Bobosharif S. answered • 03/08/18

Mathematics/Statistics Tutor

_{1}a

_{2}a

_{3}a

_{4, }wherea

_{1}=4a+b, a

_{2}=3a+4b+c, a

_{3}=3a+4b, a

_{4}=3c]

David M.

asked • 03/08/18I don't know how to prove this

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Bobosharif S. answered • 03/08/18

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Mathematics/Statistics Tutor

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

=aa_{1}a_{2}a_{3}a_{4, }wherea_{1}=4a+b, a_{2}=3a+4b+c, a_{3}=3a+4b, a_{4}=3c]

Edward A. answered • 03/08/18

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Math Tutor, Retired Computer Scientist and Technical Communicator

Mark M. answered • 03/08/18

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Mathematics Teacher - NCLB Highly Qualified

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.

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Edward A.

03/08/18