https://hoc24.vn/hoi-dap/question/68302.html
Goi số đã cho là \(A\) ta có:
\(A=4a+3\)
\(=17b+9\)
\(=19c+3\)
\(\Rightarrow A+25=4a+3+25=4a+28=4\left(a+7\right)\)
\(=17b+9+25=17b+34=17\left(b+2\right)\)
\(=19c+13+25=19c+38=19\left(c+2\right)\)
\(\Rightarrow A+25⋮4;17;19\)
Mà \(ƯCLN\left(4;17;19\right)=1\)
\(\Rightarrow BCNN\left(4;17;19\right)=4.17.19=1292\)
\(\Rightarrow A+25⋮1292\)
\(\Rightarrow A+25=1292k\left(k=1;2;3...\right)\)
\(\Rightarrow A=1292k-25=1292k-1292+1267\)
\(=1292\left(k-1\right)+1267\)
Do \(1267< 1292\) nên \(1292\) là số dư trong phép chia
\(\Rightarrow A\div1292\) dư \(1267\)
Vậy số đó chia \(1292\)\(\) dư \(1267\)
