- #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 ##.