- #1

Math100

- 773

- 219

- Homework Statement
- (Brahmagupta, 7th Century A.D.) When eggs in a basket are removed ## 2, 3, 4, 5, 6 ## at a time there remain, respectively, ## 1, 2, 3, 4, 5 ## eggs. When they are taken out ## 7 ## at a time, none are left over. Find the smallest number of eggs that could have been contained in the basket.

- Relevant Equations
- None.

Let ## x ## be the smallest number of eggs.

Then

\begin{align*}

&x\equiv -1\pmod {2}\equiv 1\pmod {2}\\

&x\equiv -1\pmod {3}\equiv 2\pmod {3}\\

&x\equiv -1\pmod {4}\equiv 3\pmod {4}\\

&x\equiv -1\pmod {5}\equiv 4\pmod {5}\\

&x\equiv -1\pmod {6}\equiv 5\pmod {6}\\

&x\equiv 0\pmod {7}.\\

\end{align*}

Note that ## lcm(2, 3, 4, 5, 6)=60 ##.

This means ## x\equiv -1\pmod {60}\equiv 59\pmod {60} ##.

Now we have ## x=59+60m ## for some ## m\in\mathbb{N} ##.

Thus ## x=59+60(1)=119\implies x\equiv 0\pmod {7} ##.

Therefore, the smallest number of eggs that could have been contained in the basket is ## 119 ##.

Then

\begin{align*}

&x\equiv -1\pmod {2}\equiv 1\pmod {2}\\

&x\equiv -1\pmod {3}\equiv 2\pmod {3}\\

&x\equiv -1\pmod {4}\equiv 3\pmod {4}\\

&x\equiv -1\pmod {5}\equiv 4\pmod {5}\\

&x\equiv -1\pmod {6}\equiv 5\pmod {6}\\

&x\equiv 0\pmod {7}.\\

\end{align*}

Note that ## lcm(2, 3, 4, 5, 6)=60 ##.

This means ## x\equiv -1\pmod {60}\equiv 59\pmod {60} ##.

Now we have ## x=59+60m ## for some ## m\in\mathbb{N} ##.

Thus ## x=59+60(1)=119\implies x\equiv 0\pmod {7} ##.

Therefore, the smallest number of eggs that could have been contained in the basket is ## 119 ##.