# The power of the number nine – is it just magic or is it true?

The power of the number nine – is it just magic or is it true?

Most people do not 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 decimal number system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. That may not seem like much, but it’s magic for the nine times table. For each product in the multiplication table by nine, the sum of the digits in the product add up to nine. Let’s go down 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 a product, such as 27 , the sum gives nine, i.e. 2 + 7 = 9. Now let’s expand on this thought. Can a number be said to be evenly divisible by 9 if the sum of the digits of that number is nine? How about 673218? The sum of the digits is 27, which is 9. The answer to 673218 divided by 9 is 74802 even. Does this work every time? It looks like that. Is there an algebraic expression that could explain this phenomenon? If it is true, there will be a proof or theorem that explains it. Do we need it to use it? Of course not!

Can we use magic 9 to check large 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 tells us it’s not wrong. What I mean is that if the digit sum of your answer was not 9, then you would know that your answer was wrong.

Well, that’s all well and good if your numbers are such that the sum of their digits is nine, but what about the rest of the number, the ones that don’t add up to nine? Can magic nines help me regardless of which numbers I’m a multiple of? You 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 digit sum of 76 is 13, added back is 4. So the remainder of 9s to 76 is 4. The sum of the digits of 23 is 5. That makes 5 the remainder of 9s to 23. At this point multiply the two remainders by 9s, i.e. 4 times 5, which equals 20, whose digits sum to 2. This is the remainder of 9s we’re looking for when we add the digits of 1748. Of course, the sum of the digits is 20, which adds up again to 2. Try it yourself with your worksheet with multiplication problems.