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Most people don’t realize the full power of the number nine. First, it is the largest single digit in the base ten number system. The digits of the base ten number system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. It may not sound like much, but it’s magic for the multiplication table of nine. For each product of the multiplication table of nine, the sum of the digits in the product adds up to nine. Let’s go through the list. 9 times 1 equals 9, 9 times 2 equals 18, 9 times 3 equals 27, and so on for 36, 45, 54, 63, 72, 81, and 90. When we add the digits of the product, like 27, the sum adds up to nine, that is, 2 + 7 = 9. Now let’s expand on that thought. Could a number be said to be divisible by 9 if the digits of that number add up to nine? How about 673218? The digits add up to 27, which equals 9. The answer to 673218 divided by 9 is 74802 pairs. Does this always work? It looks like it is. Is there an algebraic expression that can explain this phenomenon? If it is true, there would be a proof or a theorem that explains it. Do we need this, to use it? Of course not!
Can we use Magic 9 to check big multiplication problems like 459 by 2322? The product of 459 times 2322 is 1,065,798. The sum of the digits of 459 is 18, which is 9. The sum of the digits of 2322 is 9. The sum of the digits of 1,065,798 is 36, which is 9.
Does this prove that the statement that the product of 459 times 2322 equals 1,065,798 is correct? No, but it does tell us that it’s not bad. What I mean is that if the sum of the digits in your answer had not been 9, then you would have known that your answer was wrong.
Well, this is all very well if your numbers are such that their digits add up to nine, but what about the rest of the number, those that don’t add up to nine? Can the magic nines help me no matter what numbers I am a multiple of? I bet you can! In this case we pay attention to a number called the remainder of 9s. Let’s take 76 times 23, which equals 1748. The sum of digits in 76 is 13, the sum again is 4. So the remainder of 9s to 76 is 4. The sum of digits of 23 is 5. That makes let 5 be the remainder of 9s of 23. At this point, multiply the two remainders of 9, that is, 4 by 5, which is equal to 20 whose digits add up to 2. This is the remainder of 9 that we are looking for when we add the digits of 1748. Sure enough, the digits add up to 20, added again is 2. Try it yourself with your own multiplication problem worksheet.
Let’s see how you can reveal a wrong answer. How about 337 times 8323? Could the answer be 2,804,861? It seems correct, but let’s apply our test. The sum of the digits of 337 is 13, the sum again is 4. So the remainder of the 9 of 337 is 4. The sum of the digits of 8323 is 16, the sum again is 7. 4 times 7 is 28, which is 10, the sum again is 1. The remainder of the 9 in our answer to 337 times 8323 must be 1. Now we add the digits of 2,804,861, which is 29, which is 11, the sum again is 2. This tells us that 2,804,861 is not the correct answer to 337 times 8323. And it sure isn’t. The correct answer is 2,804,851, whose digits add up to 28, which is 10, the sum again being 1. Be careful here. This cheat only reveals one wrong answer. It is not a guarantee of a correct answer. Know that the number 2,804,581 gives us the same sum of digits as the number 2,804,851, but we know that the last one is correct and the first one is not. This trick does not guarantee that your answer is correct. It’s just a small guarantee that your answer is not necessarily wrong.
Now for those who like to play around with math and math concepts, the question is how much of this applies to the largest digit in any other base number system. I know that the multiples of 7 in the base 8 number system are 7, 16, 25, 34, 43, 52, 61, and 70 in base eight (see note below). All of its digit sums add up to 7. We can define this in an algebraic equation; (b-1)*n = b*(n-1) + (bn) where b is the base number and n is a digit between 0 and (b-1). So, in the case of base ten, the equation is (10-1)*n = 10*(n-1)+(10-n). This resolves to 9*n = 10n-10+10-n which equals 9*n equals 9n. I know this seems obvious, but in math, if you can get both sides to solve the same expression, that’s fine. The equation (b-1)*n = b*(n-1) + (bn) simplifies to (b-1)*n = b*n – b + b – n which is (b*nn) which is equal to (b-1)*n. This tells us that multiples of the largest digit in any base number system act the same as multiples of nine in the base ten number system. It’s up to you to find out if the rest is true too. Welcome to the exciting world of mathematics.
Note: The number 16 in base eight is the product of 2 times 7 which is 14 in base ten. The 1 in the base 8 number 16 is in the 8s position. Therefore, 16 in base 8 is calculated in base ten as (1 * 8) + 6 = 8 + 6 = 14. The different base number systems are another whole area of mathematics worth investigating. Recalculate the other multiples of seven in base eight in base ten and see for yourself.