Long Division Shortcut (Part 2)

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In the last post we looked at the problem of 926/18, and we simplified it to 463/9, so we could make it a short division problem.

What if the problem had been 927/18?
Both numbers are not even this time, so it is not readily apparent if they have common factors.

If you know how to factor (if you don’t, you can get a lesson at The Pretty Good Guide to Prime Factorization at MathMojo.com.) then you factor both of these numbers by 9.

Here’s a hint: If a number can be crunched to 9 or 0, then nine is a factor of that number. If you want to know more about crunching, I refer you to “The See-Say-Write Method of Speed Addition“.

There are also many hints you can find about how to determine if numbers are divisible by other numbers. MathMojo will eventually cover this in depth, but I’m sure you can find info if you google “divisibility rules.”

Ok, so let’s factor 927/18.

Using short division by 9, we get 103/2. How easy is the problem now? Just cut 103 in half in your mind and get 51 remainder 1. But remember, like in the last post,
If you factor a division problem before you solve it, you must multiply the remainder by that factor after you are done.

So the answer to 927/18 is 51 remainder 9, (not 1).

Go ahead and check it. Remember how? If not, check out this post.

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