Find the smallest number of eggs

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The smallest number of eggs in the basket is determined to be 119, based on a series of modular equations. The equations reveal that when eggs are removed in groups of 2, 3, 4, 5, and 6, specific remainders are left, while removing them in groups of 7 results in none remaining. The least common multiple of 2, 3, 4, 5, and 6 is 60, leading to the conclusion that x is congruent to 59 modulo 60. By solving for x with the condition of being divisible by 7, the final result is confirmed as 119. This problem is attributed to Brahmagupta, who is known for various mathematical challenges.
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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 ##.
 
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Math100 said:
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 ##.
Correct.

I didn't know that Brahmagupta had more problems associated with him. I only knew a geometric problem:
problem 13 in
https://www.physicsforums.com/threads/math-challenge-september-2019.976793/
solution on page 380f. in the solution manual (last attachment)
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/
 
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