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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 is the largest single digit in the base ten number system. The numbers in the base ten number system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. It may not seem like much, but it’s magic for the multiplication table of nines. For each product of the multiplication table by nine, the sum of the digits of the product totals 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 the product , such as 27, the sum totals nine, that is, 2 + 7 = 9. Now, let’s extend this thought. Could we say that a number is also divisible by 9 if the digits of this number add up to nine? How about 673218? The digits add up to 27, which add up to 9. The answer to 673218 divided by 9 is 74802 even. Does it work every time? It seems so. Is there an algebraic expression that could 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 large multiplication problems like 459 times 2322? The product of 459 times 2322 is 1065798. 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 1065798 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 that tells us it’s not wrong. What I mean is that if your digit sum of your answer hadn’t been 9, then you would have known your answer was wrong.
Well, that’s fine if your numbers are such that their digits add up to nine, but what about the rest of the number, the ones that don’t add up to nine? Can the magic nine help me regardless of the number of which I am a multiple? You bet it’s possible! In this case, we pay attention to a number called the remainder 9s. Let’s take 76 times 23 which is equal to 1748. The sum of the digits over 76 is 13, added again is 4. Therefore the remainder of the 9’s for 76 is 4. The sum of the digits of 23 is 5. That makes 5 the remainder of the 9s by 23. At this point, multiply the two remainders of 9, i.e. 4 times 5, which equals 20 whose digits add up to 2. This is the remainder of 9 that we are looking for when we add the digits 1748. Of course the digits add up to 20, added again is 2. Try it yourself with your own multiplication problems worksheet.
Let’s see how this can reveal a wrong answer. How about 337 times 8323? Could the answer be 2,804,861? This seems correct but let’s apply our test. The sum of the digits of 337 is 13, added again is 4. Thus, the remainder of the 9s of 337 is 4. The sum of the digits of 8323 is 16, added again is 7. 4 times 7 is 28, which is 10 , added again is 1. The remainder of the 9 in our answer to 337 times 8323 must be 1. Now let’s add the digits of 2,804,861, either 29, or 11, added together 2 again. This tells us that 2,804,861 is not the correct answer to 337 times 8323. And of course it is not. The correct answer is 2,804,851, the digits of which add up to 28, or 10, added back to 1. Be careful here. This trick only reveals one wrong answer. This is no 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, yet we know that the latter is correct and the first is not. This trick does not guarantee that your answer is correct. It’s just a little reassurance that your answer isn’t necessarily wrong.
Now, for those who like to play with math and mathematical concepts, the question is how well this applies to the largest digit of any other base number system. I know that the multiplications 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 their 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). Thus, 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 equals 9*n equals 9n. I know it sounds obvious, but in math, if you can get the same expression on both sides, 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 multiplications of the largest digit in any base number system act the same as multiplications of nines in the base ten number system. Whether the rest is equally true is for you to find out. Welcome to the exciting world of mathematics.
Note: The number 16 in base eight is the product of 2 times 7, or 14 in base ten. The 1 in base 8 number 16 is in position 8s. Therefore, 16 in base 8 is calculated in base ten as (1 * 8) + 6 = 8 + 6 = 14. Different base number systems are a whole other area of mathematics worth studying. Recalculate the other multiples of seven in base eight in base ten and check them for yourself.
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