gọi ab là số cần tìm,a,b\(\in N\)*
Theo đề ra ,ta có:\(\left\{{}\begin{matrix}\left(10a+b\right)\left(a+b\right)=405\\\left(10b+a\right)\left(a+b\right)=486\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}11ab+10a^2+b^2=405\left(1\right)\\11ab+10b^2+a^2=486\left(2\right)\end{matrix}\right.\)
lấy (1)-(2)<=>10(a2-b2)-(a2-b2)=-81
<=>9(a2-b2)=-81
<=>a2-b2=-9<=>(a+b)(a-b)=-9
vì a,b\(\in\)N* nên a+b>a-b;(a+b),(a-b)\(\in\)Z
*TH1:\(\left\{{}\begin{matrix}a+b=3\\a-b=-3\end{matrix}\right.\)
=>a=0,b=3
*TH2:\(\left\{{}\begin{matrix}a+b=9\\a-b=-1\end{matrix}\right.\)
=>a=4,b=5
*TH3:\(\left\{{}\begin{matrix}a+b=1\\a-b=-9\end{matrix}\right.\)
=>a=-4,b=5
vậy số cần tìm là 45