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?
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