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

I don't know how to prove this

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Bobosharif S. | Mathematics/Statistics TutorMathematics/Statistics Tutor
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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
=aa1a2a3a4, wherea1=4a+b, a2=3a+4b+c, a3=3a+4b, a4=3c]
 
Mark M. | Mathematics Teacher - NCLB Highly QualifiedMathematics Teacher - NCLB Highly Qualif...
4.9 4.9 (169 lesson ratings) (169)
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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.

Comments

Mark, 123123 / 7 appears to be 17589. Is this really a counterexample?
Edward A. | Math Tutor, Retired Computer Scientist and Technical CommunicatorMath Tutor, Retired Computer Scientist a...
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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?