Most people don't realize the full power of the number nine. First, it is the largest single number in the ten number system. The numbers of the ten number system are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. This may not sound like much but it is the magic of the nine times multiplication table. For each product of the nine multiplication table, the sum of the digits in the product is 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 numbers of the product, such as 27, the sum adds up to nine, i.e. 2 + 7 = 9. Now let's expand this thought. Can it be said that a number is equally divisible by 9 if the digits of that number add up to nine? What about 673218? Numbers add up to 27, which add up to 9. The answer to 673218 divided by 9 is even 74802. Does this work every time? It seems so. Is there an algebraic expression that explains this phenomenon? If this is true, then there will be evidence or theory to explain this. Do we need this to use it? of course not!

Can we use magic 9 to check large multiplication problems like 459 by 2322? 459 times 2322 equals 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 459 times 2322 equals 1,065,798? No, but she tells us it's not wrong. What I mean is that if the sum of your answer number is not 9, you will know that your answer was wrong.

Well, that's all well and good if your numbers are such that their digits go to nine, but what about the rest of the number, the ones that don't go to nine? Can magic threads help me no matter what numbers i am multiple? You bet you can! In this case, we pay attention to a number called the remainder 9. Let's take 76 times 23 which equals 1748. The sum of the digits in 76 is 13, and their sum again is 4. Hence the 9s remainder of 76 is 4. The sum of the number 23 is 5. This makes the remaining 5 9s of 23. At this point, multiply the two remainders of the 9s, that is 4 by 5, which equals 20 the number of digits we add to 2. This is the sum of the 9s that we add up again. Try it out for yourself with your own multiplication problems worksheet.

Let's see how you can reveal the wrong answer. How about 337 times 8323? Could the answer be 2,804,861? Sounds correct but let's apply our test. The sum of 337 is 13, and its sum again is 4. So the remainder of digit 9 of 337 is 4. The sum of the digits 8323 is 16, whose sum again is 7. 4 times 7 equals 28, which equals 10, whose sum again is 1. The remaining 9s of our answer to 337 in 8323 must be 1. Now let's add the numbers from 2,804,86 1, which again means that the output is 2.86, which equals 11.861. 323. It certainly is not. The correct answer is 2,804,851, whose digits add up to 28, which equals 10, and their sum again is 1. Be careful here. This trick only reveals the wrong answer. There is no confirmation of the correct answer. Know that the number 2,804,581 gives us the same sum of numbers as the number 2,804,851, but we know that the latter is true and the former is not. This trick does not guarantee that your answer is correct. It's just a simple confirmation that your answer is not necessarily wrong.

Now for those who like to play with mathematics and mathematics concepts, the question is how much of this applies to the largest number in any other basic number systems. I know that multiples of 7 in the base 8 number system are 7, 16, 25, 34, 43, 52, 61, and 70 in the base eight (see note below). All of their digits 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 number between 0 and (b-1). So in the base ten case, 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 this sounds obvious, but in math, if you can get both sides to solve for the same expression, 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 equals (b-1) * n. This tells us that multiplying the largest number in any base number system works in the same way as multiplying nine in the base-ten number system. Whether the rest of that is also true is up to you to find out. Welcome to the exciting world of mathematics.

Note: The number 16 to the eight is the product of 2 times 7, which is 14 to the tenth. The digit 1 is in the base 8 number the digit 16 is in the 8th position. Hence 16 in the base 8 in the base ten is calculated as (1 * 8) + 6 = 8 + 6 = 14. The different base number systems are a whole other area of mathematics worth studying. Recalculate the other multiples of seven into the base eight to the base ten and check for yourself.