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.
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 sum of the digits of 337 is 13, summed again is 4. So the remainder of 9 to 337 is 4. The sum of the digits of 8323 is 16, summed again is 7. 4 times 7 is 28, which is 10, summed again is 1 .The remainder 9s of our answer to 337 times 8323 should be 1. Now let’s add the digits of 2,804,861, which is 29, which is 11, added back to 2. This tells us that 2,804,861 is not the correct answer to 337 at 8323. And it certainly isn’t. The correct answer is 2,804,851, whose digits add up to 28, which is 10, and the sum is again 1. Be careful here. This trick only reveals a wrong answer. This 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 second is correct and the first is not. This trick is not a guarantee that your answer is correct. This is just a little reassurance that your answer is not necessarily wrong.
Now for those who like to play around with math and mathematical 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 the base eight system (see note below). The sum of all their digits is 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 solves 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 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 multiplications of the largest digit in any base number system work the same way as multiplications of nines in base ten number system. Whether the rest is also true is up to you to find out. Welcome to the exciting world of mathematics.
Note: The number 16 to base eight is the product of 2 times 7, which is 14 to base ten. The 1 in the number 16 of base 8 is in the 8s position. Therefore, 16 in base 8 calculates to base ten as (1 * 8) + 6 = 8 + 6 = 14. Different base number systems are a whole other area of mathematics worth exploring. Recalculate the other multiples of base eight to base ten and check for yourself.
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