Giải:
Ta có: \(\overline{abcabc}:\overline{ab}=10010\)
\(\Rightarrow\overline{abc}.1001=10010.\overline{ab}\)
\(\Rightarrow\left(10.\overline{ab}+c\right).1001=10010.\overline{ab}\)
\(\Rightarrow10010.\overline{ab}+1001c=10010.\overline{ab}\)
\(\Rightarrow1001c=0\Rightarrow c=0\)
Vậy c = 0
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ta có
abcabc:ab=10010
=>(ab.10000+c.1000+ab.10+c.1):ab=10010
=>(ab.(10000+10)+c.(1000+1):ab=10010
=>(ab.10010+c.1001):ab=10010
=>\(\left(ab.10010+c.1001\right).\dfrac{1}{ab}=10010\)
=>ab.10010.\(\dfrac{1}{ab}+c.1001.\dfrac{1}{ab}\)=10010
=>10010+c.1001.\(\dfrac{1}{ab}\)=10010
=>c.1001.\(\dfrac{1}{ab}\)=10010-10010=0
=>c.1001=0
=>c=0
vậy c=0
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