- #1
Math100
- 779
- 220
- Homework Statement
- Find an integer having the remainders ## 2, 3, 4, 5 ## when divided by ## 3, 4, 5, 6 ##, respectively. (Bhaskara, born ## 1114 ## ).
- Relevant Equations
- None.
Let ## x ## be an integer.
Then ## x\equiv 2\pmod {3}, x\equiv 3\pmod {4}, x\equiv 4\pmod {5} ## and ## x\equiv 5\pmod {6} ##.
This means
\begin{align*}
&x\equiv 2\pmod {3}\implies x+1\equiv 3\pmod {3}\implies x+1\equiv 0\pmod {3},\\
&x\equiv 3\pmod {4}\implies x+1\equiv 4\pmod {4}\implies x+1\equiv 0\pmod {4},\\
&x\equiv 4\pmod {5}\implies x+1\equiv 5\pmod {5}\implies x+1\equiv 0\pmod {5},\\
&x\equiv 5\pmod {6}\implies x+1\equiv 6\pmod {6}\implies x+1\equiv 0\pmod {6}.\\
\end{align*}
Observe that ## lcm(3, 4, 5, 6)=60 ##.
Thus ## x+1\equiv 0\pmod {60}\implies x\equiv -1\pmod {60}\implies x\equiv 59\pmod {60} ##.
Therefore, the integer is ## 59 ##.
Then ## x\equiv 2\pmod {3}, x\equiv 3\pmod {4}, x\equiv 4\pmod {5} ## and ## x\equiv 5\pmod {6} ##.
This means
\begin{align*}
&x\equiv 2\pmod {3}\implies x+1\equiv 3\pmod {3}\implies x+1\equiv 0\pmod {3},\\
&x\equiv 3\pmod {4}\implies x+1\equiv 4\pmod {4}\implies x+1\equiv 0\pmod {4},\\
&x\equiv 4\pmod {5}\implies x+1\equiv 5\pmod {5}\implies x+1\equiv 0\pmod {5},\\
&x\equiv 5\pmod {6}\implies x+1\equiv 6\pmod {6}\implies x+1\equiv 0\pmod {6}.\\
\end{align*}
Observe that ## lcm(3, 4, 5, 6)=60 ##.
Thus ## x+1\equiv 0\pmod {60}\implies x\equiv -1\pmod {60}\implies x\equiv 59\pmod {60} ##.
Therefore, the integer is ## 59 ##.