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 may not sound like much, but it’s magic for the multiplication table of nine. For each product in the multiplication table of nine, the sum of the digits of the product adds up to nine. Let’s lower 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 , since 27, the sum adds up to nine, that is, 2 + 7 = 9. Now let’s extend that 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 all the time? It looks like it is. Is there an algebraic expression that can explain this phenomenon? If true, there will be a proof or theorem to explain it. Do we need this to use it? Of course, no!

Can we use 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 prove that the statement that the product of 459 times 2322 equals 1,065,798 is correct? No, but it tells us that it is not bad. What I mean is that if the digit sum of your answer had not been 9, then you would have known that your answer was wrong.

Well this is fine if your numbers are such that your 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 the numbers I have? You can bet it can! In this case, we pay attention to a number called remainder of 9. Let’s take 76 times 23 which is equal to 1748. The sum of digits in 76 is 13, added again is 4. Therefore, the remainder of 9 to 76 is 4 The digit sum of 23 is 5. That makes 5 the remainder of 9 of 23. At this point, multiply the two remainders of 9, that is, 4 by 5, which is equal to 20 whose digits add up to 2. This is the remainder of 9 that we are looking for when we add the digits of 1748. Sure enough, the digits add up to 20, added back is 2. Try it yourself with your own multiplication problem worksheet.

Let’s see how you can reveal a wrong answer. How about 337 times 8323? Could the answer be 2,804,861? It seems correct, but let’s apply our test. The digit sum of 337 is 13, added again is 4. So the remainder of 9 from 337 is 4. The digit sum of 8323 is 16, added again is 7. 4 times 7 is 28, which is 10, added Again it is 1. The remainder of 9 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 again is 2. This tells us that 2,804,861 is not the correct answer to 337 by 8323. 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. It 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 last one is correct and the first one is not. This trick does not guarantee that your answer is correct. It is just a small guarantee that your answer is not necessarily wrong.

Now for those who like to play around with math and math concepts, the question is how much of this applies to the largest digit in 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 your 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). So in the base ten case, the equation is (10-1) * n = 10 * (n-1) + (10-n). This solves for 9 * n = 10n-10 + 10-n which is equal to 9 * n is equal to 9n. I know this seems like a no-brainer, 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 multiplication of the largest digit in any base number system works the same way as multiplication of nine in the base ten number system. If the rest is also true, it is 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 8 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 in base eight in base ten and check for yourself.