Math Is Fun Forum

drpalmer1

Member

Registered: 2008-06-07

Posts: 1

Dividing by 3

what if you need to know if you can reduce a large fraction down? Well one way to know if a fraction is divisible by 3 is to add up individual digits of the whole. If they add up a number that is divisible by 3 the large number is also divisible by 3. example. 1242 is divisible by 3 because 1+2+4+2=9 and 3 into 9 goes 3 times. chances are the number will also be divisible by 6 or 9 and sometimes both if it is divisible by 3. example 99 (9+9=18) is divisible by 3. 3 into 18 goes 6 times. But 99 is also divisible by 9. 9 x 11= 99.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Dividing by 3

This type of rule is available for all numbers.  Some are particularly ugly, such as for 7.  Others, such as 11, are nice.  Every one is derived through the use of modulus.

For 11, the difference between the even and odd digits is divisible by 11 if and only if the number itself is divisible by 11.  So for 1015817:

1 - 0 + 1 - 5 + 8 - 1 + 7 = 11, and hence 1015817 is divisible by 11.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Dividing by 3

How would you derive the rule for 7 then?

The only rule for 7 that I know of is to double the last digit, then take the result away from the rest of your number.

eg. 637259

63725 - (2x9) = 63707. This is divisible by 7, and so the original number also is.
Obviously I cheated there, and to do it properly you'd need to iterate the step until the number gets reduced enough. Each iteration reduces the order of the number by 1, so you'd need to iterate the method maybe n-2 times for your answer. It'd be easier to just try dividing by 7 directly.

A rule for 7 that's similar to the rules for 3 or 11 would make things a lot quicker.

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Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Dividing by 3

Take a number with digits:

Now write it in decimal notation:

Now reduce each coefficient by modulo 7.  There will be a pattern, but it's an ugly one.  Then that equals 0 (modulo 7), the number must be divisible by 7.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."