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?