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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 digit in the base ten number system. The digits in the base ten number system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. It might not sound like much, but it’s magic for the nine times table. For each product in the nine times table, 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 product, such as 27, the sum adds up to nine, that is, 2 + 7 = 9. Now let’s expand this thought. Could you say that a number is divisible by 9 if the digits of that 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 this work every time? It seems so. 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 the 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 show that the statement that the product of 459 times 2322 equals 1,065,798 is correct? No, but he tells us 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 incorrect.
Well, that’s all well and good 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 which numbers I am multiple? Bet it can! In this case we pay attention to a number called remainder of 9s. Let’s take 76 times 23, which is equal to 1748. The sum of the digits of 76 is 13, added back is 4. So the remaining 9s of 76 is 4. The sum of the digits of 23 is 5. That makes 5 be the 9s remaining in 23. At this point multiply the two remainders of 9, that is, 4 by 5, which is equal to 20 whose digits add to 2. This is the remainder of 9 that we are looking for when we add the digits of 1748. Sure the digits add up to 20, added back is 2. Try it yourself with your own multiplication problems worksheet.
Let’s see how it can reveal an incorrect answer. How about 337 times 8323? Could the answer be 2,804,861? That sounds about right, but let’s apply our test. The sum of digits of 337 is 13, added back is 4. So the remainder of 9 from 337 is 4. The sum of digits of 8323 is 16, added back is 7. 4 times 7 is 28, which is 10, added back is 1. The remainder of 9s from our answer to 337 times 8323 must be 1. Now add the digits of 2,804,861, which is 29, which is 11, added back is 2. This tells us that 2,804,861 is not the correct answer to 3237 times. And it sure isn’t. The correct answer is 2,804,851, whose digits add up to 28, which is 10, added back is 1. Be careful here. This trick only reveals an incorrect answer. There 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, but we know that the second is correct and the first is not. This trick does not guarantee that your answer will be correct. It’s just a little reassurance that your answer isn’t necessarily wrong.
Now, for those who like to play around with mathematical and mathematical concepts, the question is how much of this applies to the largest digit of 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 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 in 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 resolves to 9*n = 10n-10+10-n which is equal to 9*n is equal to 9n. I know this seems obvious, but in math, if you can get both sides to solve the same expression, that’s good. 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 the multipliers of the largest number in any base number system act the same as the multipliers of nine in the base ten number system. If the rest is also true, it’s up to 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, which is 14 in base ten. The 1 in 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. Different base number systems are another area of mathematics worth investigating. Recalculate the other multiples of seven from base eight to base ten and check for yourself.
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