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The Magic Cafe Forum Index :: Magical equations :: Why does this work (0 Likes) Printer Friendly Version

Good to here.
drkptrs1975
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Take any two digit number, to see if one number is a multiple of another. Take the tens digit and muliply it by the number you want to see go into it. Then subtract that from the original number. Here is the equation.

10*a+b, and number c. We want to know if 10*a+b is a multiple of c, So if 10*b+b-a*c is a muliple of c, then 10*a+b is a multiple of c.


Here is an example. Take 56, want to know if it is a multiply of 7. So it is 7*5 is 35 subtract that from 56 is 21. 21 is a multiple of 7 so that would make 56 is a multiple of 7.

I cannot figure out why this works, nor can I find a counter example.
landmark
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Well drkptrs, here is why it works:

Let's use your example of 56 being a multiple of 7. What does that mean? It means a bunch of 7s will add up to 56. Or another way to say that is that you can subtract a whole bunch of 7s from 56 and not have a remainder.

So first you subtracted 7*5 and saw what you had left. Well now since you have 21 left, you'll be able to keep subtracting 7s without a remainder. So 7 is a multiple of 56. You see there's nothing special about that 5 in the tens place you used. It could have been any number that resulted in a product less than the original number. You could have used, say, 6. Then 7*6 =42. But 56-42 = 14. And since 14 is a multiple of 7, then the original number must be a multiple of 7.

Hope this makes things a little more clear,
Jack Shalom
Heinz Weber
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It doesn't matter which multiply of c you subtract from the original number (no need to be a, each number would work) As long as you subtract 'whole' c's from the original number the property of being a multiply of c (or not) cannot change.

I thought I can say it clearer ;-)

Heinz