Is 8303 divisible by 17?
March 16, 2008 am31 6:19 am
If yes, we can add multiples of 17, and the sum will still be multiples of 17.
Assume 8303 is 17n.
Then 8303 + 17 = 17(n+1)
8320 = 17(n+1)
Divide both sides by 10.
832 = 17k (where k = (n+1)/10 )
850 – 832 would be 50(17) – 17k, or 17(50-k)
that’s 18 = 17(50-k).
no good. k is not an integer, n is not an integer, 17 doesn’t divide 8303.
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JD2718 
I figured you were doing something like dividing both sides by 10 in the last post, but I was too underslept to figure it out. ;-)
I really like the method here of adding a multiple to get a ten to drop out. That’s pretty.
I use this trick (add or subtract multiples of the potential factor to get an easier number to factor) all the time, and as you have, I use it recursively (I factor car number plates and train car numbers and phone numbers and …).
When the trick first occurred to me, it sped up factoring a lot, and I rued the time I had wasted doing things the hard way. The things where you use the factors of 1001 is another handy trick that I also use, though perhaps I don’t use it as often as I could 0 – it usually only occurs to me to use it when the number is six digits long, even though it obviously works in other cases.
Most of my skills have slowed as I age. But this one, depending on cleverness/experience, it has sharpened. I find myself often combining tricks in a single example.
I also check if some substring of a phone number will divide the rest of the number…