Fear Factor? Never Fear Factoring Again!

I'm about to share a secret with you that can instantly reduce the amount of time spent on math tests and homework where you need to find the greatest common factor (GCF) of two whole numbers, such as when reducing fractions to simplest form. This is a methodology/short-cut that I came up with long ago, but never saw described in any textbook, much to my surprise. The basic idea is actually quite simple:

Given two whole numbers A and B where A > B, the GCF of A and B must also be a factor of their difference: A-B.

Here's an example. You need to reduce the fraction 84/105 to simplest form. It might take you a minute or two to figure out all of the factors of these numbers, but you can get it by simply subtracting 105-84 = 21. The number 21 divides evenly into both 84 and 105, so that also happens to be the greatest common factor! Using this method, the final answer of 4/5 comes quickly and easily.

Just to be clear, the difference is not always the GCF, but the GCF is always a factor of the difference. When simplifying the fraction 85/106, you will also get 21 as the difference, but that is not the GCF. In this case, 85/106 is already in simplest form, as the GCF of those two numbers is 1. So, here, the difference of 21 merely helps to narrow down the possibilities, as you only need to check 3 and 7 (the prime factors of 21) and find that neither of those divide evenly into 85 or 106. Since none of the factors of the difference work, this fraction can't be reduced.

You can also subtract out the lower number multiple times if necessary, and it still works! For the fraction 91/343, the difference is 343-91=252. Doing this again gets 252-91=161. And finally: 161-91=70. The prime factors of 70 are 2, 5, and 7... so this greatly narrows down the search of common factors for 343 and 91. Since 2 and 5 clearly don't work, you are left with 7, which does divide evenly into both numbers, leaving you with the answer: 13/49.

Note that subtracting multiple times like this is actually the same operation as finding the "remainder" when dividing, so my original statement can be further refined:

Given two whole numbers A and B where A > B, the GCF of A and B must also be a factor of the remainder when dividing A by B.

I have one more short-cut to share here: if you have a calculator, you can quickly find the remainder by dividing A by B, subtracting out the whole number (leaving just the decimal), and then multiplying by B again. Using the same example as above, dividing 343 by 91 yields a long decimal number that starts with 3.76923... so you then subtract 3 from that to leave just the decimal part. Multiplying that result by 91 gets you the remainder of 70! (Note that the decimal calculation may be slightly inaccurate, so you could end up with 70.000001 or 69.999999... but it should still be obvious what the remainder would be, since we are talking about whole numbers here.)

I hope this helps!!


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