Q&A #5879

Reducing fractions with large denominators

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From: Gail (for Teacher2Teacher Service)
Date: Mar 25, 2001 at 13:44:44
Subject: Re: Reducing fractions with large denominators

One thing that works well for my students is for them to know ways to
recognize if a number has a certain factor...   for example, if I wanted to
tell if a number had 2 as a factor (which means I can divide 2 evenly into
the number), all I have to do is look to see if the number is even.  If it
is odd, then I know I won't be able to divide it by 2 evenly.

If the number ends with 0 or 5, then I know 5 is a factor.  And if the
number ends in 0, I know 10 is a factor.

If I can add all the digits, and the sum is evenly divisible by 3, then the
whole number is divisible by 3...   like this 1,234,560   1+2+3+4+5+6+0 =
21, and 3 divides into 21 evenly, so it will also divide 1,234,560 evenly.

So, what good is this?  Well, one of the ways students can "reduce" or
simplify fractions is to look for a factor that will divide evenly into both
the numerator and the denominator...   We call that a "common" factor.

 Take 48 / 120   for example...

120 has 5 as a factor, but 48 doesn't, so I can't use 5 to make the
fraction simpler.

120 has 3 as a factor, because 1+2+0 is 3..   And so does 48 because 4+8=12
so I could divide both by 3 and get a new simplified fraction:   16 / 40

16 and 40 both have a factor of 2 (because they are both even... )

Now my fraction is simplified to 8/20 and I can use the factor of 2 again...

Now it is 4/10, and I can use the factor of 2 one more time to get 2/5.

Of course, I can use any factor that is common to both numbers, and the
larger a factor I use, the less steps I have to take...  I hope this helps.

 -Gail, for the T2T service

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