Divisibility of n by 7: Elementary Proof

[JFo]

Remove the last digit from a number and subtract twice this digit from the new (shorter) number. Show that the original number is divisible by 7 iff this difference is divisible by 7.

I have only the division algorithm and the fact that the integers are closed under addition/multiplication/subtraction to work with plus elementary arithmetic. By the way, all numbers are in base 10

Heres what I've got:
let n be an integer. By the division algorithm I can find an integers c,d s.t. n = 10c + d where 0<= d < 10

The number 10c is n with the last digit removed. We need to show 10c + d is divisble by 7 iff 10c - 2d is divisible by 7

Suppose 10c + d is divisble by 7. Then 10c + d = 7m for some integer m.

using the division algorithm we can find integers e,f st. d = 7e + f where
0<= f < 7

given the restrictions that 0<=d<10 we must have e = 0 or 1.

Here's where I'm stuck. Any suggestions?

 

[Muzza]

The number 10c is n with the last digit removed.

No, 10c is n with the last digit replaced with a zero. You want to consider c - 2d.

This proof can be done like so (using various elementary divisibility properties):

c - 2d is divisible by 7
<=>
10c - 20d is divisibly by 7
<=>
10c - 20d + 21d is divisibly by 7
<=>
10c + d is divisible by 7
<=>
n is divisible by 7.

If you had modular arithmetic in your inventory, this type of problem would be routine.

 

[JFo]

Thank you muzza! I really need to get rid of this extra chromosome!