Question 174065: Your class has just completed a unit on divisibility rules. One of the students ask why divisibility by numbers other than 3 and 9 can not be tested by dividing the sum of the digits by the tested number. How do you respond?
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website!
I should think the answer is obvious since the student just completed a unit on the divisibility rules. That is like a student just finishing a unit on the pathagorean theorem and that same student asking why doesnt the theorem work on a circle...WHY because it just doesnt . I dont know who comes up with some of these questions....This one is useless imo
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but here is an attempt at an explanation:
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You can see why if you go back to how the rule of adding the digits to test for divisibility by 3 (or 9) was derived:
First, if several different numbers are all divisible by 3, then their sum is also divisible by 3.
Second, any number multiplied by a number that is divisible by 3 is also divisible by 3.
Third, say you have a 3 digit number 'xyz'. This can be written as:
100x + 10y + z
= (99x + x) + (9y + y) + z
= 99x + 9y + (x + y + z)
which is x+y+z=99x+9y
The point of that is that if we subtract out the sum of the digits from a number, we're always left with something that's divisible by 9 (and therefore also divisble by 3) because it's in the form (99*x + 9*y), or for larger numbers 999 times something, 9999 times something, etc. But not so for other numbers.
You're probably familiar with the rule for testing divisibility by 2 or 5: just look at the rightmost digit. It's really the same principle but with less algebraic manipulation. Again as with the 3-digit number 'xyz', you have:
100*x + 10*y + z
Of course the 100*x and 10*y parts are divisble by 2 and 5, so all you have to do is pull out the last digit 'z' and test its divisibility.
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