 108 is a multiple of 9 and it works again.