The Power of the Number Nine – Is It Just Magic Or Is It Real

Most people don’t realize the full power of the number nine. First it’s 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. That may not seem like much but it is magic for the nine’s multiplication table. For every product of the nine multiplication table, the sum of the digits in the product adds up to nine. Let’s go down the list. 9 times 1 is equal to 9, 9 times 2 is equal to 18, 9 times 3 is equal to 27, and so on for 36, 45, 54, 63, 72, 81, and 90. When we add the digits of the product, such as 27, the sum adds up to nine, i.e. 2 + 7 = 9. Now let’s extend that thought. Could it be said that a number is evenly divisible by 9 if the digits of that number added up to nine? How about 673218? The digits add up to 27, which add up to 9. Answer to 673218 divided by 9 is 74802 even. Does this work every time? It appears so. Is there an algebraic expression that could explain this phenomenon? If it’s true, there would be a proof or theorem which explains it. Do we need this, to use it? Of course not!

Can we use magic 9 to check large multiplication problems like 459 times 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 statement that the product of 459 times 2322 is equal to 1,065,798 is correct? No, but it does tell us that it is not wrong. What I mean is if your digit sum of your answer hadn’t been 9, then you would have known that your answer was wrong.

Well, this is all well and good 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 magic nines help me regardless of what numbers I am multiple? You bet you it can! In this case we pay attention to a number called the 9s remainder. Let’s take 76 times 23 which is equal to 1748. The digit sum on 76 is 13, summed again is 4. Hence the 9s remainder for 76 is 4. The digit sum of 23 is 5. That makes 5 the 9s remainder of 23. At this point multiply the two 9s remainders, i.e. 4 times 5, which is equal to 20 whose digits add up to 2. This is the 9s remainder we are looking for when we sum the digits of 1748. Sure enough the digits add up to 20, summed again is 2. Try it yourself with your own worksheet of multiplication problems.

Let’s see how it can reveal a wrong answer. How about 337 times 8323? Could the answer be 2,804,861? It looks right but let’s apply our test. The digit sum of 337 is 13, summed again is 4. So the 9’s remainder of 337 is 4. The digit sum of 8323 is 16, summed again is 7. 4 times 7 is 28, which is 10, summed again is 1. The 9s remainder of our answer to 337 times 8323 must be 1. Now let’s sum the digits of 2,804,861, which is 29, which is 11, summed again is 2. This tells us that 2,804,861 is not the correct answer to 337 times 8323. And sure enough it isn’t. The correct answer is 2,804,851, whose digits add up to 28, which is 10, summed again is 1. Use caution here. This trick only reveals a wrong answer. It is no assurance of a correct answer. Know that the number 2,804,581 gives us the same digit sum as the number 2,804,851, yet we know that the latter is correct and the former is not. This trick is no guarantee that your answer is correct. It’s just a little assurance that your answer is not necessarily wrong.

Now for those who like to play with math and math concepts, the question is how much of this applies to the largest digit in any other base number systems. I know that the multiplies 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 their digit sums add up to 7. We can define this in an algebraic equation; (b-1) *n = b*(n-1) + (b-n) 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 solves to 9*n = 10n-10+10-n which is equal to 9*n is equal to 9n. I know this looks obvious, but in math, if you can get both side to solve out to the same expression that’s good. The equation (b-1)*n = b*(n-1) + (b-n) simplifies to (b-1)*n = b*n – b + b – n which is (b*n-n) which is equal to (b-1)*n. This tells us that the multiplies of the largest digit in any base number system acts the same as the multiplies of nine in the base ten number system. Whether the rest of it holds true too is up to you to discover. 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. Hence 16 in base 8 is calculated in base ten as (1 * 8) + 6 = 8 + 6 = 14. Different base number systems are whole other area of mathematics worth investigating. Recalculate the other multiples of seven in base eight into base ten and verify them for yourself.