Prove that the six-digit whole number abc,abc is divisible by 7?

Question

I don't know how to prove this

Answer 1

abcabc=100,100*a+10,010*b+1001*c.

abcabc/7=(100,100/7) * a+(10,010/7)*b+(1001/7)*c

             = 14300*a +1430b +143c

 

[ The rest is not necessary

=(1*10,000+4*1000+3*100+0*10+0*1)*a+

                 +(1*1000+4*100+3*10+0*1)*b+

                                 (1*100+4*10+3*1)c =

=10,000a+(4a+b)1000+(3a+4b+c)100+(0 a+3b+4c)10+3c

=aa₁a₂a₃a_4, wherea₁=4a+b, a₂=3a+4b+c, a₃=3a+4b, a₄=3c]

 

Answer 2

David, what are the factors of abcabc ? abc is one factor, what is the other? Since abc might or might not be divisible by 7, is the other factor divisible by 7?

Answer 3

Take the last digit (ones), double, subtract from original.

If result is divisible by 7, original is.

Try 123,123.

It is a counter example.