prime factorization of 624

I was deriving thejamesmachine's method for determining if a number is divisible by 7.  Here's how he expressed it

For seven:

Take the last digit, double it, and subtract it from the rest of the number. If the answer is divisible by 7 (including 0), then the number is also divisible by 7:

861:86-1x2=84

84/7=12

.

624  =

312 * 2  =

156 * 2 * 2 =

78 * 2 * 2 * 2 =

39 * 2 * 2 * 2 * 2

13 * 3 * 2 * 2 * 2 * 2 =

2^4 * 3 * 13

http://web2.0calc.com/questions/what-will-go-into92

Check the link out

Thanks, thejamesmachine for that link....I didn't know the "trick" for testing 7 as a factor.........

{Now....I want to see if I can figure out WHY that works.......LOL!!! )

Yes Chris, I agree, that would be a good challange  :)

I had not heard of it before james put it on the forum either .   Thanks James

Is N exactly divisible by 7?

Let N = 10n + a    where n and a are integers and a is in the units position.

Construct M = n - 2a   (i.e. twice the units value subtracted from what's left, where what's left is considered to be a number in its own right).

If M is exactly divisible by 7 then we can write  n - 2a = 7m  where m is an integer.

So n = 7m + 2a

In that case N = 10(7m + 2a) + a  or  N = 70m + 21a

So N/7 = 10m + 3a and since m and a are integers the resulting value of N/7 is an integer.  i.e. N is exactly divisible by 7 with no remainder.

.

I have forgotten what you are proving Alan :/

Thanks Alan :)

Thanks for that proof, Alan.....

Welcome for the method!