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This is a homework problem that I'm confused about. I understand the solution until it says "With the Chinese Remainder Theorem and some computation this shows that $n \equiv 301 \pmod{420}.$ I'm not sure how to use Chinese Remainder Theorem and how to get this, any explanation is appreciated.

A woman with a basket of eggs finds that if she removes either 2, 3, 4, 5, or 6 at a time from the basket, there is always one egg left over. If she removes 7 eggs at a time from the basket, there are no eggs left over. If the basket holds up to 500 eggs, how many eggs does she have?

Solution. Let $n$ be the number of eggs in the basket. We know that $n$ is $1$ mod $2,3,4,5,6$. This means $n-1$ is $0$ mod $2,3,4,5,6$ and is divisible by all of those numbers. Since the LCM of $2,3,4,5,6$ is $60$, $n-1$ is a multiple of $60$, and $n \equiv 1 \pmod{60}$. Finally, we know that $n \equiv 0 \pmod{7}$. With the Chinese Remainder Theorem and some computation this shows that $n \equiv 301 \pmod{420}$. So the possible values of $n$ are $301, 721, 1141, \ldots$ and the only one less than $500$ is $\boxed{301}$

Andrew Chin
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    The Chinese Remainder theorem shows existence abstractly...here you can actually produce the number, so you don't really need it. Do you understand the argument that leads to $n\equiv 1 \pmod {60}$? – lulu Jul 21 '20 at 17:22
  • yes, I understand that –  Jul 21 '20 at 17:23
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    Ok, so we know that $n\in {61,121,181, \cdots, 481}$. Now we are also given that $7,|,n$ so you just need to search that list to find a multiple of $7$. Again, CRT guarantees existence, but since you end up solving it explicitly, you don't need to rely on the abstract result. – lulu Jul 21 '20 at 17:25

3 Answers3

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You can just try the positive integers that are congruent to $1$ modulo $60$ until you find one that is also divisible by $7.$ The first few $61,121,181,241$ don't work, but $301=7\cdot 43$ does work. I believe the function of CRT here is to let you know that this is the unique solution that is a positive integer that is less than or equal to $60\cdot 7=420.$

Favst
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Solving $\begin{cases} n\cong1\bmod{60}\\n\cong0\bmod7\end{cases}$ with the Chinese remainder theorem yields the following (using Bezout coefficients): since $2\cdot 60-17\cdot7=1$, we get $n=0\cdot{120}-1\cdot{119}=-119\bmod{420}$.

Now since $0\le n\lt500$, we get $n=301$.

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You know $n \equiv 1 \pmod{60}$ and $n \equiv 0 \pmod{7}$, so by the Chinese Remainder Theorem you know there is a solution and, since $60$ and $7$ are coprime, that solution is unique modulo $420$.

It will be equivalent to one of $1,61,121,181,241,301,361\pmod{420}$ and the only case which is $0 \pmod{7}$ is $301$

Henry
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