Math Series Part V: Dividing Big Numbers
Eleven is a prime number, and as such cannot be broken down into smaller digits whose product result in eleven. There are a few observations that we can make, for example, that allow us to see whether any number greater than eleven is in fact divisible by eleven. Indeed, the criterion for a number, written in decimal base, to be divisible by 11 is for us to take the alternating sum of the digits that make up the number. If the sum is a number, positive or negative, that is perfectly divisible by 11, then so is the entire number. If not, then this sum represents our remainder.
If we take, for example, the number 76,720 and try to determine its remainder in a more general manner, we can achieve this by breaking the number down into segments that are divisible by 11 and then subtracting those segments from the original number. For example, we know that 66,000 is perfectly divisible by 11, and so we can subtract: 76,720 – 66,000 = 10,720. Likewise, we know that 1,100 is divisible by 11: 10,720 – 1,100 = 9,620. If we continue in this fashion: 9,620  8,800 = 820; 820 – 770 = 50; 50 – 44 = 6. Therefore, 76,720 divided by 11 has a remainder of 6. We divided the number by 11 (6,000 + 100 + 800 + 70 + 4 = 6,974) times. Hence, 76,720/11 is 6,974 + 6/11, and we can use this simple method of segmentation to find the remainder of any large number that is divided by 11, and indeed by most numbers.
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In Part I of this series, we look at the fascinating Möbius band  Mathematical Wonders Series: Part II
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Yet, if we return to our opening paragraph, we can see that by taking the alternating sum, i.e. that of 7 – 6 + 7 – 2 + 0 = 6, we get the same remainder we found above. This technique helps us greatly in determining whether we have divided our number correctly to get the appropriate remainder.
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10=0
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It is by drawing on the modular arithmetic system that we are also able to use modulo 11 and the 11clock to prove some multiplications and divisions by eleven. If we look at, to start with, the multiplication table of the decimal clock numbers (as seen in the table), we can see that certain numbers in the decimal system give us certain remainders. For example, 2,578 = 2 * 1000 + 5 * 100 + 7 * 10 + 8 * 1, which means that in base 10 our remainder is 8 as 10=0. However, if we look at the table below for base 11, we can rewrite the number 2,578 in base eleven in order to find out whether it is perfectly divisible by 11. Thus, by observing that 10 in mod. 11 is 10, we also notice that this is equivalent to 1, as all we have to do is take one step back on our 11clock to reach the same position. Hence, we see that 2578 = 2 * (1)^{3} + 5 * (1)^{2} + 7 * (1) + 8 * 1 = 2 +5 – 7 + 8 = 4 in mod. 11. 2,578 is, therefore, not directly divisible by 11; we get a similar result by using our alternating summation system mentioned above without even needing to convert to mod. 11. All we need to do is draw from the conversion equation from mod. 10 to mod. 11: For numbers in the form a*10^{n} + b*10^{n+1} + c*10^{n+2}, with n=0, we simply replace 10 by (1) in order to get the alternating summation needed.
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Through this, we can now use the base 11 modular system to verify our results for large multiplications of decimalbased numbers. Take, for instance, 2,015 * 321. The decimal result to this is 2,015 + 40,300 + 604,500 = 646,815. This can also be written as 6 * 10^{5} + 4 * 10^{4} + 6 * 10^{3} + 8 * 10^{2} + 1 * 10 + 5. Although 10=0 mod. 10, it is also equal to 1 in mod. 11. Hence, in mod. 11, 646,815 = as 6 * (1)^{5} + 4 * (1)^{4} + 6 * (1)^{3} + 8 * (1)^{2} + 1 * (1) + 5. Therefore, 646,815 = 6 + 4  6 + 8  1 +5 = 4 mod. 11.
We verify that 2,015 * 321 is indeed 646,815 by comparing the multiplication of 2,015 * 321 in mod. 11: (2 * 10^{3} + 1 * 10 + 5) * (3 * 10^{2} + 2 * 10 + 1) mod. 10 = (2 * (1)^{3} + 1 * (1) + 5) * (3 * (1)^{2} + 2 * (1) + 1) mod. 11 = (2 – 1 + 5) * (3 +  2 + 1) mod. 11 = (2 * 2) = 4 mod. 11. As these two results match, we have proved by 11 that the result of the multiplication is correct.
We end this mod. section by giving a proof by 11 of a threedigit number divided by a twodigit number that results in a nonzero remainder. Let’s consider 111 / 12 = 9.25 = 9 + 2 * 10^{1} + 5 * 10^{2} mod. 10. In mod. 11, 9.25 = 9 + 2 * (1)^{1} + 5 * (1)^{2} = 9  2 + 5 = 12. If we take the same steps for the division in mod. 11, we see that (1 * 10^{2} + 1 * 10 + 1) / (1 * 10 + 2) mod. 10 = (1 * (1)^{2} + 1 * (1) + 1) / (1 *(1) + 2) mod. 11 = (1 – 1 + 1) / (2 – 1) mod. 11 = 1. As 12 is equal to 1 in mod. 11, we have proved by 11 that 111 / 12 is indeed 9.25.
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