có vì chữ số tận cùng của a là 7
còn b là 8
A + B khong chia het cho 7
vi A tan cung la so 7
B tan cung la so 4
\(A=7.7.7...7=\left(7^4\right)^{25}=\left(\overline{...1}\right)^{25}=\overline{....1}\)
\(B=8.18.28.38...998.1008\rightarrow B\)có 101 thừa số
\(\Leftrightarrow B=\overline{...6}x\overline{...6}x\overline{...6}x...x\overline{...6}.1008=\overline{....6}.1008=\overline{....8}\)
\(\Rightarrow A+B=\overline{....1}+\overline{.....8}=\overline{......9}⋮̸5.\)
Bài giải
\(A=7\text{ x }7\text{ x }7\text{ x }...\text{ x }7=7^{100}=\left(7^4\right)^{25}=2401^{25}=\overline{\left(...1\right)}^{25}=\overline{\left(...1\right)}\)
\(B=8\text{ x }18\text{ x }28\text{ x }38\text{ x }...\text{ x }998\text{ x }1008\)
\(B=\left(8\text{ x }18\text{ x }28\text{ x }38\right)\text{ x }\left(48\text{ x }58\text{ x }68\text{ x }78\right)\text{ x }...\text{ x }\left(968\text{ x }978\text{ x }988\text{ x }998\right)\text{ x }1008\)
\(B=\overline{\left(...6\right)\text{ }}\text{ x }\overline{\left(...6\right)}\text{ x }...\text{ x }\overline{\left(...6\right)}\text{ x }\overline{\left(...8\right)}\)
\(B=\overline{\left(...6\right)}\text{ x }\overline{\left(...8\right)}\)
\(B=\overline{\left(...8\right)}\)
\(\Rightarrow\text{ }A+B=\overline{\left(...1\right)}+\overline{\left(...8\right)}=\overline{\left(...9\right)}⋮̸5\)
